Monty Hall Problem - Numberphile
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- Опубліковано 26 вер 2024
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This video features Lisa Goldberg, an adjunct professor in the Department of Statistics at University of California, Berkeley.
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Third option: you intentionally switch to the open door to guarantee that you get the goat
Made me laugh
This sounds like something Dwight would say
@@1Bam712 yeah~!
I mean who wouldn't want a goat
I agree, I want the goat too!
I've heard explanations for this game too many times but I've only found a few which are super easy to understand, this is my preferred "Explanation":
Say you choose 1 door out of 3. There's a 1/3 chance you get a car, and a 2/3 chance you get a zonk. This is without the switching nonsense, just barebones and I'm sure we can all agree on it.
Now let's say you chose a zonk, fair enough, it's a 2/3 chance, more likely to show up. Now Monty opens a door with a zonk in it. There are only two zonks in the game, the one you've more likely selected, and the one Monty just opened. Therefore, the third door _must_ have the car.
Let's go back a step, and go on the assumption you chose the car first try. Now Monty will open one of the other two doors, with a zonk, of course. If you switch you'll get the other zonk, no matter what. This is why it's not a foolproof strategy.
Buuut, with this logic, if you choose a car first (1/3 chance) and then switch, you'll always get a zonk. And, if you choose a zonk first (2/3 chance) and then switch, you'll always get the car. This, is why switching is the better option.
Therefore, if you stay, you'll get a 1/3 chance of getting the car, but if you switch you'll have a 2/3 chance of getting the car. That's what the video says in the end anyways, but I found this explanation far easier to comprehend xD
No! I found the explanation in the video much simpler and easier (and quicker) to understand!
jeez, chill out. It's not like OP said that his explenation is THE BEST one.
this is a much better explanation. why are ppl explaining it with nonsense like concentrated probability? this comment is brilliant!
I've spent the last 15 minutes reading and watching explanation but this is the one that makes the most sense.
@Kieron Rana Why is the video confusing at all? You choose at 1/3 chance, leaving 2/3 chance the car is behind the other doors. When one of those other doors is eliminated, that 2/3 chance still applies to the unopened door. What's confusing at all about that?
Before you comment saying its 50/50, read this.
Scenario 1: You pick the car. Monty shows a goat. You switch to the other goat. You lose.
Scenario 2: You pick goat 1. Monty shows the other goat. You switch to the car. You win.
Scenario 3: You pick goat 2. Monty shows the other goat. You switch to the car. You win.
So as you can see, you only lose by switching if you picked the car the first time around. And the chance that you picked the car first time around is 33%.
You win by switching if you picked a goat the first time around. And the chance that you picked a goat first time around is 67%.
The reason its not 50/50 after a goat is revealed is because you made your first selection BEFORE the goat was shown to you. You made your first choice when there was still 3 doors so there is a 33% chance you picked the car. Revealing where a goat is does not change the fact that you still picked your door when there was 3 doors to choose from.
4/8/20 Edit:
I’ve seen a lot of comments saying I have forgotten the other three scenarios in my explanations, which I have defined below.
Scenario 4: You pick goat 1. Monty shows goat 2. You don’t switch. You lose.
Scenario 5: You pick goat 2. Monty shows you goat 1. You don’t switch. You lose.
Scenario 6: You pick the car. Monty shows you one of the goats. You don’t switch. You win.
As you can see, Scenario’s 4-6 show that sticking with your original choice only wins 33% of the time. The reason you don’t combine the perecentages of Scenarios 1-3 with Scenarios 4-6 is because you add more then one independant variable to the experiment. The independant variable is whether you switch or not. Everything else must remain the same. There’s always 2 goats 1 car, you always get first pick, Monty always shows a goat, you are always offered the option to switch.
If you want to analyze the results of switching, there are only 3 scenarios to analyze (1-3.)
If you want to analyze the results of not switching, there are only 3 scenarios to analyze (4-6.)
But you cannot group Scenarios 1-6 together and analyze the results of all 6 percentages because you have two different independant variables that change halfway through the experiment.
Imagine you have two different plant foods, A and B, were trying to figure out which plant food makes your plant grow faster. You hve two identical plants, but lets say you also watered one twice as much as the other. You have no way of knowing whether it was the water, the plant food, or both that made the plants grow at the rates they did. If you’re thinking to yourself “Why would you water one plant twice as much as the other? It should be the same water?” Then that is perfect. That means you understand that for the Monty Hall problem, you can’t combine the switching and not switching scenarios into one large experiment.
i realised that the reason i didn't get it was because this problem only works when the game show host opens a door with a goat behind. if the door picking was random then another outcome would be that the car would be revealed instantly and you lose
"i realised that the reason i didn't get it was because this problem only works when the game show host opens a door with a goat behind. if the door picking was random then another outcome would be that the car would be revealed instantly and you lose"
The rules of the problem requires of the host to know where the prize is, must reveal a goat from a door not chosen, and offer the contestant an option to switch.
If the host does not know where the prize is and opens a door with a goat, then the staying and switching chances are equal.
Yes exactly, I think that might be why it's so hard to get one's head around
This rather makes sense that what's explained in the video and "concentrating probability"
METUBE
Switching wins if the contestant chose a door with a goat. 2 out of 3 doors have goats. So the chances to win by switching are 2/3.
If you pick the car and switch you always lose. But if you pick a goat and switch you always win, since Monty is forced to eliminate the other goat from the game. But because you are twice more likely to pick a goat than a car in the first place, this means that you are twice more likely to be in a situation where switching wins you the game.
Perfectly explained. 2/3 chance of winning now, rather than 1/3. Must switch. Thanks!
This is the most succinct and clear answer I've seen, thank you 😁🙏🏻
Yes, this is correct because 2/3 of the time Monty would be forced to avoid the car.
@lloydburden2341
Monty is always going to eliminate a goat so you are actually only picking between a goat and a car. It’s fifty-fifty.
@JamesBond-st4qu That is incorrect. Yes, you are always going to be picking between one goat and one car, but the odds aren't going to be 50:50.
I always sum it up as: "In order to lose by switching, you have to have picked right at first. So you will lose by switching 1/3 of the time. So switch."
This is brilliant.
Better way of putting it is your original chance was 33.3% but then it's 50/50 once being told which door is wrong. So your chance goes up by 16.6% if you change door thats how it becomes a 66.6% chance of being right
@@nathanream6483 No thats not a better way to put it..
@nathanream6483 No. It's never 50/50. Your door never rises from 33% to 50%. You selected your door when there were three doors. When a goat was revealed, your door was not an option, so it didn't constitute a new game. It all depends on how the goat was revealed, by accident, or by intentionally avoiding the car.
To lose a game, only game i know is go for broke!
How darE YOU DETECTIVE DIAZ I AM YOUR SUPERIOR OFFICER *inhales* B O N E ? ! !
Haha, I watched this video because I just saw that episode of Brooklyn Nine-Nine. 😂
Ryan Williams I me too! I finally get that Kevin is right
Same lol
BOOOOOONNNEEEE
I’m on that same episode
The explanation with 100 doors got rid of any misunderstandings I had about the answer of this problem. Thank you!
Yeah, I couldn't get rid of the fact that door 1 and 2 is also a 2/3 chance. The 100 doors easily made me realize my mistake, you picked the door randomly without any knowledge of the problem, where as the host knows so it's easier to see that you gain another 1/3 by switching.
Yeah it was like the deal or no deal show
@@raymonddurkin it's totally false logic, why would 1 door change its chance and not the other door, just because you "chose" it. All of the doors chances change equally and intent does not change maths lol. If you choose door 1, door 2 opens and 3 stays shut how can it suddenly have a better chance than if you chose door three and door 2 opened and 1 was left behind. The only thing changing is the persons intent, which can not change the outcome of maths.
@@knifeyonline Did you even watch the video?
@@mefit8725 yes and I was already aware of this theory. I've known about it for decades...
I didnt understand it until the 1 in 100 part. That made it incredibly clear to me.
Bruteforce the possibilities, remember the constant is that the host always removes a Wrong answer.
NO SWITCH
If you're not gonna switch, forget the theatre of the host opening another door, you locked in at 1/3 without the influence of the constant.
WITH SWITCH
1) Pick Right, either Wrong is removed, switch to the other = lose
2) Pick Wrong1, Wrong2 is removed, switch to Right = win
3) Pick Wrong2, Wrong1 is removed, switch to Right = win
2/3 chances to win with switch.
@@Slothacious small doubt bro in case 1 of urs for it appears as two cases
Pick right one wrong 1 opened and switch u lose
Pick right one wrong 2 opened and switch u lose
How the heck I can get more probability while switching plz bro
@@mahakalgaming719 The probabilities of the two possibilities you described are 1/3x1/2=1/6. So the probability of picking the car is still 1/6+1/6=1/3.
Nice try though, we get similar explanations from others who have no understanding of probability all the time.
But it's not 1 in 100 it's 3. This is fake it's the same odds for the remaining 2
@@jdbaker82 It's not. When you pick a door, it's very unlikely you get the correct door. There are 2 other doors remaining, there is a higher chance it's in one of those 2, right? Well the host eliminates one of them, so it is much more likely for the car to be in the other one. That doesn't mean it will always be, it just means most of the time it will be.
Try this - I made an extra video explaining it in my own words... ua-cam.com/video/7u6kFlWZOWg/v-deo.html
Thank you, that's a fantastic explanation. This original video critically failed to explain that Monty cannot open your door, which you instead emphasized.
+Numberphile You didn't explain it well enough. The reason why this works is because when you start the game. It's statistically unlikely that you will pick the correct door. So when every door but the correct door opens, that in term has a "slightly larger chance of being the correct door because you were unlikely to pick it in the first place. In term this chance is not at 66.666666...% chance. it's more around 40 (I didn't calculate it, I just know it's a bit bellow 66.666..%)from my estimates. The reason why it shows itself as such is because there are only 3 variables.
+Chase Boyd No, it is definitely higher than 40%. If it was only 40% than that would mean that you have a better chance of winning the car by staying as opposed to switching. which is wrong. I curious to know why you think "it's a bit below 66.666..%".
+Chase Boyd What are you talking about? Take three playing cards and try the game yourself. See? 67% win rate by swapping, not 40%. The probability of winning by swapping is exactly equal to the probability of picking a goat with your first pick.
+Chandrakant Sharda
In a word, NO.
"The math thing isn't the problem, you just need to bone."
HHHHHOOOW DARE YOU DETECTIVE DIAAAZZ
WhAt GrOsS tHoSe ArE oUr DaDs
Roooossaaa. And that's what really solved Kevin's and Ray's problem. Nine-nine!
Night shifts drifting you and Kevin apart...
Should I teach you kindergarden statistics
Sincerely,
Raymond Holt.
This is the first time I've ever seen the Monty Hall problem explained on a larger scale eg 1/100 vs 99/100. When you inflate to that scale, you get a gut feeling that the odds are in your favor to switch because the chance you picked the car on your first try is 1/100 which are extremely low odds. Extremely effective way to present the problem, and one I intend to use in future. Thank you!
It doesn't chance the fact that if one wrong is always deleted and you can chance your answer the probability of being right is 50/50 either way. But If you would play The game again you would win More Times by answering differently than before or by chancing to the other door
@@juliasrouvali1924 If you stay with your initial choice your odds are 1:100, or put mathematically P = 0.01. The probability the car is behind the two doors is 100%, or 1.0 .. the probability that it's behind the other door is Po = 1.0 - P = 1.0 - 0.01 = 0.99
@Vini Chenzo How it' has more chance? When the other door is opened, the choice is now reduced to two doors right? Then it must be 1/2 which is 50%. I don't understand the logic please explain.
@@user-ft5jp1ot2h Why would you open another door in the first place? Just because the author of the puzzle told you to?
@Vini Chenzo oh. I get it now bro. Thanks for the explanation 😊
Short explanation:
1. You have a statistical FACT that your first choice will be 67% of time wrong. If you chose door A, it's most probably wrong.
2. It means 67% probability of the right choice is divided between door B and door C together.
3. Host shows you it is not B.
But B + C were 67%, so if B=0%, we can count C as whole 67%.
4. So if B is out and A is most probably wrong, it can be just C.
Funny how it works.
You can think of it also this way:
A and B and C make together 100% probability, each one makes 33% of it.
A + B = 67%
A + C = 67%
B + C = 67%.
If A is right choice only in 33% of time, B + C must be right choice in 67% of time.
You chose A. You know it's most probably wrong. B+ C are right for 67%. Now you know B is out. It means C is right for 67%.
Mathematically after the host reveal I'd write it like A + B + C = 33 + 0 + 67. But at the beginning you don't know B is zero.
Even if you chose another letter, if you chose B, it would be only 33% time correct. So 67% correct probablity is A + C. Host shows you A is out = 0%. So C is 67% and B is 33% right, which one you choose?
Simplest explanation that I have, since some other explanations in the comments are confusing
The door the host opens is always a goat.
So if you choose a car at the start, you will switch to a goat.
And if you choose a goat, you will switch to a car. (Since the second goat door was opened by the host)
Therefore, you have a 1/3 chance of choosing a car at first, and switching into a goat.
But a 2/3 chance to choose a goat at first, but switch to a car.
switching can save you 15% or more on car insurance
Wait....switching will save you 15% or more on car insurance if you get the goat, not the car.
LOL
smartest comment on here
switching to or from? that's the question
I never knew I wanted this comment. I love it.
Great explanation! However if i may add. i would postulate, math aside, that the reason the probability concentrates on the final door vs reverting back to splitting between the final and the initial door is because Monty's pick is a dependent probability and not an independent probability (very important point here). For example, if Monty were allowed to pick "your" door and show you that you picked the goat, your chances don't increase. and remain 50/50 as his choice of door is no longer dependent on your initial choice.
This explains some of my confustion
Great reply, this is what makes it make sense now.
great point. So the most important question becomes: Does the host's revelation logic ever hypothetically reveal the contestant's door to be a zonk? If so, then your 1/3rd chance of being right the first time goes up to 50/50 after his reveal (even if he randomly ends up revealing a door that was not your original pick). That all said, the show might get clever and reveal a zonk door when they are trying to make you switch with the illusion that you are playing the odds. In such a case, your actually better off staying with your original pick. It all goes back the most important question mentioned above. I hadn't heard anyone take it to that discussion before.
@@axiom167 "I hadn't heard anyone take it to that discussion before."
That's because it would no longer be the MHP in the first place.
Nope! If you have a 1 in 3 chance of picking the winning door, and then he opens your door to show you a goat, then you have a 1 in 2 chance of picking the winning door if you choose to pick another door, which almost certain you will pick a new door. Your chances of choosing the winning door now went from 1 in 3, to 1 in 2, still better odds every time you switch.
Another way to look at it is from Monty's perspective:
Monty can only open a door that you didn't pick, and he can only open a door that has a Zonk behind it.
There's a 1 in 3 chance that you picked the car. So there's a 2 in 3 chance that you picked a Zonk. So 2 out of 3 times, Monty has only one door that he's allowed to open. Because one is your Zonk and the other is the car.
@@gutenbird you're right. The premise is that you're at a point where he picked a goat. It's not important whether he picks one every time. Because we're focusing on this specific scenario only.
@@Motorsheep This is not correct. If for example he randomly opened a door and just by chance it resulted to have a goat, then at that point the probabilities for the two remaining ones would be 1/2, so no advantage by switching.
To understand why, notice that the 2/3 are the total games in which the player starts picking a goat, regardless of what the host does next. If so, if he acts randomly, then he has 1/2 chance to reveal the other goat and 1/2 chance to reveal the car, so half of those times he would reveal one and the other half the other. Therefore, when he reveals a goat, you cannot say that all the 2/3 are still possibilities, but only the half in which that occurs.
If you played a lot of times, like 900, the results would be like:
1) In 300 games (1/3 of 900) you start selecting the car door. In all of them the host will reveal a goat because the other doors only have goats.
2) In 600 games (2/3 of 900) you start selecting a goat door.
2.1) In 300 of them the host reveals the other goat.
2.2) In 300 of them the host accidentally reveals the car.
So, he only reveals a goat and offers the switch 600 times (cases 1 and 2.1), from which you win by staying in 300 (case 1), that are 1/2 of 600, and by switching also in 300 (case 2.1).
In contrast, in the standard Monty Hall game you would have had the opportunity to win by switching in all the 600 games that you started picking a goat door.
@@RonaldABG I cannot find fault with your logic. So I was right initially and the increased chance of winning by switching applies only if you know that monty is 66.6667 per cent likely to have been forced to pick the door he picked.
your explanation made sense. after reading so many explanations, watching countless videos, i just had to think from his pov
great explanation Motorsheep. it makes more sense to me now that i know to keep in mind rules that confine the host
Many people get confused as they think the contestant picks a door hoping for a car, then get left wiyth a fifty fifty chance of winning so swapping does not matter.
The chance was 66% a door you picked was a goat, and a car in the other, not a 50;50 chance of either.
programed a simulation on my TI-83 plus. Ran a sample of 500 games, where the choice was always switched after a "goat" was removed. switching won 67.6%.
+djsnowman06
sure you dont mean 66.7?
+Dawning Knight Eh, not necessarily - 500 iterations allows for a little deviance from the true probability.
nope it was the result i got. 67.6%. within 1 percent deviance actually surprised me.
+Dawning Knight nah,probability in real life is NOT accurate.Say you flip a coin100 times.you have a 50% chance of heads or tails.But it does not mean you will get exactly 50 heads and 50 tails. if you run a sample stimulation of more games, you would get closer to 66.667%
Guys I was just pointing out what could have been a typo, as 66.7 is more likely that 67.6
I DISAGREE.
In the game, you are effectively given TWO chances to choose your "winning door".
Choice 1 is between 1, 2 and 3 and you have a 1/3 chance of picking the correct door. So you pick door 1 with a 1/3 chance of winning.
You are then asked to choose again, after Monty opens door 2, and you have to choose door 1 or 3 and you make this choice by either staying with door number 1 or switching to door number 3. So you switch, or stick, but the odds remain at 50% chance of door 1 or door 3.
The odds of 1/3 are now irrelevant now you know that door 2 is not a winner.
The odds of picking the correct door on the second choice are ALWAYS 50% or 1/2, EVEN if you had a million doors, and Monty opened 999,998 of the other doors, the second choice is the same risk.
UNLESS you are assuming that these doors are all analogous to Schrodinger's cat, where the car AND the zonk are both behind ALL doors - until you open it, when it is forced to choose and present itself as containing either the car, or the zonk.
Another way to see it is to look at all the possibilities. At first there are three choices (Door 1, Door, 2, Door 3) and then you get two choices (Switch or Stay.) Therefore, there are six possible scenarios:
Car --> Stay = Car
Goat1 --> Stay = Goat
Goat2 --> Stay = Goat
Car --> Switch = Goat
Goat1 --> Switch = Car
Goat 2 --> Switch = Car
As you can see, always choosing to switch makes the probability of any of the first three outcomes 0% and results in winning a car 2 out of 3 times.
Chris Champagne After you chose a door and one of the other two doors is open, you will see one of the goats and you will be asked to either switch or stay, so you are left with four possible scenarios:
Car --> Stay = Car
Goat --> Stay = Goat
Car --> Switch = Goat
Goat --> Switch = Car
So it doesn't matter whether you switch or stay, it's a fifty-fifty chance.
But... On second thought, since there are two goats and one car, you are most likely to chose a door that has a goat, and that's why it's better to switch I guess.
U53i2 Your second thought is correct. 2/3rds of the time you will choose a goat initially, thereby forcing the host to reveal which of the two remaining doors is not a car, which means the other door is. Only 1/3rd of the time will you choose the car initially and switch to a goat regardless of which door the host reveals.
Chris Champagne wow. your explanation is the best one I have seen yet. I was trying to think of this in a numerical sense but i could never wrap my head around this concept. Now i finally get it. Thank you!
U53i2 No need for your second thought here. Your table is correct. Recall from algebra: probability = # of successful outcomes / # of possible outcomes.
Although INITIALLY there was a 2/3 probability you chose a goat, after a goat is eliminated (which is always going to happen the way the problem is described) then you have two possible outcomes (goat or car) and two choices (stay or switch) as you diagrammed. The extra goat is not a choice now (unless you're VERY adamant about "proving" this silly logic) so there is one "success" (presumably car) out of two possibilities (there are only one goat and one car remaining).
Stated another way, the 2/3 probability doesn't "shrink" to the extra unchosen door, part of it stays with the shown goat and that possible outcome is eliminated from the denominator (# of outcomes shrinks from 3 to 2). It is possible to write a program to demonstrate to those who don't understand programming that the video's logic is sound -- you just assume it is when you write the rules for the demonstration.
If your first pick is Goat A and you do not swap, you get Goat A.
If your first pick is Goat B and you do not swap, you get Goat B.
If your first pick is the car and you do not swap, you get the car.
You will only win 1 time out of 3 times if you don't swap.
Incredible how this comment section is still filled with people confidently denying this explanation. I get not quite understanding the math, that's fine. But it takes quite an ego to assume you know better than all mathematicians, rather than assume you are missing something.
Well, once they successfully convince them they are smarter than actual mathematicians, everything is possible for them.
They become Superman in their eyes, but Superdumb in other peoples eyes.
This is making me have a mental breakdown.
It's very simple, just imagine it on a bigger scale. Like they did at the end
Your not picking the one door your picking two. Think of it this way instead of opening a door and asking you to switch he could have said now you can switch to pick both of the other doors and if the prize is behind one of them you win. Its the same thing but you can see the math better that way.
Extremely simple
First time you see this concept, it does that to you, even great mathematicians had this reaction :).
But the problem is the remaining 2/3 chance can be distributed among 2 doors including the door u have already opened
I had a lot of trouble understanding this and for many years couldn't get past the whole "but you get to choose from 2 doors, so it's 50/50" idea.
The way I understood it was when someone made a thread on some imageboard and one of the ways they explained it was:
"Think about the problem as though Monty offers to give you the two doors you didn't choose. You can keep the one door you initially chose, or you can take the other two doors instead, which do you choose?"
Simply put.
Every time you choose a goat and switched you win the car
Every time you choose a car and switched you win the goat
There is a 66.6% chance of choosing a goat on the first pick.
Therefore switching would win you the car 66.6% of the time.
this is genious :)
You beat me to it. I have known about this "problem" for years and the way you describe it is exactly the way I do as the simplest explanation.
Your comment convinced me way better than the video! ^_^
TheRu5tyNaiL A most impressively simple explanation!
TheRu5tyNaiL Your way of describing the overall outcome is a neat summary and in the sense that you are able to make countering assertions by using the same words with slightly different rearrangements it appears to be balanced. Therefore it would appear to be convincing on that basis of being balanced or plainly spoken. What you say is simply put. But this is a situation where Occam's Razor does not make it correct.
I would ask you to consider how when you choose the car, and you can't know when you do, the further possible outcomes exist: A choice between Car and Goat A or a choice between the car and goat B. These, two, separate choice outcomes form only half of the decision space.
When you "choose a goat" in the first place, you choose either Goat A or Goat B. By Monty then having the knowledge about these, he eliminates the other goat. That is, if you choose one, Goat A, your choice will be between the car and Goat B, Similarly for Goat B.
So "Choosing a goat in the first round" leads to a pair of binary choices.So "Choosing a car in the first round" also leads to a pair of binary choices.
Thus, the entire decision space, or set of possible outcomes is four fold and two of each are between the separate goats.
If you say: "What about switching?", I will say that the appearance of having somewhere to switch to or from is an illusion. At the time when one is making the second choice of door, one is choosing between a car and a goat from an even number of balanced opportunities.
Easier for me to see like this:
Assuming you select door 1 and DON'T switch.
If the prize is actually behind door 1: you win
If the prize is actually behind door 2: you lose
If the prize is actually behind door 3: you lose
Assuming you select door 1, and you DO switch:
If the prize is actually behind door 1: you lose
If the prize is actually behind door 2: you win
If the prize is actually behind door 3: you win
It's hard to intuitively understand how just changing your answer each time can improve your chances of winning, but it's plain to see here which strategy is better.
It's so simple but people will still refuse to look at it that way.
The Middle East version of the show features 2 cars and 1 goat.
LOL
This is a very underrated comment.
Sorry, 2x4wd and 1xcamel
The Chinese version has 1 car and 2 dogs.
3 Cars, 2 are vbieds rigged with 1000lbs of high explosive.
I've always thought it was easier to think of it this way ... lets say the car is behind door 1.
If you pick and stay. 1 = win, 2 = lose, 3 = lose
Pick and swap,
1 = lose (because you'll swap to 2 or 3, whichever is not opened),
2 = win (because 3 will be opened, so you can only swap to 1),
3 = win (because 2 will be opened and you can only swap to 1)
Who needs "concentrating probabilities" if you have logic :D
Some people have asked what happens to the solution if the host opens the door with the car behind it. So to clear this up, Steve Harvey will never be the host of this game.
cool thing is, i think it doesnt change the strategy, the case where "Steve Harvey" opens the car door by mistake means you couldve done nothing to get the car now, as we couldnt choose the "opened" door. Thus, this case increases our chance of losing, but the strategy remains optimal.
"A horrible mix up, still a great night"!
Cannot stop laughing at this comment
Darkjet Productions
Imagine there are 1 million doors instead of 3.
You pick a door and your selection has a 1 in a million chance of being right. After all there are a million to choose from. So you pick one in a million.
Now the host starts opening doors, it doesn't matter if he doing it blindly or with perfect knowledge and is avoiding the Prize Door.
Nothing will change as far as your odds of winning are concerned as he opens 100 doors, 100,000 doors or 999,998 doors.
Your odds do not increase because you always knew that there were 999,999 doors that didn't have Prize behind them ... and now all he's doing is showing your what you already knew. He's verifying that 999,998 don't have the Prize. You always knew that.
You original choice is still a really, really, really long shot.
Your door still as a 1 in a million chance.
And that other unopened door has a 999,999 in a million chance.
Swap!
Brilliant way of making this statement.
Idk why people are arguing about such a simple thing. If you choose door number 1, Monty is now left with door number 2 and 3. Ofcourse he will always choose the door with the goat. Suppose if he opens door number 3. Now you can ask this question - why didn't he choose door 2? You can now safely assume that since Monty didn't choose door 2, there is a high chance that it contains the car. On the other hand, you have zero information on whether the first door that you chose has the car or not. So yeah it's always better to switch.
People are arguing, I think, because its not clear that Monty will always open a door with a goat. If his opening of the door is a choice (rather than a necessity), the odds depend on the reason of the choice.
The simplest explanation:
1. If you switch, you get the "opposite" of what you first picked.
2. 2/3 of the time you initial pick will be a goat.
3. That's it.
Actually, you missed a step:
3. No matter whether you first picked a goat or a car, you will always be offered an opportunity to switch.
(This rules out the possibility that the host may open doors at random.)
I stand corrected Mr Tenner, thanks.
No problem, Mr. Bodey. :-)
Ok, but lets say you choose door 1, initially, doors 1, 2 and 3 have a 1/3 chance of being the correct ones, say he opens door 3 and its a goat, why is it correct to assume that there is a 1/3 chance for door 1 and 2/3 chance for door 2 instead of 1/2 for both doors.
There are now only 2 possible outcomes, why would one, given the new information, have a higher chance than the other?
And I'm not saying you're wrong, everyone seems to agree the explanation on the video was correct, just trying to figure out why my logic is flawed.
@@victorf6672 Remember the host cannot reveal your door and neither the one that has the car. Since you pick a goat door in 2 out of 3 times, the host is who leaves the car hidden in the other door he does not reveal in those same 2 out of 3 times.
Imagine you play 900 times. We suppose the car is randomly placed in the 3 doors, so in about 1/3 of the time it should appear in each door: in about 300 times in door 1, in about 300 times in door 2 and in about 300 in door 3. The same with the other contents. Suppose you always pick door 1.
1) In 300 games, door 1 has the car (yours). The host can reveal any of the other two because they both have goats.
1.1) In 150 of them, the host reveals door 2.
1.2) In 150 of them, the host reveals door 3.
2) In 300 games, door 2 has the car. The host is forced to reveal door 3.
3) In 300 games, door 3 has the car. The host is forced to reveal door 2.
You asked about when the host reveals door3. That could only be case 1.2) or 2), which is a total of 450 games. In 150 you win by staying (case 1.2) and in 300 you win by switching (case 2). Now note that 150 represents 1/3 of 450 and 300 represents 2/3 of 450.
I understand the 1/3 and 2/3 thing. What i don't understand is: why is the 2/3 happening to the other door and not to the one i chose first. And one more question: what would happen if i did not choose a door and only choose after one was opened. Whould that then mean both doors still have the same chance?
The rules prevent Monty from touching the door you chose first, so it remains as it is (1/3) and you gain no further information about it. The other door remains unopened either because it is the car (2/3 of the time) or the second goat (1/3 of the time), and you know it cannot be the only goat as Monty would have opened it.
And if you only choose after Monty reveals a goat, it's 50/50 so you're right - same chance.
Because you always get what your first pick is if you do not swap.
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.
Let's say there are two groups of doors.
1. Your door: 1/3 to be the car, 2/3 to be the goat.
2. Other two doors: 2/3 to be the car, 4/3 to be the goat (100% to have 1 + 1/3 to have another).
Now remove 1 goat (and 1 door) from the 2nd group.
1. Your door: 1/3 to be the car, 2/3 to be the goat.
2. Other door: 2/3 to be the car, 1/3 to be the goat.
If you have flashcards, 1-10, you pick one randomly and your partner removes 8, He tells you to choose the card that says number 1.. Would you stay with your first choice or choose the card remaining? obviously you can tell that both cards are not 50/50 to be number 1, Although it "Looks" like this is the case, its simply not due to the fact that your original choice percentage can not change due to other parameters changing, Yours is a constant The control group, very small chance of getting lucky and picking the right number, the others are the variable
It confused people because its on such a small scale.. The math is the same with unlimited doors... imagine theres 1 million red matches, and 1 blue, your told to grab one at random, and then all except yours and 1 match are removed.. Do you honestly believe you have equal % chance you picked the blue, over the last match remaining?
I think i understand now, why the higher chance is always with the other side: because monty would never open the door with the car. If he opened doors at random, the chances after opening would be 50/50, given he didn't open the door with the car already. But because he can only open doors with zonks, the chances for the car to be behind the other door is higher. I think that is what is confusing most people: the chance is biased by monty's actions.
"Probability doesn't kick in, do I have to teach you college level statistics?"
I don't know. Do I have to teach you high school level statistics?
@@Duxxmachina Do I have to teach you 8th grade statistics?
@@arlo4639 If you’ll excuse me, I have to leave you both a snide email about kindergarten statistics
@@denchdwarf5949 the maths isn't the problem they just need to bone due to the night shift
Do i have to teach you highschool level statistics 😂
Think of it as the number of choices each has. When the contestant starts, there are three equally valid choices. 1/3 Now it's Monty's turn. He can't pick the door that you picked, and he can't pick the door that has the car. 1/3 of the time both of those are the same. 2/3 of the time, they are different. If they are different, Monty only has one door he can open. So, 2/3 of the time, the door he doesn't open has the car.
My (easier to understand) explanation:
Three equally likely scenarios.
1: You picked the door with the car. Monty shows you either of the zonks. Switching will give you a zonk (the other one).
2: You picked the door with zonk A. Monty shows you the other zonk(B). Switching will give you the car.
3: You picked the door with zonk B. Monty shows you the other zonk(A). Switching will give you the car.
In 2 out of 3 scenarios switching gives you the car!
But following this logic wouldn't you have to say:
1. You picked the door with the car. Monty shows you zonk(B). Switching will give you zonk(A).
2. You picked the door with the car. Monty shows you zonk(A). Switching will give you zonk(B).
@@BranoneMCSG Each of your two examples have a probability of 1/6. So there is still a 1/6+1/6=1/3 probability of picking the door with the car.
@@BranoneMCSG that is not correct. Because picking the car is one singular choice, and a choice in which switching loses. Just because he can eliminate either goat doesn't mean you are twice as likely to lose. You're only losing in 1 scenario, the one where you pick the car.
But you do not know the zonk is in the door you first picked!! Your saying it from an outside perspective, the car could easily be in the door you picked the first time, as at the time your on the show making a decision, it is still 50/50
@@Jykobe491 But there are 2 zonks whereas only 1 car, so to start with, you already have a 2/3 chance of having originally picked a zonk
Quickest explanation for everyone who doesn't get it (aside from "test it yourself"):
If you pick a goat, you win. Because switching will get you the car. 2 goats out of 3, here's your chances.
Monty opening a door is superfluous misdirection.
He could just as well ask: "Do you want the content behind your chosen door or the content behind both the other two doors?"
+Mattias Sollerman
That wouldn't be much of a problem though would it?
+Mattias Sollerman That's a very different riddle and it has an easy answer: never switch. Opening the door is essential.
+busTedOaS nope same thing still taking 2 doors over 1 lol
Yes, my point was that like most riddles, the problem disappears when you formulate it 'properly'.
You should switch as the likelihood of the car being behind one of the other two doors is twice as high as it being behind your chosen door.
+Mattias Sollerman I love the way you put it. It makes the decision to switch seem as beneficial as it gets.
Since there is one winning and two losing doors, when picking a door there is one chance of winning and two of losing. Or 1/3 winning and 2/3 losing.
Switch door chances:
1) you pick the right door (car), you switch and lose
2) you pick the wrong door (goat 1), Monty opens the goat 2 door, you switch and win
3) you pick the wrong door (goat 2), Monty opens the goat 1 door, you switch and win
So, by switching, you have 2 chances of winning and 1 of losing (or 2/3 winning and 1/3 losing).
Keep door chances:
1) you pick the right door (car), you keep and win
2) you pick the wrong door (goat 1), Monty opens the goat 2 door, you keep and lose
3) you pick the wrong door (goat 2), Monty opens the goat 1 door, you keep and lose
So, by keeping, you have 1 chance of winning and 2 of losing (or 1/3 winning and 2/3 losing).
After years of wrestling with this problem, I thought I'd finally got it - that I truly understood it and could even explain it to others.
Then I met the *Betrand's Box* puzzle and utterly failed to transfer my knowledge of Monty Hall.
Seriously, check out Betrand's Box paradox on wikipedia. It's cool :)
What perplexes me the most isn't that some people don't get it. That's understandable, since the problem IS very counterintuitive. What perplexes me is the fact that, even when every major maths site will give you the right answer (switching is the winning strategy), and even experimentation through computer simulations will give you 2/3 chances of winning if you switch and 1/3 if you don't. They STILL go with "nah, there's no chance I've got it wrong, I'm just smarter than the entire world."
J Thorsson Because mathematicians know much more about math than you.
math has limits, the human ego has no limits
2 things are infinite, the universe and human stupidity
I think the key element that people miss in understanding this is that the game show host categorically cannot open the door with the car in it. That's the key that keeps the probability to 2/3 as he knows where the car is. This isn't the case in games like deal or no deal where the banker doesn't know what's in the box.
You hit the nail on the head
Yes! Everyone tries to apply the monty hall logic to deal or no deal and it's not the same
James Young what made sense to me was remembering the fact that the host can’t pick the door that the contestant is behind, so revealing the door with the zonk does absolutely nothing with regards to the probability for his door (in fact, it 100% expected that the host will open a door with a zonk), it is still 1/3 as it was initially. However the other closed door was not “protected” in the reveal and the revelation of the zonk points to the door has a 50% chance of being a car.
@@Kvadraten376 "the host can’t pick the door that the contestant is behind"
What? now we have cars, goats and contestants behind doors?
jorgensenmj You missed the point. If a contestant picks a door hiding a goat, then the host, by rule of the show, will need to open the door hiding the other goat. In this sequence of events, switching gives the win. However, you will have chosen the goat 2/3 of the time, so switching gives the win 2/3 of the time.
When you first pick a door (say, Door 1), there’s a 1/3 chance that you picked the car and a 2/3 chance that you picked a goat.Monty knows where the car is, so he opens one of the remaining doors (say, Door 3) to show a goat. This is important because it doesn't change the fact that there was a 2/3 chance you picked a goat from the start.If you stick with Door 1: You have a 1/3 chance of having picked the car originally.
If you switch: You now have a 2/3 chance of winning the car because Monty has effectively eliminated one of the wrong doors for you. The odds that you picked a goat originally were 2/3, so switching increases your chance of getting the car.
This is so awesome because I understand it yet my mind refuses it so much haha.
Yeah I keep understanding it then saying to myself but it's still 50/50
@@oliverhargadon2316 It is 50/50, there are gonna be other cards that are gonna be opened
@@oliverhargadon2316 The odds are 50/50 the whole time
@@aizawatakeru4184 there's a 33% you choose the car, but then you say there's a 50% to win if you switch. Where did the remaining 17% go ?
I really like this explanation - because what's more likely, that you chose the correct door out of 100 doors on the first try? Or that you didn't? (and if you didn't then Monty just showed you which one is the right one) - so yes, it is better to switch!
Insert *thank-you.gif*
But there was not 100 doors from the start
@@nuclearchris832 From 3:28.
Aaron, that was an elegant explanation. Succinct. Nice job.
@@nuclearchris832
Doesnt matter.
This is an EASY explanation.
You have 3 ways to play.
1: You pick Goat A, Monty shows Goat B, switch and you WIN
2: You pick Goat B, Monty shows Goat A, switch and you WIN
3: You pick the car, Monty either shows Goat A or B, you switch and LOSE
2 times out of 3 you win.
+Brian Bucher Exactly!
easiest explanation ive come across
You ignored the ways to play where you DONT switch.
1: You pick Goat A, Monty shows Goat B, switch and you WIN
2: You pick Goat A, Monty shows Goat B, dont switch and you LOSE
3: You pick Goat B, Monty shows Goat A, switch and you WIN
4: You pick Goat B, Monty shows Goat A, dont switch and you LOSE
5. You pick the car, Monty either shows Goat A or B, switch and LOSE
6: You pick the car, Monty either shows Goat A or B, dont switch and WIN
You win 3 times, you lose 3 times. So its 50%.
@skunkonejpc - exactly. I can't believe I had to scroll this far down for this. The problem can be simplified very easily with critical thinking. The initial guess is redundant and has no bearing on the outcome. Same with the first zonk reveal. The problem doesn't even start until after the zonk reveal, where you have to choose 1 out of 2 doors.
+Brett Cameron
Don't get confused too. You can see in skunkonejpc's diagram that switching wins more times. One thing is the probability of winning making a random selection, and another probability is when you select a specific option. If you have got information, you can know if one is more likely than the other.
The first choice matters because it gives you information about which is more likely to be the correct. Since from the host point of view when selecting which door to reveal is the same as selecting which door not to reveal, let's say his closed door is "his selection". In this way, at first basically you both are deciding which doors will remain closed for the second part of the game. You randomly select one and then the host selects another one.
The difference between you two is that you chose randomly but the host knew the positions and must reveal a goat, meaning that he must leave the car hidden. So, all times you got a goat door at first, he purposely selects the car door to leave it hidden. It is more likely that you got a goat door (2/3), so it is more likely that the host's closed door is which has the prize and then it is better to bet in his option.
The fact that there are two doors left has nothing to do with who was more likely to select the correct one.
for anyone confused still
think of it like this,
the only reason its not a 50/50 is because he lets you choose a box before he reveals one,
since you choose a box, just because he now reveals one doesnt actually change the fact you had a 33% chance to land on the box you are on.
now for the complicated part.
You had a 66% chance to be wrong, and a 33% chance to be right, he now halfs your chances of being wrong by removing one of the incorrect answers, theres a 33% chance we are right still, but now there are only 2 answers, meaning the other answer must have a 66% chance of housing the right answer.
"in order to lose by switching, you must have chosen right at the beginning, so you will lose 1/3 of the time by switching, so you should switch" - someone in the comments
Thanks helped me alot
I actually didn't have to go through the video just because of this comment. Thanks!
Wrong, the revelation of data changed your percentage, because it changed the overall number of outcomes in the game. So many people make that mistake because they don't know how to separate "ideas" from "inherent representation" of what is happening.
"theres a 33% chance we are right still"
@@bobcatgaze the loopholes you're going through to disprove something that isnt wrong is funny
My biggest question is that couldn’t you view it the other way, of grouping the one he shows you and the one you picked having a 66% chance, so staying with yours has a 66% chance and switching has a 33%?
Another take on explaining why it works:
What if you implemented the following strategy: always choose door 1, never switch. You will get the car 1 time out of 3 (i.e. every time the car is behind door 1).
Now amend that strategy to always pick door 1 AND to always switch. You will miss the car when it's behind door 1 (1 in 3 times), but you will get the car every time it's behind door 2 OR door 3 (2 in 3 times), because when either of those is the case Monty will open the other, and the option to switch to will be the car.
That is true but chances are always 50/50 if you can play only once. Getting correct door out of 3 three Times in a row with choosing The same door everytime gives you only 0.333×0.333×0.333 is something like 3% chance of being right but by quessing different number everytime gives you 33% chance of being right
@@juliasrouvali1924 - WOW! You make no sense at all!
If you play this game just one time, you have a 1/3 chance of your initial guess being right...PERIOD! Now granted, the chance of being right three times in a row is slim, but that has NOTHING to do with the initial problem! The initial problem is simply stating that by ALWAYS switching, one has a 2/3 chance of winning, as compared to only a 1/3 chance of winning by never switching.
NOTE: RIP Monty Hall
1921-2017
He died in his house in California.
T_T
did he die in the hall ?
when ambulance entered the house they saw 3 doors as well so they played the game one last time hahahah
Which door did he die behind?
He chose the wrong door.
If people want a "logical but easily understood reason", then I usually tell them: The first door, you picked. But the second door, *Monty* picked. He picked it with full knowledge of what doors have and don't have the prize. (It's clearly illustrated with the 100-doors explanation; Monty picked the 98 doors that he knew didn't have the prize.) So switching your decision basically means you're using Monty's knowledge instead of just your own blind guess.
Thinking about it from the perspective of Monty and the problem simplifies.
Once the contest chooses a door, Monty must choose a door that does not have the car. 2/3 of the time Monty does not have a choice on which door he chooses, because in one of the non-chosen doors, there is a car and he has to avoid that one. 1/3 of the time he does not have to avoid any doors, because the car is in the door that has already been chosen and he can simply open up either one of the door.
Monty is forced to not open the right door with 2/3 probability. Thus we open the door that he has avoided.
I went on excel and created a simulation of this problem. Along the top I put in the following headers:
Actual | Selected | Shown | Changed | Stayed
I then created a table with the following formulas for each header:
=RANDBETWEEN(1, 3) //for Actual
=RANDBETWEEN(1, 3) //for Selected
=IF([@Actual][@Selected],6-([@Actual]+[@Selected]),IFS([@Actual]=1,RANDBETWEEN(2,3), [@Actual]=2, CHOOSE(RANDBETWEEN(1,2),1,3),[@Actual]=3, RANDBETWEEN(1,2))) //for Shown (note that the IFS() function is new to Excel 2016)
=([@Actual]=(6-([@Selected]+[@Shown])))*1 //for Changed
=([@Actual]=[@Selected])*1 //for Stayed
---------------------------
I then ran the simulation 10,000 times. I added up all of the Changed and got 6,637 wins and added up the Stayed and got 3,363. According to this sim, changing will result in a win 66.37% of the time while not changing will result in a win 33.63% of the time.
---------------------------
That is assuming you want the car. If you want the goat, then after one of the doors are opened, just select the opened door and you will win the goat 100% of the time.
Makai Post
You’re a brilliant person who knows more than the normal average. How did you learn about building tetrahedrons?
@@alliseeisW
What does the _normal average_ know?
This comment section is classic confusion between probability vs possibility. It’s like saying there’s a 50% chance that Leprechauns exist, because either they do, or they don’t.
I am not a native english speaker and i have just learned that thoose are two different words. What is the diffrence in between?
You may help by giving the answer:
There is a road splits up to two diffrent roads: A and B. Than the road B splits up to road B1 and B2 after a few miles. After a few more miles they all reunite together.
What is the possibility/probability of me using the road B1 to go to the reuniting point? (when i see a road splitting, i am equally likely to choose one)
@@sodr7440
In your example, you have 3 possibilities: route A, route B1 and route B2. So you can say that any single route represents 33.3% of the possibilities. But the probabilities depends on how much each possibility is used. I mean, if 95% of the people use route A, 4% use route B1 and 1% use route B2 (and if I can consider you a normal driver), then there is only 1% of probabilty of you using B2, in spite of the fact that it represents 33.3% of the possibilities.
@@tomasaguirre4654 Thanks!
@@sodr7440 , that's because Josh Lau has no idea what he's talking about. It's about PROBABILITY vs CONDITIONAL PROBABILITY. People don't know the difference between PROBABILITY (you have 1/3 chance of initially picking the correct door) vs CONDITIONAL PROBABILITY (the host has shown you an incorrect door - based on this new information what is the chance that the other door is correct). Conditional Probability re-calculates odds based on new information provided.
For people having trouble with this :
Imagine you pick door A and the host says, you want me to tell whats behind door A? or do you prefer to know whats behind both door B and C?
You switch to door B and C, the host opens a door with a donkey for dramatic tension, and finally opens the last door with your prize.
This is the exact same game you played, it's not the idea of knowing theres a donkey behind one of the doors, its the fact you have been offered two doors instead of one and choose the better of both.
If you choose to stay, your first pick must be the car to win. if you choose to switch, your first pick must be a goat to win. The probability of your first pick to be a car is 1/3, and to be a goat is 2/3. So it’s 2/3.
This problem isn't as complicated as people (including statisticians make it). You don't need conditional probability. Monty knows which doors have the goats and will always show you one, independent of your initial pick. It doesn't change its probability so it's not a 50-50 chance when you're down to 2 doors.
All that matters is the first choice of doors which breaks down to flipping a weighted coin with outcomes:
1. Car (p = 1/3), switching gets goat
2. Goat (p = 2/3), switching gets car
A simple demonstration for why you should switch:
Suppose door number 3 had the car.
- if you chose 1 and he revealed that there is a goat behind door 2, then switching would let you win
- if you chose 2 and he revealed that there is a goat behind door 1, then switching would let you win
- if you chose 3 and he revealed that there is a goat behind door 1, then switching would make you lose
As you can see, in two of the three cases, switched made you win
+Justin Scheepers NOPE, you forgot one case: if you chose 3 and he revealed that there is a goat behind door 2, then switching would make you lose. So it should be 2/4
+Cuong Pham Ok, well you could also say that for choosing 1 and him revealing 3 with a goat, where switching would make you win. All you're doing by "adding" new scenarios is increasing the fraction, which can again be broken down back in 1/3.
I'm not saying this in a mean way, but there are flaws in everyone's reasoning.
Woof Bork, your reasoning doesn't work because if you've already picked a goat, only one door can have the other goat. Only in the certain case that Cuong described (where you pick a car at start) means both the other doors have a goat. So Cuong is right that there are *only* 4 possibilities.
However, Cuong, I think you'll mistaken that possibility isn't the same as probability. There's a 1/3 chance you pick door 1 and reveal a goat on door 2. There's a 1/3 chance you pick door 2 and reveal a goat.
That means the total leftover 1/3 chance is when you pick door 3. So picking a door in goat 2 is now 1/6 and picking a door in goat 3 is now 1/6. If it was 1/3 and 1/3 instead, then you would have 4 1/3s but that's impossible since the sum of the percentages must equal 1.
Cuong Pham - You are likely to choose each door 1/3 of the time. if you choose door 1 and the car is behind door 3, the host will open door 2 100 percent of the time, making the probability of the host opening door 2, 1/3 of the time. if you pick door 2 and the car is behind door 3, the host will open door 1, 100 percent of the time, making the host opening door 1, 1/3 of the time. However if you pick door 3, and the car is behind door three, the host will open door 1, 50 percent of the time and door 2, 50 percent of the time with 1/6 chance of each.
oke simple question what is your change after he revealed a door you have 2 doors left one of them has a goat .
you forget that after he opens the door that the chance changes because you go from 3 doors to choose from to 2 doors to choose from
The best explanation of this problem I have seen so far. thank you!!
Thank you so much for the full explanation. I've seen this problem presented so many times in so many videos, but every time they fail to properly explain why the chances on the other door raise to 2/3, and that's what truly matters on this explanation. Many thanks again.
@georgerivera278 The most transparent explanation imo:
1. If you pick the car first and switch, you always lose, because the door you switch to has to be a goat.
2. If you pick a goat first and switch, you always win, because the door you picked is a goat, and the door opened by the host is a goat.
You can easily verify the logic of these. Now, how likely are you to pick the car first, and how likely are you to pick a goat first?
Since you have a 2/3rds chance of choosing a goat, that means you have a 2/3rds chance of winning when you switch due to #2.
It's confusing because the way humans think makes it seem like the act of switching is raising the probability of the other door from 1/3rd to 2/3rds. In reality though, the odds aren't changing and are defined completely by whether you pick a goat or a car first. It's just that the host showing you a goat makes picking a goat always win if you switch.
It doesn’t raise to 2/3 tho. There’s a vital error in the reasoning here and this videos animation illustrates the error perfectly. After Monty shows us the car isn’t behind door 2, doors 3 and 1 still remain unknown so the 2/3 chance gets spread across both remaking doors. 1 and 3. So each door only ever has a 1/3 change.
In the animation you literally see the 2/3 bracket reduce to only cover door 3 after Monty opens door 2. But there’s still door one sitting there unopened. That 2/3 bracket should cover doors 3 AND 1
There are more goats than cars.
You've probably already chosen one of the goats since there are more goats than cars.
Remembering that you've already probably chosen a goat, Monty reveals, with 100% certainty, where the other goat is.
So you've probably already chosen a goat, and you now know where the other goat is with 100% certainty.
Therefore, the door you haven't chosen probably has the car.
wrong sorry - again flawed logic
@@paulatkins9675 How so?
I understand it thanks to you
@@paulatkins9675 nope, he is correct
@@paulatkins9675 then ur logic is imperfect.
Wow! This is the clearest explanation of this concept I’ve heard. Excellent example with the 100 doors.
Instead of thinking about it like "The host will always open the door with the goat",
think about it like this "The host will deliberately keep the door with the car closed"
Now you can see that if you switch to the closed door, 2/3 times it will be closed because it has a car behind it.
@@gutenbird What exactly are you trying to prove? If the host doesn't know where the car is and opens another door that reveals a goat then it's 50/50. If you want to include the ability of picking the car if the host reveals it then you must include that same ability of switching to the revealed goat in the original problem as well.
@@gutenbird If he reveals a goat without knowing where the car is it's 50/50.
@@klaus7443 It was late and I was a bit tired. Let me clarify for myself so I feel better about my explanation. The host doesn’t need to know from the beginning where the car is. But after the contestant selects the host will then have to obtain knowledge of what is behind the other 2 doors and eliminate one of the goats. As with attempting to pick a random card in the deck of 52 and then the dealer looking at the deck and turning over 50 of the other cards. The dealer didn’t need to know the location of the card up front or whether the person had originally selected that card but afterwards he would have to turn over 50 cards that aren’t that card while himself looking through the deck. You are correct to say if he didn’t look and 50 other cards came out, it would be equal. He can simply peak after the first pick is made. Sorry.
Chance of it being behind the first door = 1/3
Chance of it being behind the door which was opened = 0
So the chance of it being behind the remaining door must be 2/3.
Because probabilities must add up to 1.
Correct.
Ah yes when I watched the first five minutes of this B99 episode, it was 50/50 for me for 20 minutes until I realized I had no idea how the game show worked.
Well, here's my explanation.
Lets suppose we have 2 persons - Alex and Ben.
--------------------------------------------------------
Alex ALWAYS stays on his initial choice.
Ben ALWAYS switches.
--------------------------------------------------------
Lets see under what conditions they win
Alex only wins if his initial choice is a car. If he picks a goat he loses.
If Ben's initial choice is a goat he wins. (Monty will open another door with a goat and Ben will switch to a car)
Did you get it?
--------------------------------------------------------
Alex picks the car = he wins
Ben picks a goat = he wins
--------------------------------------------------------
The probabilty for Alex to pick the car is 1/3
The probabilty for Ben to pick a goat is 2/3
Ask if you have questions.
Your comment made the most sense of any of those explanations I have seen.
Better explanation then movie itself, should be an accepted answer :)
The reason I get confused is because I see it as 50/50. Once Monty shows you a door he asks you if you want to switch. To me this overrides the original choice and presents you with a new one. 1 or 3. Now the choice isn't ⅓ because there aren't 3 options. It seems to me ½ because it's between 2 doors.
Let's say you're on a "Who wants to be a Millionaire" show.
1) You've chosen an answer and then you use the "50 50" lifeline. Computer eliminates 2 incorrect answers, this MAY include your initial choice.
Q: Will there be a difference if you swap your choice after that?
Common sense tells us that there will be no difference, and that's right. Computer eliminates answers NOT depending on our choice, thus making two remaining options equal, e.g. 50 - 50 percent.
Sounds true, right?
Now~
2) Computer eliminates 2 incorrect answers, NOT touching your initial choice.
Now ask yourself. Will the 2 remaining options (your initial choice and one other) be 50 - 50?
What is your common sense telling you?
The thing is, whats behind the door is hidden so basically Alex might win because he originally picked the right choice which is the car, him being exposed to whats behind door number 2 still doesn't give him as much information as this theory explains because the prize is hidden regardless.
For those who still didn't get the intuition for this truth, think like this... the only way you win on swapping is by picking a wrong door, what is the probability of picking a wrong door? 2/3
Finally, this is what just did it for me.
@@christopherclark7460 Same lol
Got it ! Thanx
But how do you know you picked the wrong door?
@higorss I simply don't know which ones are the wrong doors, but I can calculate the odds of picking a wrong door from the doors, which is 2 out of 3.
This means 2 out of 3 times I pick the wrong door.
So, 2 out of 3 times I will win on swapping cuz the other wrong door (Apart from the wrong door I picked) will then be revealed by the host, leaving me the right door, right?
I'm hoping that you got this.
I remember reading about the Monty Hall Problem in a book about numbers and probabilities some years ago and I found the explanation of the answer fascinating.
On the actual game show of Let's Make a Deal, though, this strategy is somewhat less effective, because the actual Monty Hall did not always offer a choice to switch on the show, and if his producers were breathing down his neck, he may have been required to offer more switches in disadvantageous positions (i.e. suppose the player has picked the car, but Monty, knowing the car has been picked, now offers the switch because the other choice is worse - whereas in situations where the player did ONT pick the car, the producers tell him "just show them the Zonk they picked"). But that sort of next-level gaming theory is well above my pay grade.
And yet you seem to understand the problem better than most of the lemmings who just state that switching is "always" better.
Who’s here after watching B99?!
“BONE?!!!”
Nine nine!
That’s exactly why I’m here! 😂
NINE NINE!!
Noine-noine
NINE NINE
😄🙋🏻♂️
If you don't believe that the people in the video are right; try to disprove this:
If you initially guess a car and then switch, you lose.
If you initially guess a zonk and then switch, you win.
There are more zonks than cars and therefore switching increases your chances.
Before you respond or discard this, try to ACTUALLY come up with a scenario where this isn't true.
Initial guess door 1 (car). He reveals door number 2 (zonk). You switch to door number 3 (zonk) = lose
Initial guess door 1 (car). He reveals door number 3 (zonk). You switch to door number 2 (zonk) = lose
Initial guess door 2 (zonk). He reveals door number 3 (zonk). You switch to door number 1 (car) = win
Initial guess door 3 (zonk). He reveals door number 2 (zonk). You switch to door number 1 (car) = win.
It is in your best interest to switch every time, seeing as how the probability of initially selecting the car is 33% versus the 66% chance of selecting a zonk. but as a programmer, allowing for every contingency still gives a 50/50% chance of getting either a zonk or a car.
+Travis Washeck
You mess sub-scenarios with main scenarios.
Let's play a game.
Roll a dice.
1 2 = win
3 4 = roll again, any number = win.
5 6 = lose
Here are all the possibilities.
1
2
3 - 1
3 - 2
3 - 3
3 - 4
3 - 5
3 - 6
4 - 1
4 - 2
4 - 3
4 - 4
4 - 5
4 - 6
5
6
You win in 14 scenarios out of 16. Does this means your winning chance is 14/16 = 87.5%?
NOPE.
Your winning chance is 66.666...% since you will win if your first roll is 1 2 3 or 4 and will lose if it is 5 or 6.
+Travis Washeck that's what I thought. It's a 50/50 chance after they reveal the zonk
Xealous
That's wrong.
Let's say you always stick 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.
The reason I don't think this paradox is right is because those are all *seperate* events. One event doesn't effect the other, so you can't come up with probable odds when you combine events that are completely seperate.
For the idiots:
In the Monty Hall problem, you have 1/3 chance of getting a car, and 2/3 of not getting a car. That is, in your first pick. Monty only reveals one door that doesn't have the car, then you have to choose: swap or not?
Here's the logical part that you don't understand. You have a 2/3rds chance of getting a "not car". If so, then Monty opens another door that is also a "not car". If you luckily get the 2/3rds chance of picking that "not car", when you swap, you automatically get the car.
---
For example, let "o" be the door with a "not car" behind it and "c" the door with a car behind it.
The layout becomes something like this:
o o c
Let's say I choose door number one, I'll mark it with a capital x
X o c
Monty reveals the other door with an "o", meaning you can already see one of the doors with an "o" in it (revealed door in quotation)
X"o"c
Then if you swap, you get the c on door 3, and you get the car!
-----
Same thing for the other o.
o X c
"o"X c
The same thing happens, swap the X and you get the car!
-----
On the other hand, if you get the 1/3rd chance of getting a car on your first pick, Monty will open a door with a "not car". If you swap, you automatically get a "not car".
Let's get the layout from above.
o o c
If I chose the c on door 3, you now get
o o X
Monty will either reveal the left or right "not car". Let's say the right one.
o"o"X
When you switch the X, you get the not car.
-----
Summary: There are 3 doors. If you swap, you have 2 possibilities to win, and 1 to lose, therefore automatically making the winning chance 2/3, and the losing chance 1/3.
-----
Do you get it now? Or do I have to be more detailed?
if you initially select one of the zonks, then swapping gurantees you win. probability of randomly selecting one of the zonks initially, is 2/3. probability you lost if you swap, is the probability you randomly selected the correct door at the start, which was 1/3. probability of winning without swapping, is still the probability you randomly chose the correct door initially, which is 1/3.
(if you initially selected one of the zonks, then he removes one of the zonks, the only thing left to swap to is the correct door)
I look at it a different way. When the door get's opened, there's a 1/2 chance you already picked the right one, not 1/3 chance. As one door is eliminated. If they ask you to choose again, it's not a 1/3 chance vs a 1/2 chance, but a 1/2 chance vs a 1/2 chance. Therefor both doors give you the same chance for winning.
well daniel you dont look at it a different way but the wrong way.
Best explanation of the Monty hall problem I've ever seen
Easiest way to understand: When you first chose, you probably chose the wrong door, since it was a 1/3 probability. With Monty revealing one door, you get a second chance to choose. And since you most likely chose the wrong door the first time (highest probability), switching increases your chances.
And ypu can choose your door a second time :)
50/50
That theory would work, if he wouldnt open a door with a goat in case uve picked the one with the car by first try. Chances of winning this game r always 50/50 cause in the end u choose between 2 doors. At the point ur asked if ud like to switchy, ur basically picking one out of 2. One door of the 3 cant be counted in the probabilty "calculation", it just belongs to the concept of the show, as making people doubt their choice brings more drama to it.
@@viktoria8299 It's best not to think about the probability of what's behind the doors after the reveal. Instead, think about the probability of you either picking a zonk or car on the first guess.
You're more likely to have picked a zonk. If you've picked a zonk and then switch doors after the reveal, then you're guaranteed to get the car. That means by switching, you're more likely to get the car. Because that's the more likely senario - not the more likely door.
@@viktoria8299 the game works since Monty always reveals a goat from the remaining doors in any scenario. 1/3 games you picked the car and Monty leaves a goat, sticking wins. 2/3 games you picked a goat, Monty leaves the car, switching wins.
The third door not being in play doesn’t mean it’s 50/50. What you picked in the first round always carries over as your switch choice and what you didn’t pick carries over as your switch choice. 1/3 games you picked the car initially and 2/3 games you picked a goat, you can’t separate the rounds because what you picked 100% determines the outcome of both options.
still doesn't make sense to me. because your chance of the first door being it hasn't changed? you pick door 1, which has 1/3 chance to be car. door 2 is opened and is zoink, so you know door 3 has 1/2 chance of being a car, since now you can compare it only with door 1. but that doesn't mean that it changed anything about door 1 or 3. you could say the same for door 1 right? so why switch when its eithere 1/2 or the other 1/2, it will be 1/2 either way and we dont know which half like wtf ????
Here's a slightly different way to put it. Regardless of which strategy you choose - switch or stick, there are three possible outcomes for each (car, goat1, goat2):
Switch strategy: Three possible outcomes. Remember Monty knows where the goats are.
#1 You picked the car. Monty shows you one of the goats. (he chooses which to show you - does not matter) You switch to the other goat. You LOSE.
#2 You picked goat1. Monty shows you goat2. You switch to the car. You WIN!
#3 You picked goat2. Monty shows you goat1. You switch to the car. You WIN!
Stick strategy. Three possible outcomes. Remember Monty knows where the goats are.
#1 You picked the car. Monty shows you one of the goats. (he chooses which to show you - does not matter) You stick. You WIN!
#2 You picked goat1. Monty shows you goat2. You stick. You LOSE!
#3 You picked goat2. Monty shows you goat1. You stick. You LOSE!
Switch strategy: WIN 2/3 Stick strategy: WIN 1/3 (stays the same as the original odds since you do not benefit from Monty's knowledge of where the goats are.)
Now I get it, thank you dculp
In my quest for the simplest explanation of this, I've come up with the following attempt. This game is equivalent in probabilities to the following variant:
There are three doors, one with a car, two with zonks. You are given the following options:
1. Select one random door. If it has the car, you get the car.
2. Select two random doors. If either has the car, you get the car.
What is your best strategy? Option 1 or option 2?
This is all there is to it. Whether the host opens one of your set of two doors, or does a strip tease, plays the 5th symphony, or anything else, is just a distraction.
Brilliant. This is indeed a perfect analysis.
To everyone who thinks it's 50/50:
The only reason why it isn't 50/50 is because the chance of picking a goat when you have 3 doors and 2 goats to choose from is 66%. Since the host is forced to reveal the other goat you have a 66% chance of getting the car when you switch. If there was no such rule then yes, it would be 50/50. Same with the 100 doors. The host will have 99 doors and you have 1 to open. His chance of getting the car is 99% and your chance is 1%. The host is forced to keep the winner for last so that's his remaining door. The key is that it's the host opening the 98 doors not you.
I see it like this. G is goat (1&2) and C is car. An arrow is change and a full stop is stop.
These are all the things that can happen:
G1 > C
G2 > C
C > G
Changing doors will win 2/3 times
G . G
G . G
C . C
Remaining put will win 1/3 times
an easier explanation as to why it works to always switch is that you have a higher probability of getting it wrong in the beginning. so when he removes the wrong answer out of the 2 remaining its more likely for the final to contain the winner
When you think about not playing the game once, but multiple times, it suddenly makes sense. Because if you play 100 times, and always choose and stick to door #1, you'll win on average 33 times, and lose 67 times. But when you choose door #1 and then switch to the door that wasn't opened by the host, you'll win 67 games and only lose 33 (which is the number of times, your initial guess of door #1 was correct).
now this one makes sense. Thanks😊😊
I was skeptical 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.
For the generalized problem with N doors and G goats revealed, the solution is as follows:
Chance of winning by remaining with your original choice = 1/N, the same as the chance of winning if the goat was never revealed in the first place.
Collective chance of winning by switching to any one of the remaining doors = 1 - 1/N, the complimentary probability.
Number of remaining doors = N - G - 1
Divide the collective probability of winning by switching, by the number of remaining doors, to get the probability of winning by switching to an individual door: (1 - 1/N)/(N - 1 - G), which simplifies to (N-1)/(N*(N - 1 - G))
@carultch That seems ridiculously complicated. The only reason the "other" door has a 2/3 chance of having the door is because Monty would not reveal the car. Two-thirds of the time, Monty would be forced to pick the door he did because the car was behind the other door. It's as simple as that. I don't think your complicated math actually demonstrates that.
When Monty asks the contestant, "Keep door number 1 or switch for number 3?"...
What I want to know is what are the chances that *Monty* will keep the car? At that moment he knows whether staying or switching will result in giving away a car, the only thing he doesn't know is which of the 2 alternatives the contestant will choose. At that moment *his* chances of keeping the car are 1/2. But the contestant chooses "switch" so *her* chances of getting the car are 2/3. Where did the extra 1/6 of a car come from?
I'ma condense this to 3 lines that hopefully you non-math people can understand:
- If you choose incorrectly and then switch, you will win.
- You have a higher chance of choosing incorrectly (2/3)
- Therefore, you have a higher chance of winning if you switch (2/3)
It ain't rocket science, people. It would only be 50/50 if Monty exposed a Zonk BEFORE you chose one and you were left with two options from the get-go, but as it stands, you choosing one influences which one he exposes, and chances are best that he has exposed the only other losing option, leaving you with best odds if you switch.
I always thought it must be 50:50. But with 100 doors it finally makes sense. I think it is not about the chances that you pick the right one in the first place, but the wrong ones. Chances are so bad and he cannot open the winner door. Omg I feel enlighted. Thanks for that.
Summary is what you chose is probably wrong, plus the fact the host will always reveal another wrong one, further confirming your wrong initial choice, so the remaining is most probably the right one so always switch to that to maximize winning.
Best non visual breakdown & explaining further, you only have 1/3 chance of choosing the car door, so switching has a bigger winning rate of 2/3. the host will ALWAYS remove a goat door which gives the change of choice (switching) an additional 1/3 (total of 2/3) compared to your initial choice of 1/3. this solution only works if the host ALWAYS removes a goat door. if the host doesn't open any doors then this will truly be a 1/3 chance of winning regardless if the host asks you to change your choice or not.
The key thing to understand is that, upon choosing a door, you're more likely to lose. The prize is more likely to be behind one of the other two doors. That's the crux of the whole thing. By opening one of the other two doors to reveal a goat, the host practically tells you that the prize is most likely behind the other closed door.
For all the people not getting it and thinking the chance becomes 50 / 50, understand this simple point: the chance you have from the first door is less than 50% and this *does not change* (if you think about it this would be impossible as no subsequent events can change what's already behind it). Then from simple mathematics, if you have two doors and you know one has a less than 50% chance of winning, the other must have a greater than 50% chance of winning.
The brute force method is easiest for me to visualize. With three doors and three prizes (car, goat 1, goat 2), we can get the following possible combinations:
1. *C* G1 G2
2. *C* G2 G1
3. G1 *C* G2
4. G1 G2 *C*
5. G2 *C* G1
6. G2 G1 *C*
Whether I pick door 1, door 2, or door 3, there are only two out of six possible combinations that result in a win. For each of those six possible combinations of doors and prizes, I can open one of the remaining doors and eliminate one of the remaining prizes. That means for whatever door I pick, there are 12 possible outcomes that are equally likely to occur. In this example, I have picked door one and _X_ is the door that is opened:
1a. *C* _X_ G2 or 1b. *C* G1 _X_
2a. *C* _X_ G1 or 2b. *C* G2 _X_
3a. G1 _X_ G2 or 3b. G1 *C* _X_
4a. G1 _X_ *C* or 4b. G1 G2 _X_
5a. G2 _X_ G1 or 5b. G2 *C* _X_
6a. G2 _X_ *C* or 6b. G2 G1 _X_
Again, each of these 12 outcomes is equally likely. If I randomly eliminate door two, then I have two winning combinations if I stay with door one and two winning combinations if I switch to door 3. My odds are still 1:3 in either case. If I randomly eliminate door three, then I have two winning combinations if I stay with door one and two winning combinations if I switch to door two. Again, my odds have not improved. This is because there is are cases where the car is eliminated when a door is opened (3a, 4b, 5a, and 6b). So why do my odds increase when we restrict opening a door to goats only? The same twelve options happen with the same frequency as without the restriction, but the possible outcomes where a car is eliminated are now swapped:
1a. *C* _X_ G2 or 1b. *C* G1 _X_
2a. *C* _X_ G1 or 2b. *C* G2 _X_
3a. G1 *C* _X_ or 3b. G1 *C* _X_
4a. G1 _X_ *C* or 4b. G1 _X_ *C*
5a. G2 *C* _X_ or 5b. G2 *C* _X_
6a. G2 _X_ *C* or 6b. G2 _X_ *C*
Now it becomes clear (at least to me). For the 12 possible outcomes after I pick a door and a goat has been revealed, sticking with door #1 has only four ways to win (out of 12). However, limiting open doors to goats only means that in _every_ instance where a goat is under door one, switching will get me the car. My odds of winning by sticking with door 1 are 4:12 (1:3). My odds of winning by switching are 8:12 (2/3).
Edited for formatting
@@deinemutter5257 Did you not read the post? The first set of outcomes is if I randomly pick in a situation and then a random door is opened and eliminated. Those are the 12 possible outcomes. The second set, however, shows how the odds change because of the reason you just stated.
Holy cow! That's a lot of work when you can just say there was a 2/3 chance the host had to open the door he did because the car was behind the other door.
You are taking a lot of effort to try to say that if you have 1/3 odds picking the right door out of 3, you should switch to one of the other doors because your 2/3 odds of winning are over there. That is false.
I’m not reading all that
At the outset, there is a 33.3% chance that your choice is correct (1 out of 3), which also means there is a 66.7% chance (2 out of 3) that you are wrong. When Monty eliminates one of the doors you didn't choose by opening the door with the donkey, now those same 2 in 3 odds fall on the one unopened door you didn't choose. That's why you always switch.
but why do they fall on one unopened door?
@@dovydaslevanavicius9050 They can't go anywhere else. Remember, your choice has a 1 in 3 chance of being correct which means that whatever remains has a 2 in 3 chance.
Why can't people understand?
I'll try explain it myself then.
Say you picked a goat door at first, Monty would open the other goat door, then if you switch you win.
100% of the time of you pick a goat door you win.
You picking a goat door initially = Win. No matter what.
With me so far?
Good
Now, say you picked the car door at first, Monty would open a goat door, then if you switch you lose.
Still with me?
If you switch when you picked a goat door initially, you always win
If you switch when you picked a car door initially, you always lose.
The chances of picking a car door initially is 1/3.
The chances of picking a goat door initially is 2/3.
So 2/3 of the time, if you switch you will win.
That's all there is to it.
Now, reply to this and tag all the stupid people who don't understand.
Ooops I'm tagging you. You don't understand. Nice try though.
Wow this is probably the best explanation of the reversing of the probability.
No I still think it is 50/50. This whole 2/3 theory is based on the overall possibilities of what could happen if you switch/stay on each door hypothetically. Yes, if you look at the overall "what if's" if you switch/stay from Door A,B, or C, then you would come out with the results of a 2/3 chance of success if you switch. But when you are actually physically playing the game, you have already chosen a door when it's time to make the stay/switch decision. This whole "what if's" of door A,B,C becomes irrelevant. You have chosen a door. At this point in the decision, the only knowledge you have is the car is in your door or the other door. That is purely the only knowledge you have. Based on that knowledge, you have a 50% chance of winning and a 50% chance of losing. Whether played one time or 45083097302857290857 times, it is still 50% chance of winning or losing.
Thanks.
You're an idiot
I don't understand why this happens. I understand that you had a 1/3 chance of getting the car on your first try. But after he opens the door with the zonk, you only have 2 choices (not three):stay or switch. If you only have 2 options, that would make either choice just as likely. Shouldn't the odds changing affect each option equally, and not just the one you didn't choose?
How's this sound: 3 doors, 2 goats, 1 car. You pick door A. Monty gives you a choice, you can take what's behind door A, or alternatively you can take what's behind both doors B and C, and remove the goat by yourself.
You get 2 doors, doubling your chance of the car. You just have to remove a goat.
Connor Stickels Sounds like a different game: pick 1 door or pick 2 doors
HumptyDumptyOakland it's really the same, you're choosing between taking what's behind your first door or you switch to the better of the other two.
It's a different game but it works in the same way.
Connor Stickels Brilliant! That's a fantastic way to look at it - it's like Monty knows and will open all the goat doors for you - before or after doesn't really matter. You can keep A or take the other 99 doors. If you switch to the 99 doors, Monty will open 98 goat doors, leaving you with the last door, for better or worse (almost always better).
*****
"2 doors 1 with a car and one with a goat."
Let's say the game start with 1 door (1 car).
After you pick that one and only door, the host add a goat door.
Right now it's exactly *2 doors 1 with a car and one with a goat*, the same as your reasoning.
Does this means your chance to win if you stay is 50% and your chance to win if you swap is 50%?
Totally wrong.
"the odds are you'll choose a door randomly either swap or stay and you have 1/2 chances that the car is behind one of the two"
Why must randomly pick a door?
Just pick the last door.
You are not comparing between staying and swaping but you claim swaping and staying has the same result?
You use a very bad logic.
Best way to understand, but correct me if I'm wrong. Door 1 has a 1/3 chance of being correct. Doors 2 and 3 (as a set) have a 2/3 chance of being correct. Monty showing you the result of either (because he knows the answer) doesn't actually change that possibility. So switching still leaves you with that same 2/3 chance of the set.
" Door 1 has a 1/3 chance of being correct. Doors 2 and 3 (as a set) have a 2/3 chance of being correct."
Yes.
"Monty showing you the result of either (because he knows the answer) doesn't actually change that possibility."
Well, if he shows you the car that definitely changes it but if he deliberately reveals a goat, it doesn't change the fact that if he had the car behind either of his two doors, he still has it.
Hank is right, but a fundamental rule of the game is that Monty knows where the car is, and will ALWAYS reveal a goat. This is an important factor.
Because of this, the fundamental odds do not change. At all times, there is a 1/3 chance the car is in the door you originally picked, and a 2/3 chance it is in a door you did NOT originally pick. The only thing Monty is changing is the number of doors representing those odds. At the beginning of the game, there is a 2/3 chance the car is in a door you did NOT pick, and 2 doors representing those odds - - or 1/3 each. Makes sense. After Monty reveals a goat door, he doesn't change the odds that the car is in a door you did NOT pick, but he DOES change the number of doors representing those odds - - now there's only 1 door left - representing that 2/3rds chance.
Now, if the host were acting randomly, and randomly revealed a goat, then it would be 50/50. The reason is because if you chose a goat to begin with, the host could potentially randomly open the car door. This would obviously ruin the game. This would happen 1/3rd of the time. So here are the scenarios :
You choose the car - host reveals goat - you switch to a goat - LOSE
You choose a goat - host reveals goat - you switch to a car - WIN
You choose a goat - host reveals car - GAME IS RUINED
Ergo, of the scenarios where the host RANDOMLY reveals a goat, you have a 50/50 shot.
This isn't how the Monty Hall problem works though. Monty will ALWAYS reveal a goat; NEVER a car. Ergo, there is ALWAYS a 1/3 chance you have the car - and a 2/3 chance the car is in the other remaining door.
Thank you all for your patience and dedication to teaching.
That is classy of you. Regards, and take care.
In order for you to lose after switching you must have guessed the right door and that can only happen at 1/3 probability
This is the most intuitive and simplest explanation of this problem I’ve ever seen. Kudos.
First time I heard this problem, I was certian "the math" was bunch of bs. Years later I got interest in probability and I returned to this problem to solve it, it came just after hearing the rules. Too many people tries to be "street start" regarding science. My comment might not have much sense, but to truly undertand what you're saying, I needed years of education.
Kudos for nicely summing it up
@@ac4694 You chose a door having 33% chance to win.. Then the host opens a door.. You flip a coin and chose a random door.. Now sometimes you chose the same door as the one you started. You switched to the same door.
@@mariomario4676 you should practice some reading XD
@@ac4694 Elaborate please 🥺
Well this really works, I wrote a program which simply chooses a random door and then switches to the one left closed.
Here are result (no of wins).
When switching doors: 48752
When not switching doors: 26486
Of course it works.
We do not believe in probabilities, but we are calculating them.
This is more interesting in terms of human psychology than the actual math. When there are 3 doors your ego fools you. Anything more than 3 and you can see it clear as day because your ego isn't there anymore.
Exectly. It's an introduction into 'game theory' where odds change based on deliberate actions rather than random chance. One of the most complex and interesting fields in mathematics.
Bruh i just saw this after watching dl sneaky and meteos talk about it
SAME HAHA
yea gained respect for meteos and DL for not flaming sneaky too hard haha
*Best explanation:* There are two possibilities:
(A) *1/3* chance that the car hides behind the door you picked. Both other doors are zonks, one of which has been opened. *By switching you lose the car.*
(B) *2/3* chance that the car is behind one of the other two doors, one of which has now been opened, revealing no car. *By switching you win the car.*
Hence, by switching you have a 2/3 chance of winning the car.
I like that. I think the most intuitive way to look at it is this: You pick one door, then Monty offers you the chance to instead have BOTH the other two doors. Of course you jump at the chance because you understand owning two doors is better than owning one door.
could you not claim there was a 2/3 chance it was behind door 1 or door 2?
Let's put it this way - - if you choose to switch, you will ALWAYS be switching to the OPPOSITE of what you picked. If your initial pick was a car, you will switch to a goat. If your initial pick was a goat, you will switch to the car. You will NEVER switch from one goat to the other goat - - because if your initial choice is a goat, the host will reveal the OTHER goat - leaving only the car, so you will ALWAYS be switching to the car if your initial choice is a goat.
GOAT A
GOAT B
CAR
Your first choice is GOAT A, host reveals GOAT B, you switch to CAR - WIN
Your first choice is GOAT B, host reveals GOAT A, you switch to CAR - WIN
Your first choice is CAR, host reveals either GOAT A or GOAT B, you switch to a goat - LOSE
Switching wins 2 out of 3 times. The only way you LOSE by switching is if you chose the CAR to begin with. It's the only scenario where you lose by switching. The odds that you chose the CAR to begin with are 1 in 3 - ergo, the only way you lose by switching is 1 in 3 - meaning, you win 2 out of 3 times by switching.
I mostly agree. One thing you get wrong is to use the odds of picking the car (1/3) as the precondition. The actual precondition is "picking car AND host offering switch", which can actually be anything, doesn't have to be 1/3 at all...
@@insignificantfool8592 The host always offers the switch.. it's part of the problem, and the host offering the switch changes nothing about the odds you picked the car.
@@willoughbykrenzteinburg then we talk of different problems. I talk about the problem stated in the the letter to Parade and often repeated in a similar fashion elsewhere. In those cases, it is not said that the host _always_ offers the switch.
@@insignificantfool8592 Then they are presenting the problem incorrectly....simple as that. Offering the switch is a fundamental element of the problem....
@@willoughbykrenzteinburg I'd argue the problem that is known as the Monty Hall problem is what you would call the incorrect one.
Do you have any example where the problem is presented correctly?
Let's say door 1 contains the car behind it and doors 2 & 3 are not winners. If you do not switch, then we can all agree that you have a 1/3 chance of winning. However, if you do switch, then there are three possible scenarios.
SWITCH SCENARIO 1:
1.) You pick door 1.
2.) He reveals door 2 or 3.
3.) You switch to the other door that he did not reveal.
4.) You lose.
SWITCH SCENARIO 2:
1.) You pick door 2.
2.) He is forced to reveal door 3, because door 1 has the car behind it.
3.) You switch to door 1.
4.) You win.
SWITCH SCENARIO 3:
1.) You pick door 3.
2.) He is forced to reveal door 2, because door 1 has the car behind it.
3.) You switch to door 1.
4.) You win.
As can be seen in the 3 'switch scenarios' that you have control over, you have a 2/3 chance of winning if you switch.
These are all the possible scenarios but they dont matter because he will give you the one thats zonk. So it comes down to 50%
fidodido665 yikes, you didn’t read his comment in the slightest
Here's another way to view it. You cannot know which door you picked, but let's for argument's sake assume either
A: You picked the wrong door, i.e. a door with a goat.
B: You picked the right door, i.e. a door with a car.
If you're in case A then he's literally showing you the right door by opening the other wrong door.
If you're in case B and he opens another door, he is only confusing your choice. If you choose the other door, you're taking the wrong door and you'll probably be very pissed.
So with the suggested logic you'll get a car in case A and you do not get a car in case B. However case A is the most probable, with a chance of 2/3, and case B is the most improbable, with a chance of 1/3.
Basically, since you most likely chose the wrong door in the first place (because there’s two goats and one car), once Monty gets rid of a bad door, that doesn’t change the fact that you probably chose the wrong door in the first place, so you should switch it.
One of the simplest explanations. People try to interject all kinds of variables to debunk this. But math does not lie.