Too many people are saying the same thing, so I just made this video to respond to 99.9% of comments disagreeing... If you think it's obvious that the host will always open exactly 1 door after you select yours, and that the host will never reveal a car, or your door, please watch this video. I can PROVE that it's not obvious, try the challenge if you're so confident. ua-cam.com/video/3AJVaeysml4/v-deo.html
The situation is that "decisions without consequences" and "choices with practical benefits" are confused and regarded as having the same meaning. It has been clearly decided. If there are really two choices, shouldn't we get both things? In fact, the rule changes from one of three choices to one of two, misleading you into thinking that you have already chosen once, but in fact you cannot get any of the so-called choices at all. When you finally make the real choice, in order to prove how smart and knowledgeable you are, of course it should be counted as 2/3 when there are only two choices. This is a testament to the Goebbels effect. If scholars and the educational community do not correct the extension of Monty Hall's problem and still believe that 2/3 is correct, then what is the point of educational scholarship? ? It was ruined.
Maybe thinking this way....... Try to understand it from the perspective of the host. The player provides his prediction to the host. The host opens a pair of doors with sheep, leaving only two pairs of doors, one sheep and one car, for the player to choose. If the player Predicted the door of the car, what will the host do? of couse Want the player to change their choice. The host asks you to calculate the odds with misleading you already have once choice before. you swapping your choice, 2/3 will win. On the contrary, the player predicted the door with the sheep. What will the host do to make the player still choose the door with the sheep? you can tell. ........ Dawyer's door problem, calculate the chance of the host winning.
1) not a paradox 2) Monty never reveals a prize 3) Monty always reveals a door containing a goat You will never trade a goat for a goat. Ever - because Monty is REQUIRED BY RULE to reveal a goat door. 1/3rd of the time, you will have chosen a car. Monty will reveal a goat. You will switch to the other goat. 2/3rds of the time, you will have chosen a goat. Monty will reveal the other goat. You will switch to the car.
@@asafoster7954 No. That part of being less smart is on you guys. This 2/3 probability to win if you switch is a trick played on your mind. Here is what takes away. Monty will always reveal a goat. If so this eliminates your first decision. Now say it was door A that was taken. You are now left with door B or C. If you chose B before that means: Stay with B: 50% chance to win. Switch to C: 50% chance to win. I leave it open to suggest behind witch door the prize is hidden. Why? Because as the participant making your choices YOU do not know which door does hold the prize. You can only accumulate your winning chance when you assume the door that holds the prize. You can simulate it mechanically. But it will not play out like this in reality. The prize can be behind door A in four consecutive shows or five times behind door C. And you must not forget that it isn't constantly you flipping through ten games of the Monty Hall show. To maybe equal up to this 2/3 winning chance throughout all games. You play maybe once and you either win or lose. And that is all there is. Now toss me some more calculations if you want. But reality hold your butt.
@@asafoster7954your the dumb one. If you actually pay attention to the video you realize that he admits that the Monty hall problem is correct as long as you are told that the host is systematically avoiding the car. The Monty hall problem is based off of assumptions If I word the question slightly different and mention the fact that the host is not picking cars than you would want to switch
if you're looking at this like a logic problem, then you need to understand that there are 3 endings. in the first ending, the game show host picks the car, and you lose. in the second ending, the game show host shows you that you picked the car, and you win. in the third ending, the game show host shows you that you didn't pick the car, and you lost. if you're talking about the probability of getting the car, and you're including the first ending in your math? you really shouldn't be. if the first ending happens BEFORE you get to make a choice. it eliminates you BEFORE YOU MAKE YOUR CHOICE. how can something that PREVENTS YOU FROM MAKING A CHOICE IN THE FUTURE factor into the probability of the choice? it can't. which means we need to eliminate the first ending in our math, because it unequivocally CANNOT effect the choice; all it can do is prevent you from making the choice to begin with because, again, IT ENDS YOUR CHANCE TO MAKE THE CHOICE.
There are 2 possible outcomes... win a car or you win a goat. It's literally that simple. People have overcomplicated this just because they didn't want to piss off people pushing a female role model.
Stand strong, dude. The drones are explaining the probabilities of which door Monty opens, not the probability of getting the car. It is that simple. Bottom line is it is either ⅓ vs. ⅓ or 50/50. 🤙
It cannot be 1/3 & 1/3 or 2/3 & 2/3, probability has to add up to one, I encourage you to get a die, and four pieces of paper, write "winner" one one piece of paper, and "1", "2", and "3" on the others, roll the die, on a 1 or 2, place the winner paper beneath the 1 paper, 3 and 4 for 2, and 5 and 6 for the 3 paper. Then roll to choose a paper again, then remove a piece of paper without the winner paper. Then switch the paper you chose, mark whether you got the winner paper, repeat many times and see how many win.
@@rat-prophetistfordism8344 I appreciate your reply. Correct, probabilities have to add up to one, which means each door can ONLY have a 1/3 probability for the car to be behind and 1/3 probability for the car to NOT be behind. At no point in time does ONE door possess a 2/3 probability for the player to not get the car. Also, at no point in time does the player have a 1/3 probability to pick the door Monty eliminates. The 2/3 hoax wants you to believe that this is about picking one door from three, and the rest doesn't matter. Player has TWO SEQUENTIAL PICKS, with Monty's elimination of a no-car door being RANDOM to the player. The Monty Hall Problem asks, does the player have a higher probability to get the car by picking AGAIN the same door he initially picked, or pick the remaining door that Monty did not eliminate with his SECOND SEQUENTIAL PICK. And yes, the simulations showing 2/3 no-car probability are all FLAWED. Most are done in Python, which is a compiled and interpreted accelerator. That means you have to prompt and code all the rules BEFORE each function and re-prompt and re-code EVERY STEP of the simulation along the way to get the correct result. No one disputes player's 1/3 probability to get the car. Isolate the simulation to find player's probability to NOT GET THE CAR with his second sequential pick with two doors left and you will find the correct answer: Probability of not picking the car in Pick A: P (not car in Pick A) = ⅔ Conditional Probability of not picking the car in Pick B, given Pick A was not the car: After the host eliminates a door that does not have the car, the scenario reduces to two doors: one that the player initially picked (not the car), and one that wasn't picked (which could be the car) If the player initially picked a door without the car (probability ⅔), there is a ½ chance that the car is behind the remaining door after the host's reveal. Therefore, the probability of not picking the car in Pick B, given that Pick A was not the car, is: ½ Overall Probability (Law of Total Probability): Combining these probabilities using the law of total probability: P (not car in Pick B) = P (not car in Pick B∣not car in Pick A) ⋅ P (not car in Pick A) Substitute the values we have: P (not car in Pick B) = ½ ⋅ ⅔ = ⅓ Therefore, the probability that the player does not get the car after TWO SEQUENTIAL PICKS is ⅓. Final probabilities after player’s second sequential pick is a 50/50 guess: Probability of getting the car: ⅓ Probability of not getting the car: ⅓
The crux of the confusion is that Hall doesn't ever choose the prize in his reveal. It's not about doors, it's a logic problem about three different scenarios. Scenario triggered by the choice of either the good door, or either of the two bad doors. Choose the good door, and he can reveal either of the remaining ones. If you switch, you will necessarily fail. But if you choose either of the other two doors, he must eliminate the other bad door. So in either of those scenarios, switching gives you the prize. Two good scenarios, one bad, if you switch. Two bad scenarios, one good, if you don't. 2/3 odds for switching. They are simply scenarios that always play out, not "doors". It took me a bit, originally, too, but that's because I didn't quite get the problem. If it actually were about the doors, the superficial stats, then round two probability in a vacuum is 50/50. But knowing how each scenario plays out, how everything is selected and locked in changes those odds.
I was typing up something similar, but say Mr. Dan's Comment, and thought a reply might be better than a new post. At 2:20, you say "nowhere does it state that the host picking a car is not a possibility." It does, though. "The host, who knows what's behind the doors..." As Mr. Dan says, the host will not open a door with the prize and then ask the player--who has just been revealed to have lost--if they want to switch to the other losing prize.
You're talking about form then like with the horses. Those who fully grasp their form are in the realm of certainty not probability.. The show had form obviously , but I never watched it,so my realm would be probability only and purely.
@@philip5940 I have never seen the show, either. I am going entirely off the wording in the problem statement you seemingly did not notice or deliberately ignored. The form, as you put it, is stated right there. You assumed each choice was independent and random. However, it is stated the host knows what is behind each door. So, if nothing else, the host's choice is no longer truly random. Therefore the probabilities you assigned to the second pick are essentially arbitrary.
@@BoomStickization No you are not going by the wording at all. We are never informed that the host must actively chose a door that has a goat behind it. It would have been really easy to state that. But they don't, probably as their urge to make the paradox appear as fancy as possible did one up on themselves. Because the wording is so specifically design to confuse and unnatural that it is hard to see this failed attempt was not intentional. We are only told that he knows what is behind the doors and that in the instance of the show we are "supposed to imagine" unfolded the event that the door he opened had a goat behind it. At no point do they state that the host HAS to open a door with a goat behind it, which is an absolute necessity for the paradox to emerge. As a result of this the answer to the question is either "insufficient information is given" or "1/2" since everything about a stochastic process like throwing a dice, drawing a lottery ticket, opening a door etc is universally assumed to be a uniform distribution among each individual side, ticket or door etc. We need to clearly be informed of the contrary to change that. But here, we never are, despite how easy it would have been to do so. Instead they put a lot of semantic effort in all sorts of other information to make the paradox appear fancy, but missed the very core of the paradox in the process, resulting in that they are not even describing a situation where the paradox emerges. Also much of the aftermath was a smoke screen as they then tried to pretend that the information that the host HAS to open a door was already stated in the initial phrasing, when clearly it was not. The fact that the played down the importance of that missing information just made people even more confused afterwards as it is the very mechanism that makes the paradox emerge in the first place. For instance running a computer simulation where the assumption that he has to open a door with a goat behind it is implemented, is such an obvious smoke screen to ignore the fact that it was never stated in the initial problem. And such a program (a Monte Carlo simulation) would almost be nonsensical as the majority among the ones who managed to write the code without copy/pasting would realize that the probability of winning by swapping would trivially be 2/3 under those assumptions.
@@dataandcolours I am going to leave this conversation after this response. Your statement seems to boil down to "the writer is clearly and deliberately trying to deceive us," and I do not see that, at all. I agree that the conditions could be made more explicitly clear. There may be assumptions in the statement that are not as clear as the author intended. But that is far from being "specifically designed to confuse," or an "active smokescreens." The contestant even being offered a choice implies there is a chance for them to win. Thus, the host must have revealed a losing door. This is a reasonable assumption in the context of an American game show. So, the question is a conditional probability problem: "is there an advantage is the contestant switching if the host revealed a losing door?"
i arived at 1/2 in a different way suppose lets say there are 3 doors a b c for first circumstance assume car is behind a all possible events are 1) you chose a and he opens b you switch - goat 2) you chose a he opens b you dont switch - car 3 ) you chose a he opens c you switch -goat 4) you chose a he opens c you dont switch- car 5) you chose b , he opens c you switch- car 6) you chose b , he opens c you dont switch - goat 7) you chose c he opens b you dont switch - car 8) you chose c he opens b you switch - goat all these can be multiplied and divided by 3 cause similar outcomes would be if car was behind b or c when you switch no of times you win is 2/4 no of times you dont switch and win is 2/4 now multiplying 3 to both numerator and denominator for the outcomes would lead to 6/12 or 1/2
The whole point of the host opening a door, is to confuse the player. If he opens the door with the car, why would he ask the player if he wants to switch? The host will always open the door with a goat
But we should never have to assume things like that, especially as complex of an assumption like that. Things like that needs to be clarified in the initial question. But they never are (especially not in how the question was asked in Parade Magazine). We are not intended to play private detectives here. Because if we go down that rabbit hole, when do we even stop? * Why did the host even open a door and asked if the player wanted to switch if the player had chosen a door with a goat behind it. According to game theory it would then be in the interest of the host to just open the door the contestant chose and declare "You lost!" * But why didn't the contestant just look through the keyhole? * Maybe the information about the number of cars are false. * Maybe the host only ask if the contestant wants to switch if his/her initial pick was a door with the car behind it. This is why the initial phrasing of the question makes this question so dumb. It really does everything it can to describe an instance of the show, as doing so is the only way to camouflage all the rather important aspects that needs to be established in the first place for the paradox to emerge. This makes this version of "Three Prisoner Dilemma" a rather pointless example of a combination of sloppiness, intentional deceit, psychology and a bit of conditional probability. A mixture that is outright de-educational as the intent seem to be on mesmerizing rather than educating.
They gotta make a show and engage a little stagecraft. They could simply start with only two choices, not three to speed it all up but it ain't in interests of entertainment. If it starts with two choices , then everyone would agree probability is 50/50 . However when he starts at three choices then brings it down to two choices , it then happens that gullible folk are purseuded to believe that some phantom effect of a pre-existing ⅔ probability still exists for the now changed independent new conditions . But this stuff emerged during the Uri Geller era when folk also believed in spoon bending and his other scams . Pyramid power too , it was endless .
@@philip5940 "However when he starts at three choices then brings it down to two choices , it then happens that gullible folk are purseuded to believe that some phantom effect of a pre-existing ⅔ probability still exists for the now changed independent new conditions ." If you look at how the Monty Hall Problem is _intended_ to be worded, where the host always acts to reveal a goat from the remaining two doors (knowing where the car is), then the second 2-door scenario is _provably_ (and easily so) dependent on the first choice. Merely declaring it to be independent simply because you don't want to think doesn't change the reality here.
If the host will ALWAYS open the door with the goat, that means the chance is not 2/3 or 66.666% of the goat being behind door number 2, but its 100%. You said: ALWAYS. Always = 100%. So what you said is literally against what the correct answer was that i heard about.
If I were the game show, I would instruct the host to show the goat if they picked the car and show the car if they picked the goat. I would want to give away fewer cars. Why would I give the contestant a second chance if they got the first pick wrong? Knowing this, as a contestant I wouldn't switch. I am curious to see what the stats on the monte hall show turned out to be.
Exactly. Considering we are given no information at all on the rules the host have to obey we simply cannot answer the question as it was composed in Parade Magazine and you reasoning exemplifies this perfectly. The phrasing was not just "slightly sloppy", it was probably deliberately vague. I mean just the start "Suppose you are on a game show...". How is that not introducing us to consider an instance of the show, hence perfectly distancing us from being told how the host must act (the rules)? :)
@@gnlout7403 Yes the show actually worked in all sort of mysterious ways. However we obviously cannot add outside information to the question given. And because of this the question as phrased in Parade Magazine is absurdly ambiguous and the most natural way of interpreting it from a logical standpoint actually makes the result the 1/2 vs 1/2 scenario. It's only if we add a whole lot of additional information that the 1/3 vs 2/3 situation emerges.
@@dataandcolours no. Again. The show never worked like this. And we've been over this already. You botched logical Moralitys objection. He's not confused about what you think he's confused about. And you can do the problem as stated with the information you have. Hope that helps
@@gnlout7403 Please refrain from claiming I have claimed things without clarifying what I have claimed. I hope you understand that just comes off as very passive aggressive in a weak attempt to settle some scores. The intent of discussing here is simply to point out how massively poor the original wording was and how correct people are at pointing that out.
Monty hall problem is a great example of real world problem. Most problems that people see in text books are problems with fair dice. Monty hall problem is one with loaded dice (which is always what happens in real world). Monty hall picks door ensuring that car door is never opened. Thats the most important part!
Too many people complicate this. Here is the simplest proof: - Your first choice is probably (66%) a goat. - The host reveals the other goat (Fact. 100% chance that he does this.) - Thus, the remaining door is probably (66%) not a goat.
@@jacobinozznope. Your door stays one third. The remaining door is 2/3rds. Test it with 100 doors. Or a deck of cards. How often do you win by staying?
OK if someone can help me with this: The moment the host asks ''Do you want to open door number 2?'' is it not true that Monty opening door number 3 is completely irrelevant to the chances? Because the moment there are only two doors left, he has no choice. He HAS to ask (If that is the assumption that he HAS to ask btw. And second assumption he can never ask you if you want to open the door you chose yourself) if you want to open the only other door left. There is no choice. So the whole ''Oh he picks the one with the goats'' Is completely irrelevant at this point because he is not PICKING any door cause there is only one door left. If door number 2 has a goat? He HAS to ask the question if you want to choose that door be opened. And if the door has a car, he has to ask the same question also. He only has one choice and therefor your Him knowing what is behind the door is completely irrelevant. And even if the hosts has the ability to ask if you want to open door number one. Or the ability to not ask any questions at all --> All of this can be done to manipulatie you and he can in his head choose a door randomly. Either making you believe he wants you to open the door he is asking about. Or either making you believe that he does not want you to open the door he is talking about. And even those two choices is a matter of manipulation of wich you have no control over. From the moment i saw this paradox the only thing that comes up in my head is that its 50/50. Because the QUESTION of the host is entirely irrelevant. That seems to be the real confusion of all of this. Is that people believe the previous choices of the host affect the last choice. And thats wrong because the last choice is not a choice anymore. EDIT: Ok i was completely wrong lmao. This one fucked with my head. Now i finaly get it.
Where you messed up is here - 'it doesn't say the host picking the car isn't a possibility'. It's not about what's possible in other games, it's about the game you are presented with. In this scenario, you pick, the host, knowing what's behind the doors, opens a door with a goat. Then he offers the switch. He doesn't open a door with a car in the scenario you are asked about, therefore that's not an actual option in the probability tree.
I am not talking about other games, I am talking about this game. If you roll a 6 with dice it doesn't mean a 4, or 1 wasn't possible at the time. The odds that you would get that 6 is still 1/6th, unless it's impossible to roll certain numbers.
@@logicalmorality4646 this game is not dice. It is a specific problem. The Monty Hall problem. He doesn't pick the car in this scenario, yet you include it as a possibility.
@@logicalmorality4646 Why you did you outline that you're talking about the Monty Hall problem in the title, when you're not? Instead you're just talking about straight up probability.
@@logicalmorality4646 Certain things are eliminated by the rules. In your dice example, if someone changes dice and replaces 6 by 5 then your dice can never throw 6. The trick is to know that dice is loaded. In this game show, the host can NEVER reveal the car because at that point the game becomes pointless. So he will always have to reveal a goat. The host does not chooses doors with equal probablity, in fact by the rules of this game he will always chose goat door. Otherwise the game is meaningless.
To simplify, I deal three cards, one of which is an ace. You select one card. I have the two remaining. I am twice as likely to have the ace. I remove a card that isnt an ace, but Im still twice as likely to have been dealt the ace. The odds of my having the ace didnt change by removing a none ace card.
I was skeptical about this paradox until I wrote this: There are 3 options: I select a car, then the host reveals a goat and the remaining door must contain a goat. I select a goat, then the host reveals another goat and the remaining door must contain a car. I select a goat, then the host reveals another goat and the remaining door must contain a car. In 2 of 3 cases the remaining door contains a car... so I should switch to it.
But there is also a two thirds chance that the car was behind either the door he opened or the door you chose so you should stick with your choice by that logic or admit that the entire concept you are using is wrong. Why artificially tie the two doors you didn't choose together? Probability doesn't work that way. QED
@@TimBee100 He never opens the price door. If you wanna change the monty hall problem so that the price can be revealed its still always more advantageous to switch.
Lets say the first player quits after hall removes a loser and asks the player if they want to switch, And a new player comes in facing two doors with a winner and a loser and is asked to continue the first players game not knowing which pick is original , and which pick is a switch. Without changing anything that has happened, the second play will be asked which of the two doors would you like to pick. there odds of winning will be 50 / 50 because one has a car and one does not have a car.
Actually no. Their odds of winning will depend on which door they pick. If they pick the original contestant's door, they will have a 1/3 chance of getting the car. It is the doors that have the odds, not the player. What you are thinking of is the statistical average of many such games where the first player leaves and a new player guesses. It is no different than the first player flipping a coin to decide to switch or stay. The odds for one game will depend on the door but the statistics will average out to 50/50.
@@Hank254 A second player is faced with two doors one has a winner and the other is a loser. They are asked to pick one of the two doors. They might pick the first players original pick or the other. They don't know. Try it. Two doors. One winner .
@@immrnoidall Yes, all of that is correct. What I am explaining is that the doors already have different probabilities of having the car. Take a deck of cards and two coffee cans. Pick a random card from the deck and put it in can A. Put the other 51 cards in can B. You are trying to say that if a person came in after you did that, and they chose can A, they would have a 50-50 shot at picking the 'winning' card. That is simply not correct. They have a 1/52 chance if they pick can A and a 51/52 chance if they pick can B. The cans have different probabilities regardless of what the person saw.
Ok, apparently this video attracts traffic from people that want to prove their 50/50 probability is right. But keep in mind the video poster serves a different math problem than what majority believe talking about. One where the host can also reveal doors with cars and make you lose prematurely. In other words, if you've picked a door and the host decides to show you a car in another door, there is no point in asking to switch. You already lost and that's the end. The video poster believes this could happen in the game show, because the phrasing doesn't explicitly say it can't. While I agree it doesn't explicitly say so in the phrasing, I still think logically it's not what the game show is about, here's why: 1) It makes no sense to involve randomness, because it phrases "the host, who knows what's behind the doors..." and the host picks. Doesn't it seem odd to explicitly phrase the host knows all the doors contents and the host proceeds to roll a dice for his door pick? What is the point of saying he knows all the contents in this context? 2) Let's say it's not random, and the host, who knows what's behind all the doors, picks deliberately. This is incredibly unfair to the contestants, because this means the host controls the show and decides whenever a contestant should lose or not. 3) In addition to this the phrase "...opens another door, say no.3, which has a goat" should be read as "...opens another door, which always has a goat, say no.3" Therefore I think these game rules apply by logically reading the math problem: 1) The host opens another door every show. 2) The door the host picks will always have a goat. 3) The host always asks if you want to switch. based on this ruleset, switching will always give you a 2/3 chance of winning.
"1) It makes no sense to involve randomness, because it phrases "the host, who knows what's behind the doors..." and the host picks. Doesn't it seem odd to explicitly phrase the host knows all the doors contents and the host proceeds to roll a dice for his door pick? What is the point of saying he knows all the contents in this context?" Exactly.
But here is the point. If the intent actually is to clarify how the scenario is unfolding with the assumptions that creates the 2/3 vs 1/3, the question is not just slightly poorly phrased, it's almost perfectly taylor made to be systematically unclear an ambigious. A perfection to such an extent that it is generally hard to believe it was not intentional. Which is why the only useful lesson from this is "Ask stupid and deliberatley poorly phrased questions => get a meaningless debate." As a result, it's absolutely not a good question for educational purposes as it mixes up several other confusing aspects as psychology and extremely poorly phrasing to a subject that many already find confusing without all this additional and deliberately misleading nonsense. That is why it's important to point out why they actually went so far overboard when they phrased the initial question, that they unambiguously managed to describe a scenario were we cannot establish an asymmetric redistribution of the probabilities. It could perhaps serve as a valuable lesson to never ask questions in that manner.
Nailed it. It’s clearly discussed as a math problem and it’s quite obvious that it implied the host will always reveal a goat and always offer a switch. As you say, it makes no sense to point out that the host knows where the car is but could then randomly reveal it. Why do people have to complicate things?
It's 50/50 Following mathematical logic, It would be 66% but If would exist a strict pattern. If Monty show would follow this "pattern" a lot of people would have won the car. No all the times the car is put in the same place. According to the math, a bullet would never hit the target because you can devide infinitely the distance, for instance, 1meter left, half meter, 1/4 meter 1/10000000000 and never would hit the object.
I find it quite strange that people believe it is always better to switch mathematically even though the odds do not seem to change all that much with just three doors, especially if the host cannot pick the prize door. What I see: (Legend: X=revealed; 0=unopened door; 1=chosen door) *All* possible choices if [0][1][0] is the correct answer, [1][0][X] incorrect [0][1][X] correct [X][0][1] incorrect [X][1][0] correct So no matter if the door is switched or not, you will always have a 50/50 chance of winning if you switched for just that scenario. What is flawed with what is given above if what seems to be the consensus believes that there are higher odds of winning if you switched?
If the host can not reveal the car, then those odds change from 1/6 you pick car host reveals goat 2 1/6 you pick car host reveals goat 1 1/3 you pick goat 1 host reveals goat 2 1/3 you pick goat 2 host reveals goat 1 0/3 you pick goat 1 host reveals car 0/3 you pick goat 2 host reveals car Because remember the two losing switches are a subset of you having selected the car. They should add up to 1/3 probability total as there is a max 1/3 chance you can pick the car. On the other hand if the host can't reveal the car, then if you select either goat you have won for sure by switching because there is no subsets which would normally exist if he revealed randomly. (normally 50% of the time you would lose by switching if you picked a goat, because half the time you would be trading a goat for another goat)
@@logicalmorality4646 I'm not certain about your logic. Two you have of the above six possibilities are precluded because he can't reveal a car. Now the possibilities are: 1/4 you pick car host reveals goat 2. 1/4 you pick car host reveals goat 1. 1/4 you pick goat 1 host reveals goat 2. 14 you pick goat 2 host reveals goat 1 Two you're right, two you're wrong. 50%.
If he can't pick the car, and you select goat 1 for example, then the probability of him revealing goat 2 is now 100% correct? So it goes from 1/6 to 1/3
With one event removed , the odds have now gone from 1:3 to 1:2 for chances of having the car . Either event choice is a 50/50 chance of being the car in other words . The psychology behind the question would make me stick with original event choice . But he could already have factored for me calculating a psychological strategy and be a step ahead , l know .
It's never 1/6. It's 1/3 stay, 2/3 switch as long as the host must open a door with a goat. Here are the possibile scenarios You. Prize Door. Don’t. Do 1. 1. Win. Lose 1. 2. Lose. Win 1. 3. Lose. Win 2. 1. Lose. Win 2. 2. Win. Lose 2. 3. Lose. Win 3. 1. Lose. Win 3. 2. Lose. Win 3. 3. Win. Lose Scenario 1: You pick door 1 and the prize is actually behind door 1. In this case, Monty will open either door 2 or 3 and show you that nothing is behind one of those doors. If you stay with door 1, you win. Scenario 2: You pick door 1 and the prize is actually behind door 2. In this case, Monty must open door 3 and show you that nothing is behind it. If you stay with door 1, you lose. Scenario 3: You pick door 1 and the prize is actually behind door 3. In this case, Monty must open door 2 and show you that nothing is behind it. If you stay with door 1, you lose.
You have one try, one door. The House has two tries, two doors. The house then gets rid of one bad door. IF the house had the one good door, they will always get rid of their bad door thus leaving their good door for you to trade with. ( I'm using "House" as a casino reference.)
I would say concealed choices and prize and booby prize . Forget car , forget doors, forget goats . People are too easily bamboozled by superficial irrelevant details .
Yes if you make all of those additional assumptions it would be beneficial to switch. If you explain the scenario correctly and concise there aren't really that many that falls for it. But for some reason it almost never is explained in a concise way. The problem is that many of the necessary conditions for the paradox to emerge is not stated in the question in Parade Magazine. Instead we are jus being described an instance of the game show and what events that unfolded then. We are never informed why the events unfolded as they did, how the host must act nor how the game works in general. If anything the phrasing in Parade Magazine is a complete train wreck of a question. It's as if every semantic option they made was to maximize the confusion, but they far overdid it and no longer even described a scenario where we can establish that the intended paradox emerges.
I don’t know that this “debunks” the Monty hall problem, but it’s a good reminder to clearly lay out all the assumptions and factors when describing it.
In other words, 2/3 of the time you will have initially picked a goat door. The host reveals the other goat door. So 2/3 of the time, by switching, you gain the car door.
@@logicalmorality4646 Hall, if he knows, could always be knocking off a goat door. There is always one available. Your odds were one of three of getting the car, and I can't see an advantage to switching. Your odds now are not one of three, but one of two. There is no harm in switching, as the odds are even. It would just be a matter of luck. What dos Monty have to do with it?
@@logicalmorality4646 It's clearly stated in the original problem - a letter from a reader - that the host KNOWS what is behind each door. No randomness is implied.
@@jasonthomas5118 It's been thoroughly demonstrated in physical tests that switching definitively doubles your chances of winning. Of course it's also been proven mathematically and graphically. My favorite is simply that when you picked a door, you divided the set of all doors into two sets. The probability of the car being in your one door set is 1/3, the probability it is in the 2 door set is 2/3, the sum of each door's probability. That is fixed for the two sets. Opening one of the goat doors reveals that door had a zero probability. Therefore, the set's probability resides entirely in the unopened door in that set, and it is 2/3ds.
This has became an analysis of how the question was worded rather than Math. This was also the debate among professors back in those days as they said the intend of the host is important.
Among other things, you apparently misread the original problem. Nothing in it implies randomness. It states he will open a door which has a goat. (In the phrase, "opens another door, say No. 3, which has a goat" I think you're interpreting the appositive phrase "say No. 3" as if it implies a random choice, but it's just labeling the door for future reference.) Of course any "debunking" of the problem has to explain away the abundant experimental evidence that, indeed, switching doubles your chances of winning the car. Good luck with that.
William Ivey and Marilyn Von Savant are correct because it is stating in the problem that a goat is revealed. Therefore, the probability begins at that point not prior to that point, which you are suggesting Logical Morality. In fact, your own probability tree answers the problem for you. The problem begins at the second column of your probability tree because it states that we know a goat is revealed and at that point we have only three options: 1) switch off the car on to a goat (goat 2 was revealed), 2) switch off goat to a car (goat 2 was revealed), switch off of goat to a car (goat 2 was revealed). That means two out of the three options favor you switching. Therefore, you should certainly switch.
@@philb1208 As I explained in the video, the host could reveal the car 99.9% or even 100% of the time whenever you pick a goat.... and everything stated in the problem could still be true. If that were the case you should never switch. In fact this is actually the most logical way to play too... 1) Do you agree the host wants the contestant to lose? (Usually the casino/carnival wants the player to lose) 2) If the player selects the car then you should give them the option to switch... They have won for sure, but if you give them the option to switch then they might do it and lose... 3) If the player picks the goat then you should reveal the car. This way the player has no option to switch, they can only trade a goat for another goat.
@@logicalmorality4646 hes right on the math, the problem is everyone is thinking about the first choice its the second choice that actually determines what the out come is because he always reveals the goat, the real trick is realizing there were NEVER THREE CHOICES. the reason is that since he will never reveal a car and he knows were the goats and car are he will choose to show you the goat at that time it can still be behind either do that part has not changed, the problem is that people are thinking about it wrong
@@Chaosmage42 based on the wording of the problem you don't have that information. All you know is that he revealed a goat once ever. You don't know why.
@@gnlout7403 no because he would never open the door with the car. It would be pointless to ask you if you want to change doors in this scenario. He always opens a door with a goat. If there's no host, the solution which this video provides would be right. But there is a host. So you're wrong.
@@iliyastoyanov3302 you said the host is trying to make you lose. I'm saying his motives are irrelevant if he's bound to open a door with a goat, which you seem to agree with
I admit my post of nine hours ago was wrong. I did this using cards. Two jokers and one ace. Shuffle the cards and put them face down and pick one at random, or choose them in order, 1, 2, 3, each time. It came out overwhelmingly in favor of switching. So the adult thing is to admit I was wrong. The key is apparently Monty knocking out a dead door.
No, there's a flaw in your method. If you pick one at random you're not following the problem. Monty knows there's an incorrect choice which he must reveal. In your example you have to select one, leave it face down, and lift one of the remaining two. In this game, you have a one in three chance of being correct. If one of the deselected ones is revealed to be correct, it's no longer the Monty Hall problem. If it is incorrect, you go on to game two, which I cannot overemphasize. In game two the odds are 50/50. Anything else is logically erroneous.
I think you didn't watch the video. Yes what you are saying is correct, the problem is that the problem does not have any rule where the host avoids revealing a car, has to reveal one door, can't reveal your door, etc...
@@kurtbader9711 "In your example you have to select one, leave it face down, and lift one of the remaining two." No you don't. You can turn over the card that was picked and record it as a win or loss by staying.
I don't think you _debunked_ the Monty Hall Problem, but you did point out the crucial assumption that is often missing from the problem statement, and your mathematics is completely correct here.
@@gnlout7403 The crucial assumption that is _often_ missing (but not always) is that the host intentionally revealed a goat from among the two remaining doors, regardless of what the contestant's original door was. You can get around this by using words like "always" - i.e., the host "always" opens a door with a goat and "never" opens the door the contestant originally picked. But this implies a repeated thing, which isn't necessary. One could also say that the host decided on following an algorithm beforehand where they would select reveal a goat door from among the two unpicked doors, regardless of what the contestant's original door was. The reason these need to be specified is because of variants of the Monty Hall Problem which appear identical from the contestant's perspective, including: Ignorant Monty. The host follows the following algorithm: After the contestant chooses a door, the host randomly selects one of the remaining two doors and reveals it. If it's the car, the host says, "Oh, too bad. You lose." and the game ends. If it's a goat, the host offers the contestant the option to stick or switch. (If the contestant is given the option to stick or switch, switching wins with probability 1/2.) Monty Hell. The host follows the following algorithm: If the contestant picked a goat door, the host reveals the contestant's door and says, "Oh, too bad. You lose." and the game ends. If the contestant originally picked the car, then the host reveals one of the two other doors and gives the contestant the option to stick or switch. (If the contestant is given the option to stick or switch, switching wins with probability 0.) Angelic Monty. The host follows the following algorithm: If the contestant picked the car, the host reveals the contestant's door and says, "Wow! Congratulations! You won!" and the game ends. If the contestant originally picked a goat, then the host reveals the other goat and gives the contestant the option to stick or switch. (If the contestant is given the option to stick or switch, switching wins with probability 1.) None of these three variants are the _actual_ Monty Hall Problem. But if people don't specify the strict rules the hosting is acting under, the contestant could be in the "stick or switch" subcase of any of these situations, just knowing the host revealed a goat from the remaining two doors. The Monty Hell and Angelic Monty scenarios further show that the host's knowledge is not enough (since the host must know the car's location in these two scenarios as well). So the host must be acting in a very specific way to get 2/3 as the probability of winning by switching if given the choice to stick or switch.
@@MuffinsAPlenty you are correct in that you don't need him to act this way every time. You are only given one scenario. You weren't told what he might do the next game. Still the fact that you are told he knows what's behind the doors and chooses to open a door with a goat means he is not picking randomly. You should be able to calculate the odds from this information
@@gnlout7403 It depends on how it's worded! The wording that the video gives does not say that the host "chose" to reveal a goat. It simply said that the host knew where the car was _and_ opened a door with a goat. I think it's reasonable to assume the host intentionally revealed a goat, but that is only _reasonable_ not _logical._ But even then, if you make the reasonable (albeit not logical) assumption that the host's knowledge implied that the host intentionally chose that goat door, you still can't eliminate the possibility of being in Monty Hell or Angelic Monty.
@@MuffinsAPlenty I disagree. If he knows and opens a door with a goat, the assumption is that's it's purposeful, not random. If it said he didn't know what's behind the doors it'd be random. I have no problem with the wording BTW as long as Monty chose the door intentionally, aka, not randomly, intentions are irrelevant.
I choose a door = 33% chance i'm right. No matter what, one of the other two doors is wrong. Opening the door that's wrong just reminds us that it's the case. The door you chose still had a 33% chance.
If your pick is wrong, switching will automatically get you the car. You pick incorrectly 2/3 of the time so switching gets you the car 2/3 of the time.
@@logicalmorality4646 I think you misunderstand me. If I pick a goat, the host will always pick a goat meaning if I switch it'll always be the car. The problem is I don't know that I've picked a goat.
@@Hank254 ... It says "the host who knows what's behind the doors, opens another door, say door #3 which has a goat" It doesn't say he always reveals a goat no matter what all the time. It just tells you in 1 instance he reveals a goat. He could reveal a car 99.999% of the time and this would still be true.
So, you have just decided to be nitpicky about host purposely avoiding the car? Well, this is usually mentioned in the problems, so I don't know where did you find the version where this isn't mentioned
My thinking was the same as yours but it turns out that I was missing an important piece of information, I did the maths again and it turns out that you SHOULD switch every time. When the host opens a door with a goat behind, that in fact does change the probability function. If you choose not to switch: Choice Outcome G1 G1 G2 G2 C C Pr(C) = 1/3 If you switch: Choice Outcome G1 C G2 C C G1 or G2 Pr(C) = 2/3 In other words, the probability of getting the car if you don't switch is 1/3 and the probability of getting a goat is 2/3. If you've chosen a goat and you switch, you WILL get the car. Since choosing a goat is more likely to happen, you should always switch.
@@logicalmorality4646 I did. Your calculations at 1:39 assume that nothing can change the probability function which is false. Pr(A) is not the same as Pr(A|B) unless Pr(B) is certain. The probability of event A can change given event B. So once you've picked your first door, Pr(C) = 1/3 and Pr(G) = 2/3 Now the host reveals a goat, and you have the option to switch. If you don't switch you have a 1/3 chance of winning the car. If you switch, this is what happens: Pr(G1|C) = 1/2 Pr(G2|C) = 1/2 Pr(G1|G2) = 0 Pr(G2|G1) = 0 Pr(C|G1) = 1 Pr(C|G2) = 1
@@thebassboar3980 I don't think you watched it until now but that's not important hahaha. Think hard about this... -Suppose the host is biased and reveals the car 90% of the time when he has the option to... (we are assuming he won't reveal your door even if you pick the car). If this were the case and you were revealed a goat, then is it still favorable to switch? OF COURSE NOT. In the event of 2 unlikely possibilities, the least unlikely becomes the most likely scenario. In this case, you are far more likely to have been revealed a goat by selecting the car (1/3 chance), than haven got a 2/3 chance of a 10% chance of the host revealing it. According to your logic, this bias should not affect the probability... But it clearly does. Just build a simulation and you will see it instantly. Or just change the bias to 100%, and it's obvious as hell you picked the car. Does that make clear sense?
Dude, the host never reveals the car after the first door has been chosen. He always reveals a goat. That’s why the suggestion that you should always switch has been the most corroborated solution.
Super simple: your initial odds of picking the wrong door are 2 out of 3. Since the host knows where the prize is and, by definition, must remain fair and neutral, the next door he removes from your choices can't be one that hides the prize. After this new information is available, since you would have initially chosen the wrong door 66.6% of the time and now only two doors remain, switching will make you win 66.6% of the time.
@@failurefiend If there is a 2/3 chance of the firs pick being a goat, that means there is a 2/3 chance the host will have the car. There is no 50/50 in this puzzle.
@@failurefiendnope. The other door does not have the same odds as the door you picked. Test it with a deck of cards and tell me how often you win by staying.
@@failurefiend Let me just explain this in (kinda) simple terms. (I am just trying to explain how to get the car initially.) There is a 2/3 chance that you choose a door with a goat behind it, being that there are 2 goats and only one car. That statistic does not change after the host opens a door with a goat behind it, so it is mathematically impossible that you have a 50/50 chance of choosing either the goat or the car. This means you have a 66.6% chance of getting the goat if you stay, but have a 66.6% chance of getting the car if you switch. (Image credit to By Rick Block - en.wikipedia.org/wiki/File:Monty-RightCarSwitch.svg, Public Domain, commons.wikimedia.org/w/index.php?curid=68521184) upload.wikimedia.org/wikipedia/commons/thumb/4/41/Monty_Hall_Problem_-_Standard_probabilities.svg/1024px-Monty_Hall_Problem_-_Standard_probabilities.svg.png
@user-vn1lz4cq2r Yes you are giving a good description of how one can think about the actual intended problem as with for instance the much better prisoner dilemma version of it. But just read how specific and advanced you must clarify the assumptions behind, as they never are in the phrasing in the Parade Magazine. You wrote "Since the host knows where the prize is and, by definition, must remain fair and neutral, the next door he removes from your choices can't be one that hides the prize." and although we can certainly choose to make that assumption, we really shouldn't in general. Since the scenario where the paradox emerges is completely dependent of that (and that he ALWAYS opens a door), those things ought to have been explicitly explained in the question, not just vaguely implied as they are the very cornerstone for the paradox in the first place. Remove any of those two and the answer is not 2/3 for switching and 1/3 for staying. That is the entire point which is actually by far the biggest takeaway from this problem. Wee need to make a busload of assumptions that are not explicitly stated in the question to arrive at the situation where the 2/3 vs 1/3 situation occurs. How do we know that the host cannot open a door with a car behind it? We don't. If he cannot, it should have been stated. It never is, it's only described how he acted in the instance "We are supposed to be at". We must for some very bizarre reason assume that in order to be able to give a concise answer. How do we know the host always chooses a door to open giving you this option and or if he doesn't always, makes this decision independent of what is behind the door the contestant chose. We don't know either, but one of these criteria needs to be informed for the paradox to emerge. But neither are informed. Again, we need to somehow assume the former or the latter for the paradox to emerge. The way the question was phrased in Parade Magazine is a good example of "Ask a vague question, get a wide variety of answers"
@@awal7664 In that case by switching you double your chances of winning from 1/3 to 2/3. It's only 50/50 if the host opens a random door (could show the car but happens to show a goat).
You're assigning probabilities to events which cannot ever happen. The comparison with the coin toss is not accurate, because we know beforehand that the presenter will never reveal a car. That's not altering probabilities after observation, because we already knew it before the observation.
No I am not, you watched half the video. I explain at length how this information is not included in the original problem, and can not be figured out either based on the information given.
@@dataandcolours You don't understand. There is looking at this as a game vs. looking at this as a puzzle. It said the host, who knows what's behind the doors, opens another door that has a goat... Yes? It's no puzzle if he can reveal the car because it's understood the host must be allowed to switch, so standard assumptions must be followed. However, it's no game then because the contestant could just figure out to switch even before the game starts. To preserve the problem as a game, rather than a puzzle, Monty can and must be allowed to do anything that is not expressly prohibited by the rules. The letter said suppose you are on a game show ... Logical Morality is approaching the problem as if it were a real game, rather than a puzzle.
@@healthquest4823 You are undertaking some sort of Hercule Poirot investigation on how to interpret it so it can become a puzzle, instead of just reading what it actually says. This question as stated in Parade Magazine is legendarily poor, there is no way around it. Instead of trying to replicate the Three Prisoner Dilemma, they made it into mostly a misunderstanding of the conditions issue with the horrible language used. There are three further conditions that all needs to be established for the 2/3 vs 1/3 scenario to occur. It's not even close.
@@dataandcolours I simply see it as how you approach the problem as written in the letter. As a puzzle, the so called assumptions are not even really assumptions at all but necessary to preserve it as a puzzle. The host simply must not reveal the car, so saying he knows what's behind the doors isn't even necessary. To preserve the narrative as a probability puzzle, the contestant must also be allowed to switch. However, the rather ridiculous thing here is if the host has knowledge of where everything is, then it automatically becomes a non-random process, eliminating the supposed intuitive 50:50 view from even being a possibility. The question is really only which door will provide the greater chance of winning, not IF there is any reason to bother switching. The other view is to literally just read it as is - a serious question about a theoretical game. In that case, Monty Hall is free to do anything not expressly prohibited in the narrative. Then the 2/3 probability cannot be assessed because he didn't have to do anything given in the narrative beyond the initial set up, giving the contestant his or her a 1/3 chance to win. There simply is no solution because it is non-random. The problem with the video is that LM declared the actual answer to be 50:50, which is not correct.
Yes I agree that it should be stated that the host will ALWAYS open a door with a goat. If this is the case then yes you should switch. I think it's not mentioned because most people will just assume that to be the case - cuz - well -- that's how the game shows operate. If he opened a door with the car its not as exciting then is it?
That rule is actually stated in the setup but this guy here ignores the rule and somehow says there is a chance he randomly choices a car. The setup says he picks a goat. So he will never choice the car according to the setup rules.
No it really doesn't make sense that that would happen. Follow this logic... 1) Does the host want you to win or lose? I assume you lose. Ultimately we don't know for sure though... 2) If you picked the car, what should the host do? He should give you the option to switch and reveal a goat. Your odds now decrease from 100% win, to something less... 3) If you pick a goat what should the host do? He should reveal the car of course. That way you have no possibility to win whether you switch doors or stick with the same door. Show a flaw in that logic...
@@logicalmorality4646 It’s difficult to find a flaw in that logic because it’s not even logical. You aren’t using logic. You are using gut feelings. Also, your gut isn’t giving you good advice. The host isn’t looking for you to lose. He’s looking to make the show exciting. He needs to build the anticipation. Sure, if you pick the car initially he could simply reveal but he won’t because that is not building drama. Not only are you not good at logic, you don’t have very good instincts either.
I made two python scripts. The first is as the MHP is played. ie. you pick a door. Monty then reveals a goat. Guess what? You get the car 2/3 of the time if you swap in that scenario. Why? Because your initial odds are 1/3 of choosing the car and there's 2/3 chances it's behind one of the other 2 doors. Monty then shows you a goat which now means that last door has 2/3 odds of being the car. Why? Because you are opening TWO doors, not one. This means 2/3 of the time it's best to pick TWO doors. In the second script I played it blind. ie. you choose a door, and then swap it randomly for one of the other doors. Guess what? You only win 1/3 of the time. Why? Because you're only choosing ONE door. What Monty is essentially doing is offering you the choice of one chance to win, or two chances. When you swap you are always swapping your one door, for the other two doors. Of course the odds are better because two chances are better than one.
After watching you last night, my son (PhD in Statistics) and I ran a 1000 cycle program in R simulating the Monty Hall Paradox. It did show the odds of winning by switching your choice is approximately 66%, not 50%.
That depends entirely on how you set it up. If you systematically avoid the car, you'll get 66% for switching. If you open the doors randomly, you'll get 50% for switching.
@@niemand7811 No. That's true only if you reveal the goat randomly, which means you will also reveal the car 33% of the time. If you don't want to reveal the car so avoid doing so by using your knowledge of "what's behind the doors," then the odds for the contestant increase to 66% by switching.
You pick a door. the chance you got it correct is 1/3. It's more likely that the car is behind the doors you didn't pick because it's 2 to 1. This is fairly obvious. By switching you are basically choosing both of the other doors. When you think of it this way it's just plain obvious.
I watched your video, and after carefully considering it, I'd like to respectfully give a rebuttal, and would love to hear your response. Your key argument (at 6:15) is that "the problem is leaving out a key assumption: the host is not opening doors randomly, he is systematically avoiding the car". So here's the counter argument: the problem says "the host opens another door...which has a goat". So to be clear, the host has opened a door with a goat, and that is stated in the problem. Whether or not (as you say) "he is systematically avoiding the car" is irrelevant. The context of the problem is framed as a single game. It's not a sampling of multiple games. It doesn't matter if the host "could have" or "could not have" done something in the past, because the host has already opened a door with a goat. It doesn't matter why they opened the door with a goat either: they very well could have opened the door with a car but that is irrelevant because they didn't. In this situation, switching is better.
Sure! Yeah I can't believe how emotional people get on this. Maybe I am right maybe I am wrong hahaha. Let's discuss it. I had a very very long debate with Calvin over this, eventually he made a simulation to test it. But the short answer is, yes it does matter if he could have. The host has revealed a goat. Now we must ask, what are all possible paths to this scenario occurring? "How could we be in the situation we find ourselves in?" is a good way to phrase it... path 1) you select car (1/3) host reveals goat A (1/2) path 2) you select car (1/3) host reveals goat B (1/2) path 3) you select goat A (1/3)host reveals goat B (1/2) path 4) you select goat B (1/3) and host reveals goat A (1/2) What we are really comparing are the odds of path 1+2 vs path 3+4... If the host could reveal a car, then the above odds are correct. However if the host must avoid the car, then in path 3, and 4, the odds change from 1/2 to 2/2. Thus 3+4 becomes more likely than 1+2
"However if the host must avoid the car" The host already avoided the car, so yes by definition, the host must avoid the car. This is not a statistical sampling of games, it's a theoretical single-instance scenario where you've already sat down, chosen a first door, and the host has already shown a goat. Let's add up the odds of your paths: path 1) 1/6 path 2) 1/6 path 3) 1/3 path 4) 1/3 So: - There's a 2/3 chance you're in path 3 or path 4. I.e., 2/3 chance you and the host both chose a goat - If you both chose a goat, the remaining door is a car. But this is all overcomplicating a much simpler way to explain it: "your first choice was probably (2/3) a goat, and they've revealed the only other goat, so the remaining door is probably a car!" (P.S. your step 3 and step 4 say "1/2" where they should say "100%", because if you chose a goat then there is 100% chance the host will reveal the other goat.)
@@jay31415 No the system the host uses clearly changes the odds. Suppose you were dealt the 5 of spades randomly... What are the odds of that happening? 1/52. Now suppose you learn the dealer won't deal red cards... Well now it's 1/26. Despite having already been dealt the 5 of spades, we have to update our calculation. The probability itself doesn't change, but now our knowledge of it has changed and is more accurate and closer to the real probability function. Thus we have to modify our calculation.
@Logical Morality But, the host already dealt a 5 of spades. So, if we update our calculation as you say, then we derive that the odds of this happening is 100%, because it already happened. What is the chance it will rain today? Where I'm at, 100% , because it already rained this morning. Rain tomorrow, I do not know, maybe 50%. But when after tomorrow, it will collapse to either 100% or 0%.
Too many people complicate this. Here is the simplest proof: - Your first choice is probably (66%) a goat. - The host reveals the other goat (Fact. 100% chance that he does this.) - Thus, the remaining door is probably (66%) not a goat.
Nah, this dude is wrong. The Monty Hall Problem is solved already. You should switch. Switching gets you the opposite of what you first picked, since there are only 2 doors left. 2 out of 3 times, you picked a goat initially. So 2 out of 3 times, switching doors means you go from goat to car.
@@jacobinozz The original poorly worded problem does not produce a 2/3 distribution. Only when the following is established do we get 2/3... 1) The host always reveals exactly 1 door after you select yours, this door can not be your door, and if there is a car behind it he will not open it. 2) The goats/cars are randomized 3) You want to win the car I think assumptions #2, and #3 are fair enough... Sure... But #1 most definitely isn't. These are very specific rules that only belong to this very specific game. So the question is, were these rules established in the problem or not? Clearly not. There is no possible way in hell you can get rule #1 and the 3 critical conditions it states to build a 2/3 probability based on the wording of the original problem.
It all condenses down to two concealed choices. A booby prize and valued prize . The probability in a randomised situation is 50/50 . It's always 50/50 . Hocus pocus and mumbo Jumbo has no effect.
@@philip5940 nope. If the host knowingly opens a door with a goat - as opposed to randomly - you double your odds by switching. There are tons of simulators and explanations on the internet that can help you understand it.
If you use cards and just randomly pick 1, 2, or 3 it is close to 70% better chance of winning if you switch. I didn't believe it either until I tried it over and over again.
I wanted to mention that I actually played this with my niece. I took three of her plastic toy French fries and placed a dot on one. I placed them on a glass table with the dot down and asked her to scramble them around, then look under the table to see where the dot was. I then picked one and moved it to the side without looking at it. Next I told her to flip one of the remaining two over that didn't have the dot. Now I could decide to switch the one I first chose with the one left. We did this over and over, and I kept track of the results on a nearby whiteboard. The times I switched clearly showed a 2/3 advantage. When I looked at what I chose first, I only won 1 in 3 times. The thing is, there was no other practical way to do this. Saying well there was nothing in the letter that excluded showing the car, as you pointed out, means my niece could have exposed the French fry with the dot if she had wanted, at times when she had it. What was I to tell her..."You can just go ahead and reveal the dot every now and then"? That would just have made it impossible for me to have the opportunity to switch during those times. So saying Savant was wrong is really stretching things. To get any particular answer to the question in the letter, you must follow her assumptions.
For a coin it's an indisputable 50/50 flip probability. Throw a coin a 100 times and you'll get results of around ⅔ & ⅓ fairly often. Throw it a thousand times and it might start to level at 50/50 . Maybe could throw a thousand coins up all at once to speed things up . Maybe..
@Philip I just did it, and it only took 16 flips to get fifty-fifty results. Heads pulled ahead at first with 7 heads to only 1 tail. Then tails started coming up more often, becoming equal at flip 16. Any flip of a coin is 50:50...period. The first two flips could just as easily been 50:50.
@Philip A coin has only two sides. The odds are factually 50:50. How soon that bears itself out could be from the very beginning of any serious of flips. The outcome does not dictate the probability function. The probability function dictates the outcome. You are confusing ideas. If you are trying to determine odds, then you need a large sampling. That's true. In this case, we already know the odds. It would be extremely unlikely that flipping a coin 100 times would result in a 2:3 outcome. That would be directly counter to the predicted odds.
This becomes subjective when one discusses the host's strategy. Unless the strategy is clearly stated factually in the rules, conclusions cannot be drawn either way, however the common interpretation is that the host deliberately avoids eliminating the car to take the game to the next phase of uncertainty. That seems reasonable to me, otherwise there would be no point in offering a second chance to pick/switch a door. Traditionally if your last selection is random, then your odds are in fact 50/50, however this does not change the odds of switching Vs not switching which is 2/3 and 1/3 respectively. Thus there are actually 3 scenarios, however if you count doing nothing at all, then your odds are 0% You always initially had a 1/3 chance of selecting the car and 2/3 chance of selecting goat. Thus by switching doors you're more likely to end up with the car, because the door you picked first up, was always more likely to be the goat and with one door and goat removed, the only remaining switch option is the car 2/3 of the time and the goat 1/3 of the time. Thus a random selection between the last two doors ( 50%), offers better odds than not switching (33.33%) at all, but not as good odds as switching (66.66%).
I argued with a close friend on this problem for several hours one time lol........ I told him I would never accept that switching my pick every time improved my chances. Its a clever illusion and nothing more
Did Monty ever reveal the "car" after the contestant picked? If he did, why would the contestant continue playing the game? I never watched the game show.
This was not a real segment on the game show... it is a math puzzle that was named for the host of a show because they use a similar format to describe it. The rules of the MHP say he will always open a door with a goat but you are correct... if he opened the door with the car, the game would have to be over and the player lost.
in probability world revealing a car is present but in the game he eliminated this probability which made you and many not understanding the outcome is 50 50
Stating that the host _always_ reveals a door, _never_ reveals the car, and _never_ reveals the contents of the constant's door is, indeed, enough to guarantee a 2/3 probability for winning by switching. But often, it's not stated in such a way that we can deduce that these always and never statements are true.
@@MuffinsAPlenty The reason it doesn't have to be stated outright is because the question is just about one instance of the game... not multiple instances. We are not told the rules and we don't need them because we are told what happened. Contestant picked a door, host knowingly opened a different door to reveal a goat and offered a switch to the remaining door. In the game described, the host did in fact follow the rules which you described. It is only when we want to consider multiple instances of the game that we need to actually state those rules so that each new game will also follow them. Multiple games are usually used in explanations so that is where the rules are discussed but the problem works fine without listing them.
@@Hank254 I don't know that the "this only happens once" argument makes much sense. If we are having an "already happened" perspective, then the concept of probability breaks down because whether switching wins has already been determined, and is not up to random chance. From that perspective, the probability of winning by switching is "either 0 or 1, but we don't have enough information to determine which is true". This is the reason that frequentist statistics uses such unnatural reasoning in concepts like confidence intervals and hypothesis testing. You can't take the probability of a parameter because the parameter is fixed value, not a random variable. I have heard someone before say that for events which have already happened, we shouldn't use the word "probability" but instead "confidence". How much confidence would the contestant have that switching wins based on the limited information they have? But even then, if you want to remove the "frequency" words like "always" and "never", you can rephrase my above reply as being in terms of the algorithm that the host employed in this specific instance when deciding which door to reveal. That algorithm only needs to be employed once. But the host many have employed an algorithm that makes it impossible for the contestant to win by switching. Or the host may have employed an algorithm that makes it impossible for the contestant to lose by switching. Or the host may have used the traditional Monty Hall Problem algorithm. Or may have just randomly opened a door. The contestant has no way of knowing which algorithm was employed and has no way of knowing how likely any of these algorithms were to be employed. So I don't know that the confidence perspective makes much sense either.
he's correct. You're told one scenario. The host can do something else with another contestant for all we know. Given what you are told, which is that the host is not selecting a losing door randomly but purposely and then offering you a switch, you double your odds by switching
@@gnlout7403 The most important aspect of Monty Hall-like scenarios is how the host decides which door to reveal. If you know how the host decides, then you can figure out whether switching is advantageous or not. If you don't know how the host decides, then you can't determine whether it is advantageous to switch or not. Whether the scenario happens once or a million times, the analysis remains the same. The law of large numbers tells us this. But the critical thing is that we _have to know how the host decided which door to reveal_ in order to come to a definitive conclusion. Saying that probability magically changes how it behaves because the scenario only plays out once is wrong.
Sadly I must admit, watching this is what convinced me the Savant solution is true. You have a 2/3 probability of picking a goat. When a choice is removed , you still have a 2/3 chance that your choice is a goat. Inversely, the only other door has a 1/3 chance of being a goat.
You are making the mistake of treating both goats as the same variable. Its two different goats though, goat A and goat B. You have a 1/3 chance of choosing goat A, a 1/3 chance of picking goat B, and a 1/3 chance of picking a car. Regard,es of what you picked there is a 100% chance either goat A or goat B will be removed as an option. If goat A is shown there is still only a 1/3 chance you picked goat B and If goat B is shown there is still only a 1/3 chance you picked goat A. Just as well you also had a 1/3 chance of having picked the car. No matter your first choice you are swapping out a 1/3 odds for another 1/3 odds from the initial pool of options. You are looking at is as there is a 2/3 chance you didn’t pick the car so inversely changing you mind is favorable. However there is also a 2/3 chance you didn’t pick goat A when goat B is revealed, in which case changing your mind is unfavorable. There is also a 2/3 chance you didn’t pick goat B when goat A is revealed, in which changing your mind is unfavorable. If changing your mind has just as much of a chance to be good(2/3) as it does to be bad(2/3) then it is fifth fifty odds. The real key to the confusion is assuming that the revealed goat, A or B, is still influencing the odds of the second opportunity to pick a car vs goat A or the car vs goat B. The reality is that goat A and B have nothing to do with one another, they are separate variables, and in the second chance one of these variables are removed making it a 1/2 chance you get a car.
@@robertdicke7249 that's not correct, and you can test this problem yourself. The actual game scenarios demonstrate this. - You. Prize Door. Don’t. Do 1. 1. Win. Lose 1. 2. Lose. L Win 1. 3. Lose. Win 2. 1. Lose. Win 2. 2. Win. Lose 2. 3. Lose. Win 3. 1. Lose. Win 3. 2. Lose. Win 3. 3. Win. Lose Scenario 1: You pick door 1 and the prize is actually behind door 1. In this case, Monty will open either door 2 or 3 and show you that nothing is behind one of those doors. If you stay with door 1, you win. Scenario 2: You pick door 1 and the prize is actually behind door 2. In this case, Monty must open door 3 and show you that nothing is behind it. If you stay with door 1, you lose. Scenario 3: You pick door 1 and the prize is actually behind door 3. In this case, Monty must open door 2 and show you that nothing is behind it. If you stay with door 1, you lose.
@@robertdicke7249 Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car you get the car. You only win 1 out of 3 games if you stay with your first pick. Basic math/logic kids understand, idiot among idiots doesn't.
The way I see... it's true that after playing the same game over and over for a ten times or more the probabilities of getting the car after switching doors increases..BUT...that doesn't mean that every single time I play the game would be more beneficial to switch...it all depends on if you had chosen the wrong or the right door at the beginning...after eliminating one pf the door..the chances of getting a car by switching it doesn't increase it's the same.
You don't need to play multiple games in order for it to double your odds by switching. One game and you still double your odds as long as the host opens a door with a goat purposely.
The problem does specifically say that the host will open a door that reveals a goat. He’ll reveal a goat every time. If you refer to the show itself, Monty hall never revealed the desirable prize. If he chose the door at random, your analysis would make sense, except that, if Monty opened the door with the car, it wouldn’t make much sense for him to then ask you if you’d like to change your choice.
It's stated in the problem that he reveals a door with a goat. And like they said before me: It wouldn't make amy sense to ask you if you want to change your choice if the prize had already been revealed.
The explaination is quiet simple : Only one car and TWO goats. So : 1/3 : you pick the car, then you must stay to win. 2/3 : you pick goat 1 or goat 2, then you must switch to win. Saying that the final choice is a 50/50 chance means that your initial choice is 1/2 to pick the car and 1/2 to pick a goat, which is mathematically wrong. Then : ALWAYS SWITCH. CQFD as we say in France :)
What the world has missed was: 1. Initially, all doors had 1/3 probability. By picking door one, the 2/3 probability favored doors 2+3. That's accurate. 2. With an important paradigm shift (revealing that 3 is a dud), you now have 'new' information to factor into re-calculating probability. 3. The probability of the car being behind doors 1 and 2, becomes 50-50 (1/2 - 1/2). Marilyn Vos Savant, with an IQ of 228, got it wrong.
This is very simple, if the 3rd choice is wrong, then disregard it(remove it completely), the you only have 2 choices with 50/50 chances. Monty hall only works on program simulation.
Incorrect. you double your odds by switching in the given scenario which is that the host knows what’s behind the doors and chooses to open a door with a goat. now, if the host did not know what was behind the doors and open the door with a goat, the two remaining doors would indeed be 50-50.
Not gonna lie, this is a pretty lame and clickbait video. It is obvious that the host will not choose a random door, risking to reveal the prize. This entire video basically comes down to: ""The question was wrongly phrased", even though in many videos (by the more popular math channels) it was stated that the host will always reveal a goat. Like others said, of course the host will not reveal the prize if you have chosen a door with a goat. Therefore you have a 2/3 chance to choose a door with a goat, forcing the host to reveal the other goat and thus you also have a 2/3 chance to win the prize if you switch.
Uhh no actually in math you are supposed to approach things logically instead of like an idiot. For example if I tell you "I deal you 2 cards, and one is the ace of spades, what is the other?".... It does not logically follow that the other card is the queen of diamonds because we are playing "English Poker Queen", and that the host is cheating and wants you to lose, and that is the best strategy in doing so... You would need to be told that information in order to know that... Yes it's obvious what the rules of Monty Hall are... AFTER BEING TOLD THEM...
@@logicalmorality4646 you are told that the host will always reveal a goat as he is not allowed to reveal a car. Did you even watch the original show? Please understand that the host is not allowed to reveal the car. It is explicitly stated before the round. The source you're pulling this from is a little ambiguous with it's phrasing, however the original gameshow where monty hall hosted this called "let's make a deal" always included the part where the host isn't allowed to reveal the car in all the publications they've ever released.
Hey I was wondering what your answer to this problem is:There are three bags, each containing two marbles. Bag A contains two white marbles, Bag B contains two black marbles, and Bag C contains one white marble and one black marble. You pick a random bag and take out one marble, which is white. What is the probability that the remaining marble from the same bag is also white?
Oops sorry on closer inspection I think it's 2/3. 1/6 Pick bag B, draw marble 1 (remainder is white) 1/6 Pick bag B, draw marble 2 (remainder is white) 1/6 Pick bag C, draw marble 1 (remainder is black) I think that makes sense Then the rest of the scenarios we can ignore, and we can see 2/3 of these cases are white/white
Have you tried chat GTP? It's really good at mathematical and scientific logic. It's super good at this kind of stuff, I would ask it, it's like a calculator basically but can speak English. It's really bad at philosophy and politics though
You chose a weird hill to die on. You are right that the original question is not phrased well. But we know what the intention of the question was (maybe not right away but when you hear an explanation - you even present the intended interpretation in the video). You can complain all you want about the phrasing but this has nothing to do with "debunking". Also, if you want to be this pedantic about the phrasing of the question (which isn't a bad thing) then you should be consistent. You are basically complaining that people are interpreting something into the question that isn't explicitly written down and then you do the exact same thing by assuming that he choses the door randomly. Nothing in the question tells you this. This interpretation is even discouraged by stating that the host knows what's behind the doors. There could be a complex rule. It could be based on mood. There could even be situatons where he doesn't open a door at all.
In probability we assume that randomness is a default unless stated otherwise. If it's non-random then we simply just don't have enough info to make any calculation as there is a near infinite amount of non-random ways he could select. Given the information we have we can answer it for random selection, or just say the problem requires more info.
@@klaus7443 Mr.Mature is back! The sad thing is I know you aren't actually a child. I know you are at least... Let's say 18. No actual child can be this immature. What happens with humans is that mature children become mature over time, but immature children actually regress and grow in their malevolence, irrationality, and ignorance. To reach a level of immaturity this high it takes years of practice.
So, I’m thinking about this a lot. Now, I think sometimes just in concepts, without words in my mind, I know is weird, but it is like that. I don’t know how to explain it, but I have something that’s telling me it’s wrong that opinion with ‘better always switching because 33,(3) to 66,(6)’. I’m not a genius by others ‘definition and opinion(s)’, I took it like this: if after the third option’s elimination, we have 2 choices and one is double (by probability) than the other, why it’s not 25% to 75%, or 20% to 80% also 10% to 90%? How we count this, by having 2 options? How we arrive to 33,(3)-66,(6) and not to 25-75, those should be equal, but aren’t (33,3…=25 and 66,6…=75).This is a long comment 😁🫤. My point is, could be 3 options or 100, in the end is the same result, having the possibility to choose from 2 (and my mind tells me that is 50/50). The next question was for me: How probability works? What if I have 100 options divided in 2 groups of 50 options. And I must to choose one group, where I ‘think’ the price is. Now, let’s consider that we eliminate 98 of those options (the wrong ones, of course)obviously we eliminate 49 from each group, how we calculate the probability to find the price? I mean, what is the probability to choose the write group? Most ‘terrifying’ question, what is the probability in each group that if we choose a single option, that will be the price, considering that we don’t know in which group it is? 🤯🫠 Sorry for the long text, I still didn’t explain what I meant completely, because it will be too long. 😁😬 Finally, I have this intuition that is still 50/50 chances, because, in fact there is that possibility to choose too keep the first choice or not, from 2 options, because one wrong is gone. 🙃
All good, hahaha. Sorry I don't quite fully understand what you're saying. But in summary... -It depends on the rules of the game -If host must not reveal car, must reveal exactly 1 door each time after you select, and the door must not be your door... Then the odds become 2/3 -In original problem none of these rules were stated, so the answer is 50/50 by default or unknowable.
no. You're not told rules of a game, only what the host did this one time. Under these conditions, which is knowledge of the doors and the purposeful opening of a single losing door from the remaining two, you double your odds by switching. What he did before or does after with another contestant is not relevant to this scenario.
Or if only the rules of the MHP stated that the host can't show a car... _"Version 1 (Classic Monty). You are a player on a game show and are shown_ _three identical doors. Behind one is a car, behind the other two are goats._ _Monty Hall, the host of the show, asks you to choose one of the doors. You_ _do so, but you do not open your chosen door. Monty, who knows where the_ _car is, now opens one of the doors. He chooses his door in accordance with_ _the following rules:_ _1. Monty _*_always opens a door that conceals a goat._* _2. Monty never opens the door you initially chose._ _3. If Monty can open more than one door without violating rules one and_ _two, then he chooses his door randomly._ _After Monty opens his door, he gives you the options of sticking with your_ _original choice, or switching to the other unopened door. What should you_ _do to maximize your chances of winning the car?_
Usually this is "verified" by using a certain algorithm where the host always opens one of the two remaining doors with a goat behind it. You could well write a program that randomly opens one of the remaining to doors. Of all the cases that look like the MH problem (discounting the ones where the door with the car was opened) switching doors wins 50% of the time. You could also write a program where a second door is only opened if your first choice was the door with the car. In that case switching doors never wins in cases that look like MH. I will refrain from providing the programs, as the math is pretty clear.
@@Stubbari your problem description is much more accurate than the original. I like it. I would also add "the contestant knows / you know that these rules are applied in your game".
@@insignificantfool8592 except whenever it's actually tested, it's one third if you stay and two thirds if you switch. But there are simulators online if you Google them. Check out the Mythbusters episode of them testing it. But if you know you only pick the winning door one in three times with your initial pick (we would agree on this), then you'd only win one in three games if you stick, logically. No test necessary, but if you disagree, you would test it
@@gnlout7403 I explained three algorithms. The first one gets the 1/3, 2/3 result, the second 1/2, 1/2 and the third one 1 , 0. If you wish we can play a few rounds of the game where winning is 50/50 if you switch. This hash value is good for up to ten games: d09eaa8b6cbae2cf74b92bcd4c8b72e4 There are three doors. What's your pick (123)
Every time goat is picked the last door will have to be car. Since there's a 2 in 3 chance of picking goat, that's the chance of switching every time bringing you the car.
Not if the host reveals randomly. If he reveals doors randomly and your strategy is to always switch then... 1/6 you pick car host reveals goat 1 (loss) 1/6 you pick car host reveals goat 2 (loss) 1/6 you pick goat 1 host reveals car (neutral) 1/6 you pick goat 2 host reveals car (neutral) 1/6 you pick goat 1 host reveals goat 2 (win) 1/6 you pick goat 2 host reveals goat 1 (win)
@@logicalmorality4646 If the the host reveals a car. The door that is picked in the beginning and the the remaining door is automatically 0/6 or 0%. Which why is we don't ever consider the Host revealing the car. Sticking to the the restrictions of the question "and the host, who knows what is behind the doors, opens another door, say No. 3, which has a goat" "another door" meaning, the host will never pick the door the player has selected "say No. 3, which has a goat" the host has chosen Door No. 3 and a goat is there
@@goonin3590 No it isn't that makes no sense. If it's possible for him to reveal the car (and there is no rule saying it's not) then it certainly does factor into our math. It makes it half as likely that you would pick a goat, and he would reveal a goat.
@@logicalmorality4646 No, because it is part of the game and if he has done this bit with the car 50 other times or whatever and the contestant knows the setup is always to reveal a goat from the other 2, he will realize it’s not random.
the presentation of your video sucks and your math is wrong. your notation is wrong: 1/6 should be placed behind the tree branches, not at the second stage of the tree, since this place is reserved for the branch probability which is 1/2. of course, probability is modelled due to certain circumstances, for example the described game. in the reality of the game, there is no case "car is revealed by host" and therefore, the probability of this event is zero, thus plays no role.
There are 2 possible scenarios, one where you pick a car and one where you pick a goat. If you pick the car you have a win if you stay, if you pick a goat, you win if you swap. You naturally have a 2/3 chance of picking the goat so to account for that you’re better off deciding to swap. That’s all it is. A lot of people overcomplicate this to a level that’s just straight up unnecessary and makes it look more questionable than it is. There are 2 scenarios, one scenario is more likely than the other, so account for the more likely scenario. That’s it, and what is the most likely scenario? Picking a goat and having a goat revealed. That’s why you should always swap.
@@skrypto2652 sure. But I think you have to learn to differentiate nonsensical from plausible scenarios. How many game shows do you know where the host keeps the prize for himself? I don't know any. How many game shows do you know where the contestant picks something and gets what he picked. That's literally all of them.
You brought up an impossible scenario, the host opening your chosen door, that’s not how the game works and if you watched the video or read my comment properly you’d know that. The way it works is simple enough for you to understand so try, you choose one out 3 options, and here’s the part you missed, the host then reveals another door that is a goat(he always reveals a goat because he knows what’s behind all the doors, it’s not random) he then asks if you want to stick with your original choice or switch to the other door that hasn’t been revealed for both of which you still don’t know what it is, that’s kind of why your asked if you want to switch or not genius.
@@pwcinla I think you must have missed the point of the video. The way the problem was phrased in Parade Magazine, the alleged paradox doesn't even emerge as you need to do a busload of additional assumptions that was not stated in the question. It's only when you clearly establishes how the host always have to act (and in a very specific way) that the paradox emerges in the first place. So the Parade Magazine article appears to actually have created a Monty Hall Paradox Paradox judging by how many that see to be confused by it.
@@dataandcoloursthe host doesn't 'always' have to act any way. You are given a scenario in which the host knows what's behind the doors and chooses to open a door with a goat. In this specific scenario, do you double your odds by switching? Yes, you do.
@@gnlout7403 You may say so. But if the host can act any way he wants it becomes impossible to calculate the conditional probability and the question instantly becomes obviously impossible to answer. Why did he "choose" to open a door with a goat? In the "specific" scenario you give as an example it is obviously impossible to calculate the conditional probability, so the answer would be "Impossible to give an concise answer". Again you are scratching of what I like to call the Monty Hall Paradox Paradox: The way Parade Magazine presented it and consequently argued have actually created a lot of misconceptions that is not even about the intended paradox in the first place, but amongst others the misconception that it would not matter what rationale/reason the host had for doing what he did. From that perspective the entire circus has probably unfortunately had a net negative effect on education, especially if it's brought up in a school scenario were it steals time from other more useful scenarios.
@@dataandcolours he made a mistake when doing the video which is why he's getting flack. First, you're not told what the host is going to do in future games. You're only given one specific scenario. You're also not told his motivations, so you shouldn't add them into the scenario. Logical Morality ran the problem as if the host could, in the future, open a door with car. That's different from the one scenario you are given.
Based on everything staying in fixed positions behind doors and only playing odds the door you pick will have you at 33.3 chance to win car or 66.6 you will not win the car. Monty knowing what’s behind the doors opens one with goat. The mistake is thinking down to two doors you are down to 50/50 odds. Actually you are in exactly the same position as you started 33.3/66.6. Giving me the chance to give up my 33.3 door for a 66.6 door will be done in a second.
if monty can only open one door, that means you have a 50% chance. istg people purposefully don't know how to present this problem and are then somehow shocked when someone with a background in math says it's 50%. the way the problem is presented, it's 50%.
In the Monty Hall problem, the host is guaranteed to choose a door with a goat behind it. He cannot choose the door you chose. 2/3 of the time, you guess wrong, by switching, 2/3 of the time, you get a car. You misread the Monty Hall problem and failed to debunk it. Remove this video from youtube.
Say there is a got and a sheep instead of two goats, and you pick the door with the car. There is two different scenarios under this condition. The host revealing the goat and the host revealing the sheep. Which means there’s is still just a 2/4 chance once one door is revealed.
@@Noname61574 as long as the host must open a door with a goat, you double your odds by switching. This dude made this video not understanding this point, but he 'corrects' it later in the comments. Here are the scanarios - You. Prize Door. Don’t. Do 1. 1. Win. Lose 1. 2. Lose. Win 1. 3. Lose. Win 2. 1. Lose. Win 2. 2. Win. Lose 2. 3. Lose. Win 3. 1. Lose. Win 3. 2. Lose. Win 3. 3. Win. Lose
@@gnlout7403 making it still a 1/2 chance if the entire premise of this illusion is that one was already revealed before the switch, extremely redundant
@@Obs23456 one door wasn't revealed before the switch. That would produce different odds. the chart is the list of nine possible scenarios and outcomes for switching Vs staying. Do you not agree with the chart? If so, where is it wrong? Then tell me how you are able to pick the winning door with your first pick out of any more than two doors.
Everything about this depends on how you define "The Monty Hall Paradox". There are 2 scenarios that are discussed, and the definition in each scenario makes all the difference and allows 2 answers to be true depending entirely on what problem you are solving. These things are the same in both scenarios: -There are 3 doors -You choose 1 -The host opens 1 of the 2 doors that you did not pick This is where "The Monty Hall Paradox" changes into 2 different math problems. Math problem 1: The host chooses a door that he knows has a goat behind it. Math problem 2: The host randomly chooses one of the two remaining doors. Math problem 1 is the scenario presented on the show. This is evident because one of the options presented to contestants on the show is NOT "Do you want to take the prize behind the door that Monty opens". This is not an option because there is a 100% certainty that Monty is not going to open the door with the car behind it. So no matter what your choice is initially, Monty is going to open a door with a goat behind it. This is a constant. Because of this, your choice from the beginning is always 50/50 because the choice is not between 3 doors, it is between 2. Your choice is irrelevant, because no matter what you choose, what happens next is that a door will be opened that ALWAYS has a goat. In reality, there are only ever 2 choices ON THE SHOW--goat or no goat. "The Monty Hall Paradox" is an awful name for the paradox because the math problem that everyone is solving with the result that you should switch doors is NOT the situation that you are presented with on the actual Monty Hall show. In reality, The Monty Hall Show does not present the circumstances that are used as a starting point in "The Monty Hall Paradox". If you define the scenario as "The host will always pick a goat", then the odds are 50/50.
You started off well acknowledging the math problem, but then still went with 50/50. The reason it's 67/33 is because you can take advantage of the host! 1. You have to point to 1 out of 3 doors. There is 2/3 chance you select a goat. 2. Monty always reveals a second door with a goat. 3.There is a higher chance you both picked doors with goats at this point, because if you selected goat A, he has to show goat B. If you selected goat B, he has to show goat A. This means the last door having the car is most likely. 4. A different way to sum it up ,the only way you lose with a switch is if you switch out the car door in step 1. The chance of having picked a car in step 1 is 1/3. So it's important to know for the math problem you make a choice out if 3 doors BEFORE Monty makes it two. Monty can only remove ONE specific door 2/3 of the time!!
Logical Morality, after going through your video step by step, I see that you negated... and revealed a goat..., as a rule barring the car from being revealed; but you did not do the same for...who knows what's behind the door. This leaves you with a non-random game, where you never have to reveal the car, but could if you wanted to. The only way to achieve 50:50 odd is by revealing the car 1/3 of the time. The only way to prove Marilyn Savant wrong is to be consistent and reject all her standard assumptions, making it a randomized game, that yields 1/3:1/3:1/3 (winning the car, getting the goat, and blowing the game before a switch can be made).
I just have one question. At 2:10 you talk about the probability function and the host's likelihood of picking either the goat or the car. You say even though the host picked a goat he still could have picked the car because nothing in the description says he can't pick the car. But if he knows where everything is, how can he even make a random pick? The is no "could have" anymore that I can imagine. He would always have prior knowledge and would have to intentionally pick either the car or a goat if the contestant happened to get a goat.
It's equivalent to saying "a card dealer has a marked deck of cards how will he deal?". He could choose to shuffle and deal them randomly, or he could deal them non-randomly a million different ways depending on his goals/strategy. Why does having a marked deck necessitate cheating? My argument in the video isn't that he will deal them randomly for sure, it's that he may or may not deal them randomly, there is no information to eliminate either possibility... Additionally if his method is non-random, there is no rational reason to conclude that the only non-random method of selection he could pick is to systematically avoid the car, reveal only 1 door every time, give the option to switch every time, not select your door, etc... Thus we can either answer the question for randomness, or just say it's a stupid and pointless question that lacks key information to answer...
@@logicalmorality4646 But, to get 50-50 odds, we can't leave it up to the host to decided to not cheat because that still makes the game non-random, period. We have to be guaranteed randomness by not allowing the host to know where the goats and the car are.
@@logicalmorality4646 Curt and Hank are the biggest trolls in the world. I have never seen such a thing on any other thread. When the correct reason is explained, most people will just say, "Oh, I get it now." Your video is excellent. Savant clearly was not answering the question in the letter. Even Monty Hall himself wrote a letter of disagreement to Steve Selvin who incorrectly solved the problem in 1975. As Monty explained, if you remove a goat, it just leaves two doors left with 50:50 odds (of course that is only true you could open the contestant's door too). It's just starts over with a new game. Monty went on to explain no game ever was played like that and no game ever had 2/3 odds. Only the rule, the car cannot be revealed, leads us to know the car has a 2/3 chance of being behind the other door since the host must refrain from revealing it two-thirds of the time.
@@healthquest4823 "Only the rule, the car cannot be revealed, leads us to know the car has a 2/3 chance of being behind the other door since the host must refrain from revealing it two-thirds of the time.." That's absolutely hilarious. First you say the car cannot be revealed then you say the host must refrain from revealing it two-thirds of the time....in the SAME sentence yet!
@logicalmorality4646. the decision tree as well as the clip coin example are wrong: At 1:40, you mention the "line 3" (blue circle) is not possible, but has a 1/6 chance. That's wrong. If you choose goat 1, then the host MUST choose goat, therefore the propabilities are 0 (car) and 2/6 (!) (goat). The same argument holds for case 3. Therefore the chance that the host picks a goat is 1/6 + 1/6 + 2/6 + 2/6 = 1 (for sure, the host can always pick a goat), but the probabilities that the remaining door is a car 0 + 2/6 + 2/6 = 2/3. BTW that's the core of the Monty Hall problem. Your decision tree as sketched at 1:400 shows all possibilities when the host can pick ANY door (this means he could choose the car also). This is not Monty Hall. In Monty Hall, two possibilities can not occur, i.e. their probability is 0. If you simple "leave them out", then the decision tree does not sum up to 1 (!), but to 4/6. When calculating correctly, the probabilities are (1/6 + 1/6) + (0 + 2/6) + (0 + 2/6) = 2/6 + 2/6 + 2/6 = 1.
Dude, you're ignoring rules for the host. 1) Host must pick a door that the player did not pick and open it, 2) Host must always pick a door with a goat to open, 3) Host must offer you the choice of switching to the open door. If those rules are not followed then yes, its not to your advantage to switch, but game shows have rules to build excitement and ratings and such.
No you got it backwards. As a puzzle, there must be strict rules the host must follow...or there is no definitive answer. Logical Morality is approaching this as written as a game and can do anything not expressly prohibited in the narrative. There wouldn't be any tension in a game if you followed rules always leading to the same conclusion - just switch to get better odds.
@@healthquest4823 dude,you just said the host must follow strict rules, and I stated the rules. If the rules are ignored then the probability can be anything. Logical Morality is ignoring generally implied rules for this game/puzzle.
@@darkthedrummerful You're starting to understand. The probability indeed CAN be anything then. Yes! That is what you want for a game. If the contestants can figure out what to do - always switch - then it's no good for a game. Even Monty Hall himself wrote a letter saying that no such game was ever played like that, nor could it be. They never give odds to the contestants that in theory could be calculated to improve their chances of winning. A TV game is not a puzzle that is supposed to be figured out. It is a non-random crap shoot, controlled by the people in charge. They can do whatever they want for the sake of making the game fun to watch by changing their own rules anytime as they wish. LM is describing the letter literally as written. The first thing it said was, "Suppose you are on a game show... " Savant could have and should have responded she could not answer the question as written. She then should have stated she can turn it into a probability puzzle, though, by making a few assumptions. In that case, you would improve your chances of winning from 1/3 to 2/3 by switching.
@@darkthedrummerful Did you know that John F. Kennedy's lowest approval rating was 7 points higher than Donald Trump's highest approval rating? LOL 56% to 49% Haha, incredible.
50/50 odds in this scenario are contained to an individualistic instance of the problem, mathematics takes in the percentile chance through more tests than any one person in this position could reasonably be expected to participate in. Meaning that the odds statistically will always be 66.7% if you switch, but the odds of any one person in the experiment, playing the game exactly one time, will always be a 50/50 gamble of being within the 33.3% that lose by switching, or the 66.7% that win by switching...The key word in this problem is statistics. You can't cherry pick a single person's instance when calculating probability.
I think you watched the first half. Given that the host systematically avoids revealing the car, has to reveal 1 door each time, will not reveal your door, etc... Then yes it becomes 2/3. If you play those rules it's 2/3. These rules are not part of the problem though. That's a different problem. What happened in Monty Hall is like if I asked you to list all prime numbers between 1 and 20, and you say "13 is the only correct answer because it's the only prime number between 1-20 that is larger than 12 and smaller than 14"... Well being between 12 and 14 was not part of the problem, that's a different problem. So that answer is in fact incorrect. The 2/3 solution to Monty Hall adds a bunch of rules that are not part of the problem.
@@logicalmorality4646 How many times is the game intended to be played in this context? Once or multiple, if once then the probability only takes place in a single instance meaning in essence that you are only arguing that the contestant or player after having chosen one door and another being revealed to be wrong leaves the contestant's choice as a 50/50 whether they are going to be more likely or not to win. If multiple, then you are not appropriately accounting for the fact that no matter what two ways you look at it, no single instance has any true basis on the scope of probability, probability by its nature requires recurring patterns to bare any meaning. More to the question, did you conduct a large scale mathematical experiment, because by any logic if you did, it is likely that the sheer scope of the problem would show that looking at it from any singular data point renders it irrelevent.
@@AxOwLynx00777 I don't understand what you mean... Whether you flip a coin once or 100000 times the probability of getting a heads is still 50% on any toss if it's a fair coin. Probability doesn't change based on the amount of trials you run... If you are doing a Baysian probability then you are trying to determine the probability by the results, where as normally we are estimating the results by the probability. In neither case does the probability itself change, in Baysian we simply just don't know what the probability is and are gaining evidence to try to uncover it. Yes he built a simulation and ran like 1000 trials or something, you can ask him the specifics. I think you are half reading/half watching everything too.
@@AxOwLynx00777 While it is true that the difference in probabilities is small with only three doors, Logical Morality is absolutely right that even a single data point follows the probability. If you want a more stark demonstration, you can use a deck of cards which would be a 52-door Monty Hall problem. You will clearly see it is not 50/50 when you follow the rules. In fact, if you can pick the 'winner' 1 time out of 20 I would be surprised.
@@logicalmorality4646 You're coin analogy isn't a valid counter argument, it doesn't account for the initial third element, you are correct in stating probability doesn't change only in as much as you don't understand the angle mathematicians approach the problem from and your approach is mathematically incorrect because no matter what way you look at it, the door you are offered to switch to is always going to be more likely to be correct, in which playing this game once means your probability accounts for the one door you choose initially versus the two you don't. Your initial choice from the information given will always be 33.3% likely to be correct and the door that remains for you to switch to accounts that your initial choice meant you were 66.7% likely to be incorrect. Meaning of the 2 doors you don't initially choose it is more likely one of them is correct, so when one of those doors is revealed to be incorrect, all probability of both doors falls to the last door, giving it a 66.7% chance versus all the initial choices of 33.3%. Whether or not you win or lose these probabilities remain the same, you are only "more" likely to win if you switch, not guaranteed, basic law of probability, something being more likely only works if you run the experiment enough times for a pattern to develop, you are pushing the problem to the final stage and not accounting for the third element that is the door that is revealed by the host still had a probability to begin with
Dude, the Monty Hall Problem is not incorrect. You must be misunderstanding something. Suppose you have 100 doors. Your first random choice has a 1/100 chance of winning. Monty then eliminates all but one other door. That one door has a 99/100 chance of winning. I think you're assuming that Monty is just randomly choosing the other door. He's not. If you were to pick the winning door at first, then Monty would pick the other door at random. But if you initially pick a losing door, then Monty will pick the winning door, and now you have to decide to whether you want to switch or not. It's definitely better to switch.
FYI, you made a typo in your comment, you omitted the word "Monty will not not pick the winning door". In fact, you are the one assuming he cannot open a door that contains a car. Also, you are assuming he cannot open the door the contestant chose. Also, you are assuming the host must open a door at all or that his propensity to choose to do so is independent on what is behind the door the contestant chose. None of this information is given and that is why the Monty Hall Paradox, at least as stated in Parade Magazine, is embarrassingly ambiguous at best, flat-out wrong if being frank. Your reasoning is all good, but again you are the one making a busload of assumptions that simply aren't stated in the question. I am aware that people often makes all this assumptions as that is the only way to actually make the paradox emerge in the first place. But in reality that is a very bad habit and few sufficiently points out how bad the initial phrasing actually is. This causes a weird scenario where it becomes hard to distinguish between who misinterpreted the initial conditions or who actually fell for the paradox itself.
@@NDAsDontCoverIllegalActs I think you're not the understanding the problem. Nobody is suggesting starting with two doors. The point of changing the problem to a large number of doors is to demonstrate the absurdity of not switching since your odds of picking the prize and not switching are incredibly small, the larger the number of doors.
@@NDAsDontCoverIllegalActs maybe I'll ask the question again. How often do you pick the ace of diamonds from a deck of cards? Since you object to the larger number, use three cards. One's the ace of diamonds. How often do you pick it? That's your answer as to both why you switch and why this game differs from a two door problem.
I understand your argument, but i think the assumptions are valid. Your goal is to win the game show, while the game shows goal is for you to lose. Yeah, they could literally spell it out for you within the problem, but they dont need to because the rules or goals within the game show are clear: you try to win, they try to prevent you from winning. They spell it out enough because they state youre on a game show, which has understood goals, and then state the host knows where the car is. There is 0% probability the host will reveal the car to you so it can be eliminated from the probability chart.
Actually in that case it makes sense to never switch doors. 1) If you select the car you have a 100% chance of winning, so the host will give you the option to switch doors, this way your odds are less than 100% as you may choose to switch. 2) If you select a goat the host will not give you the option to switch because you have already lost Given the host wants you to lose, knows where the car is, and there is no rule stating he must reveal a door, this is a perfectly rational interpretation.
The Monty Hall Problem is statistical bullshit because they are two separate cycles. Imagine this... Imagine you don't choose a door the first time (because that is essentially what you are doing by being offered a choice you can change). So Monty removes one choice and allows you to pick one of two doors. You can pick the one you originally picked the first round... or you can change... Either way you are making a choice between two doors. The fallacy is that you choice exists between 3 doors... The choice isn't made there. The choice is made when you get down to 2 because that is when you actually choose your door and have to deal with the consequences.
Except that's not at all what you are doing. Can you answer this variation of the MHP. There are 2 goats and 1 car behind 3 doors. I choose a random door. Then you get to looo behind the remaining 2 doors and choose 1. What are the odds of me getting the car and what are the odds of you getting the car?
Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car you get the car. You only win 1 out of 3 games if you stay with your first pick. Basic math/logic kids understand, idiot among idiots doesn't.
My issue with the problem is that the question is misleading, as it puts you in a scenario and says door X has a higher chance (2/3rd's) than the first door chosen, when door X (the last door that the host didn't open) is a hypothetical door to begin with. Since the answer assumes a strategy of choosing the 2/3rds door, it is assumed before even playing: Which door is two thirds? The opposite/last door that the host doesn't pick. But which door is that? Either of the two doors. But which one? Either, or both at the same time? It's easy to see that the reason why it works is because you're treating both of the two doors (the other two that the contestant didn't initially choose) as a group. So in other words, the answer is no different than saying that a dice has 1/6th chance of landing on any given number, or that you have a 1/52 chance of picking the ace of spades out of a deck of cards. Since the Monty Hall answer is given as a strategy or procedure assumed before even playing, no specific door actually has 2/3rd's odds, contradicting what we already know, that all doors have 1/3rd odds. The question has an implied contradiction within it.
Until Monty has made his move, there is no "two thirds door." The door that has probability 2/3 has that probability because it is the door that Monty kept once the player's initial pick imposed a constraint upon him. That is, the player chooses door A, and now door A is off limits to Monty. Monty can only consider door B and door C. Then, with knowledge of the prize's location, and deliberately avoiding the untimely reveal of the prize, he keeps one door closed and opens the other. So Monty's final door is the best of those two (B and C), and that has better odds than just a single door on its own which was not the beneficiary of such an unfair selection process.
I think you make an interesting point. You already know beforehand that one of the two doors that you are not going to pick is in effect 2/3. What you really have is 1/3:2/3 or 2/3. You are just waiting to know which one to pick. It's not really even a game anymore. It's cheating. I might also point out that statisticians will say odds cannot redistribute. The unopened unchosen door doesn't actually have 2/3 odds. It's more like you have grained inside knowledge and, as above, are cheating by learning one of the two doors you didn't pick isn't the prize. Would anyone say the doors changed their odds if you had an accomplice who signaled you - Hey psssst, door 2 only has goats, because he managed to sneak a peek?
Since in the original letter, with no assumptions being made, it states the host knows what's behind the doors, randomness is not a possibility. And only random opening of the doors can yield 50:50 odds between switching and not switching. It's fine to do the problem as a straight math problem, but it is not presented as such. As written, word for word, it is a non-random game with no answer. With Savant's assumptions it becomes a puzzle. Doing it as a straight math problem might have been nice as a lead in, but the letter just isn't written where that particular approach can be justified as giving the correct objective answer.
@@Hank254 No random opening of the doors can be read from it to justify 50:50 odds since it clearly says the host knows what's behind the doors. Opening the doors randomly while at the same time knowing what's behind them would be a contradiction. How could I randomly pick which cup the ball might be under if I already know where the ball is?
@@Hank254 I've not had any problems here. I've been arguing this point with Logical Morality for a long time saying 50:50 is incorrect. He said he was solving the problem as a math problem where randomness is the default. Fine, but the letter doesn't present the problem as a math problem. It's presented as a game show where the host has knowledge of where everything is. LM says at 6:15 what they are leaving out is "The host is not opening doors randomly; he is systematically avoiding the car." However, leaving that out does not mean he IS opening doors randomly and is not systematically avoiding the car. Leaving that out only means we don't know what he is doing...it's a non-random game, with no particular answer. LM will answer we can assume randomness because in matematics concerning odds, randomness is the default. OK, but we cannot ignore that it states the host knows what's behind the doors, which is pretty counter to any assumption of his opening doors randomly. For this to be solvable as a math problem, it would have had to say, and the host opened one of the other doors which happened to have a goat. Then we could take it to be a random opening. Even if it had just left out the part "who knows what's behind the doors" and just said, " ...and the host opens another door which has a goat," that would have been enough to allow for randomness. But, that's just not the case.
@@Hank254 One cannot open doors randomly to deliberately avoid the car. That is a direct contradiction. The game show could not possibly have allowed the opening of doors randomly because that would have risked exposing the car. The letter was not presented as a math problem. The game show in reality and as described in the letter was a non-random game, controlled by management. There is nothing in the letter that leads either to a math problem or even a puzzle. As written it has no answer.
@@Hank254 The odds can be written 50/50 or 50:50, or just 1/1 or 1:1, but the probability is written as 1/2... . Since probability is always written as a fraction of wins to tries, it's better to use a colon to show odds: 1:1. For the probability of 2/3 (you have a two in three chance of winning), you have the odds of 2:1 (two to one chances of winning over losing). 50-50 is actually an idiom so should be spelled out: fifty-fifty.
The problem is well understood doooooooooooooooooooode. In the end you make a decision between the two remaining doors in the game show context. You decide whether you stay or switch. So the thing is you get a 50/50chance to either win or lose as YOU as the game show participant wouldn't know whether your prior choice does contain the prize or the other goat. You can only calculate this 2/3 win chance by switching IF you assume the knowledge behind which door the prize remains. And how many times do you have to participate in the game show to realize your actual 2/3 winning streak? Do you understand now how unlikely that is?
@@niemand7811 Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Switching means the opposite. It's just basic math/logic kids understand. Sadly, it's far too hard for idiots.
Pov: you don't understand Monty Hall Problem Wait its probably worse you do understand the problem but still goes on do math just like the people who don't understand the problem. You do know that host knows where the car is, but you are still taking one of the three scenario as host reveling the car.🤡🤡
If you want to inform what action the host must make, it is easily done by stating that the host always opens a door among the doors that the host did not pick and that he always chooses a door with a goat behind it. Had they just clarified that the 2/3 Vs 1/3 would actually emerged. But instead they describe an instance and what happened at that instance. We are just told that the host knows what behind the doors. Implying that means we have the information that he will always open a door among the ones you did not pick which always contains a goat is about as far from Mathematics you can get. It's like saying "I have a bicycle and it is 12km to my work" and believe we have established I ride it to work. If the intent is to convey information, convey it directly and concise!
@@tbd5330 This is why we articulate questions so they are concise. Not so you have to make social assessment as a result of someone avoiding stating the conditions. The only thing that is assumed in questions regarding probabilities is that stochastic variables with a clear even distribution among the outcomes like a dice, lottery tickets, opening of boxes/doors, flipping of a coin can be assumed to be evenly distributed, unless clearly stated otherwise. It is also common to add unimportant information in questions to test whether a person can differentiate between what is important for the question. "Boris, a professional casino player from Bulgaria have one red and three yellow normal die. If he throws all at once, what is the probability the sum of the sum of the red dice and the sum of the yellow die will be 21?" Perfectly fine question. If they wanted to inform that the default even distribution from a door opening are not present in the scenario, they need to clearly inform the reader of that. They never do. And requiring the reader that we must assume he must without us stating so, is unbelievably inconsistent with logic. If we start of making assumptions like that, where do we even stop??? "Clearly it's common sense the goats would smell and probably make noises, surely the contestant would have gotten some clue where the goats are, especially if the host opena the middle door as you can then see at what direction that goat is trying to communicate with the other 🐐" :) Also, not only do we need to establish he must open a door that contains a goat. We must be informed the doors he must chose from is a proper subset of the unopened doors with a cardinality above the number of prizes. We are never informed about any of this either. What's interesting is how clearly the question is not generating the asymmetrical redistribution of probabilities many assume it does and especially how almost all people claiming so doesn't even understand how extremely concise and specific the scenario needs to be for it to emerge, which is why the entire circus is demonstrably not beneficial from an educational point of view, but obvious outright counter-productive. It's not even close. It fails to establish any of 3 of the conditions that needs to all be met at the same time to get the 2/3 Vs 1/3 scenario. The three prisoner dilemma has all the mathematical conditions that many alleges this question has, yet without all the epic fails at communicating them. Therefore it is superior in every way and should get the focus instead.
@@tbd5330 That's not even what I said, it' the wrong question to ask as I demonstrated. Nor is it even sufficient to clarify he always chooses a door with a goat behind it to make the 2/3 vs 1/3 emerge, we also need to be informed he cannot open the door you chose and that he can choose freely among the now clarified two doors he can choose among in the scenario you pointed at the car door, as once the default behaviour of a stochastic variable like opening a door are established to not be present here, there is a very high requirement to unambiguously establish all the conditions present. But they dont't even get close to establish the super specific scenario needed for the 2/3 Vs 1/3 scenario.
It's nonsense because the context is wrong. The choice to keep the same door also becomes a 1 in 2 choice. The paradox comes because people assert the probability of the previous state and compare that with the probability of the new state.
@@healthquest4823 Yes but it's nonsense because what is actually happening here, is that people are attributing a previous state, that no longer exists, to the present state. The new state is generated in isolation of the previous state. I'm going to code this, to prove that it is nonsense. The whole thing is just a mental paradox but nobody seems to have thought about it for more than one second. The joys of group think lol.
@@Martin-jg5le No, that's where you are going wrong. The new state is not formed in isolation of the previous state. When the goat is revealed, the initial choice is off limits. Its odds are locked at 1/3. No goat from that door could've been revealed. The two choices are not independent. The first choice constrains the two other doors to be an independent set having 2/3 odds, and if we learn one is a goat, the only door left gets to keep all the 2/3 odds to itself.
@@healthquest4823 That's blatantly incorrect because the new choice is decided based on a new set of variables and is independent of the previous choice. What people are doing here, is using the previous state to determine the likelyhood of the door being correct but using the new state to determine the other door. When in reality you have to either use the previous state or the present state, to determine the likelyhood of both doors. Look at it this way, apply the same logic to the unselected door, using the previous state and you'll soon realise that it has the same odds, in the negative. This is just one of those interesting questions that somehow glitches people's brains.
@@Martin-jg5le No it is fact. You just can't understand it. The correct answer has been know ever since 1975 when the problem was first published with its solution. You can easily set this up to play, as I have, and you absolutely win 2/3 of the time by switching.
@@onesem9399 then explain how the computer is wrong. If you had 100 doors and you chose 1 of them, the chance is 1/100. If 98 doors were taken away and you were left with the door with 1/100 or the other oprtion, it would be the other option.
@@Tantheman.if the door u picked did not get out then it would be a new problem where u pick between 2 doors since u can change ur answer then it would be 50 50
The claim in the video at 6:00 that the answer is clearly 50:50, seems wrong to me. He states it's because the problem is leaving out a key assumption, which is "the host is not opening doors randomly, he is systematically avoiding the car." So if you put that in, you get the host is systematically avoiding the car, and that leads to 2/3 for switching. So, while it's true that if you leave that assumption out, you don't get the 2/3 for switching, you also don't get 50:50. To get 50:50 he must BE opening the doors randomly. It is Logical Morality who is assuming that if it isn't stated the host is systematically avoiding the car, he must be opening the doors randomly. For Logical Morality to be making any sense, as I see it, he would have had to say: It's clearly 50:50 because a key assumption was left out, which is the host is not systematically avoiding the car, he is opening the doors randomly.
I did explain that in the video hahaha, I said "randomness is the default" which it is, just like base 10 is the default in arithmetic If the host chooses to cheat or has a bias, it must be stated otherwise the question is simply unanswerable... Which I said it very well may be. If there is an answer it's 50/50... Otherwise there is just no answer.
@@logicalmorality4646 There is no default position. You just read the narrative, and so long as nothing he does had to have been done that way, the we can only conclude anything he did could have been otherwise. So it's non-random, in that the host was not barred from doing anything otherwise. The ONLY way for it to be random is if something is stated saying he had to open a door randomly.
@@healthquest4823 Still at it eh? Arguing is bad for your mental health, nor are you going to prove anything by ignoring what I say, presenting the same argument I have already addressed, or lying about what you said... Your assessment of my motives are way off as well. The motive should be clear, it's in theme with most the videos on this channel... To sew seeds of distrust in science populism.
@@logicalmorality4646 I have never lied about a thing. I reposted my description of the French fry game I played and my comments about Savant. I deleted it to not make you look bad at a time I was defending you against the other posters here.
Here's the thing. You are assuming that everyone has misread the problem. I think you have misread the problem, and I can see how. The key words here are "the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat." A clearer reading of the problem is if you remove the words "say No. 3", the problem then becomes worded "and the host, who knows what's behind the doors, opens another door, which has a goat." These two statements are functionally identical, as the words "say No.3" is just providing an example of a door that has a goat behind it. Right there, the problem is assuming that the host will never reveal the car, making it correct that you always want to switch. Even logically, a game show host will never reveal the car on round 1, because that just doesn't make very compelling TV, so it is in the Host's interest to always reveal a goat first, then ask you if you would like to switch.
As I explained in the video, you can't conclude the following information from "the host knows where the car is" 1) The host must reveal exactly 1 door each time you play, how do you know he can't open zero? 2) The host systematically avoids revealing the car (illogical assumption, he very well may reveal the car and then not even give you the option to switch) 3) The host can not open your door I dealt you the 2 of spades and know what the other cards are... What card will I deal you next? It's comical that anyone can answer this question, the correct answer is "it depends on the rules of the game, and the intentions of the host, both of which are not stated"... It's only obvious rules 1,2,3 are there AFTER you have been told them.
@@logicalmorality4646 Ok, I'll bite and answer each one of those points. 1) It is literally stated in the problem. "You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat." It straight up tell you that the host will always open another door. 2) Refer to my original comment, which you seem to have completely ignored. The problem again literally states that the door Hall will open "has a goat". 3) Again actually stated in the wording of the problem "You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door" It tells you that the host will always pick another door, meaning, not your door.
Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 ouf of 3 games if you stay with your first pick.
Here is what happen if the host open the door randomly. Your pick, Host reveal, The last door C G1 G2 stay = win, switch = lose C G2 G1 stay = win, switch = lose G1 C G2 stay = lose, switch = lose G1 G2 C stay = lose, switch = win G2 C G1 stay = lose, switch = lose G2 G1 C stay = lose, switch = win However, in MHP, scenario #3 and #5 doesn't exist. Since the host cannot reveal the car in scenario #3, he has to reveal G2 instead. Since the host cannot reveal the car in scenario #5, he has to reveal G1 instead. So it becomes C G1 G2 stay = win, switch = lose C G2 G1 stay = win, switch = lose G1 G2 C stay = lose, switch = win G1 G2 C stay = lose, switch = win G2 G1 C stay = lose, switch = win G2 G1 C stay = lose, switch = win
Let's say the host always reveals Goat A. If your first pick is Goat A, you get Goat A and are not in a MH situation at all, so we discard that game. The two remaining situations are indeed covered by the MH problem, let's see what happens: If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. So you win one out of two games, as expected.
Your premise is off because the goats are irrelevant and you're only concerned about the car. The problem would be better as an escape problem with two locked jail cells and one unlocked. Trust me you're better off switching every time.
2:20 "Nowhere does it state that the host picking the car is not a possibility". This is the crux of your incorrect conclusion. The game is setup such that the host does, indeed, know where the car is, and is forced to show the player one of the two goats.
Yeah, only that in most descriptions of the problem, it isn't. Most notably, the letter to Parade, that made this problem famous, does not mention the host to be forced to do anything.
@@insignificantfool8592 _"Most notably, the letter to Parade, that made this problem famous, does not mention the host to be forced to do anything."_ Except, of course, he is forced to play the game where he opens another door and shows a goat.
@@Stubbari yes, also from your perspective. If you're to guess my number between 1 and 10 and you know I'm forced to pick an even number, your chances are 1 and 5. If you don't know I'm forced to do so, chances are 1 and 10 even if I do pick an even number.
I thought about this and this isn't true, assuming randomness is involved (which i believe isn't, I think the original context describes pretty clearly that monty has to reveal a goat, but whatever). Let's just assume Monty can reveal either a car or a goat. Since he knows all doors, u don't know how likely he wants to show the car and make you lose. Keep in mind that if your initial pick is a car, you force Monty's hand to reveal a goat. In other words, it might be more likely that your initial pick was a car if he shows a goat. Making you more likely to lose. The answer is that if randomness is involved in monty's pick, then you just don't know.
Are there in fact three "live" doors. With two goats, there is always a goat for Monty to knock off, if he in fact knows. So there are always, each time this is done, only two "live" doors. How can the odds be other than fifty-fifty. The three doors are only for the original choice, not for the switch choice. Assuming Monty knows, I would say so what? He can and will always knock out a goat, at least the way this problem is posited, with the effect of eliminating a door. If, if fact, this is how the program works, Monty's knowledge seems to me to be irrelevant. If he is picking at random and could pick the car, even this problem is not a problem. I do think the odds improve from one out of three in the original choice to fifty-fifty in the switch choice.
Monty’s knowledge isn’t necessary upfront but at least after the contestant makes his initial choice, the host will have to find out what is behind the doors and reveal one of the goats. If the contestant has watched the show before and realizes that no matter what happens, Monty always reveals a goat from the other 2 doors, he will know it isn’t random. Therefore, staying with the initial pick will represent the initial 1 out of 3 picks and the switch will represent 2 out of the 3 other picks. So at the beginning before being offered the switch your pick is 1/3 and the one he doesn’t show you represents 2/3s of the chances.
Oops I thought this guy got it but I guess not. The odds do not change in this game no matter the choices because there is literally only 1 car. 1 car. 3 choices.This later simplifies down to 1 car, 2 choices. That's because the host always eliminates the 3rd choice which is always a goat. You are always left with 2 choices in the end, and one is a car. You are always left with a 50/50 chance no matter what.
Nope. Your pick always remains a one in three shot. Imagine picking a specific card from a deck of cards, or just increase the doors, and it'll become clear. The other door wins the rest of the time.
@@gnlout7403 those are just odds though, generic odds and they have nothing to do with whether you should actually change doors in the game. Of course the game changes when you make it 100 *doors but even then the same concept still applies. Perhaps if you're on this game show for 100 appearances and you're trying to capitalize on as many wins as you can then maybe going with the generic odds is a good bet. If you're dealing with a generic 2/3 then your final choice between two doors is more or less 50/50. It's more like zero to 100% in reality but 50/50 is the best way to say it.
@@gnlout7403Well again, that's just general odds. In reality you could win 10 times in a row by "not switching" before you lose once. That's how random chance works.
@@Magneticitist you could. You could win by staying with 100 doors but statistically you would only do so one percent of the time. The other 99 times you'd win by switching
I don't think your explanation is terrific, but I'm still 100% convinced that your conclusion is correct. The problem with all the original "YOU SHOULD ALWAYS SWITCH" scenarios is that both the math and the illustrations (and also the programs that run thousands of simulations showing that switching wins 2/3 of the time) completely ignore the fact that by having the host pick one of the doors with a goat, you've retroactively CHANGED THE ORIGINAL PROBABILITY. When that one bad choice is eliminated, then the chances that you picked correctly the first time effectively changes from one 1/3 to 1/2! Every other video I've watched on this subject simply ignores this fact.
I am a "philosopher", not a mathematician hahaha, but I gave it a shot. If you have any tips to explain it better let me know! Yes people somehow seem to be incapable of looking at the fact that the original problem did not include the following... "The host must always reveal exactly 1 door after you select, the door he reveals must not be your door, and he can't reveal the car"...
@@logicalmorality4646 I hate to tell you this, but after a great deal of thought and argument, I have come to the conclusion that your argument is INCORRECT. Originally I thought the fact that the game essentially restarts after the host reveals one of the bad outcomes meant that it becomes a 50-50 choice and that the initial choice then becomes moot. Unfortunately, that's actually not the case. The argument that made me change my mind is what I call the ten thousand door scenario. Imagine the same exact game, except that there are 10,000 doors. You get to pick one. The host reveals 9,998 doors without a prize. Now it seems that you now have a 50-50 shot so it doesn't matter if you switch. But the problem is you can't escape the fact that the chances that you picked the correct door in the first place was 1 in 10,000. Even after all those other choices are eliminated, you're still left with the unchangeable fact that the chances that you picked the correct door at first is vanishingly small. Increasing the number of doors in the game helps to see that logically, the chances that the one door out of 9,999 the host (who already knows the location of the one prize) did not reveal is far more likely to be the actual location of the prize. Obviously it's still possible that your initial choice was correct, but it's still only 1 in 10,000 because that's what the odds were when you made that choice. Getting to pick again doesn't change that fact.
@@stevecollins3496 That probability is offset by the fact it's nearly impossibly unlikely the host would reveal 9998 doors randomly and manage to miss the car. The odds are still 50/50 with 10,000 doors given the host is randomly opening them. Scenario 1) You picked the car (1/10000 chance) and then it doesn't matter which doors the host opens, the remaining door is a goat no matter what. Scenario 2) You picked a goat (9999/10000 chance) host reveals every door except the car (1/9999 chance) 9999/10000 x 1/9999 = 1/10000 Scenario 1, and scenario 2 have the same the odds... Switching is just as likely to make you win as losing. Switching is the same odds as staying the same... I don't think you thought about it, I think you just parroted a pop video hahaha.
@@logicalmorality4646But that doesn't follow the rules of the "Monty Hall Problem." In that scenario, it is stipulated that the host does not open the door you chose OR the door with the car.
Dude you got this all wrong! The Host of COURSE and ALWAYS opens a door with a goat?? Pretty obvious isn't it otherwise he would have stolen the car from the person playing the game? That leads to Scenario1 you chose the wrong door Scenario prob: 2/3 Prob that you win in this scenario if you stick with you choice:0 . From the other 2 alternatives: 1 car and 1 Goat, the talkmaster eliminates the Goat: Probability that you win when you change the doors: 100%*66%(scenario prob)=66% overall that you win compared to 1/3 that you chose the right door right from the start, twice as much...
Too many people are saying the same thing, so I just made this video to respond to 99.9% of comments disagreeing... If you think it's obvious that the host will always open exactly 1 door after you select yours, and that the host will never reveal a car, or your door, please watch this video. I can PROVE that it's not obvious, try the challenge if you're so confident.
ua-cam.com/video/3AJVaeysml4/v-deo.html
What if I want a goat? 🐐
I talked about that in the problem lol. It wouldn't matter in this case because it's 50/50
No one said you can't choose the opened door
You have every right to!
Yes. A goat is adorable and will love you. A car is an annoying chore to own
Then you'll get the car every time
The situation is that "decisions without consequences" and "choices with practical benefits" are confused and regarded as having the same meaning. It has been clearly decided. If there are really two choices, shouldn't we get both things? In fact, the rule changes from one of three choices to one of two, misleading you into thinking that you have already chosen once, but in fact you cannot get any of the so-called choices at all. When you finally make the real choice, in order to prove how smart and knowledgeable you are, of course it should be counted as 2/3 when there are only two choices. This is a testament to the Goebbels effect. If scholars and the educational community do not correct the extension of Monty Hall's problem and still believe that 2/3 is correct, then what is the point of educational scholarship? ? It was ruined.
@@dawyer Thank u so much for how well u articulated this explanation. It blows my mind how quickly people dismiss this obvious fact.
Maybe thinking this way....... Try to understand it from the perspective of the host. The player provides his prediction to the host. The host opens a pair of doors with sheep, leaving only two pairs of doors, one sheep and one car, for the player to choose. If the player Predicted the door of the car, what will the host do? of couse Want the player to change their choice. The host asks you to calculate the odds with misleading you already have once choice before. you swapping your choice, 2/3 will win. On the contrary, the player predicted the door with the sheep. What will the host do to make the player still choose the door with the sheep? you can tell.
........ Dawyer's door problem, calculate the chance of the host winning.
1) not a paradox
2) Monty never reveals a prize
3) Monty always reveals a door containing a goat
You will never trade a goat for a goat. Ever - because Monty is REQUIRED BY RULE to reveal a goat door.
1/3rd of the time, you will have chosen a car. Monty will reveal a goat. You will switch to the other goat.
2/3rds of the time, you will have chosen a goat. Monty will reveal the other goat. You will switch to the car.
Exactly haha this guy is not very smart lol
@@asafoster7954 No. That part of being less smart is on you guys. This 2/3 probability to win if you switch is a trick played on your mind.
Here is what takes away. Monty will always reveal a goat. If so this eliminates your first decision. Now say it was door A that was taken.
You are now left with door B or C. If you chose B before that means: Stay with B: 50% chance to win. Switch to C: 50% chance to win. I leave it open to suggest behind witch door the prize is hidden. Why? Because as the participant making your choices YOU do not know which door does hold the prize. You can only accumulate your winning chance when you assume the door that holds the prize. You can simulate it mechanically. But it will not play out like this in reality. The prize can be behind door A in four consecutive shows or five times behind door C. And you must not forget that it isn't constantly you flipping through ten games of the Monty Hall show. To maybe equal up to this 2/3 winning chance throughout all games. You play maybe once and you either win or lose. And that is all there is. Now toss me some more calculations if you want. But reality hold your butt.
@@niemand7811 just ask chat GPT to explain it to you at a 5th grade level. Sounds like that's what you need.
@@niemand7811I Agree, for me it's more like a trick rather that real math.
@@asafoster7954your the dumb one. If you actually pay attention to the video you realize that he admits that the Monty hall problem is correct as long as you are told that the host is systematically avoiding the car.
The Monty hall problem is based off of assumptions
If I word the question slightly different and mention the fact that the host is not picking cars than you would want to switch
if you're looking at this like a logic problem, then you need to understand that there are 3 endings.
in the first ending, the game show host picks the car, and you lose. in the second ending, the game show host shows you that you picked the car, and you win. in the third ending, the game show host shows you that you didn't pick the car, and you lost.
if you're talking about the probability of getting the car, and you're including the first ending in your math? you really shouldn't be.
if the first ending happens BEFORE you get to make a choice. it eliminates you BEFORE YOU MAKE YOUR CHOICE. how can something that PREVENTS YOU FROM MAKING A CHOICE IN THE FUTURE factor into the probability of the choice? it can't. which means we need to eliminate the first ending in our math, because it unequivocally CANNOT effect the choice; all it can do is prevent you from making the choice to begin with because, again, IT ENDS YOUR CHANCE TO MAKE THE CHOICE.
There are 2 possible outcomes... win a car or you win a goat. It's literally that simple. People have overcomplicated this just because they didn't want to piss off people pushing a female role model.
Stand strong, dude. The drones are explaining the probabilities of which door Monty opens, not the probability of getting the car. It is that simple. Bottom line is it is either ⅓ vs. ⅓ or 50/50. 🤙
Oh its you again. Hey.
Are you part of the cyber bullying team on here spreading the propaganda as well? If not, how are you buddy?@@Drunken_Incel-bw7sl
@Tristan
I appreciate your explanation. Most people assume the winning chance for the game show participants from the view of the game show host.
It cannot be 1/3 & 1/3 or 2/3 & 2/3, probability has to add up to one, I encourage you to get a die, and four pieces of paper, write "winner" one one piece of paper, and "1", "2", and "3" on the others, roll the die, on a 1 or 2, place the winner paper beneath the 1 paper, 3 and 4 for 2, and 5 and 6 for the 3 paper. Then roll to choose a paper again, then remove a piece of paper without the winner paper. Then switch the paper you chose, mark whether you got the winner paper, repeat many times and see how many win.
@@rat-prophetistfordism8344 I appreciate your reply.
Correct, probabilities have to add up to one, which means each door can ONLY have a 1/3 probability for the car to be behind and 1/3 probability for the car to NOT be behind. At no point in time does ONE door possess a 2/3 probability for the player to not get the car.
Also, at no point in time does the player have a 1/3 probability to pick the door Monty eliminates. The 2/3 hoax wants you to believe that this is about picking one door from three, and the rest doesn't matter.
Player has TWO SEQUENTIAL PICKS, with Monty's elimination of a no-car door being RANDOM to the player.
The Monty Hall Problem asks, does the player have a higher probability to get the car by picking AGAIN the same door he initially picked, or pick the remaining door that Monty did not eliminate with his SECOND SEQUENTIAL PICK.
And yes, the simulations showing 2/3 no-car probability are all FLAWED. Most are done in Python, which is a compiled and interpreted accelerator. That means you have to prompt and code all the rules BEFORE each function and re-prompt and re-code EVERY STEP of the simulation along the way to get the correct result. No one disputes player's 1/3 probability to get the car. Isolate the simulation to find player's probability to NOT GET THE CAR with his second sequential pick with two doors left and you will find the correct answer:
Probability of not picking the car in Pick A:
P (not car in Pick A) = ⅔
Conditional Probability of not picking the car in Pick B, given Pick A was not the car:
After the host eliminates a door that does not have the car, the scenario reduces to two doors: one that the player initially picked (not the car), and one that wasn't picked (which could be the car)
If the player initially picked a door without the car (probability ⅔), there is a ½ chance that the car is behind the remaining door after the host's reveal.
Therefore, the probability of not picking the car in Pick B, given that Pick A was not the car, is: ½
Overall Probability (Law of Total Probability):
Combining these probabilities using the law of total probability:
P (not car in Pick B) = P (not car in Pick B∣not car in Pick A) ⋅ P (not car in Pick A)
Substitute the values we have:
P (not car in Pick B) = ½ ⋅ ⅔ = ⅓
Therefore, the probability that the player does not get the car after TWO SEQUENTIAL PICKS is ⅓.
Final probabilities after player’s second sequential pick is a 50/50 guess:
Probability of getting the car: ⅓
Probability of not getting the car: ⅓
The crux of the confusion is that Hall doesn't ever choose the prize in his reveal.
It's not about doors, it's a logic problem about three different scenarios. Scenario triggered by the choice of either the good door, or either of the two bad doors. Choose the good door, and he can reveal either of the remaining ones. If you switch, you will necessarily fail. But if you choose either of the other two doors, he must eliminate the other bad door. So in either of those scenarios, switching gives you the prize.
Two good scenarios, one bad, if you switch. Two bad scenarios, one good, if you don't. 2/3 odds for switching. They are simply scenarios that always play out, not "doors". It took me a bit, originally, too, but that's because I didn't quite get the problem.
If it actually were about the doors, the superficial stats, then round two probability in a vacuum is 50/50. But knowing how each scenario plays out, how everything is selected and locked in changes those odds.
I was typing up something similar, but say Mr. Dan's Comment, and thought a reply might be better than a new post.
At 2:20, you say "nowhere does it state that the host picking a car is not a possibility." It does, though. "The host, who knows what's behind the doors..." As Mr. Dan says, the host will not open a door with the prize and then ask the player--who has just been revealed to have lost--if they want to switch to the other losing prize.
You're talking about form then like with the horses. Those who fully grasp their form are in the realm of certainty not probability.. The show had form obviously , but I never watched it,so my realm would be probability only and purely.
@@philip5940 I have never seen the show, either. I am going entirely off the wording in the problem statement you seemingly did not notice or deliberately ignored.
The form, as you put it, is stated right there. You assumed each choice was independent and random. However, it is stated the host knows what is behind each door. So, if nothing else, the host's choice is no longer truly random. Therefore the probabilities you assigned to the second pick are essentially arbitrary.
@@BoomStickization No you are not going by the wording at all. We are never informed that the host must actively chose a door that has a goat behind it. It would have been really easy to state that. But they don't, probably as their urge to make the paradox appear as fancy as possible did one up on themselves. Because the wording is so specifically design to confuse and unnatural that it is hard to see this failed attempt was not intentional. We are only told that he knows what is behind the doors and that in the instance of the show we are "supposed to imagine" unfolded the event that the door he opened had a goat behind it. At no point do they state that the host HAS to open a door with a goat behind it, which is an absolute necessity for the paradox to emerge.
As a result of this the answer to the question is either "insufficient information is given" or "1/2" since everything about a stochastic process like throwing a dice, drawing a lottery ticket, opening a door etc is universally assumed to be a uniform distribution among each individual side, ticket or door etc. We need to clearly be informed of the contrary to change that.
But here, we never are, despite how easy it would have been to do so. Instead they put a lot of semantic effort in all sorts of other information to make the paradox appear fancy, but missed the very core of the paradox in the process, resulting in that they are not even describing a situation where the paradox emerges.
Also much of the aftermath was a smoke screen as they then tried to pretend that the information that the host HAS to open a door was already stated in the initial phrasing, when clearly it was not. The fact that the played down the importance of that missing information just made people even more confused afterwards as it is the very mechanism that makes the paradox emerge in the first place.
For instance running a computer simulation where the assumption that he has to open a door with a goat behind it is implemented, is such an obvious smoke screen to ignore the fact that it was never stated in the initial problem. And such a program (a Monte Carlo simulation) would almost be nonsensical as the majority among the ones who managed to write the code without copy/pasting would realize that the probability of winning by swapping would trivially be 2/3 under those assumptions.
@@dataandcolours I am going to leave this conversation after this response.
Your statement seems to boil down to "the writer is clearly and deliberately trying to deceive us," and I do not see that, at all.
I agree that the conditions could be made more explicitly clear. There may be assumptions in the statement that are not as clear as the author intended. But that is far from being "specifically designed to confuse," or an "active smokescreens."
The contestant even being offered a choice implies there is a chance for them to win. Thus, the host must have revealed a losing door.
This is a reasonable assumption in the context of an American game show.
So, the question is a conditional probability problem: "is there an advantage is the contestant switching if the host revealed a losing door?"
i arived at 1/2 in a different way suppose lets say there are 3 doors a b c for first circumstance assume car is behind a
all possible events are
1) you chose a and he opens b you switch - goat
2) you chose a he opens b you dont switch - car
3 ) you chose a he opens c you switch -goat
4) you chose a he opens c you dont switch- car
5) you chose b , he opens c you switch- car
6) you chose b , he opens c you dont switch - goat
7) you chose c he opens b you dont switch - car
8) you chose c he opens b you switch - goat
all these can be multiplied and divided by 3 cause similar outcomes would be if car was behind b or c
when you switch no of times you win is 2/4 no of times you dont switch and win is 2/4 now multiplying 3 to both numerator and denominator for the outcomes would lead to 6/12 or 1/2
You’ve misread the problem. 🙄
The whole point of the host opening a door, is to confuse the player. If he opens the door with the car, why would he ask the player if he wants to switch? The host will always open the door with a goat
But we should never have to assume things like that, especially as complex of an assumption like that. Things like that needs to be clarified in the initial question. But they never are (especially not in how the question was asked in Parade Magazine). We are not intended to play private detectives here. Because if we go down that rabbit hole, when do we even stop?
* Why did the host even open a door and asked if the player wanted to switch if the player had chosen a door with a goat behind it. According to game theory it would then be in the interest of the host to just open the door the contestant chose and declare "You lost!"
* But why didn't the contestant just look through the keyhole?
* Maybe the information about the number of cars are false.
* Maybe the host only ask if the contestant wants to switch if his/her initial pick was a door with the car behind it.
This is why the initial phrasing of the question makes this question so dumb. It really does everything it can to describe an instance of the show, as doing so is the only way to camouflage all the rather important aspects that needs to be established in the first place for the paradox to emerge. This makes this version of "Three Prisoner Dilemma" a rather pointless example of a combination of sloppiness, intentional deceit, psychology and a bit of conditional probability. A mixture that is outright de-educational as the intent seem to be on mesmerizing rather than educating.
They gotta make a show and engage a little stagecraft. They could simply start with only two choices, not three to speed it all up but it ain't in interests of entertainment.
If it starts with two choices , then everyone would agree probability is 50/50 . However when he starts at three choices then brings it down to two choices , it then happens that gullible folk are purseuded to believe that some phantom effect of a pre-existing ⅔ probability still exists for the now changed independent new conditions . But this stuff emerged during the Uri Geller era when folk also believed in spoon bending and his other scams . Pyramid power too , it was endless .
thats why the problem is stupid it wasnt explained properly
@@philip5940 "However when he starts at three choices then brings it down to two choices , it then happens that gullible folk are purseuded to believe that some phantom effect of a pre-existing ⅔ probability still exists for the now changed independent new conditions ."
If you look at how the Monty Hall Problem is _intended_ to be worded, where the host always acts to reveal a goat from the remaining two doors (knowing where the car is), then the second 2-door scenario is _provably_ (and easily so) dependent on the first choice. Merely declaring it to be independent simply because you don't want to think doesn't change the reality here.
If the host will ALWAYS open the door with the goat, that means the chance is not 2/3 or 66.666% of the goat being behind door number 2, but its 100%. You said: ALWAYS. Always = 100%. So what you said is literally against what the correct answer was that i heard about.
If I were the game show, I would instruct the host to show the goat if they picked the car and show the car if they picked the goat. I would want to give away fewer cars. Why would I give the contestant a second chance if they got the first pick wrong? Knowing this, as a contestant I wouldn't switch. I am curious to see what the stats on the monte hall show turned out to be.
Exactly. Considering we are given no information at all on the rules the host have to obey we simply cannot answer the question as it was composed in Parade Magazine and you reasoning exemplifies this perfectly. The phrasing was not just "slightly sloppy", it was probably deliberately vague. I mean just the start "Suppose you are on a game show...". How is that not introducing us to consider an instance of the show, hence perfectly distancing us from being told how the host must act (the rules)? :)
The actual show never worked like this. He only offered money for your door.
@@gnlout7403 Yes the show actually worked in all sort of mysterious ways. However we obviously cannot add outside information to the question given. And because of this the question as phrased in Parade Magazine is absurdly ambiguous and the most natural way of interpreting it from a logical standpoint actually makes the result the 1/2 vs 1/2 scenario. It's only if we add a whole lot of additional information that the 1/3 vs 2/3 situation emerges.
@@dataandcolours no. Again. The show never worked like this.
And we've been over this already.
You botched logical Moralitys objection. He's not confused about what you think he's confused about.
And you can do the problem as stated with the information you have.
Hope that helps
@@gnlout7403 Please refrain from claiming I have claimed things without clarifying what I have claimed. I hope you understand that just comes off as very passive aggressive in a weak attempt to settle some scores. The intent of discussing here is simply to point out how massively poor the original wording was and how correct people are at pointing that out.
Monty hall problem is a great example of real world problem. Most problems that people see in text books are problems with fair dice. Monty hall problem is one with loaded dice (which is always what happens in real world). Monty hall picks door ensuring that car door is never opened. Thats the most important part!
Too many people complicate this. Here is the simplest proof:
- Your first choice is probably (66%) a goat.
- The host reveals the other goat (Fact. 100% chance that he does this.)
- Thus, the remaining door is probably (66%) not a goat.
Yeah it's really quite simple. It's not a paradox at all.
P (A+B) = 66%
P (A+C) = 66%
P (C+B) = 66%
So , the door you choose plus the door opened is also 66%
@@jacobinozznope.
Your door stays one third. The remaining door is 2/3rds.
Test it with 100 doors. Or a deck of cards.
How often do you win by staying?
Can i ask ?
A = 33%
B = 33%
C = 33%
If they chose A and the B is goat, doesnt it should be
A = 33% + (33%/2)
C = 33% + (33%/2)
@@NDAsDontCoverIllegalActs agreed. Im saying the answer becomes obvious when you increase the number of doors.
OK if someone can help me with this: The moment the host asks ''Do you want to open door number 2?'' is it not true that Monty opening door number 3 is completely irrelevant to the chances?
Because the moment there are only two doors left, he has no choice. He HAS to ask (If that is the assumption that he HAS to ask btw. And second assumption he can never ask you if you want to open the door you chose yourself) if you want to open the only other door left. There is no choice. So the whole ''Oh he picks the one with the goats'' Is completely irrelevant at this point because he is not PICKING any door cause there is only one door left.
If door number 2 has a goat? He HAS to ask the question if you want to choose that door be opened. And if the door has a car, he has to ask the same question also. He only has one choice and therefor your Him knowing what is behind the door is completely irrelevant.
And even if the hosts has the ability to ask if you want to open door number one. Or the ability to not ask any questions at all --> All of this can be done to manipulatie you and he can in his head choose a door randomly. Either making you believe he wants you to open the door he is asking about. Or either making you believe that he does not want you to open the door he is talking about. And even those two choices is a matter of manipulation of wich you have no control over.
From the moment i saw this paradox the only thing that comes up in my head is that its 50/50. Because the QUESTION of the host is entirely irrelevant. That seems to be the real confusion of all of this. Is that people believe the previous choices of the host affect the last choice. And thats wrong because the last choice is not a choice anymore.
EDIT: Ok i was completely wrong lmao. This one fucked with my head. Now i finaly get it.
Where you messed up is here - 'it doesn't say the host picking the car isn't a possibility'.
It's not about what's possible in other games, it's about the game you are presented with.
In this scenario, you pick, the host, knowing what's behind the doors, opens a door with a goat. Then he offers the switch.
He doesn't open a door with a car in the scenario you are asked about, therefore that's not an actual option in the probability tree.
I am not talking about other games, I am talking about this game. If you roll a 6 with dice it doesn't mean a 4, or 1 wasn't possible at the time. The odds that you would get that 6 is still 1/6th, unless it's impossible to roll certain numbers.
@@logicalmorality4646 this game is not dice. It is a specific problem. The Monty Hall problem.
He doesn't pick the car in this scenario, yet you include it as a possibility.
@@logicalmorality4646 Why you did you outline that you're talking about the Monty Hall problem in the title, when you're not? Instead you're just talking about straight up probability.
@@logicalmorality4646 Certain things are eliminated by the rules. In your dice example, if someone changes dice and replaces 6 by 5 then your dice can never throw 6. The trick is to know that dice is loaded. In this game show, the host can NEVER reveal the car because at that point the game becomes pointless. So he will always have to reveal a goat. The host does not chooses doors with equal probablity, in fact by the rules of this game he will always chose goat door. Otherwise the game is meaningless.
To simplify, I deal three cards, one of which is an ace. You select one card. I have the two remaining. I am twice as likely to have the ace. I remove a card that isnt an ace, but Im still twice as likely to have been dealt the ace. The odds of my having the ace didnt change by removing a none ace card.
I was skeptical about this paradox until I wrote this: There are 3 options:
I select a car, then the host reveals a goat and the remaining door must contain a goat.
I select a goat, then the host reveals another goat and the remaining door must contain a car.
I select a goat, then the host reveals another goat and the remaining door must contain a car.
In 2 of 3 cases the remaining door contains a car... so I should switch to it.
oh thank you, i was so confused until i read this
granted, i haven't actually watched the video yet
@@__sarik Because there are two goats :)
But there is also a two thirds chance that the car was behind either the door he opened or the door you chose so you should stick with your choice by that logic or admit that the entire concept you are using is wrong. Why artificially tie the two doors you didn't choose together? Probability doesn't work that way. QED
@@TimBee100 read the first option. Also the door he reveals can never be the car or what would the point to the game be?
@@TimBee100 He never opens the price door. If you wanna change the monty hall problem so that the price can be revealed its still always more advantageous to switch.
Lets say the first player quits after hall removes a loser and asks the player if they want to switch, And a new player comes in facing two doors with a winner and a loser and is asked to continue the first players game not knowing which pick is original , and which pick is a switch.
Without changing anything that has happened, the second play will be asked which of the two doors would you like to pick. there odds of winning will be 50 / 50 because one has a car and one does not have a car.
Actually no. Their odds of winning will depend on which door they pick. If they pick the original contestant's door, they will have a 1/3 chance of getting the car. It is the doors that have the odds, not the player. What you are thinking of is the statistical average of many such games where the first player leaves and a new player guesses. It is no different than the first player flipping a coin to decide to switch or stay. The odds for one game will depend on the door but the statistics will average out to 50/50.
@@Hank254 A second player is faced with two doors one has a winner and the other is a loser. They are asked to pick one of the two doors. They might pick the first players original pick or the other. They don't know. Try it. Two doors. One winner .
@@immrnoidall Yes, all of that is correct. What I am explaining is that the doors already have different probabilities of having the car. Take a deck of cards and two coffee cans. Pick a random card from the deck and put it in can A. Put the other 51 cards in can B. You are trying to say that if a person came in after you did that, and they chose can A, they would have a 50-50 shot at picking the 'winning' card. That is simply not correct. They have a 1/52 chance if they pick can A and a 51/52 chance if they pick can B. The cans have different probabilities regardless of what the person saw.
@@Hank254 lol. 51? No . Sorry. I said nothing of the sort.
Now pay attention. There are 2 doors. One is a winner. And one is not a winner.
@@Hank254 In the actual TV game show, if you first picked one of the 2 losers, more often than not, you just lost, no switch was offered.
Ok, apparently this video attracts traffic from people that want to prove their 50/50 probability is right. But keep in mind the video poster serves a different math problem than what majority believe talking about. One where the host can also reveal doors with cars and make you lose prematurely. In other words, if you've picked a door and the host decides to show you a car in another door, there is no point in asking to switch. You already lost and that's the end. The video poster believes this could happen in the game show, because the phrasing doesn't explicitly say it can't.
While I agree it doesn't explicitly say so in the phrasing, I still think logically it's not what the game show is about, here's why:
1) It makes no sense to involve randomness, because it phrases "the host, who knows what's behind the doors..." and the host picks. Doesn't it seem odd to explicitly phrase the host knows all the doors contents and the host proceeds to roll a dice for his door pick? What is the point of saying he knows all the contents in this context?
2) Let's say it's not random, and the host, who knows what's behind all the doors, picks deliberately. This is incredibly unfair to the contestants, because this means the host controls the show and decides whenever a contestant should lose or not.
3) In addition to this the phrase "...opens another door, say no.3, which has a goat" should be read as "...opens another door, which always has a goat, say no.3"
Therefore I think these game rules apply by logically reading the math problem:
1) The host opens another door every show.
2) The door the host picks will always have a goat.
3) The host always asks if you want to switch.
based on this ruleset, switching will always give you a 2/3 chance of winning.
Completely agree, very good explanation.
"1) It makes no sense to involve randomness, because it phrases "the host, who knows what's behind the doors..." and the host picks. Doesn't it seem odd to explicitly phrase the host knows all the doors contents and the host proceeds to roll a dice for his door pick? What is the point of saying he knows all the contents in this context?"
Exactly.
But here is the point. If the intent actually is to clarify how the scenario is unfolding with the assumptions that creates the 2/3 vs 1/3, the question is not just slightly poorly phrased, it's almost perfectly taylor made to be systematically unclear an ambigious. A perfection to such an extent that it is generally hard to believe it was not intentional.
Which is why the only useful lesson from this is "Ask stupid and deliberatley poorly phrased questions => get a meaningless debate."
As a result, it's absolutely not a good question for educational purposes as it mixes up several other confusing aspects as psychology and extremely poorly phrasing to a subject that many already find confusing without all this additional and deliberately misleading nonsense. That is why it's important to point out why they actually went so far overboard when they phrased the initial question, that they unambiguously managed to describe a scenario were we cannot establish an asymmetric redistribution of the probabilities. It could perhaps serve as a valuable lesson to never ask questions in that manner.
Nailed it. It’s clearly discussed as a math problem and it’s quite obvious that it implied the host will always reveal a goat and always offer a switch.
As you say, it makes no sense to point out that the host knows where the car is but could then randomly reveal it.
Why do people have to complicate things?
It's 50/50 Following mathematical logic, It would be 66% but If would exist a strict pattern. If Monty show would follow this "pattern" a lot of people would have won the car. No all the times the car is put in the same place. According to the math, a bullet would never hit the target because you can devide infinitely the distance, for instance, 1meter left, half meter, 1/4 meter 1/10000000000 and never would hit the object.
I find it quite strange that people believe it is always better to switch mathematically even though the odds do not seem to change all that much with just three doors, especially if the host cannot pick the prize door.
What I see:
(Legend: X=revealed; 0=unopened door; 1=chosen door)
*All* possible choices if [0][1][0] is the correct answer,
[1][0][X] incorrect
[0][1][X] correct
[X][0][1] incorrect
[X][1][0] correct
So no matter if the door is switched or not, you will always have a 50/50 chance of winning if you switched for just that scenario.
What is flawed with what is given above if what seems to be the consensus believes that there are higher odds of winning if you switched?
If the host can not reveal the car, then those odds change from
1/6 you pick car host reveals goat 2
1/6 you pick car host reveals goat 1
1/3 you pick goat 1 host reveals goat 2
1/3 you pick goat 2 host reveals goat 1
0/3 you pick goat 1 host reveals car
0/3 you pick goat 2 host reveals car
Because remember the two losing switches are a subset of you having selected the car. They should add up to 1/3 probability total as there is a max 1/3 chance you can pick the car. On the other hand if the host can't reveal the car, then if you select either goat you have won for sure by switching because there is no subsets which would normally exist if he revealed randomly. (normally 50% of the time you would lose by switching if you picked a goat, because half the time you would be trading a goat for another goat)
@@logicalmorality4646 I'm not certain about your logic. Two you have of the above six possibilities are precluded because he can't reveal a car. Now the possibilities are:
1/4 you pick car host reveals goat 2.
1/4 you pick car host reveals goat 1.
1/4 you pick goat 1 host reveals goat 2.
14 you pick goat 2 host reveals goat 1
Two you're right, two you're wrong. 50%.
If he can't pick the car, and you select goat 1 for example, then the probability of him revealing goat 2 is now 100% correct? So it goes from 1/6 to 1/3
With one event removed , the odds have now gone from 1:3 to 1:2 for chances of having the car . Either event choice is a 50/50 chance of being the car in other words . The psychology behind the question would make me stick with original event choice . But he could already have factored for me calculating a psychological strategy and be a step ahead , l know .
It's never 1/6. It's 1/3 stay, 2/3 switch as long as the host must open a door with a goat.
Here are the possibile scenarios
You. Prize Door. Don’t. Do
1. 1. Win. Lose
1. 2. Lose. Win
1. 3. Lose. Win
2. 1. Lose. Win
2. 2. Win. Lose
2. 3. Lose. Win
3. 1. Lose. Win
3. 2. Lose. Win
3. 3. Win. Lose
Scenario 1: You pick door 1 and the prize is actually behind door 1.
In this case, Monty will open either door 2 or 3 and show you that nothing is behind one of those doors. If you stay with door 1, you win.
Scenario 2: You pick door 1 and the prize is actually behind door 2.
In this case, Monty must open door 3 and show you that nothing is behind it. If you stay with door 1, you lose.
Scenario 3: You pick door 1 and the prize is actually behind door 3.
In this case, Monty must open door 2 and show you that nothing is behind it. If you stay with door 1, you lose.
You have one try, one door. The House has two tries, two doors. The house then gets rid of one bad door. IF the house had the one good door, they will always get rid of their bad door thus leaving their good door for you to trade with. ( I'm using "House" as a casino reference.)
I would say concealed choices and prize and booby prize . Forget car , forget doors, forget goats . People are too easily bamboozled by superficial irrelevant details .
Yes if you make all of those additional assumptions it would be beneficial to switch. If you explain the scenario correctly and concise there aren't really that many that falls for it. But for some reason it almost never is explained in a concise way.
The problem is that many of the necessary conditions for the paradox to emerge is not stated in the question in Parade Magazine. Instead we are jus being described an instance of the game show and what events that unfolded then. We are never informed why the events unfolded as they did, how the host must act nor how the game works in general. If anything the phrasing in Parade Magazine is a complete train wreck of a question. It's as if every semantic option they made was to maximize the confusion, but they far overdid it and no longer even described a scenario where we can establish that the intended paradox emerges.
I don’t know that this “debunks” the Monty hall problem, but it’s a good reminder to clearly lay out all the assumptions and factors when describing it.
In other words, 2/3 of the time you will have initially picked a goat door. The host reveals the other goat door. So 2/3 of the time, by switching, you gain the car door.
If he systematically avoids revealing the car yes. If he randomly opens no. The wording from Parade magazine suggests randomness.
@@logicalmorality4646 Hall, if he knows, could always be knocking off a goat door. There is always one available. Your odds were one of three of getting the car, and I can't see an advantage to switching. Your odds now are not one of three, but one of two. There is no harm in switching, as the odds are even. It would just be a matter of luck. What dos Monty have to do with it?
@@logicalmorality4646 It's clearly stated in the original problem - a letter from a reader - that the host KNOWS what is behind each door. No randomness is implied.
@@jasonthomas5118 It's been thoroughly demonstrated in physical tests that switching definitively doubles your chances of winning. Of course it's also been proven mathematically and graphically.
My favorite is simply that when you picked a door, you divided the set of all doors into two sets. The probability of the car being in your one door set is 1/3, the probability it is in the 2 door set is 2/3, the sum of each door's probability. That is fixed for the two sets. Opening one of the goat doors reveals that door had a zero probability. Therefore, the set's probability resides entirely in the unopened door in that set, and it is 2/3ds.
@@logicalmorality4646 "The wording from Parade magazine suggests randomness."
ROFLMAO!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
This has became an analysis of how the question was worded rather than Math. This was also the debate among professors back in those days as they said the intend of the host is important.
Among other things, you apparently misread the original problem. Nothing in it implies randomness. It states he will open a door which has a goat. (In the phrase, "opens another door, say No. 3, which has a goat" I think you're interpreting the appositive phrase "say No. 3" as if it implies a random choice, but it's just labeling the door for future reference.)
Of course any "debunking" of the problem has to explain away the abundant experimental evidence that, indeed, switching doubles your chances of winning the car. Good luck with that.
I put all my responses under 1 comment to make it more organized.
William Ivey and Marilyn Von Savant are correct because it is stating in the problem that a goat is revealed. Therefore, the probability begins at that point not prior to that point, which you are suggesting Logical Morality.
In fact, your own probability tree answers the problem for you. The problem begins at the second column of your probability tree because it states that we know a goat is revealed and at that point we have only three options: 1) switch off the car on to a goat (goat 2 was revealed), 2) switch off goat to a car (goat 2 was revealed), switch off of goat to a car (goat 2 was revealed). That means two out of the three options favor you switching. Therefore, you should certainly switch.
@@philb1208 As I explained in the video, the host could reveal the car 99.9% or even 100% of the time whenever you pick a goat.... and everything stated in the problem could still be true. If that were the case you should never switch. In fact this is actually the most logical way to play too...
1) Do you agree the host wants the contestant to lose? (Usually the casino/carnival wants the player to lose)
2) If the player selects the car then you should give them the option to switch... They have won for sure, but if you give them the option to switch then they might do it and lose...
3) If the player picks the goat then you should reveal the car. This way the player has no option to switch, they can only trade a goat for another goat.
@@logicalmorality4646 hes right on the math, the problem is everyone is thinking about the first choice its the second choice that actually determines what the out come is because he always reveals the goat, the real trick is realizing there were NEVER THREE CHOICES. the reason is that since he will never reveal a car and he knows were the goats and car are he will choose to show you the goat at that time it can still be behind either do that part has not changed, the problem is that people are thinking about it wrong
@@Chaosmage42 based on the wording of the problem you don't have that information. All you know is that he revealed a goat once ever. You don't know why.
The host always knows where the car is. The host revealing the car would be pointless to the gameshow, so the probability of this happening is zero.
Exactly. He is trying to make you loose.
the host?
Well, if he "always" opens a door with a goat, his motives would be irrelevant.
But this is a single scenario problem. There is no always.
@@gnlout7403 no because he would never open the door with the car. It would be pointless to ask you if you want to change doors in this scenario. He always opens a door with a goat. If there's no host, the solution which this video provides would be right. But there is a host. So you're wrong.
@@iliyastoyanov3302 you said the host is trying to make you lose.
I'm saying his motives are irrelevant if he's bound to open a door with a goat, which you seem to agree with
@@gnlout7403 yeah basically. he has the inside info from the team of the show.
I admit my post of nine hours ago was wrong. I did this using cards. Two jokers and one ace. Shuffle the cards and put them face down and pick one at random, or choose them in order, 1, 2, 3, each time. It came out overwhelmingly in favor of switching. So the adult thing is to admit I was wrong. The key is apparently Monty knocking out a dead door.
No, there's a flaw in your method. If you pick one at random you're not following the problem. Monty knows there's an incorrect choice which he must reveal. In your example you have to select one, leave it face down, and lift one of the remaining two. In this game, you have a one in three chance of being correct. If one of the deselected ones is revealed to be correct, it's no longer the Monty Hall problem. If it is incorrect, you go on to game two, which I cannot overemphasize. In game two the odds are 50/50. Anything else is logically erroneous.
I think you didn't watch the video. Yes what you are saying is correct, the problem is that the problem does not have any rule where the host avoids revealing a car, has to reveal one door, can't reveal your door, etc...
@@kurtbader9711 "In your example you have to select one, leave it face down, and lift one of the remaining two."
No you don't. You can turn over the card that was picked and record it as a win or loss by staying.
@@logicalmorality4646 the problem literally says that in this scenario a goat is revealed
I don't think you _debunked_ the Monty Hall Problem, but you did point out the crucial assumption that is often missing from the problem statement, and your mathematics is completely correct here.
What assumption is missing?
@@gnlout7403 The crucial assumption that is _often_ missing (but not always) is that the host intentionally revealed a goat from among the two remaining doors, regardless of what the contestant's original door was.
You can get around this by using words like "always" - i.e., the host "always" opens a door with a goat and "never" opens the door the contestant originally picked. But this implies a repeated thing, which isn't necessary. One could also say that the host decided on following an algorithm beforehand where they would select reveal a goat door from among the two unpicked doors, regardless of what the contestant's original door was.
The reason these need to be specified is because of variants of the Monty Hall Problem which appear identical from the contestant's perspective, including:
Ignorant Monty. The host follows the following algorithm: After the contestant chooses a door, the host randomly selects one of the remaining two doors and reveals it. If it's the car, the host says, "Oh, too bad. You lose." and the game ends. If it's a goat, the host offers the contestant the option to stick or switch. (If the contestant is given the option to stick or switch, switching wins with probability 1/2.)
Monty Hell. The host follows the following algorithm: If the contestant picked a goat door, the host reveals the contestant's door and says, "Oh, too bad. You lose." and the game ends. If the contestant originally picked the car, then the host reveals one of the two other doors and gives the contestant the option to stick or switch. (If the contestant is given the option to stick or switch, switching wins with probability 0.)
Angelic Monty. The host follows the following algorithm: If the contestant picked the car, the host reveals the contestant's door and says, "Wow! Congratulations! You won!" and the game ends. If the contestant originally picked a goat, then the host reveals the other goat and gives the contestant the option to stick or switch. (If the contestant is given the option to stick or switch, switching wins with probability 1.)
None of these three variants are the _actual_ Monty Hall Problem. But if people don't specify the strict rules the hosting is acting under, the contestant could be in the "stick or switch" subcase of any of these situations, just knowing the host revealed a goat from the remaining two doors. The Monty Hell and Angelic Monty scenarios further show that the host's knowledge is not enough (since the host must know the car's location in these two scenarios as well). So the host must be acting in a very specific way to get 2/3 as the probability of winning by switching if given the choice to stick or switch.
@@MuffinsAPlenty you are correct in that you don't need him to act this way every time. You are only given one scenario. You weren't told what he might do the next game. Still the fact that you are told he knows what's behind the doors and chooses to open a door with a goat means he is not picking randomly. You should be able to calculate the odds from this information
@@gnlout7403 It depends on how it's worded! The wording that the video gives does not say that the host "chose" to reveal a goat. It simply said that the host knew where the car was _and_ opened a door with a goat. I think it's reasonable to assume the host intentionally revealed a goat, but that is only _reasonable_ not _logical._
But even then, if you make the reasonable (albeit not logical) assumption that the host's knowledge implied that the host intentionally chose that goat door, you still can't eliminate the possibility of being in Monty Hell or Angelic Monty.
@@MuffinsAPlenty I disagree. If he knows and opens a door with a goat, the assumption is that's it's purposeful, not random. If it said he didn't know what's behind the doors it'd be random.
I have no problem with the wording
BTW as long as Monty chose the door intentionally, aka, not randomly, intentions are irrelevant.
I choose a door = 33% chance i'm right. No matter what, one of the other two doors is wrong. Opening the door that's wrong just reminds us that it's the case. The door you chose still had a 33% chance.
this might actually the best explanation i have found
Go to Wikipedia. It is THOROUGHLY explained. I thought it was 50/50 as well, but the Genius is right. It's a TRULY REMARKABLE problem. ✌
If your pick is wrong, switching will automatically get you the car. You pick incorrectly 2/3 of the time so switching gets you the car 2/3 of the time.
No it won't there is 6 possible outcomes, 2/6 you win by switching, 2/6 you lose, and 2/6 are neutral (you trade a goat for a goat)
@@logicalmorality4646 I think you misunderstand me. If I pick a goat, the host will always pick a goat meaning if I switch it'll always be the car. The problem is I don't know that I've picked a goat.
@@marctspence You can't get that information from the information given in Parade magazine. There is no rule saying that.
@@logicalmorality4646 Sure there is, the Parade article specifically states the host reveals a goat after the contestant picks a door.
@@Hank254 ... It says "the host who knows what's behind the doors, opens another door, say door #3 which has a goat"
It doesn't say he always reveals a goat no matter what all the time. It just tells you in 1 instance he reveals a goat. He could reveal a car 99.999% of the time and this would still be true.
So, you have just decided to be nitpicky about host purposely avoiding the car?
Well, this is usually mentioned in the problems, so I don't know where did you find the version where this isn't mentioned
It's the original version that popularized it.
My thinking was the same as yours but it turns out that I was missing an important piece of information, I did the maths again and it turns out that you SHOULD switch every time. When the host opens a door with a goat behind, that in fact does change the probability function.
If you choose not to switch:
Choice Outcome
G1 G1
G2 G2
C C
Pr(C) = 1/3
If you switch:
Choice Outcome
G1 C
G2 C
C G1 or G2
Pr(C) = 2/3
In other words, the probability of getting the car if you don't switch is 1/3 and the probability of getting a goat is 2/3. If you've chosen a goat and you switch, you WILL get the car. Since choosing a goat is more likely to happen, you should always switch.
Watch the video before commenting.
@@logicalmorality4646 I did. Your calculations at 1:39 assume that nothing can change the probability function which is false. Pr(A) is not the same as Pr(A|B) unless Pr(B) is certain. The probability of event A can change given event B.
So once you've picked your first door, Pr(C) = 1/3 and Pr(G) = 2/3
Now the host reveals a goat, and you have the option to switch. If you don't switch you have a 1/3 chance of winning the car. If you switch, this is what happens:
Pr(G1|C) = 1/2
Pr(G2|C) = 1/2
Pr(G1|G2) = 0
Pr(G2|G1) = 0
Pr(C|G1) = 1
Pr(C|G2) = 1
@@thebassboar3980 I don't think you watched it until now but that's not important hahaha. Think hard about this...
-Suppose the host is biased and reveals the car 90% of the time when he has the option to... (we are assuming he won't reveal your door even if you pick the car).
If this were the case and you were revealed a goat, then is it still favorable to switch? OF COURSE NOT. In the event of 2 unlikely possibilities, the least unlikely becomes the most likely scenario. In this case, you are far more likely to have been revealed a goat by selecting the car (1/3 chance), than haven got a 2/3 chance of a 10% chance of the host revealing it. According to your logic, this bias should not affect the probability... But it clearly does. Just build a simulation and you will see it instantly. Or just change the bias to 100%, and it's obvious as hell you picked the car.
Does that make clear sense?
Dude, the host never reveals the car after the first door has been chosen. He always reveals a goat. That’s why the suggestion that you should always switch has been the most corroborated solution.
And why would the host reveal the car? If that were the case the game wouldn’t make any sense
ua-cam.com/video/C4vRTzsv4os/v-deo.htmlsi=_8UD3iJ2PIuQgOwl
48 secs video on Numberphile2 explaining this so easily 😂
Super simple: your initial odds of picking the wrong door are 2 out of 3. Since the host knows where the prize is and, by definition, must remain fair and neutral, the next door he removes from your choices can't be one that hides the prize. After this new information is available, since you would have initially chosen the wrong door 66.6% of the time and now only two doors remain, switching will make you win 66.6% of the time.
But the other un opened door also had the same chance of being wrong. You're just describing a 50/50 state using the language of thirds
@@failurefiend If there is a 2/3 chance of the firs pick being a goat, that means there is a 2/3 chance the host will have the car. There is no 50/50 in this puzzle.
@@failurefiendnope. The other door does not have the same odds as the door you picked.
Test it with a deck of cards and tell me how often you win by staying.
@@failurefiend Let me just explain this in (kinda) simple terms. (I am just trying to explain how to get the car initially.) There is a 2/3 chance that you choose a door with a goat behind it, being that there are 2 goats and only one car. That statistic does not change after the host opens a door with a goat behind it, so it is mathematically impossible that you have a 50/50 chance of choosing either the goat or the car. This means you have a 66.6% chance of getting the goat if you stay, but have a 66.6% chance of getting the car if you switch. (Image credit to By Rick Block - en.wikipedia.org/wiki/File:Monty-RightCarSwitch.svg, Public Domain, commons.wikimedia.org/w/index.php?curid=68521184) upload.wikimedia.org/wikipedia/commons/thumb/4/41/Monty_Hall_Problem_-_Standard_probabilities.svg/1024px-Monty_Hall_Problem_-_Standard_probabilities.svg.png
@user-vn1lz4cq2r Yes you are giving a good description of how one can think about the actual intended problem as with for instance the much better prisoner dilemma version of it. But just read how specific and advanced you must clarify the assumptions behind, as they never are in the phrasing in the Parade Magazine. You wrote
"Since the host knows where the prize is and, by definition, must remain fair and neutral, the next door he removes from your choices can't be one that hides the prize."
and although we can certainly choose to make that assumption, we really shouldn't in general. Since the scenario where the paradox emerges is completely dependent of that (and that he ALWAYS opens a door), those things ought to have been explicitly explained in the question, not just vaguely implied as they are the very cornerstone for the paradox in the first place. Remove any of those two and the answer is not 2/3 for switching and 1/3 for staying.
That is the entire point which is actually by far the biggest takeaway from this problem. Wee need to make a busload of assumptions that are not explicitly stated in the question to arrive at the situation where the 2/3 vs 1/3 situation occurs.
How do we know that the host cannot open a door with a car behind it? We don't. If he cannot, it should have been stated. It never is, it's only described how he acted in the instance "We are supposed to be at". We must for some very bizarre reason assume that in order to be able to give a concise answer.
How do we know the host always chooses a door to open giving you this option and or if he doesn't always, makes this decision independent of what is behind the door the contestant chose. We don't know either, but one of these criteria needs to be informed for the paradox to emerge. But neither are informed. Again, we need to somehow assume the former or the latter for the paradox to emerge.
The way the question was phrased in Parade Magazine is a good example of
"Ask a vague question, get a wide variety of answers"
If the host has to eliminate a losing door, you win 2/3rds of the time if you switch.
If he chooses randomly, it's 50 50.
Dawg did you watch the video?
@@awal7664 What about the video?
@@Stubbari switching is always 50/50 because the host will NEVER show the car, he always shows a goat
@@awal7664 In that case by switching you double your chances of winning from 1/3 to 2/3.
It's only 50/50 if the host opens a random door (could show the car but happens to show a goat).
@@Stubbari you got it backwards
You're assigning probabilities to events which cannot ever happen. The comparison with the coin toss is not accurate, because we know beforehand that the presenter will never reveal a car. That's not altering probabilities after observation, because we already knew it before the observation.
No I am not, you watched half the video. I explain at length how this information is not included in the original problem, and can not be figured out either based on the information given.
Nope he isn't. That claim is based on making tons of assumptions that are never there in the initial question. In mathematics we don't do that!
@@dataandcolours You don't understand. There is looking at this as a game vs. looking at this as a puzzle. It said the host, who knows what's behind the doors, opens another door that has a goat... Yes? It's no puzzle if he can reveal the car because it's understood the host must be allowed to switch, so standard assumptions must be followed. However, it's no game then because the contestant could just figure out to switch even before the game starts. To preserve the problem as a game, rather than a puzzle, Monty can and must be allowed to do anything that is not expressly prohibited by the rules. The letter said suppose you are on a game show ... Logical Morality is approaching the problem as if it were a real game, rather than a puzzle.
@@healthquest4823 You are undertaking some sort of Hercule Poirot investigation on how to interpret it so it can become a puzzle, instead of just reading what it actually says. This question as stated in Parade Magazine is legendarily poor, there is no way around it.
Instead of trying to replicate the Three Prisoner Dilemma, they made it into mostly a misunderstanding of the conditions issue with the horrible language used. There are three further conditions that all needs to be established for the 2/3 vs 1/3 scenario to occur. It's not even close.
@@dataandcolours I simply see it as how you approach the problem as written in the letter. As a puzzle, the so called assumptions are not even really assumptions at all but necessary to preserve it as a puzzle. The host simply must not reveal the car, so saying he knows what's behind the doors isn't even necessary. To preserve the narrative as a probability puzzle, the contestant must also be allowed to switch. However, the rather ridiculous thing here is if the host has knowledge of where everything is, then it automatically becomes a non-random process, eliminating the supposed intuitive 50:50 view from even being a possibility. The question is really only which door will provide the greater chance of winning, not IF there is any reason to bother switching.
The other view is to literally just read it as is - a serious question about a theoretical game. In that case, Monty Hall is free to do anything not expressly prohibited in the narrative. Then the 2/3 probability cannot be assessed because he didn't have to do anything given in the narrative beyond the initial set up, giving the contestant his or her a 1/3 chance to win. There simply is no solution because it is non-random. The problem with the video is that LM declared the actual answer to be 50:50, which is not correct.
If the contest is a farmer who hates cars and needs a goat then he should not switch
🤣 wth
Yes I agree that it should be stated that the host will ALWAYS open a door with a goat. If this is the case then yes you should switch. I think it's not mentioned because most people will just assume that to be the case - cuz - well -- that's how the game shows operate.
If he opened a door with the car its not as exciting then is it?
That rule is actually stated in the setup but this guy here ignores the rule and somehow says there is a chance he randomly choices a car. The setup says he picks a goat. So he will never choice the car according to the setup rules.
It is always stated
No it really doesn't make sense that that would happen. Follow this logic...
1) Does the host want you to win or lose? I assume you lose. Ultimately we don't know for sure though...
2) If you picked the car, what should the host do? He should give you the option to switch and reveal a goat. Your odds now decrease from 100% win, to something less...
3) If you pick a goat what should the host do? He should reveal the car of course. That way you have no possibility to win whether you switch doors or stick with the same door.
Show a flaw in that logic...
@@logicalmorality4646 It’s difficult to find a flaw in that logic because it’s not even logical. You aren’t using logic. You are using gut feelings. Also, your gut isn’t giving you good advice. The host isn’t looking for you to lose. He’s looking to make the show exciting. He needs to build the anticipation. Sure, if you pick the car initially he could simply reveal but he won’t because that is not building drama. Not only are you not good at logic, you don’t have very good instincts either.
@@gutenbird Even a 5 year old child can understand logic that basic. Really sad, did you drop out of elementary school?
I made two python scripts.
The first is as the MHP is played. ie. you pick a door. Monty then reveals a goat. Guess what? You get the car 2/3 of the time if you swap in that scenario. Why? Because your initial odds are 1/3 of choosing the car and there's 2/3 chances it's behind one of the other 2 doors. Monty then shows you a goat which now means that last door has 2/3 odds of being the car. Why? Because you are opening TWO doors, not one. This means 2/3 of the time it's best to pick TWO doors.
In the second script I played it blind. ie. you choose a door, and then swap it randomly for one of the other doors. Guess what? You only win 1/3 of the time. Why? Because you're only choosing ONE door.
What Monty is essentially doing is offering you the choice of one chance to win, or two chances. When you swap you are always swapping your one door, for the other two doors. Of course the odds are better because two chances are better than one.
After watching you last night, my son (PhD in Statistics) and I ran a 1000 cycle program in R simulating the Monty Hall Paradox. It did show the odds of winning by switching your choice is approximately 66%, not 50%.
That depends entirely on how you set it up. If you systematically avoid the car, you'll get 66% for switching. If you open the doors randomly, you'll get 50% for switching.
@@healthquest4823 Thanks for saying it out loud. At the end I remain with a 50% chance to chose the right or the wrong door. Nothing else.
@@niemand7811 No. That's true only if you reveal the goat randomly, which means you will also reveal the car 33% of the time. If you don't want to reveal the car so avoid doing so by using your knowledge of "what's behind the doors," then the odds for the contestant increase to 66% by switching.
Authority fallacy
@@entpboss5285 entp boss fallacy
You pick a door. the chance you got it correct is 1/3. It's more likely that the car is behind the doors you didn't pick because it's 2 to 1. This is fairly obvious. By switching you are basically choosing both of the other doors. When you think of it this way it's just plain obvious.
I watched your video, and after carefully considering it, I'd like to respectfully give a rebuttal, and would love to hear your response.
Your key argument (at 6:15) is that "the problem is leaving out a key assumption: the host is not opening doors randomly, he is systematically avoiding the car".
So here's the counter argument: the problem says "the host opens another door...which has a goat". So to be clear, the host has opened a door with a goat, and that is stated in the problem. Whether or not (as you say) "he is systematically avoiding the car" is irrelevant. The context of the problem is framed as a single game. It's not a sampling of multiple games. It doesn't matter if the host "could have" or "could not have" done something in the past, because the host has already opened a door with a goat. It doesn't matter why they opened the door with a goat either: they very well could have opened the door with a car but that is irrelevant because they didn't.
In this situation, switching is better.
Sure! Yeah I can't believe how emotional people get on this. Maybe I am right maybe I am wrong hahaha. Let's discuss it.
I had a very very long debate with Calvin over this, eventually he made a simulation to test it. But the short answer is, yes it does matter if he could have. The host has revealed a goat. Now we must ask, what are all possible paths to this scenario occurring? "How could we be in the situation we find ourselves in?" is a good way to phrase it...
path 1) you select car (1/3) host reveals goat A (1/2)
path 2) you select car (1/3) host reveals goat B (1/2)
path 3) you select goat A (1/3)host reveals goat B (1/2)
path 4) you select goat B (1/3) and host reveals goat A (1/2)
What we are really comparing are the odds of path 1+2 vs path 3+4... If the host could reveal a car, then the above odds are correct. However if the host must avoid the car, then in path 3, and 4, the odds change from 1/2 to 2/2. Thus 3+4 becomes more likely than 1+2
"However if the host must avoid the car"
The host already avoided the car, so yes by definition, the host must avoid the car. This is not a statistical sampling of games, it's a theoretical single-instance scenario where you've already sat down, chosen a first door, and the host has already shown a goat.
Let's add up the odds of your paths:
path 1) 1/6
path 2) 1/6
path 3) 1/3
path 4) 1/3
So:
- There's a 2/3 chance you're in path 3 or path 4. I.e., 2/3 chance you and the host both chose a goat
- If you both chose a goat, the remaining door is a car.
But this is all overcomplicating a much simpler way to explain it: "your first choice was probably (2/3) a goat, and they've revealed the only other goat, so the remaining door is probably a car!"
(P.S. your step 3 and step 4 say "1/2" where they should say "100%", because if you chose a goat then there is 100% chance the host will reveal the other goat.)
@@jay31415 No the system the host uses clearly changes the odds. Suppose you were dealt the 5 of spades randomly... What are the odds of that happening? 1/52. Now suppose you learn the dealer won't deal red cards... Well now it's 1/26.
Despite having already been dealt the 5 of spades, we have to update our calculation. The probability itself doesn't change, but now our knowledge of it has changed and is more accurate and closer to the real probability function. Thus we have to modify our calculation.
@Logical Morality But, the host already dealt a 5 of spades. So, if we update our calculation as you say, then we derive that the odds of this happening is 100%, because it already happened.
What is the chance it will rain today? Where I'm at, 100% , because it already rained this morning. Rain tomorrow, I do not know, maybe 50%. But when after tomorrow, it will collapse to either 100% or 0%.
Too many people complicate this. Here is the simplest proof:
- Your first choice is probably (66%) a goat.
- The host reveals the other goat (Fact. 100% chance that he does this.)
- Thus, the remaining door is probably (66%) not a goat.
I am on the same page
24 possible plays
win without changing - 25 %
win changhing - 25%
loose without changing - 25%
loose changing - 25%
Nah, this dude is wrong.
The Monty Hall Problem is solved already. You should switch.
Switching gets you the opposite of what you first picked, since there are only 2 doors left. 2 out of 3 times, you picked a goat initially.
So 2 out of 3 times, switching doors means you go from goat to car.
@@ZZubZZero I am not convinced yet.
@@jacobinozz Why not? Everything I said is correct. Also did you watch the video...?
@@jacobinozz The original poorly worded problem does not produce a 2/3 distribution. Only when the following is established do we get 2/3...
1) The host always reveals exactly 1 door after you select yours, this door can not be your door, and if there is a car behind it he will not open it.
2) The goats/cars are randomized
3) You want to win the car
I think assumptions #2, and #3 are fair enough... Sure... But #1 most definitely isn't. These are very specific rules that only belong to this very specific game. So the question is, were these rules established in the problem or not? Clearly not. There is no possible way in hell you can get rule #1 and the 3 critical conditions it states to build a 2/3 probability based on the wording of the original problem.
@@logicalmorality4646 Exactly. Other view is the probability to fail 2/3 is the same than choose the correct door+door opened
your first pick will be a goat more often than it will be a car, that is all you need to know
Exactly
It all condenses down to two concealed choices. A booby prize and valued prize . The probability in a randomised situation is 50/50 . It's always 50/50 . Hocus pocus and mumbo Jumbo has no effect.
@@philip5940 nope.
If the host knowingly opens a door with a goat - as opposed to randomly - you double your odds by switching.
There are tons of simulators and explanations on the internet that can help you understand it.
@@gnlout7403 I never suspected quantum physics extended to the macro .
@@philip5940 it's not quantum physics.
But it is a settled problem.
You can test it with cards if you don't believe it.
If you use cards and just randomly pick 1, 2, or 3 it is close to 70% better chance of winning if you switch. I didn't believe it either until I tried it over and over again.
I wanted to mention that I actually played this with my niece. I took three of her plastic toy French fries and placed a dot on one. I placed them on a glass table with the dot down and asked her to scramble them around, then look under the table to see where the dot was. I then picked one and moved it to the side without looking at it. Next I told her to flip one of the remaining two over that didn't have the dot. Now I could decide to switch the one I first chose with the one left. We did this over and over, and I kept track of the results on a nearby whiteboard. The times I switched clearly showed a 2/3 advantage. When I looked at what I chose first, I only won 1 in 3 times.
The thing is, there was no other practical way to do this. Saying well there was nothing in the letter that excluded showing the car, as you pointed out, means my niece could have exposed the French fry with the dot if she had wanted, at times when she had it. What was I to tell her..."You can just go ahead and reveal the dot every now and then"? That would just have made it impossible for me to have the opportunity to switch during those times.
So saying Savant was wrong is really stretching things. To get any particular answer to the question in the letter, you must follow her assumptions.
For a coin it's an indisputable 50/50 flip probability. Throw a coin a 100 times and you'll get results of around ⅔ & ⅓ fairly often. Throw it a thousand times and it might start to level at 50/50 . Maybe could throw a thousand coins up all at once to speed things up . Maybe..
@Philip Get a coin and flip it 100 times, and please come back with the results
@Philip I just did it, and it only took 16 flips to get fifty-fifty results. Heads pulled ahead at first with 7 heads to only 1 tail. Then tails started coming up more often, becoming equal at flip 16. Any flip of a coin is 50:50...period. The first two flips could just as easily been 50:50.
@@healthquest4823 uhuh , 16 is a small inconclusive sample. It's gotta be a big number of times.
@Philip A coin has only two sides. The odds are factually 50:50. How soon that bears itself out could be from the very beginning of any serious of flips. The outcome does not dictate the probability function. The probability function dictates the outcome. You are confusing ideas. If you are trying to determine odds, then you need a large sampling. That's true. In this case, we already know the odds. It would be extremely unlikely that flipping a coin 100 times would result in a 2:3 outcome. That would be directly counter to the predicted odds.
This becomes subjective when one discusses the host's strategy. Unless the strategy is clearly stated factually in the rules, conclusions cannot be drawn either way, however the common interpretation is that the host deliberately avoids eliminating the car to take the game to the next phase of uncertainty. That seems reasonable to me, otherwise there would be no point in offering a second chance to pick/switch a door. Traditionally if your last selection is random, then your odds are in fact 50/50, however this does not change the odds of switching Vs not switching which is 2/3 and 1/3 respectively. Thus there are actually 3 scenarios, however if you count doing nothing at all, then your odds are 0% You always initially had a 1/3 chance of selecting the car and 2/3 chance of selecting goat. Thus by switching doors you're more likely to end up with the car, because the door you picked first up, was always more likely to be the goat and with one door and goat removed, the only remaining switch option is the car 2/3 of the time and the goat 1/3 of the time. Thus a random selection between the last two doors ( 50%), offers better odds than not switching (33.33%) at all, but not as good odds as switching (66.66%).
Ah, the old change the question to fit the answer trick.
I argued with a close friend on this problem for several hours one time lol........ I told him I would never accept that switching my pick every time improved my chances. Its a clever illusion and nothing more
I don't even understand the Monty Hall puzzle
Did Monty ever reveal the "car" after the contestant picked? If he did, why would the contestant continue playing the game? I never watched the game show.
Neither have I hahhaa, all you have is the information given. Based on the information given, this is a possibility.
This was not a real segment on the game show... it is a math puzzle that was named for the host of a show because they use a similar format to describe it. The rules of the MHP say he will always open a door with a goat but you are correct... if he opened the door with the car, the game would have to be over and the player lost.
@@logicalmorality4646 you are horrible at discerning information, and even worse at comprehending what you read
in probability world revealing a car is present but in the game he eliminated this probability which made you and many not understanding the outcome is 50 50
You are wrong. The premise of the problem is that host will NEVER reveal the car. This is why he has to know which door has a car.
Stating that the host _always_ reveals a door, _never_ reveals the car, and _never_ reveals the contents of the constant's door is, indeed, enough to guarantee a 2/3 probability for winning by switching. But often, it's not stated in such a way that we can deduce that these always and never statements are true.
@@MuffinsAPlenty The reason it doesn't have to be stated outright is because the question is just about one instance of the game... not multiple instances. We are not told the rules and we don't need them because we are told what happened. Contestant picked a door, host knowingly opened a different door to reveal a goat and offered a switch to the remaining door. In the game described, the host did in fact follow the rules which you described. It is only when we want to consider multiple instances of the game that we need to actually state those rules so that each new game will also follow them. Multiple games are usually used in explanations so that is where the rules are discussed but the problem works fine without listing them.
@@Hank254 I don't know that the "this only happens once" argument makes much sense.
If we are having an "already happened" perspective, then the concept of probability breaks down because whether switching wins has already been determined, and is not up to random chance. From that perspective, the probability of winning by switching is "either 0 or 1, but we don't have enough information to determine which is true".
This is the reason that frequentist statistics uses such unnatural reasoning in concepts like confidence intervals and hypothesis testing. You can't take the probability of a parameter because the parameter is fixed value, not a random variable.
I have heard someone before say that for events which have already happened, we shouldn't use the word "probability" but instead "confidence". How much confidence would the contestant have that switching wins based on the limited information they have?
But even then, if you want to remove the "frequency" words like "always" and "never", you can rephrase my above reply as being in terms of the algorithm that the host employed in this specific instance when deciding which door to reveal. That algorithm only needs to be employed once. But the host many have employed an algorithm that makes it impossible for the contestant to win by switching. Or the host may have employed an algorithm that makes it impossible for the contestant to lose by switching. Or the host may have used the traditional Monty Hall Problem algorithm. Or may have just randomly opened a door. The contestant has no way of knowing which algorithm was employed and has no way of knowing how likely any of these algorithms were to be employed. So I don't know that the confidence perspective makes much sense either.
he's correct. You're told one scenario.
The host can do something else with another contestant for all we know.
Given what you are told, which is that the host is not selecting a losing door randomly but purposely and then offering you a switch, you double your odds by switching
@@gnlout7403 The most important aspect of Monty Hall-like scenarios is how the host decides which door to reveal. If you know how the host decides, then you can figure out whether switching is advantageous or not. If you don't know how the host decides, then you can't determine whether it is advantageous to switch or not.
Whether the scenario happens once or a million times, the analysis remains the same. The law of large numbers tells us this.
But the critical thing is that we _have to know how the host decided which door to reveal_ in order to come to a definitive conclusion. Saying that probability magically changes how it behaves because the scenario only plays out once is wrong.
Sadly I must admit, watching this is what convinced me the Savant solution is true. You have a 2/3 probability of picking a goat. When a choice is removed , you still have a 2/3 chance that your choice is a goat. Inversely, the only other door has a 1/3 chance of being a goat.
You are making the mistake of treating both goats as the same variable. Its two different goats though, goat A and goat B. You have a 1/3 chance of choosing goat A, a 1/3 chance of picking goat B, and a 1/3 chance of picking a car. Regard,es of what you picked there is a 100% chance either goat A or goat B will be removed as an option. If goat A is shown there is still only a 1/3 chance you picked goat B and If goat B is shown there is still only a 1/3 chance you picked goat A. Just as well you also had a 1/3 chance of having picked the car. No matter your first choice you are swapping out a 1/3 odds for another 1/3 odds from the initial pool of options.
You are looking at is as there is a 2/3 chance you didn’t pick the car so inversely changing you mind is favorable. However there is also a 2/3 chance you didn’t pick goat A when goat B is revealed, in which case changing your mind is unfavorable. There is also a 2/3 chance you didn’t pick goat B when goat A is revealed, in which changing your mind is unfavorable. If changing your mind has just as much of a chance to be good(2/3) as it does to be bad(2/3) then it is fifth fifty odds.
The real key to the confusion is assuming that the revealed goat, A or B, is still influencing the odds of the second opportunity to pick a car vs goat A or the car vs goat B. The reality is that goat A and B have nothing to do with one another, they are separate variables, and in the second chance one of these variables are removed making it a 1/2 chance you get a car.
@@robertdicke7249 that's not correct, and you can test this problem yourself. The actual game scenarios demonstrate this. -
You. Prize Door. Don’t. Do
1. 1. Win. Lose
1. 2. Lose. L Win
1. 3. Lose. Win
2. 1. Lose. Win
2. 2. Win. Lose
2. 3. Lose. Win
3. 1. Lose. Win
3. 2. Lose. Win
3. 3. Win. Lose
Scenario 1: You pick door 1 and the prize is actually behind door 1.
In this case, Monty will open either door 2 or 3 and show you that nothing is behind one of those doors. If you stay with door 1, you win.
Scenario 2: You pick door 1 and the prize is actually behind door 2.
In this case, Monty must open door 3 and show you that nothing is behind it. If you stay with door 1, you lose.
Scenario 3: You pick door 1 and the prize is actually behind door 3.
In this case, Monty must open door 2 and show you that nothing is behind it. If you stay with door 1, you lose.
You are correct, Kevin
@@robertdicke7249
Assume you stay with your first pick.
If your first pick is Goat A, you get Goat A.
If your first pick is Goat B, you get Goat B.
If your first pick is the car you get the car.
You only win 1 out of 3 games if you stay with your first pick.
Basic math/logic kids understand, idiot among idiots doesn't.
The way I see... it's true that after playing the same game over and over for a ten times or more the probabilities of getting the car after switching doors increases..BUT...that doesn't mean that every single time I play the game would be more beneficial to switch...it all depends on if you had chosen the wrong or the right door at the beginning...after eliminating one pf the door..the chances of getting a car by switching it doesn't increase it's the same.
You don't need to play multiple games in order for it to double your odds by switching.
One game and you still double your odds as long as the host opens a door with a goat purposely.
The problem does specifically say that the host will open a door that reveals a goat. He’ll reveal a goat every time. If you refer to the show itself, Monty hall never revealed the desirable prize. If he chose the door at random, your analysis would make sense, except that, if Monty opened the door with the car, it wouldn’t make much sense for him to then ask you if you’d like to change your choice.
If the problem requires you to watch the show, then that needs to be stated in the problem.
It's stated in the problem that he reveals a door with a goat. And like they said before me: It wouldn't make amy sense to ask you if you want to change your choice if the prize had already been revealed.
@@stefanief.687 I addressed this exact point in the video. Watch vid before commenting.
The explaination is quiet simple :
Only one car and TWO goats.
So :
1/3 : you pick the car, then you must stay to win.
2/3 : you pick goat 1 or goat 2, then you must switch to win.
Saying that the final choice is a 50/50 chance means that your initial choice is 1/2 to pick the car and 1/2 to pick a goat, which is mathematically wrong.
Then : ALWAYS SWITCH.
CQFD as we say in France :)
What the world has missed was:
1. Initially, all doors had 1/3 probability. By picking door one, the 2/3 probability favored doors 2+3. That's accurate.
2. With an important paradigm shift (revealing that 3 is a dud), you now have 'new' information to factor into re-calculating probability.
3. The probability of the car being behind doors 1 and 2, becomes 50-50 (1/2 - 1/2).
Marilyn Vos Savant, with an IQ of 228, got it wrong.
Maybe you should look at some of the other videos about the MHP. They do a better job at explaining it.
This is very simple, if the 3rd choice is wrong, then disregard it(remove it completely), the you only have 2 choices with 50/50 chances. Monty hall only works on program simulation.
Incorrect.
you double your odds by switching in the given scenario which is that the host knows what’s behind the doors and chooses to open a door with a goat. now, if the host did not know what was behind the doors and open the door with a goat, the two remaining doors would indeed be 50-50.
Not gonna lie, this is a pretty lame and clickbait video. It is obvious that the host will not choose a random door, risking to reveal the prize. This entire video basically comes down to: ""The question was wrongly phrased", even though in many videos (by the more popular math channels) it was stated that the host will always reveal a goat. Like others said, of course the host will not reveal the prize if you have chosen a door with a goat. Therefore you have a 2/3 chance to choose a door with a goat, forcing the host to reveal the other goat and thus you also have a 2/3 chance to win the prize if you switch.
Uhh no actually in math you are supposed to approach things logically instead of like an idiot. For example if I tell you "I deal you 2 cards, and one is the ace of spades, what is the other?".... It does not logically follow that the other card is the queen of diamonds because we are playing "English Poker Queen", and that the host is cheating and wants you to lose, and that is the best strategy in doing so... You would need to be told that information in order to know that... Yes it's obvious what the rules of Monty Hall are... AFTER BEING TOLD THEM...
@@logicalmorality4646 you are told that the host will always reveal a goat as he is not allowed to reveal a car. Did you even watch the original show? Please understand that the host is not allowed to reveal the car. It is explicitly stated before the round. The source you're pulling this from is a little ambiguous with it's phrasing, however the original gameshow where monty hall hosted this called "let's make a deal" always included the part where the host isn't allowed to reveal the car in all the publications they've ever released.
Hey I was wondering what your answer to this problem is:There are three bags, each containing two marbles. Bag A contains two white marbles, Bag B contains two black marbles, and Bag C contains one white marble and one black marble. You pick a random bag and take out one marble, which is white. What is the probability that the remaining marble from the same bag is also white?
Oops sorry on closer inspection I think it's 2/3.
1/6 Pick bag B, draw marble 1 (remainder is white)
1/6 Pick bag B, draw marble 2 (remainder is white)
1/6 Pick bag C, draw marble 1 (remainder is black)
I think that makes sense Then the rest of the scenarios we can ignore, and we can see 2/3 of these cases are white/white
Have you tried chat GTP? It's really good at mathematical and scientific logic. It's super good at this kind of stuff, I would ask it, it's like a calculator basically but can speak English. It's really bad at philosophy and politics though
@@logicalmorality4646 i see its because you already chose a white ball👍
@@logicalmorality4646 No I will definitely give it a try:)
You chose a weird hill to die on. You are right that the original question is not phrased well. But we know what the intention of the question was (maybe not right away but when you hear an explanation - you even present the intended interpretation in the video). You can complain all you want about the phrasing but this has nothing to do with "debunking".
Also, if you want to be this pedantic about the phrasing of the question (which isn't a bad thing) then you should be consistent. You are basically complaining that people are interpreting something into the question that isn't explicitly written down and then you do the exact same thing by assuming that he choses the door randomly.
Nothing in the question tells you this. This interpretation is even discouraged by stating that the host knows what's behind the doors. There could be a complex rule. It could be based on mood. There could even be situatons where he doesn't open a door at all.
In probability we assume that randomness is a default unless stated otherwise. If it's non-random then we simply just don't have enough info to make any calculation as there is a near infinite amount of non-random ways he could select. Given the information we have we can answer it for random selection, or just say the problem requires more info.
@@klaus7443 Mr.Mature is back! The sad thing is I know you aren't actually a child. I know you are at least... Let's say 18. No actual child can be this immature. What happens with humans is that mature children become mature over time, but immature children actually regress and grow in their malevolence, irrationality, and ignorance. To reach a level of immaturity this high it takes years of practice.
So, I’m thinking about this a lot. Now, I think sometimes just in concepts, without words in my mind, I know is weird, but it is like that. I don’t know how to explain it, but I have something that’s telling me it’s wrong that opinion with ‘better always switching because 33,(3) to 66,(6)’. I’m not a genius by others ‘definition and opinion(s)’, I took it like this: if after the third option’s elimination, we have 2 choices and one is double (by probability) than the other, why it’s not 25% to 75%, or 20% to 80% also 10% to 90%? How we count this, by having 2 options? How we arrive to 33,(3)-66,(6) and not to 25-75, those should be equal, but aren’t (33,3…=25 and 66,6…=75).This is a long comment 😁🫤. My point is, could be 3 options or 100, in the end is the same result, having the possibility to choose from 2 (and my mind tells me that is 50/50). The next question was for me: How probability works? What if I have 100 options divided in 2 groups of 50 options. And I must to choose one group, where I ‘think’ the price is. Now, let’s consider that we eliminate 98 of those options (the wrong ones, of course)obviously we eliminate 49 from each group, how we calculate the probability to find the price? I mean, what is the probability to choose the write group? Most ‘terrifying’ question, what is the probability in each group that if we choose a single option, that will be the price, considering that we don’t know in which group it is? 🤯🫠 Sorry for the long text, I still didn’t explain what I meant completely, because it will be too long. 😁😬 Finally, I have this intuition that is still 50/50 chances, because, in fact there is that possibility to choose too keep the first choice or not, from 2 options, because one wrong is gone. 🙃
All good, hahaha. Sorry I don't quite fully understand what you're saying. But in summary...
-It depends on the rules of the game
-If host must not reveal car, must reveal exactly 1 door each time after you select, and the door must not be your door... Then the odds become 2/3
-In original problem none of these rules were stated, so the answer is 50/50 by default or unknowable.
no.
You're not told rules of a game, only what the host did this one time.
Under these conditions, which is knowledge of the doors and the purposeful opening of a single losing door from the remaining two, you double your odds by switching.
What he did before or does after with another contestant is not relevant to this scenario.
If only someone tested this problem to find out how often they win or lose by staying and switching..... 🤔
Or if only the rules of the MHP stated that the host can't show a car...
_"Version 1 (Classic Monty). You are a player on a game show and are shown_
_three identical doors. Behind one is a car, behind the other two are goats._
_Monty Hall, the host of the show, asks you to choose one of the doors. You_
_do so, but you do not open your chosen door. Monty, who knows where the_
_car is, now opens one of the doors. He chooses his door in accordance with_
_the following rules:_
_1. Monty _*_always opens a door that conceals a goat._*
_2. Monty never opens the door you initially chose._
_3. If Monty can open more than one door without violating rules one and_
_two, then he chooses his door randomly._
_After Monty opens his door, he gives you the options of sticking with your_
_original choice, or switching to the other unopened door. What should you_
_do to maximize your chances of winning the car?_
Usually this is "verified" by using a certain algorithm where the host always opens one of the two remaining doors with a goat behind it.
You could well write a program that randomly opens one of the remaining to doors. Of all the cases that look like the MH problem (discounting the ones where the door with the car was opened) switching doors wins 50% of the time.
You could also write a program where a second door is only opened if your first choice was the door with the car. In that case switching doors never wins in cases that look like MH.
I will refrain from providing the programs, as the math is pretty clear.
@@Stubbari your problem description is much more accurate than the original. I like it. I would also add "the contestant knows / you know that these rules are applied in your game".
@@insignificantfool8592 except whenever it's actually tested, it's one third if you stay and two thirds if you switch.
But there are simulators online if you Google them.
Check out the Mythbusters episode of them testing it.
But if you know you only pick the winning door one in three times with your initial pick (we would agree on this), then you'd only win one in three games if you stick, logically.
No test necessary, but if you disagree, you would test it
@@gnlout7403 I explained three algorithms. The first one gets the 1/3, 2/3 result, the second 1/2, 1/2 and the third one 1 , 0.
If you wish we can play a few rounds of the game where winning is 50/50 if you switch. This hash value is good for up to ten games:
d09eaa8b6cbae2cf74b92bcd4c8b72e4
There are three doors. What's your pick (123)
Every time goat is picked the last door will have to be car. Since there's a 2 in 3 chance of picking goat, that's the chance of switching every time bringing you the car.
Not if the host reveals randomly. If he reveals doors randomly and your strategy is to always switch then...
1/6 you pick car host reveals goat 1 (loss)
1/6 you pick car host reveals goat 2 (loss)
1/6 you pick goat 1 host reveals car (neutral)
1/6 you pick goat 2 host reveals car (neutral)
1/6 you pick goat 1 host reveals goat 2 (win)
1/6 you pick goat 2 host reveals goat 1 (win)
@@logicalmorality4646 If the the host reveals a car. The door that is picked in the beginning and the the remaining door is automatically 0/6 or 0%. Which why is we don't ever consider the Host revealing the car.
Sticking to the the restrictions of the question "and the host, who knows what is behind the doors, opens another door, say No. 3, which has a goat"
"another door" meaning, the host will never pick the door the player has selected
"say No. 3, which has a goat" the host has chosen Door No. 3 and a goat is there
@@goonin3590 No it isn't that makes no sense. If it's possible for him to reveal the car (and there is no rule saying it's not) then it certainly does factor into our math. It makes it half as likely that you would pick a goat, and he would reveal a goat.
@@logicalmorality4646 No, because it is part of the game and if he has done this bit with the car 50 other times or whatever and the contestant knows the setup is always to reveal a goat from the other 2, he will realize it’s not random.
@@gutenbird Problem is you don't know that, you know what is written in the problem.
the presentation of your video sucks and your math is wrong. your notation is wrong: 1/6 should be placed behind the tree branches, not at the second stage of the tree, since this place is reserved for the branch probability which is 1/2.
of course, probability is modelled due to certain circumstances, for example the described game.
in the reality of the game, there is no case "car is revealed by host" and therefore, the probability of this event is zero, thus plays no role.
There are 2 possible scenarios, one where you pick a car and one where you pick a goat.
If you pick the car you have a win if you stay, if you pick a goat, you win if you swap.
You naturally have a 2/3 chance of picking the goat so to account for that you’re better off deciding to swap. That’s all it is. A lot of people overcomplicate this to a level that’s just straight up unnecessary and makes it look more questionable than it is.
There are 2 scenarios, one scenario is more likely than the other, so account for the more likely scenario. That’s it, and what is the most likely scenario? Picking a goat and having a goat revealed. That’s why you should always swap.
There's also the scenario where the host opens the door you picked. This complicates things more than most people like 😂
@@insignificantfool8592 there’s also the scenario where the host drives off with the car
@@skrypto2652 sure. But I think you have to learn to differentiate nonsensical from plausible scenarios.
How many game shows do you know where the host keeps the prize for himself? I don't know any.
How many game shows do you know where the contestant picks something and gets what he picked. That's literally all of them.
@@insignificantfool8592 my dude, the host always picks a door other than the one you choose, I thought you were joking and did the same.
You brought up an impossible scenario, the host opening your chosen door, that’s not how the game works and if you watched the video or read my comment properly you’d know that.
The way it works is simple enough for you to understand so try, you choose one out 3 options, and here’s the part you missed, the host then reveals another door that is a goat(he always reveals a goat because he knows what’s behind all the doors, it’s not random) he then asks if you want to stick with your original choice or switch to the other door that hasn’t been revealed for both of which you still don’t know what it is, that’s kind of why your asked if you want to switch or not genius.
Lmao It’s so funny when people don’t understand the Monty hall problem
I'm waiting for the day when the penny drops and he quietly deletes this video!
@@pwcinla I think you must have missed the point of the video. The way the problem was phrased in Parade Magazine, the alleged paradox doesn't even emerge as you need to do a busload of additional assumptions that was not stated in the question. It's only when you clearly establishes how the host always have to act (and in a very specific way) that the paradox emerges in the first place. So the Parade Magazine article appears to actually have created a Monty Hall Paradox Paradox judging by how many that see to be confused by it.
@@dataandcoloursthe host doesn't 'always' have to act any way.
You are given a scenario in which the host knows what's behind the doors and chooses to open a door with a goat.
In this specific scenario, do you double your odds by switching?
Yes, you do.
@@gnlout7403 You may say so. But if the host can act any way he wants it becomes impossible to calculate the conditional probability and the question instantly becomes obviously impossible to answer. Why did he "choose" to open a door with a goat?
In the "specific" scenario you give as an example it is obviously impossible to calculate the conditional probability, so the answer would be "Impossible to give an concise answer". Again you are scratching of what I like to call the Monty Hall Paradox Paradox: The way Parade Magazine presented it and consequently argued have actually created a lot of misconceptions that is not even about the intended paradox in the first place, but amongst others the misconception that it would not matter what rationale/reason the host had for doing what he did. From that perspective the entire circus has probably unfortunately had a net negative effect on education, especially if it's brought up in a school scenario were it steals time from other more useful scenarios.
@@dataandcolours he made a mistake when doing the video which is why he's getting flack.
First, you're not told what the host is going to do in future games. You're only given one specific scenario. You're also not told his motivations, so you shouldn't add them into the scenario.
Logical Morality ran the problem as if the host could, in the future, open a door with car. That's different from the one scenario you are given.
Based on everything staying in fixed positions behind doors and only playing odds the door you pick will have you at 33.3 chance to win car or 66.6 you will not win the car. Monty knowing what’s behind the doors opens one with goat. The mistake is thinking down to two doors you are down to 50/50 odds. Actually you are in exactly the same position as you started 33.3/66.6. Giving me the chance to give up my 33.3 door for a 66.6 door will be done in a second.
if monty can only open one door, that means you have a 50% chance. istg people purposefully don't know how to present this problem and are then somehow shocked when someone with a background in math says it's 50%. the way the problem is presented, it's 50%.
In the Monty Hall problem, the host is guaranteed to choose a door with a goat behind it. He cannot choose the door you chose. 2/3 of the time, you guess wrong, by switching, 2/3 of the time, you get a car. You misread the Monty Hall problem and failed to debunk it. Remove this video from youtube.
You don't get to decide what is suitable to be on UA-cam or not. Even if you are right you have a bad attitude.
Say there is a got and a sheep instead of two goats, and you pick the door with the car. There is two different scenarios under this condition. The host revealing the goat and the host revealing the sheep. Which means there’s is still just a 2/4 chance once one door is revealed.
@@Noname61574 as long as the host must open a door with a goat, you double your odds by switching. This dude made this video not understanding this point, but he 'corrects' it later in the comments.
Here are the scanarios -
You. Prize Door. Don’t. Do
1. 1. Win. Lose
1. 2. Lose. Win
1. 3. Lose. Win
2. 1. Lose. Win
2. 2. Win. Lose
2. 3. Lose. Win
3. 1. Lose. Win
3. 2. Lose. Win
3. 3. Win. Lose
@@gnlout7403 making it still a 1/2 chance if the entire premise of this illusion is that one was already revealed before the switch, extremely redundant
@@Obs23456 one door wasn't revealed before the switch. That would produce different odds.
the chart is the list of nine possible scenarios and outcomes for switching Vs staying.
Do you not agree with the chart? If so, where is it wrong?
Then tell me how you are able to pick the winning door with your first pick out of any more than two doors.
Everything about this depends on how you define "The Monty Hall Paradox". There are 2 scenarios that are discussed, and the definition in each scenario makes all the difference and allows 2 answers to be true depending entirely on what problem you are solving.
These things are the same in both scenarios:
-There are 3 doors
-You choose 1
-The host opens 1 of the 2 doors that you did not pick
This is where "The Monty Hall Paradox" changes into 2 different math problems.
Math problem 1: The host chooses a door that he knows has a goat behind it.
Math problem 2: The host randomly chooses one of the two remaining doors.
Math problem 1 is the scenario presented on the show. This is evident because one of the options presented to contestants on the show is NOT "Do you want to take the prize behind the door that Monty opens". This is not an option because there is a 100% certainty that Monty is not going to open the door with the car behind it.
So no matter what your choice is initially, Monty is going to open a door with a goat behind it. This is a constant. Because of this, your choice from the beginning is always 50/50 because the choice is not between 3 doors, it is between 2. Your choice is irrelevant, because no matter what you choose, what happens next is that a door will be opened that ALWAYS has a goat. In reality, there are only ever 2 choices ON THE SHOW--goat or no goat.
"The Monty Hall Paradox" is an awful name for the paradox because the math problem that everyone is solving with the result that you should switch doors is NOT the situation that you are presented with on the actual Monty Hall show. In reality, The Monty Hall Show does not present the circumstances that are used as a starting point in "The Monty Hall Paradox".
If you define the scenario as "The host will always pick a goat", then the odds are 50/50.
You started off well acknowledging the math problem, but then still went with 50/50. The reason it's 67/33 is because you can take advantage of the host!
1. You have to point to 1 out of 3 doors. There is 2/3 chance you select a goat.
2. Monty always reveals a second door with a goat.
3.There is a higher chance you both picked doors with goats at this point, because if you selected goat A, he has to show goat B. If you selected goat B, he has to show goat A. This means the last door having the car is most likely.
4. A different way to sum it up ,the only way you lose with a switch is if you switch out the car door in step 1. The chance of having picked a car in step 1 is 1/3.
So it's important to know for the math problem you make a choice out if 3 doors BEFORE Monty makes it two. Monty can only remove ONE specific door 2/3 of the time!!
You’re wrong
prove it
Logical Morality, after going through your video step by step, I see that you negated... and revealed a goat..., as a rule barring the car from being revealed; but you did not do the same for...who knows what's behind the door.
This leaves you with a non-random game, where you never have to reveal the car, but could if you wanted to. The only way to achieve 50:50 odd is by revealing the car 1/3 of the time. The only way to prove Marilyn Savant wrong is to be consistent and reject all her standard assumptions, making it a randomized game, that yields 1/3:1/3:1/3 (winning the car, getting the goat, and blowing the game before a switch can be made).
I just have one question. At 2:10 you talk about the probability function and the host's likelihood of picking either the goat or the car. You say even though the host picked a goat he still could have picked the car because nothing in the description says he can't pick the car. But if he knows where everything is, how can he even make a random pick? The is no "could have" anymore that I can imagine. He would always have prior knowledge and would have to intentionally pick either the car or a goat if the contestant happened to get a goat.
It's equivalent to saying "a card dealer has a marked deck of cards how will he deal?". He could choose to shuffle and deal them randomly, or he could deal them non-randomly a million different ways depending on his goals/strategy. Why does having a marked deck necessitate cheating? My argument in the video isn't that he will deal them randomly for sure, it's that he may or may not deal them randomly, there is no information to eliminate either possibility... Additionally if his method is non-random, there is no rational reason to conclude that the only non-random method of selection he could pick is to systematically avoid the car, reveal only 1 door every time, give the option to switch every time, not select your door, etc... Thus we can either answer the question for randomness, or just say it's a stupid and pointless question that lacks key information to answer...
@@logicalmorality4646 But, to get 50-50 odds, we can't leave it up to the host to decided to not cheat because that still makes the game non-random, period. We have to be guaranteed randomness by not allowing the host to know where the goats and the car are.
@@logicalmorality4646 Curt and Hank are the biggest trolls in the world. I have never seen such a thing on any other thread. When the correct reason is explained, most people will just say, "Oh, I get it now." Your video is excellent. Savant clearly was not answering the question in the letter. Even Monty Hall himself wrote a letter of disagreement to Steve Selvin who incorrectly solved the problem in 1975. As Monty explained, if you remove a goat, it just leaves two doors left with 50:50 odds (of course that is only true you could open the contestant's door too). It's just starts over with a new game. Monty went on to explain no game ever was played like that and no game ever had 2/3 odds. Only the rule, the car cannot be revealed, leads us to know the car has a 2/3 chance of being behind the other door since the host must refrain from revealing it two-thirds of the time.
@@healthquest4823 Poor little fella just doesn't get it. It's cute that he tries so hard though!
@@healthquest4823 "Only the rule, the car cannot be revealed, leads us to know the car has a 2/3 chance of being behind the other door since the host must refrain from revealing it two-thirds of the time.."
That's absolutely hilarious. First you say the car cannot be revealed then you say the host must refrain from revealing it two-thirds of the time....in the SAME sentence yet!
@logicalmorality4646. the decision tree as well as the clip coin example are wrong: At 1:40, you mention the "line 3" (blue circle) is not possible, but has a 1/6 chance. That's wrong. If you choose goat 1, then the host MUST choose goat, therefore the propabilities are 0 (car) and 2/6 (!) (goat). The same argument holds for case 3. Therefore the chance that the host picks a goat is 1/6 + 1/6 + 2/6 + 2/6 = 1 (for sure, the host can always pick a goat), but the probabilities that the remaining door is a car 0 + 2/6 + 2/6 = 2/3.
BTW that's the core of the Monty Hall problem. Your decision tree as sketched at 1:400 shows all possibilities when the host can pick ANY door (this means he could choose the car also). This is not Monty Hall. In Monty Hall, two possibilities can not occur, i.e. their probability is 0. If you simple "leave them out", then the decision tree does not sum up to 1 (!), but to 4/6. When calculating correctly, the probabilities are (1/6 + 1/6) + (0 + 2/6) + (0 + 2/6) = 2/6 + 2/6 + 2/6 = 1.
Dude, you're ignoring rules for the host. 1) Host must pick a door that the player did not pick and open it, 2) Host must always pick a door with a goat to open, 3) Host must offer you the choice of switching to the open door. If those rules are not followed then yes, its not to your advantage to switch, but game shows have rules to build excitement and ratings and such.
No you got it backwards. As a puzzle, there must be strict rules the host must follow...or there is no definitive answer. Logical Morality is approaching this as written as a game and can do anything not expressly prohibited in the narrative. There wouldn't be any tension in a game if you followed rules always leading to the same conclusion - just switch to get better odds.
@@healthquest4823 dude,you just said the host must follow strict rules, and I stated the rules. If the rules are ignored then the probability can be anything. Logical Morality is ignoring generally implied rules for this game/puzzle.
@@darkthedrummerful You're starting to understand. The probability indeed CAN be anything then. Yes! That is what you want for a game. If the contestants can figure out what to do - always switch - then it's no good for a game. Even Monty Hall himself wrote a letter saying that no such game was ever played like that, nor could it be. They never give odds to the contestants that in theory could be calculated to improve their chances of winning. A TV game is not a puzzle that is supposed to be figured out. It is a non-random crap shoot, controlled by the people in charge. They can do whatever they want for the sake of making the game fun to watch by changing their own rules anytime as they wish. LM is describing the letter literally as written. The first thing it said was, "Suppose you are on a game show... "
Savant could have and should have responded she could not answer the question as written. She then should have stated she can turn it into a probability puzzle, though, by making a few assumptions. In that case, you would improve your chances of winning from 1/3 to 2/3 by switching.
@@darkthedrummerful Did you know that John F. Kennedy's lowest approval rating was 7 points higher than Donald Trump's highest approval rating? LOL 56% to 49% Haha, incredible.
50/50 odds in this scenario are contained to an individualistic instance of the problem, mathematics takes in the percentile chance through more tests than any one person in this position could reasonably be expected to participate in. Meaning that the odds statistically will always be 66.7% if you switch, but the odds of any one person in the experiment, playing the game exactly one time, will always be a 50/50 gamble of being within the 33.3% that lose by switching, or the 66.7% that win by switching...The key word in this problem is statistics. You can't cherry pick a single person's instance when calculating probability.
I think you watched the first half. Given that the host systematically avoids revealing the car, has to reveal 1 door each time, will not reveal your door, etc... Then yes it becomes 2/3. If you play those rules it's 2/3. These rules are not part of the problem though. That's a different problem. What happened in Monty Hall is like if I asked you to list all prime numbers between 1 and 20, and you say "13 is the only correct answer because it's the only prime number between 1-20 that is larger than 12 and smaller than 14"... Well being between 12 and 14 was not part of the problem, that's a different problem. So that answer is in fact incorrect. The 2/3 solution to Monty Hall adds a bunch of rules that are not part of the problem.
@@logicalmorality4646 How many times is the game intended to be played in this context? Once or multiple, if once then the probability only takes place in a single instance meaning in essence that you are only arguing that the contestant or player after having chosen one door and another being revealed to be wrong leaves the contestant's choice as a 50/50 whether they are going to be more likely or not to win. If multiple, then you are not appropriately accounting for the fact that no matter what two ways you look at it, no single instance has any true basis on the scope of probability, probability by its nature requires recurring patterns to bare any meaning.
More to the question, did you conduct a large scale mathematical experiment, because by any logic if you did, it is likely that the sheer scope of the problem would show that looking at it from any singular data point renders it irrelevent.
@@AxOwLynx00777 I don't understand what you mean... Whether you flip a coin once or 100000 times the probability of getting a heads is still 50% on any toss if it's a fair coin. Probability doesn't change based on the amount of trials you run... If you are doing a Baysian probability then you are trying to determine the probability by the results, where as normally we are estimating the results by the probability. In neither case does the probability itself change, in Baysian we simply just don't know what the probability is and are gaining evidence to try to uncover it. Yes he built a simulation and ran like 1000 trials or something, you can ask him the specifics. I think you are half reading/half watching everything too.
@@AxOwLynx00777 While it is true that the difference in probabilities is small with only three doors, Logical Morality is absolutely right that even a single data point follows the probability. If you want a more stark demonstration, you can use a deck of cards which would be a 52-door Monty Hall problem. You will clearly see it is not 50/50 when you follow the rules. In fact, if you can pick the 'winner' 1 time out of 20 I would be surprised.
@@logicalmorality4646 You're coin analogy isn't a valid counter argument, it doesn't account for the initial third element, you are correct in stating probability doesn't change only in as much as you don't understand the angle mathematicians approach the problem from and your approach is mathematically incorrect because no matter what way you look at it, the door you are offered to switch to is always going to be more likely to be correct, in which playing this game once means your probability accounts for the one door you choose initially versus the two you don't. Your initial choice from the information given will always be 33.3% likely to be correct and the door that remains for you to switch to accounts that your initial choice meant you were 66.7% likely to be incorrect. Meaning of the 2 doors you don't initially choose it is more likely one of them is correct, so when one of those doors is revealed to be incorrect, all probability of both doors falls to the last door, giving it a 66.7% chance versus all the initial choices of 33.3%. Whether or not you win or lose these probabilities remain the same, you are only "more" likely to win if you switch, not guaranteed, basic law of probability, something being more likely only works if you run the experiment enough times for a pattern to develop, you are pushing the problem to the final stage and not accounting for the third element that is the door that is revealed by the host still had a probability to begin with
A car will not be revealed, period.
Dude, the Monty Hall Problem is not incorrect. You must be misunderstanding something. Suppose you have 100 doors. Your first random choice has a 1/100 chance of winning. Monty then eliminates all but one other door. That one door has a 99/100 chance of winning. I think you're assuming that Monty is just randomly choosing the other door. He's not. If you were to pick the winning door at first, then Monty would pick the other door at random. But if you initially pick a losing door, then Monty will pick the winning door, and now you have to decide to whether you want to switch or not. It's definitely better to switch.
FYI, you made a typo in your comment, you omitted the word "Monty will not not pick the winning door".
In fact, you are the one assuming he cannot open a door that contains a car. Also, you are assuming he cannot open the door the contestant chose. Also, you are assuming the host must open a door at all or that his propensity to choose to do so is independent on what is behind the door the contestant chose.
None of this information is given and that is why the Monty Hall Paradox, at least as stated in Parade Magazine, is embarrassingly ambiguous at best, flat-out wrong if being frank. Your reasoning is all good, but again you are the one making a busload of assumptions that simply aren't stated in the question. I am aware that people often makes all this assumptions as that is the only way to actually make the paradox emerge in the first place. But in reality that is a very bad habit and few sufficiently points out how bad the initial phrasing actually is. This causes a weird scenario where it becomes hard to distinguish between who misinterpreted the initial conditions or who actually fell for the paradox itself.
@@NDAsDontCoverIllegalActs I think you're not the understanding the problem. Nobody is suggesting starting with two doors.
The point of changing the problem to a large number of doors is to demonstrate the absurdity of not switching since your odds of picking the prize and not switching are incredibly small, the larger the number of doors.
@@NDAsDontCoverIllegalActs maybe just look up the problem with your Google machine if you can't visualize or logically comprehend it.
@@NDAsDontCoverIllegalActs maybe I'll ask the question again.
How often do you pick the ace of diamonds from a deck of cards?
Since you object to the larger number, use three cards. One's the ace of diamonds.
How often do you pick it?
That's your answer as to both why you switch and why this game differs from a two door problem.
@@NDAsDontCoverIllegalActs you're serious?
I understand your argument, but i think the assumptions are valid. Your goal is to win the game show, while the game shows goal is for you to lose. Yeah, they could literally spell it out for you within the problem, but they dont need to because the rules or goals within the game show are clear: you try to win, they try to prevent you from winning. They spell it out enough because they state youre on a game show, which has understood goals, and then state the host knows where the car is. There is 0% probability the host will reveal the car to you so it can be eliminated from the probability chart.
Actually in that case it makes sense to never switch doors.
1) If you select the car you have a 100% chance of winning, so the host will give you the option to switch doors, this way your odds are less than 100% as you may choose to switch.
2) If you select a goat the host will not give you the option to switch because you have already lost
Given the host wants you to lose, knows where the car is, and there is no rule stating he must reveal a door, this is a perfectly rational interpretation.
The Monty Hall Problem is statistical bullshit because they are two separate cycles. Imagine this... Imagine you don't choose a door the first time (because that is essentially what you are doing by being offered a choice you can change). So Monty removes one choice and allows you to pick one of two doors. You can pick the one you originally picked the first round... or you can change... Either way you are making a choice between two doors. The fallacy is that you choice exists between 3 doors... The choice isn't made there. The choice is made when you get down to 2 because that is when you actually choose your door and have to deal with the consequences.
Except that's not at all what you are doing.
Can you answer this variation of the MHP.
There are 2 goats and 1 car behind 3 doors.
I choose a random door. Then you get to looo behind the remaining 2 doors and choose 1. What are the odds of me getting the car and what are the odds of you getting the car?
This is a conditional probability problem, not two independent problems.
Assume you stay with your first pick.
If your first pick is Goat A, you get Goat A.
If your first pick is Goat B, you get Goat B.
If your first pick is the car you get the car.
You only win 1 out of 3 games if you stay with your first pick.
Basic math/logic kids understand, idiot among idiots doesn't.
My issue with the problem is that the question is misleading, as it puts you in a scenario and says door X has a higher chance (2/3rd's) than the first door chosen, when door X (the last door that the host didn't open) is a hypothetical door to begin with. Since the answer assumes a strategy of choosing the 2/3rds door, it is assumed before even playing: Which door is two thirds? The opposite/last door that the host doesn't pick. But which door is that? Either of the two doors. But which one? Either, or both at the same time? It's easy to see that the reason why it works is because you're treating both of the two doors (the other two that the contestant didn't initially choose) as a group. So in other words, the answer is no different than saying that a dice has 1/6th chance of landing on any given number, or that you have a 1/52 chance of picking the ace of spades out of a deck of cards. Since the Monty Hall answer is given as a strategy or procedure assumed before even playing, no specific door actually has 2/3rd's odds, contradicting what we already know, that all doors have 1/3rd odds. The question has an implied contradiction within it.
Until Monty has made his move, there is no "two thirds door." The door that has probability 2/3 has that probability because it is the door that Monty kept once the player's initial pick imposed a constraint upon him. That is, the player chooses door A, and now door A is off limits to Monty. Monty can only consider door B and door C. Then, with knowledge of the prize's location, and deliberately avoiding the untimely reveal of the prize, he keeps one door closed and opens the other. So Monty's final door is the best of those two (B and C), and that has better odds than just a single door on its own which was not the beneficiary of such an unfair selection process.
I think you make an interesting point. You already know beforehand that one of the two doors that you are not going to pick is in effect 2/3. What you really have is 1/3:2/3 or 2/3. You are just waiting to know which one to pick. It's not really even a game anymore. It's cheating.
I might also point out that statisticians will say odds cannot redistribute. The unopened unchosen door doesn't actually have 2/3 odds. It's more like you have grained inside knowledge and, as above, are cheating by learning one of the two doors you didn't pick isn't the prize. Would anyone say the doors changed their odds if you had an accomplice who signaled you - Hey psssst, door 2 only has goats, because he managed to sneak a peek?
Since in the original letter, with no assumptions being made, it states the host knows what's behind the doors, randomness is not a possibility. And only random opening of the doors can yield 50:50 odds between switching and not switching. It's fine to do the problem as a straight math problem, but it is not presented as such. As written, word for word, it is a non-random game with no answer. With Savant's assumptions it becomes a puzzle. Doing it as a straight math problem might have been nice as a lead in, but the letter just isn't written where that particular approach can be justified as giving the correct objective answer.
What's wrong with the letter?
@@Hank254 No random opening of the doors can be read from it to justify 50:50 odds since it clearly says the host knows what's behind the doors. Opening the doors randomly while at the same time knowing what's behind them would be a contradiction. How could I randomly pick which cup the ball might be under if I already know where the ball is?
@@Hank254 I've not had any problems here. I've been arguing this point with Logical Morality for a long time saying 50:50 is incorrect. He said he was solving the problem as a math problem where randomness is the default. Fine, but the letter doesn't present the problem as a math problem. It's presented as a game show where the host has knowledge of where everything is. LM says at 6:15 what they are leaving out is "The host is not opening doors randomly; he is systematically avoiding the car." However, leaving that out does not mean he IS opening doors randomly and is not systematically avoiding the car. Leaving that out only means we don't know what he is doing...it's a non-random game, with no particular answer. LM will answer we can assume randomness because in matematics concerning odds, randomness is the default. OK, but we cannot ignore that it states the host knows what's behind the doors, which is pretty counter to any assumption of his opening doors randomly.
For this to be solvable as a math problem, it would have had to say, and the host opened one of the other doors which happened to have a goat. Then we could take it to be a random opening. Even if it had just left out the part "who knows what's behind the doors" and just said, " ...and the host opens another door which has a goat," that would have been enough to allow for randomness. But, that's just not the case.
@@Hank254 One cannot open doors randomly to deliberately avoid the car. That is a direct contradiction.
The game show could not possibly have allowed the opening of doors randomly because that would have risked exposing the car. The letter was not presented as a math problem. The game show in reality and as described in the letter was a non-random game, controlled by management. There is nothing in the letter that leads either to a math problem or even a puzzle. As written it has no answer.
@@Hank254 The odds can be written 50/50 or 50:50, or just 1/1 or 1:1, but the probability is written as 1/2... . Since probability is always written as a fraction of wins to tries, it's better to use a colon to show odds: 1:1.
For the probability of 2/3 (you have a two in three chance of winning), you have the odds of 2:1 (two to one chances of winning over losing). 50-50 is actually an idiom so should be spelled out: fifty-fifty.
You dont even understand the problem dude
The problem is well understood doooooooooooooooooooode. In the end you make a decision between the two remaining doors in the game show context. You decide whether you stay or switch. So the thing is you get a 50/50chance to either win or lose as YOU as the game show participant wouldn't know whether your prior choice does contain the prize or the other goat. You can only calculate this 2/3 win chance by switching IF you assume the knowledge behind which door the prize remains. And how many times do you have to participate in the game show to realize your actual 2/3 winning streak? Do you understand now how unlikely that is?
@@niemand7811 It's about probability. The same scenario run on computers show 2/3 times the participants winning when they chose to switch
@@niemand7811
Assume you stay with your first pick.
If your first pick is Goat A, you get Goat A.
If your first pick is Goat B, you get Goat B.
If your first pick is the car, you get the car.
You only win 1 out of 3 games if you stay with your first pick.
Switching means the opposite.
It's just basic math/logic kids understand.
Sadly, it's far too hard for idiots.
Pov: you don't understand Monty Hall Problem
Wait its probably worse you do understand the problem but still goes on do math just like the people who don't understand the problem.
You do know that host knows where the car is, but you are still taking one of the three scenario as host reveling the car.🤡🤡
If you want to inform what action the host must make, it is easily done by stating that the host always opens a door among the doors that the host did not pick and that he always chooses a door with a goat behind it. Had they just clarified that the 2/3 Vs 1/3 would actually emerged.
But instead they describe an instance and what happened at that instance. We are just told that the host knows what behind the doors. Implying that means we have the information that he will always open a door among the ones you did not pick which always contains a goat is about as far from Mathematics you can get.
It's like saying "I have a bicycle and it is 12km to my work" and believe we have established I ride it to work. If the intent is to convey information, convey it directly and concise!
@@dataandcolours its common sense
If host knows where the car is placed, than why would he ever open the door with the car.
@@tbd5330 This is why we articulate questions so they are concise. Not so you have to make social assessment as a result of someone avoiding stating the conditions.
The only thing that is assumed in questions regarding probabilities is that stochastic variables with a clear even distribution among the outcomes like a dice, lottery tickets, opening of boxes/doors, flipping of a coin can be assumed to be evenly distributed, unless clearly stated otherwise.
It is also common to add unimportant information in questions to test whether a person can differentiate between what is important for the question.
"Boris, a professional casino player from Bulgaria have one red and three yellow normal die. If he throws all at once, what is the probability the sum of the sum of the red dice and the sum of the yellow die will be 21?"
Perfectly fine question.
If they wanted to inform that the default even distribution from a door opening are not present in the scenario, they need to clearly inform the reader of that. They never do. And requiring the reader that we must assume he must without us stating so, is unbelievably inconsistent with logic.
If we start of making assumptions like that, where do we even stop???
"Clearly it's common sense the goats would smell and probably make noises, surely the contestant would have gotten some clue where the goats are, especially if the host opena the middle door as you can then see at what direction that goat is trying to communicate with the other 🐐" :)
Also, not only do we need to establish he must open a door that contains a goat. We must be informed the doors he must chose from is a proper subset of the unopened doors with a cardinality above the number of prizes. We are never informed about any of this either.
What's interesting is how clearly the question is not generating the asymmetrical redistribution of probabilities many assume it does and especially how almost all people claiming so doesn't even understand how extremely concise and specific the scenario needs to be for it to emerge, which is why the entire circus is demonstrably not beneficial from an educational point of view, but obvious outright counter-productive.
It's not even close. It fails to establish any of 3 of the conditions that needs to all be met at the same time to get the 2/3 Vs 1/3 scenario.
The three prisoner dilemma has all the mathematical conditions that many alleges this question has, yet without all the epic fails at communicating them. Therefore it is superior in every way and should get the focus instead.
@@dataandcolours i don't think you need to write a wall of text to convey that "host would always open an empty door is not common sense".
@@tbd5330 That's not even what I said, it' the wrong question to ask as I demonstrated. Nor is it even sufficient to clarify he always chooses a door with a goat behind it to make the 2/3 vs 1/3 emerge, we also need to be informed he cannot open the door you chose and that he can choose freely among the now clarified two doors he can choose among in the scenario you pointed at the car door, as once the default behaviour of a stochastic variable like opening a door are established to not be present here, there is a very high requirement to unambiguously establish all the conditions present. But they dont't even get close to establish the super specific scenario needed for the 2/3 Vs 1/3 scenario.
It's nonsense because the context is wrong. The choice to keep the same door also becomes a 1 in 2 choice. The paradox comes because people assert the probability of the previous state and compare that with the probability of the new state.
The choice to keep the same door is a 1 in 2 chance if you make a random guess. So then you have 50:50 odds of having a 2/3 probability.
@@healthquest4823 Yes but it's nonsense because what is actually happening here, is that people are attributing a previous state, that no longer exists, to the present state. The new state is generated in isolation of the previous state. I'm going to code this, to prove that it is nonsense. The whole thing is just a mental paradox but nobody seems to have thought about it for more than one second. The joys of group think lol.
@@Martin-jg5le No, that's where you are going wrong. The new state is not formed in isolation of the previous state. When the goat is revealed, the initial choice is off limits. Its odds are locked at 1/3. No goat from that door could've been revealed. The two choices are not independent. The first choice constrains the two other doors to be an independent set having 2/3 odds, and if we learn one is a goat, the only door left gets to keep all the 2/3 odds to itself.
@@healthquest4823 That's blatantly incorrect because the new choice is decided based on a new set of variables and is independent of the previous choice. What people are doing here, is using the previous state to determine the likelyhood of the door being correct but using the new state to determine the other door. When in reality you have to either use the previous state or the present state, to determine the likelyhood of both doors. Look at it this way, apply the same logic to the unselected door, using the previous state and you'll soon realise that it has the same odds, in the negative. This is just one of those interesting questions that somehow glitches people's brains.
@@Martin-jg5le No it is fact. You just can't understand it. The correct answer has been know ever since 1975 when the problem was first published with its solution. You can easily set this up to play, as I have, and you absolutely win 2/3 of the time by switching.
why is the computer simulator saying otherwise?
Computers are not always right
@@onesem9399 then explain how the computer is wrong. If you had 100 doors and you chose 1 of them, the chance is 1/100. If 98 doors were taken away and you were left with the door with 1/100 or the other oprtion, it would be the other option.
@@Tantheman.if the door u picked did not get out then it would be a new problem where u pick between 2 doors since u can change ur answer then it would be 50 50
@@Tantheman.like either when u get the 98 doors out u picked one that got out or u picked one of two that might be right
@@onesem9399 It doesn't matter if the door I pick is correct or not, the probability of the door being wrong stays the same.
The claim in the video at 6:00 that the answer is clearly 50:50, seems wrong to me. He states it's because the problem is leaving out a key assumption, which is "the host is not opening doors randomly, he is systematically avoiding the car." So if you put that in, you get the host is systematically avoiding the car, and that leads to 2/3 for switching. So, while it's true that if you leave that assumption out, you don't get the 2/3 for switching, you also don't get 50:50. To get 50:50 he must BE opening the doors randomly. It is Logical Morality who is assuming that if it isn't stated the host is systematically avoiding the car, he must be opening the doors randomly.
For Logical Morality to be making any sense, as I see it, he would have had to say: It's clearly 50:50 because a key assumption was left out, which is the host is not systematically avoiding the car, he is opening the doors randomly.
I did explain that in the video hahaha, I said "randomness is the default" which it is, just like base 10 is the default in arithmetic If the host chooses to cheat or has a bias, it must be stated otherwise the question is simply unanswerable... Which I said it very well may be. If there is an answer it's 50/50... Otherwise there is just no answer.
@@logicalmorality4646 There is no default position. You just read the narrative, and so long as nothing he does had to have been done that way, the we can only conclude anything he did could have been otherwise. So it's non-random, in that the host was not barred from doing anything otherwise. The ONLY way for it to be random is if something is stated saying he had to open a door randomly.
@@healthquest4823 Still at it eh? Arguing is bad for your mental health, nor are you going to prove anything by ignoring what I say, presenting the same argument I have already addressed, or lying about what you said...
Your assessment of my motives are way off as well. The motive should be clear, it's in theme with most the videos on this channel... To sew seeds of distrust in science populism.
@@logicalmorality4646 Just answer this simple question: Is 50:50 the correct answer?
@@logicalmorality4646 I have never lied about a thing. I reposted my description of the French fry game I played and my comments about Savant. I deleted it to not make you look bad at a time I was defending you against the other posters here.
Here's the thing. You are assuming that everyone has misread the problem. I think you have misread the problem, and I can see how.
The key words here are "the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat."
A clearer reading of the problem is if you remove the words "say No. 3", the problem then becomes worded "and the host, who knows what's behind the doors, opens another door, which has a goat." These two statements are functionally identical, as the words "say No.3" is just providing an example of a door that has a goat behind it. Right there, the problem is assuming that the host will never reveal the car, making it correct that you always want to switch.
Even logically, a game show host will never reveal the car on round 1, because that just doesn't make very compelling TV, so it is in the Host's interest to always reveal a goat first, then ask you if you would like to switch.
As I explained in the video, you can't conclude the following information from "the host knows where the car is"
1) The host must reveal exactly 1 door each time you play, how do you know he can't open zero?
2) The host systematically avoids revealing the car (illogical assumption, he very well may reveal the car and then not even give you the option to switch)
3) The host can not open your door
I dealt you the 2 of spades and know what the other cards are... What card will I deal you next? It's comical that anyone can answer this question, the correct answer is "it depends on the rules of the game, and the intentions of the host, both of which are not stated"... It's only obvious rules 1,2,3 are there AFTER you have been told them.
@@logicalmorality4646 Ok, I'll bite and answer each one of those points.
1) It is literally stated in the problem. "You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat." It straight up tell you that the host will always open another door.
2) Refer to my original comment, which you seem to have completely ignored. The problem again literally states that the door Hall will open "has a goat".
3) Again actually stated in the wording of the problem "You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door" It tells you that the host will always pick another door, meaning, not your door.
By the way, it actually is a paradox. This is called a veridical paradox.
You are not talking about the Monty Hall problem.
You can't debunk something by solving a different problem
Assume you stay with your first pick.
If your first pick is Goat A, you get Goat A.
If your first pick is Goat B, you get Goat B.
If your first pick is the car, you get the car.
You only win 1 ouf of 3 games if you stay with your first pick.
Here is what happen if the host open the door randomly.
Your pick, Host reveal, The last door
C G1 G2 stay = win, switch = lose
C G2 G1 stay = win, switch = lose
G1 C G2 stay = lose, switch = lose
G1 G2 C stay = lose, switch = win
G2 C G1 stay = lose, switch = lose
G2 G1 C stay = lose, switch = win
However, in MHP, scenario #3 and #5 doesn't exist.
Since the host cannot reveal the car in scenario #3, he has to reveal G2 instead.
Since the host cannot reveal the car in scenario #5, he has to reveal G1 instead.
So it becomes
C G1 G2 stay = win, switch = lose
C G2 G1 stay = win, switch = lose
G1 G2 C stay = lose, switch = win
G1 G2 C stay = lose, switch = win
G2 G1 C stay = lose, switch = win
G2 G1 C stay = lose, switch = win
Let's say the host always reveals Goat A.
If your first pick is Goat A, you get Goat A and are not in a MH situation at all, so we discard that game.
The two remaining situations are indeed covered by the MH problem, let's see what happens:
If your first pick is Goat B, you get Goat B.
If your first pick is the car, you get the car.
So you win one out of two games, as expected.
@@insignificantfool8592
lmao
In MHP, the host cannot reveal your door.
Your premise is off because the goats are irrelevant and you're only concerned about the car. The problem would be better as an escape problem with two locked jail cells and one unlocked. Trust me you're better off switching every time.
2:20 "Nowhere does it state that the host picking the car is not a possibility".
This is the crux of your incorrect conclusion. The game is setup such that the host does, indeed, know where the car is, and is forced to show the player one of the two goats.
Yeah, only that in most descriptions of the problem, it isn't. Most notably, the letter to Parade, that made this problem famous, does not mention the host to be forced to do anything.
@@insignificantfool8592 _"Most notably, the letter to Parade, that made this problem famous, does not mention the host to be forced to do anything."_
Except, of course, he is forced to play the game where he opens another door and shows a goat.
@@Stubbari showing a door and being forced to show a door are two entirely different things. Ask any woman...
@@insignificantfool8592 Not from our perspective.
@@Stubbari yes, also from your perspective. If you're to guess my number between 1 and 10 and you know I'm forced to pick an even number, your chances are 1 and 5. If you don't know I'm forced to do so, chances are 1 and 10 even if I do pick an even number.
if monty shows a goat, switch.
I thought about this and this isn't true, assuming randomness is involved (which i believe isn't, I think the original context describes pretty clearly that monty has to reveal a goat, but whatever). Let's just assume Monty can reveal either a car or a goat. Since he knows all doors, u don't know how likely he wants to show the car and make you lose. Keep in mind that if your initial pick is a car, you force Monty's hand to reveal a goat. In other words, it might be more likely that your initial pick was a car if he shows a goat. Making you more likely to lose. The answer is that if randomness is involved in monty's pick, then you just don't know.
Are there in fact three "live" doors. With two goats, there is always a goat for Monty to knock off, if he in fact knows. So there are always, each time this is done, only two "live" doors. How can the odds be other than fifty-fifty. The three doors are only for the original choice, not for the switch choice. Assuming Monty knows, I would say so what? He can and will always knock out a goat, at least the way this problem is posited, with the effect of eliminating a door. If, if fact, this is how the program works, Monty's knowledge seems to me to be irrelevant. If he is picking at random and could pick the car, even this problem is not a problem. I do think the odds improve from one out of three in the original choice to fifty-fifty in the switch choice.
No he wouldn't. Please watch the video, I explain extensively why that is not a rational assumption you can derive from the wording of the problem.
Monty’s knowledge isn’t necessary upfront but at least after the contestant makes his initial choice, the host will have to find out what is behind the doors and reveal one of the goats. If the contestant has watched the show before and realizes that no matter what happens, Monty always reveals a goat from the other 2 doors, he will know it isn’t random. Therefore, staying with the initial pick will represent the initial 1 out of 3 picks and the switch will represent 2 out of the 3 other picks. So at the beginning before being offered the switch your pick is 1/3 and the one he doesn’t show you represents 2/3s of the chances.
Oops I thought this guy got it but I guess not. The odds do not change in this game no matter the choices because there is literally only 1 car.
1 car. 3 choices.This later simplifies down to 1 car, 2 choices. That's because the host always eliminates the 3rd choice which is always a goat.
You are always left with 2 choices in the end, and one is a car. You are always left with a 50/50 chance no matter what.
Nope.
Your pick always remains a one in three shot.
Imagine picking a specific card from a deck of cards, or just increase the doors, and it'll become clear.
The other door wins the rest of the time.
@@gnlout7403 those are just odds though, generic odds and they have nothing to do with whether you should actually change doors in the game. Of course the game changes when you make it 100 *doors but even then the same concept still applies. Perhaps if you're on this game show for 100 appearances and you're trying to capitalize on as many wins as you can then maybe going with the generic odds is a good bet. If you're dealing with a generic 2/3 then your final choice between two doors is more or less 50/50. It's more like zero to 100% in reality but 50/50 is the best way to say it.
@@Magneticitist test it yourself with three cards (typo) . If you stay, you win one third of the time. Not half the time
@@gnlout7403Well again, that's just general odds. In reality you could win 10 times in a row by "not switching" before you lose once. That's how random chance works.
@@Magneticitist you could.
You could win by staying with 100 doors but statistically you would only do so one percent of the time.
The other 99 times you'd win by switching
I don't think your explanation is terrific, but I'm still 100% convinced that your conclusion is correct. The problem with all the original "YOU SHOULD ALWAYS SWITCH" scenarios is that both the math and the illustrations (and also the programs that run thousands of simulations showing that switching wins 2/3 of the time) completely ignore the fact that by having the host pick one of the doors with a goat, you've retroactively CHANGED THE ORIGINAL PROBABILITY. When that one bad choice is eliminated, then the chances that you picked correctly the first time effectively changes from one 1/3 to 1/2! Every other video I've watched on this subject simply ignores this fact.
I am a "philosopher", not a mathematician hahaha, but I gave it a shot. If you have any tips to explain it better let me know! Yes people somehow seem to be incapable of looking at the fact that the original problem did not include the following...
"The host must always reveal exactly 1 door after you select, the door he reveals must not be your door, and he can't reveal the car"...
@@logicalmorality4646 I hate to tell you this, but after a great deal of thought and argument, I have come to the conclusion that your argument is INCORRECT. Originally I thought the fact that the game essentially restarts after the host reveals one of the bad outcomes meant that it becomes a 50-50 choice and that the initial choice then becomes moot. Unfortunately, that's actually not the case. The argument that made me change my mind is what I call the ten thousand door scenario. Imagine the same exact game, except that there are 10,000 doors. You get to pick one. The host reveals 9,998 doors without a prize. Now it seems that you now have a 50-50 shot so it doesn't matter if you switch. But the problem is you can't escape the fact that the chances that you picked the correct door in the first place was 1 in 10,000. Even after all those other choices are eliminated, you're still left with the unchangeable fact that the chances that you picked the correct door at first is vanishingly small. Increasing the number of doors in the game helps to see that logically, the chances that the one door out of 9,999 the host (who already knows the location of the one prize) did not reveal is far more likely to be the actual location of the prize. Obviously it's still possible that your initial choice was correct, but it's still only 1 in 10,000 because that's what the odds were when you made that choice. Getting to pick again doesn't change that fact.
@@stevecollins3496 That probability is offset by the fact it's nearly impossibly unlikely the host would reveal 9998 doors randomly and manage to miss the car. The odds are still 50/50 with 10,000 doors given the host is randomly opening them.
Scenario 1) You picked the car (1/10000 chance)
and then it doesn't matter which doors the host opens, the remaining door is a goat no matter what.
Scenario 2) You picked a goat (9999/10000 chance) host reveals every door except the car (1/9999 chance)
9999/10000 x 1/9999 = 1/10000
Scenario 1, and scenario 2 have the same the odds... Switching is just as likely to make you win as losing.
Switching is the same odds as staying the same... I don't think you thought about it, I think you just parroted a pop video hahaha.
@@logicalmorality4646But that doesn't follow the rules of the "Monty Hall Problem." In that scenario, it is stipulated that the host does not open the door you chose OR the door with the car.
@@stevecollins3496 No it clearly doesn't... Please watch the video before commenting lol.
Dude you got this all wrong! The Host of COURSE and ALWAYS opens a door with a goat?? Pretty obvious isn't it otherwise he would have stolen the car from the person playing the game? That leads to Scenario1 you chose the wrong door Scenario prob: 2/3 Prob that you win in this scenario if you stick with you choice:0 . From the other 2 alternatives: 1 car and 1 Goat, the talkmaster eliminates the Goat: Probability that you win when you change the doors: 100%*66%(scenario prob)=66% overall that you win compared to 1/3 that you chose the right door right from the start, twice as much...