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Game Theory Scene | 21(2008) | Now Playing - Фільми й анімація
I love how college classes in movies are only 4 minutes long.
I can’t stand how chalk boards in movie ‘’’math’’ classes are set up. They have a bunch of random unrelated graphs and equations to make what’s being presented more scary than it really is.
@@robrs8631 Also Newton's method is something you do numerically on a computer. Definitely not anything you would do on paper or on a chalk board. (Except maybe quickly explain it...)
It’s the end of the lecture guy
4 minute lecture. You can't argue Kevin Spacey doesn't try hard to be popular with the kids.
@@Mobin92 I mean not really.
If you learn numerical methods Newton Method is explained without computers, it's not about how to solve it, but how it works, and why it works.
For sure, homework will often require writing some script to solve it, but lectures - no.
At least I was taught this way.
imagine if he went to find Ben Campbell's exam and saw a "46% F"
Damn. That would have been good.
@@ThePandagansta Textbooks will dull your mind.
This isn’t Naruto. He clearly showed himself to be a standout from the other students since literally no one else could answer those questions. Someone that smart wouldn’t be failing his tests, it would’ve been funny tho.
@@adh2298 All he did was account for variable change.. any 6 year old can do that when playing battleship.
That would be better actually lmfao
The video is titled Game Theory, the class is named Nonlinear Equations, the question is asked is about Probability Theory.
yup plus nonlinear equations and the problems on the board were x^2 +1 and the quotient formula for derivatives. all of it makes no sense academically speaking
Game theory was invented to give the average mug the sense that they can gain an advantage over the house. It was invented by the house. They increase their customers, their losses go up, but so do their winnings which always outsrip their losses.
That is why this video, referencing game theory, presens a false picture.
There is a 100% certainty that he will be presented with a fifty fifty chance of winning the car, so his starting odds are not 33%, they're 50%. 66% is still an advantage, but you're never doubling your chances.
@@happinesstan No, he will be always be presented with 2 choices, but they are 66% and 33%. If he always stays with his original choice, he will win the car 33% of the time.
I guess if you're saying if he adopted a random strategy between switching and staying, then he would win the car 50% of the time, then that is true. But most people always stay, and then their chances are 33%, so half of always switching.
@@kevinrosenberg4368 Yeah, that's exactly what I'm saying. The experiment explains itself very well, but I agree that most people, lacking the information that the problem presents, would stick with their original pick. Therefore picking randomly would be a better [not best] choice. But the manner in which the choice is presented, essentially denies that opportunity.
It's a fascinating puzzle that, I think, is about more than probability.
What? All a bit " fuzzy"?
While many people explain to those who do not understand it, they often fail to explain why the probability of 33.3% is added on top of the other probability. Let's consider a simplified scenario for those who think the chances should be 50/50:
Let's say there are three doors, and behind them are like the following:
Goat - Car - Goat
Let's look at all the possibilities.
If you choose Door 1, the host must open Door 3, and if you change your choice to Door 2, you win.
If you choose Door 2 and decide to change your door, you lose.
If you choose Door 3, the host must open Door 1, and if you change your choice to Door 2, you win.
As we can see, in all three possibilities where you change your door, you win twice out of the three possibilities.
Similarly, let's consider the possibilities where you stick with your initial choice:
If you choose Door 1, you lose.
If you choose Door 2, you win.
If you choose Door 3, you lose.
We can clearly see that the strategy of changing your door gives you a higher chance of winning the prize. It's not a 50/50 scenario, but rather a 2/3 probability of winning if you switch doors. When host opens one of the remaining doors, he provides you with a new information. This information is not changin the initial probabilites but rather telling you that:
"The probability of the car being in one of the 2 doors you did not choose is 66.7% and I am opening one of these doors for you. In the beginning there was a 66.7% probability that the car was in one of these two doors, and I showed you which of these doors had a goat."
The 33.3% probability was added because of the information the host gave us. Thus, when we change our door, we have a 66.7% probability of winning.
Great explanation
After going through a number of comments, this finally makes proper sense.
That's an explanation, thanks
That was wonderfully explained... Thanks a lot.
Man, thank you, this shit haunts me for years and now I got it.. ty
I love how the class on solving polynomials became a probability class.
Of course, movies
Solving polynomials, the hell does that mean?
The class was about solving non-linear differential equations.
@@fredthechamp3475 I hate the fact that I now know what all this means after watching this movie when i was 10
dont forget he was also trying to determine if he wanted to recruit Ben, after realizing he was intelligent. It was more of a test for blackjack than a test for the specific class he was in.
And the teacher misses the obvious error.
Since there is a 100% certainty that he will be offered a 50/50 chance, his starting odds are not 33%.
Imagine registering for SOLVING FOR NON-LINEAR EQUATIONS and spending the class discussing the Monty Hall problem. I hope those kids went straight to Admissions to request a refund.
My thoughts exactly lmao such stupid scriptwriting
@@randomutubr222 I hate the way college classes are shown in movies. "Who explain Newton's method and how to use it..." - no prof is teaching like this? If that is the week's agenda, that's what THEY'LL teach. And secondly, what the fuck is the relevancy of Ben's mention of Raphson here? That wasn't the question. This is a math class, not the history of math class lmao.
@@ASOT666 in my classes i had, prof just don't have the time lol, they spend all of their time writing at the board trying to fit a course that they have less and less hours to fit in and are annoyed when you ask them questions
@@ASOT666 in bens defense the teacher went off track by saying "Tell me something i don't already know," and the scriptwriters used this to prove that Ben understood more than the basics by showing he knew the history behind the method. This was probably the best way to show he knew math without him actually doing math so the audience wouldnt get confused by technical jargon. Still terrible writing
It’s the end of the class and the prof want to end the class with something fun and interesting. To be honest prof and teachers like these are the ones that got me hooked into the class, not those teachers that only focus on the lecture.
Another way to think about it:
If you picked correctly the first time, the right move is to stay. If you picked incorrectly the first time, the right move is to switch. What was your odds of picking incorrectly the first time? 66%. So 66% of the time the right move is to switch.
E: Since people are being stubborn-
Suppose after the host opens door 3, you say "I will stay on door 1 since I don't improve my odds by switching". Then by that logic, if the host had opened door 2 instead, you also would stay on door 1 instead of switching to door 3. Therefore, by that logic, you don't even the game host to open a door at all! You just need to know that he would have gone and opened a door.
So we are left with the following: the host isn't necessary- you picking door 1 makes your odds of winning 50% regardless of what the host does. Which is absurd.
Thanks. This is the best explanation I have read so far.
This is known as the Monty Hall paradox. It's not really a paradox, it got the name from the solution being so unintuitive.
Perfect explanation
Damn. That’s an elegant explainer.
Think of it this way. Behind 1 door is freedom, behind the other 2 are shotguns that shoot you like in a Saw movie. After choosing the correct first choice, how confident are you now to stay and not change decisions when your life is on the line?
I gotta give this movie some credit, that stuff on the board is real. Some is just garbage but they actually have the correct formula for newton-raphson iterations to solve nonlinear equations
No idea why a non linear eqn Professor would randomly ask a game theory q tho ahhah
🤓
@@advayiyer6456 probably to probe to see if he was blackjack team material
@@advayiyer6456 They were on a scout, it was random for everyone but not for those 2 that wanted to test him.
it is funny tho that a non linear equations prof is asking a stats game theory questions
This is not "Game Theory" - this is "The Monty Hall Problem".
I mean it is Game Theory, but your answer is more specific. Lol
It is the theory of a game, the Monty Hall game. Not game theory indeed.
Yea, I was thinking John Nash's Game Theory which would've been inaccurate for this.. But it is generic game theory, I guess..
What is the name of the movie!?
@@farooqkelosiwang9697 Its name is "21"
i wish a nonlinear equations class was this easy in real life. one of my toughest undergrad classes right after topology
havent seen the movie but ar eyou inferring from other scenes?
@@anthonyhu6705what is topology?
Literally.@@definetheterms1236
@@anthonyhu6705 topology origami 😂😂😂
@@anthonyhu6705 tell me you dont know what topology is without telling me you dont know what topology is
Say what you will about spacy. But man oh man the guy can act.
i say: he is sexually assaulting ppl. and the ones who speak up get killed.
nuff said.
He was actually pretty bad in this scene.
LA confidential
I will say he’s a gay pedophile.
@@thebeautyofnature3616 He's not guilty
The scene doesn't explain it well, but the original Monty Hall problem factors in a few rules:
1.) The game show confirms that the host does indeed know which door contains the car.
2.) The host _cannot_ reveal the door you chose.
3.) As the point of the game is whether you will alter your decision, the host can never reveal the door with the car. This means whichever door they reveal will be a goat every time.
Let's say you initially pick Door 1. You are now met with three possible scenarios.
Scenario 1: The car is indeed behind Door 1. This means, regardless which of the other doors the host reveals, you win by staying. Of course, you lose if you switch.
Scenario 2: The car is behind Door 2. Since the host cannot reveal the car, they are forced to reveal Door 3. By staying, you lose. You win by switching.
Scenario 3: The car is behind Door 3. Since the host cannot reveal the car, they are forced to reveal Door 2. By staying, you lose. You win by switching.
By switching, you have a 2/3 chance of winning the car.
The host knowing the car's location is key. If the host doesn't know where the car is, they run the risk of potentially revealing its door. This would cause the game to end immediately. In the case where they don't reveal the car, then the probability of switching or not switching would be 50-50.
I've said something similar. If the choice of door was completely random and it was possible for the host to open a door that had the car behind it (Whoops!), then having two doors left would indeed be 50-50. But since the host will NEVER open the door with the car, them opening up a door AT ALL means nothing. You're either staying with your 1-in-3 door or switching to the 2-in-3 doors.
For some extra fun...(let's assume the car is behind door #2 and the host opened up door #3
From the original contestant's perspective, the odds of winning the car if they switch is 66%
From the host's perspective, the odds of winning the car if the contestant switches is 100%
If someone otherwise oblivious to the game walks in after door #3 was opened and only see 2 doors remaining without knowing the rules, the odds of winning the car if the contestant switches are 50%
People who are getting confused by this problem are thinking that they are in the third position instead of the first one. It's all about knowing the correct information.
For people confused, imagine if you have 100 doors, 1 of them has a car and 99 of them have a goat. Your guess accounts for 1% chance of being a car behind it, but imagine the show host (who knows where the car is) opens 98 doors (all goats) and leaves you with a choice to choose your door or switch to the one still not open, you clearly switch since there is a 99% chance the car is behind those doors
My man! Thank you for explaining
But now aren't you supposed to choose between 2 (and not 100) doors where in one of them there is the car. So isn't it a 50-50 probability that the car is behind one of those 2 doors?
@@neelarghoray5011 when you 1st chose you had a 99% chance of being wrong.. so its 99 time more likely you chose a wrong door.
By opening 98 other doors the host takes care of 98% of that chance, you switch and now you have 99% chance of being correct, if you stay its still only a 1% chance your original pick was right..
Hope that makes sense
@@neelarghoray5011
Think of it this way, if I pulled out a deck of cards told you to pick one at random, and hope it was the Ace of Spades, and then I searched through the rest of the deck and grabbed a card. Then I told you, either you picked the right card at the start, (1/52 chance), or I just picked the card right now. (51/52) chance.
What seems more likely, that you guessed correctly at the start? Or that I did, knowing what all of the cards were? It's the same logic since the host knows what's behind each door. He ALWAYS chooses the door with the goat.
would the situation change to random chance if host opens 97 doors (all goats) and you left to choose 3 doors (1 of which you can stay)?
Some great acting by Spacey
what's so great about it?
@@sebastiann3670 People often confuse convincing and charismatic acting (especially by great actors), with great acting.
@@jacobshirley3457 Explain the difference?
Wish he was my math teacher at high school.
The comment section of only UA-cam video which helped me to learn this classical problem...
By switching, you are essentially saying, "I'm betting that I started with a door with a goat." Since there was a 2/3 chance of this being true, switching increases your chances of winning the car to 2/3.
If you stick with your initial choice, you're essentially betting that you started with the car, which has only a 1/3 chance.
Your reverse-engineered solution is actually far more intuitive than the deeply-explained (but first principles-based) solutions elsewhere in the comments.
Ignore the starting odds, there is a 100% certainty that you will be offered a 50/50 chance, so your starting odds are evens. 66% is still an advantage, of course, but nt as big as doubling your chances.
The key to this that most people overlook is that the host's opening of a door was a deliberate opening of a losing door--it wasn't random, amd could never have a car. Had the host made a random opening, the probabilities wouldn't have changed, and the host might have opened the winning door. Easier to understand if you consider the extreme--a hundred doors, and the host deliberately eliminates 98 of 99 losing doors from the set you didnt choose, which set contains a 99 percent chance of having the winning door, leaving one juicy door remaining that you could switch to.
Amo esse ator e esse jeito de falar. ❤
I’m sure you do weirdo
After the host reveals one of the doors with a goat behind it, and you decide to switch, you get your initial odds of picking a goat, which were 66%. This is because if you picked a goat initially, switching doors will always land you at a car.
Best Explanation:
Scenario 1:
You initially pick the door with the car behind it (1/3 chance).
If you stick with your choice, you win.
If you switch, you lose.
Scenario 2 and 3:
You initially pick a goat (2/3 chance combined for both scenarios).
In both of these scenarios, Monty has to open the other door with a goat.
If you stick with your initial choice, you lose (because you originally chose a goat).
If you switch, you win the car.
The probability breakdown for switching vs. staying is:
Switch: Lose (1/3) vs. Win (2/3)
Stay: Win (1/3) vs. Lose (2/3)
Meaning if you switch you will always have a 2/3 chance of winning (the 1/3 chance of losing is from you switching when you already chose the door with the car)
For those who still have problems understanding/accepting this, try it with 1000 doors.
You choose 1 door and the host opens 998 doors behind which there's only goats, leaving only your door and one other left.
So, do you stick with your door and think the odds for your door just magically changed from 1/1000 to 50/50 or will you change to the only door the host hasn't opened as he most likely just showed you where the car is?
Yeah sorry that is bullshit because there are just 3 doors here. You open 998 doors but the host just open one! Pls don't try to b e stupid.
@@dianamon2727
Assume you stay with your first pick.
If your first pick is Goat A, you get Goat A.
If your first pick is Goat B, you get Goat B.
If your first pick is the car, you get the car.
You only win 1 out of 3 games if you stay with your first pick.
Switching means the opposite.
It's just basic math/logic kids understand.
Sadly, it's far too hard for idiots and their parents.
To audition Ben with the Monty Hall problem was simply genius
This entirely assumes the host would even consider opening an alternative door in the first place.... WHICH YOU DON'T KNOW.
The answer presented can only be given partial credit as it does not account for a player that wishes to win one of the goats.
I watched this in theaters for my after prom in high school. A decade and some change later, this is the type of homework my kids bring home and I feel like Barney rubble 💀😂
Just because you switch doesn't mean you win with a 100% certainty, it means if you play the game N times then the strategy (intial 33.33% + info gain from one open door 33.33% = 66.6% probability of it being there) statistically converges. Therefore, you have an edge if you played the strategy in which you switch.
Well, no, people are forgetting that both doors have a 66% of being correct if you're using the original calculation. Taking out an option throws the entire probability model out the door.
@@DBCOOPER888 correct, you're just left with 2 choices: to stick to your original choice or to switch
Think about it as you have a 66% of being wrong with your first pick. So its likely the prize is behind one of the doors you didnt pick. You want yo switch, but you dont know which door to switch too. But then the host tells you which door has a goat. So now you know which door to switch to.
@@raycon921 You're not left with two options, you are left with three options. Change your mind, don't change your mind, or tell Monty that your mind was never made up and revert to your original strategy of picking randomly.
This gives you a 50/50chance of picking the same door, or the only other door.
The presentation of the choice is deliberate in order to mask the third option.
I think a simpler explanation is this. When you first pick you have a 33% chance to be right and a 66% chance to be wrong. But when the host takes away one of those two wrong options, the original % is still correct, the goat has a 33% chance to be your first choice, meaning the same 66% chance is left with the other unopened door.
There was also a 66% chance the other door is the right choice, so it's still a 50 / 50 toss up.
@@DBCOOPER888 correct, theese idiots don't understand that sticking to you first choice is still a choice 🤣
@@DBCOOPER888 No. It becomes 33.3% chance when the switch is offered because the player is more likely to pick the goat in the first choice.
All possible outcomes and player takes the switch everytime:
1. player first picks goat #1, he will switch to the car.
2. player first picks goat #2, he will switch to the car.
3. player first picks the car, he switches to either goat.
So while the player has 33% to pick the car in the first place, if he can choose to switch doors, he actually wants to first pick one of the goats (66%) to get the car.
@@ojon12389 omg no better explanation than this
@@ojon12389genius
Спасибо за очередную полезную связку, все работает как в описании 🔥🔥🔥💯
The method described initially is known as Newton-Raphson method. So Raphson did get credit for that
Naayinde mone
@@pkmuhammedhisan Ente ponnu aashane enikku malayalam nalla pola ariyam. Nalla pole theri parayanum ariyam. Pakshe vendanne
@@rahulmathew4970 sorry bro..malayalee aano enn ariyaan veruthe irittatth vedi vecchetha...naattil evdeya..nyan Thalassery laa
@@pkmuhammedhisan vedi vachittu kondalle. Pathanamthitta
in my classes it was only Newton lol
When Ben first makes his choice, he had a 1 in 3 chance of choosing the right door... the other two doors together have all the remaining probability of being right, i.e. 2 in 3. When one of those doors is eliminated from consideration, Ben's first choice still has the same 1 in 3 chance of winning that he started with. The remainder of the system still has the 2 in 3 chance of being right.
His acting make the scene so much interesting ..
You change-Let's look at it this way. 99 people play the game and agree to share the prizes. If this is right they should all change their first pick. The car is placed behind door 1, 33 times, door 2 33 times and door 3 , 33 times. All 99 contestants pick door 1, and all change to the remaining door left. That means that 66 will win the car. In case 1 where the car is behind door 1, they all pick door 1 and switch, those 33 all lose. In case 2, where the car is behind door #2, door # 3 has to be eliminated since they don't eliminate the door you picked. You switch you win. Same thing happens in case 3 when the door is behind #3. They eliminate #2, you switch from 1 to 3 and win. So, if you just work it out over a larger population it's easy to see why you switch.
This is called the Monty Hall problem
The correct explanation is that once Ben has selected #1, there are four possibilities based on the Host's restrictions (can't open Ben's door, can't open the car door. In reverse order, they are:
1) (1/3) The car is behind #3, so the host must open #2.
2) (1/3) The car is behind #2, so the host must open #3.
3) (1/3) The car is behind #1, and now can choose however he wishes between #2 and #3. If he chooses randomly, this breaks down into:
3A) (1/6) He opens #2.
3B) (1/6) He opens #3.
What Ben ignores, is that he saw the host open #3. So cases 1 and 3A are eliminated. Of the two that remain, case 2 is twice as likely as case 3A, so two out of every three times the car will be behind #2.
But what if the host doesn't choose randomly in case 3? What if he always opens #3 if he can? (Then the chances #1 and #2 are the same.) Or if he always opens #2 if he can? (Then the car IS behind #2.) The point is that Ben's reasoning is wrong, even tho he gets the right answer. It isn't because #1 stays at 33.3%, it is because we can't assume the host chooses non-randomly.
Your comment convinced me to never take extra math classes, thank you kind internet stranger!
This is the complete and correct explanation. Others are trying to solve it purely based on probability and none of the explanation answer the why. Thanks!
@@tekudiv I appreciate the feedback. But one correction: people usually pick an _answer_ based on intuition, and choose a _solution_ that leads to that answer, and justify it because they think it is the right answer. That is what is happening here, and coincidentally it is the right answer.
This is possible in probability, but not really in other fields of mathematics, because the elements aren't always the same in different solutions. In geometry, if you have a triangle, its sides are such well-defined elements. But in probability, the outcomes you choose to consider can be different. Here you need to recognize that the choice of doors can be random.
Yeah, the example in the movie clip here doesn't fully go over the "rules" for the student to guess the correct answer, because it's an abbreviated version of the problem.
@@lluewhyn Those who think the answer is 50:50, if they think about rules at all, all believe them to be the "full rules" you refer to. That is, the host _will_ open a _goat_ door that _was_ _not_ selected by the contestant. These "full rules" have nothing to do with the controversy, except that when brought up they make the problem seem more complicated than it needs to be.
They think the answer is 50:50 because they do not see any difference between the two closed doors. If true, that would indeed make the chances 50:50. The most common explanations do not even attempt to explain that there is a difference, or what it could be.
The answer "My original door's chances can't change" is actually wrong, They can, but only if the host is, indeed, "playing a trick on you trying to use [either direction] psychology to get you to pick a goat." For example, if he always opens the highest-numbered door (did you notice that the "explained rules" don't say the doors are numbered?) that he can, then in the game as presented you do have a 50:50 chance. Because there is no difference. But if he always opens the lowest-numbered door, we can deduce, for a fact, that the car is behind #2 _because_ the only way he won't open it is if it has the car.
The answer "what if there are 100 doors" is actually better, since it isolates the one door that was left closed as being different. It doesn't explain the difference, but makes it more obvious.
imagine if Ben asked Fisher for extra help to take down Micky Rosa
After you pick 1 door, you have remaining 2 doors. Those doors are more likely to be 1 goat + 1 car, rather than 2 goats. In numbers, there is a 66,6% likelihood that the remaining two doors are 1 goat+1 car. Simply put, it is more likely that the door with the car is included in the remaining 2 doors (after you pick your choice). THEN, the host shows you a wrong door, no matter what. He must do so. Considering the above, it more likely that the remaining door is the correct one , rather than not. Therefore, SWITCH , and you have 66.6 % of winning, the same as the possibility that the remaining doors after your initial choice are 1 goat + 1 car. NOTE: all scenarios are likely. No quarentee you will win. We are searching for the most likely…
An easy way to understand this is to consider what you do if you had a choice between selecting one door or two -- clearly if you could select two doors, your chance of winning is 2/3 whereas selecting one door yields only 1/3 chance. In the scenario here, you initially select one door and then are given the opportunity to select two doors - you win if EITHER of the two doors is not a goat - there's only 1/3 chance that both of the doors are goats. The illusion is that by switching, you are selecting only one door.
I mean are you taking a test so you are allowed to check two answer boxes say C and D for example to get more of a chance to get it right? XD
@@MrLuffy9131😂same q??
Probably the most layman and simple explanation I’ve heard so far.
Using that logic there's a 66% chance either door will win, so you're still back to a toss up.
Imagine that you pick door 1 - and then you are given the option to switch to both door 2 and door 3 -- meaning that if you switch and EITHER door 2 or door 3 has the prize you win. Will you switch? Of course you would because, there's a 2/3 chance that prize is either behind door 2 or door 3 and only a 1/3 chance it is behind door 1. Telling you that the prize is not behind door 2 before giving you the choice to switch doesn't affect the 2/3 chance you get by switching.
See if someone still stuck on 50-50 see it in this way:
1) There are 3 doors and getting one correctly is 33.33%
2) Now out of 2 one is shown to be Goat door. 2 doors are left, 1 is your chosen door and another which is left.
3) Since there were 33.33% of you being correct, so the door left to be correct will now be 66.67% as both will sum up as 100%.
if one of goat door is shown or the condition favorable to chooser is changed, why the favorable probability only contributed to the left door? it should be eaqually improve the both doors. So the chosen door winning chance improves to 50%. the final winning chance is 50% to 50%. switch or not switch is the same. another example, 3 persons (A,B, C) are put in a jail, only one person can be released. now the police says C will not be out for sure. Do you think one of the left two will think he will have 66.67% chance out? of couse no, both chance of going out will be improve from 33.33% to 50%!
@@vitasino5823 The thing is that this game has as a rule that the host cannot reveal the player's choice and neither which contains the car. He must always reveal one that is not any of those two, which he can because he knows the locations. This is often not well clarified and that's why it's confusing.
If you notice, with those conditions the player's choice is a forced finalist: it will always be one of the last two regardless of if it is a bad choice or not, so the only way it could result being correct 50% of the time is if the player managed to pick the correct option 50% of the time when there were still 3 ones.
In contrast, the other that remains closed had to survive a possible elimination, because the host could have removed it in case it did not contain the prize. But as the host avoided it, its chances of being the winner increased.
So, your example with the 3 persons is not equivalent because it was not established from the start which of them could never be mentioned, even if he was not going to be out.
i remembered first time seeing this scene when I was 12 confused af about what he said. Now being in numerical analysis and major in applied stats, I understood everything he said was just basic intro stuff. Mind blowing how time flies so fast
Kevin Spacy what an actor
Not really game theory, but still a good scene (although in retrospect it makes no sense for him to be asking this question in a non-linear equations class).
Game theory is the study of mathematical models of strategic interactions among rational agents. It has applications in all fields of social science, as well as in logic, systems science and computer science. The concepts of game theory are used extensively in economics as well.
Idk, this is game theory by the sounds of it. What was your definition of game theory?
And maybe not a non-linear equations class, but considering what the movie is about it's his "test" to Ben to see if he's competent enough to join their gambling group.
@@terencetrumph9962 note “among rational agents.” Game theory is when multiple agents are making choices and those choices have effects on the overall outcome. This problem consists of one person making a choice, thus it would be categorized as choice theory or decision theory.
Also, I know what it’s about haha, just doesn’t seem like the time or place
@@j.d.kurtzman7333 I see your point, although I think the plural here refers to 1 and/or all and the focus is "strategic interactions" between the player(s) and the game, no? Otherwise me playing solitaire all by my lonesome has just been "choice theory", right?
How I thought of it was, even in 1 player games, variables are designed to act as an opposing force, therefore making a "2nd player" for you to overcome. Say that weren't the case, or I'm an idiot and just wrong, if you play rock paper scissors with a learning AI that can guess what you throw out based on patterns, does it become game theory rather than choice theory after a certain point?🤔
@@terencetrumph9962 the economics definition would not define solitaire as a “game” per se since only your decisions affect the outcome (ie there are no one player “games”). As to the AI, more of a philosophical question perhaps, or maybe a computer science one. Put it up to the Turing test, if it passes then I guess you’ve got a game
This is game theory. Switching doors after the host shows a goat is a "dominant strategy;" it maximizes your utility.
Here are your 3 possible door scenarios:
CGG
GCG
GGC
Always choose Door 1 but then reveal one of the G. You’ll see that there is only one scenario where you don’t get C if you change.
This is one of the best explanations I've seen
@@animeshadhikary7802 the easiest way would be, if you don't change, you have 1/3 chance, so if you change, you have the remaining 2/3 chance
Yeah, because the host always opens a door with a goat meaning if you switch to the other 2 doors and the car is there, you will always get the car. The other two doors have a 2/3 chance of being correct because there are two doors. Think of the two doors as one group of doors that has a probability of 2/3 and the first door you choose as a probability of 1/3.
It’s only 50-50 if you choose a door with one of the three doors open already with a goat.
@@hongieyo I always add more doors at the start of the explanation because ppl struggle with 33% and 67% for some reason. But if you have 10 doors and a 10% vs 90% chance ppl get it. You can also do 100 doors and have a 1% to 99% chance. I think they struggle with the 33% vs 67% choice because the chance to pick right from the start was already pretty big.
It doesnt matter which door you choose, you can choose any door you want.
The key point here is that host will always reveal a door where there is a goat, then you can choose whether to switch or not and the result will be the same - 66.66% chance to win the car not matter which door you initially chose.
I'm very impressed by the fact that you can use non-linear equations to get out of groomer charges...
The answer to the people who thought 50/50:
There are essentially two hidden rules to the door he will open. He will never open the door you picked because then the question whether you wanna switch wouldn't make sense: If the car is there, you will have to say no to switching, if it isn't, you'll have to say yes.
He will also never open the door with the car because that will mean that you already lost, and again the question about switching wouldn't make sense. So really, the answer lies within the fact that you are guaranteed the option to switch. This fact alone reveals that you should.
Nice explanation
I agree, but the suggestion that you begin with a 33% chance is erroneous, as it is a 100% certainty that you will be offfered the 50/50. So whilst the 66% is an advantage, it is not as great as implied.
It's called The Monty Hall Paradox
this "paradox" was mentioned also in the series Better Call Saul Season 2 Episode 4
It was also mentioned in Brooklyn Nine Nine where Captain Holt actually gets that one wrong saying the probability is 50/50 so it doesn't matter if you switch.
And ironically Monty Hall never actually offered the choice to switch doors after eliminating one on his game show.
My strategy will be to pick door A and always switch. There are three (equally likely possibilities).
A: The host opens a door, i switch, and lose
B: The Host opens C, I switch to B and win.
C: The Host opens B, i switch to C and win
I win 2/3 of the time.
I do this with my 10th grade pupils as a maths teacher. Everyone gets an idea to find the best strategy for the monty hall problem by drawing a probability tree for each strategy. Its funny how they sell it as a test to find the only genius in your class.
the way I prefer to explain this is by asking the same question but with 100 doors with a car behind one of them. If you pick a door, then I open 98 other doors all with goats, then that leaves just the one you picked and the one with the car behind, obviously you would switch because the chance you had of picking the right one first remains 1/100, therefore the chance of getting the car by switching is 99/100
I get the point of it but in all reality it is still a 50/50 chance assuming that all doors had equal chances to get a car
This is not "Game Theory" - this is "The Monty Hall Problem".. Some great acting by Spacey.
which is an application of game theory
@@ktktktktktktktNo.
@@ktktktktktktkt It is not, because the Monty Hall Problem is not a game. Or - to be more accurate - it is neither a _cooperative_ nor a _non-cooperative_ game between _rational agents_ .
@@michaelkarnerfors9545 That is under the assumption that most popular analyses of the problem make but the host can make different decisions too.
@@ktktktktktktkt Yeah: Monty never allowed the participant to change. 😁 That is an invention made from the problem.
I mastered card counting thanks to this movie. Was a loner type who liked Sci and Tech the most. Saw the movie and spent my vast free time "practice card counting". Funny a teacher wanted to learn too. (Taught him basics)
Kevin: "Naow if aye r'mooove a sihtin numbah from the equachun whether human groped or naught; taystimony says im innocent no mattah the numbah of deyad witnuhsayas yuh-on-ah"
Jury: 👏👏
You have three doors: A, B, C. B contains the car, the other two contain goats.
You have an option to choose twice. Once at the start and once after opening a door that contains a goat.
Let's say that you choose to another door after host shows a different door. Here are the possible scenarios --
{First time Choose A, Second time Choose B, Host open Door C} -> you win,
{First time Choose B, Second time Choose C, Host open Door A} -> you lose,
{First time Choose C, Second time Choose B, Host open Door A} -> you win
Probability of winning went to 67% boom!
I don't think so. There are 2 possible ways for you to lose by choosing Door B the first time. You only listed one (1). It's still 2 on 2. 50/50
@@erranti07 It doesn't matter, regardless of whichever door opens, you loose in case 2
@@nightlessbaronIt does matter. The reason you arrived at 66.67% probability is cause of the failure to account for the other possible event of losing when you choose Door B.
@erranti07 I guess I can explain it in even more simpler terms. The question is whether we should choose another door or not after deciding on the first choice. So, we want to find P(choosing another door) and P(not choosing another door). Also, P(choosing another door) + P(not choosing another door) = 1.
You have three doors: A, B, C. Also assume that you always choose door A on the first turn (you can repeat the same exercise with other 2 doors and average the results out --> you will end up with the same number).
A B C Stay Switch
Car Goat Goat Win Lose
Goat Car Goat Lose Win
Goat Goat Car Lose Win
Thus, probability of winning if we switch doors is 2/3 and probability of winning is 1/
@@erranti07haha nah nah, it’s nice to see you spent time to figure it out. It’s actually a pretty famous problem called Monty Hall problem 😊
Guys. Just because there are only two doors in the end DOES NOT mean an equal 50-50 spilt of probability. Probability is not necessarily evenly split between all choices.
Consider for example the case of a weighted die. Say it’s weighted in such a way that 70% of the time it lands with the 6 side facing up. What is the probability it would land on 6? According to the logic of many of you in the comments, since there is 6 sides, it will have a 1/6th possibility of landing on 6. This is wrong. It has a 70% chance of landing on 6 not a 1/6th chance.
Probability is not retroactive. Once you make your initial choice the probability you are right will always be 1/3. It does not matter that the host revealed a door. That CANNOT change the probability of your initial choice. It also cannot change the probability that the car was in one of the doors you did not choose. Since the probability the car was behind one of the doors you did not choose is 2/3, when a door is revealed, this probability remains with the unopened, unchosen door. You will always have a greater chance of winning by switching because the probability locks in when you make your initial choice. The probability is NOT evenly split between the remaining two doors.
Total probability theorem says otherwise,at the end of the day the goal is to get the car,ok so say the car is behind 2nd door then u make 3 cases for choosing each door and choosing switching or not switching and add it all to get the total probability to get the car,which comes out to be 50%,this thing makes no sense to me.
I have another model to support my argument,whichever door you choose,host chooses a door with a goat behind,so due to symmetry in choices,the 1st event i.e. selection of door by host is irrelevant talking in terms of final actions which is choosing the car door,due to 2 options,the probability is 50%.
@@mehulshakya1153 I don’t know how you end up with 50%. Perhaps you’re treating, in the event that you choose door 2 (with the car), the host revealing either door 1 or 2 as separate events. This would be wrong. Remember getting the car is all that matters. That is whether switching or staying leads to getting the car. What the host does in revealing a door does not matter. It does not matter if the host reveals either door 1 or door 2. In your scenario, staying always leads to a win and switching always leads to a loss. Switching or staying leading to wins or losses depends entirely on your initial choice. It never depends on what door the host reveals.
If you choose wrong at first, you win every time by switching. Since you choose wrong 2/3 of the time, there is a 2/3 chance you will get the car by switching. The only time you win if you stay would be when you initially choose correctly. You only have a 1/3 chance of choosing correctly with your first choice.
@@WilliamCacilhas oh wow the last argument you mentioned makes much more sense
What i meant was:(say car in d2)
P(car door)=P(car through d1)+P(car through d2)+P(car through d3)
=(1/3)×(1/2)+(1/3)×(1/2)+(1/3)×(1/2)
=1/2
Where 1/3 is probability of choosing a door and 1/2 is probability of either switching or not switching.there is no probability term involved because of the host because he for sure chooses a door with goat so its probability being 1.
@@mehulshakya1153 Ah I see what you meant. You're calculating the probability of the contestant choosing the door and not the probability of the car being behind the final chosen door.
In that case you would be right. The contestant can either switch or stay. This is a 50/50. The point of the problem, and what I hope came out clear in my response (I am not particularly good at explaining things), is to point out that one of these choices leads to wins more often than the other.
@@WilliamCacilhas oh no,what i meant was suppose the car was behind door 2( whichever door it is behind shouldnt matter as the problem is symmetrical wrt the car behind a door),then i found out the total probability of choosing door 2 through the process of choosing a door and then either switching or staying.then i added the probabilities of choosing door 1,switching to 2 and then choosing door 2,staying at same door and then choosing door 3 and switching to 2.i am confused as to why this method is giving the wrong anwser
Let me explain it in another way:
Make the door count 100, instead of 3. You pick a door, let's say 57. Then the host opens 98 of them and keeps door 57 and let's say door 14 closed. And he asks you if you wanna switch your choice. Would you?
Of course, you would... You had a chance of %1, now you have %99...
Ben is the kind of student, who when he gets an answer that is different than the teacher's edition, the professors rechecks the teacher's edition for errors.
Bigger plot twist to account for
You don’t want the car. You have to pay the insurance for it, and the game show host knows it.
Yet bigger plot twist - You are from the middle east.
So basically there are only two ways to win this game.
1) You pick the right door initially and not switch the door. The probability of this is 33%.
2) You pick the wrong door initially and switch the door. The probability of this is 66%.
So based on this, switching door will give you a better chance to the win the game.
Why? Not switching your door is in fact choosing the door out of two. It'd be no different if you decided to switch. You're asked which of these two doors you'd like. Staying or swapping is a new decision, not related to the original one. The odds are 50% regardless of what your first choice was since that wasn't the door that was revealed.
@@aaronanderson6958 The contents are not shuflled again for the second part. If you already had a goat behind your door before the revelation, that goat will still be there after the revelation, and the same with the car. So by staying with your door you cannot win more times than if no option was ever revealed and only the first part of the game existed.
This is better seen in the long run. If you played 900 times, you would be expected to start selecting the door that hides each of the three contents (goatA, goatB and car) in about 300 games (1/3 of 900). So in total 600 times a goat and 300 times the car.
As the host always reveals a goat from the two doors that you did not pick, in the 600 games that yours already had a goat, the revealed goat must be the second one, so the car must have been left in the switching door. Only in the 300 attempts in which you started selecting the car, the switching door will have a goat.
Therefore, despite you always end with two closed doors, which you originally picked only happens to be correct 300 times (1/3 of 900), while the other that the host had to leave closed happens to be correct 600 times (2/3 of 900).
@@aaronanderson6958 I also thought this until I realized that the hosts choice has 2 constraints, not 1: it must reveal a goat and it must not be the door you picked. The door you picked was never up for consideration to be eliminated so the chance remains 1/3. The remainder of 2/3 has to be attributed to the only other choice left.
The best explanation I've seen for this comes from the damninteresting website, written by Alan Bellows:
"In explaining the effect, it helps to increase the scale of the question. Imagine that there are 100 doors to choose from instead of three, but still just one prize. When hypothetical contestant Contessa chooses her door, she effectively divides the doors into Set A that contains her one door (1% chance of including the prize), and Set B that contains 99 doors (99% chance). Our imaginary Monty then proceeds to reveal goats behind 98 of the 99 doors in Set B, skipping over one seemingly random door. The odds that Contessa picked the winning door on her first try remain at one-in-a-hundred, so when asked if she wants to keep her original door or switch to that one other unopened door, the better answer is more obvious. Monty is essentially asking, “Do you want to keep your door and its chance of winning, or take all 99 of the other doors and their chance of winning?”
Memories of Monday, Wednesday, Friday college classes that lasted 50 minutes per class for 16 weeks, and all the homework flash back. Don't miss it.
I found most people who say the probability is 50/50 simply because there are only two choices/possibilities (two doors left, one has a goat and the other has a car), but one important thing to keep in mind is that just because there are only two choices/possibilities doesn't mean the probability is 50/50. Thank about our real life, there are so many scenarios where there are only two possibilities, such as I buy the power ball and I either win or lose, or I go to a job interview and I either get hired or not. However in neither scenario the probability is 50/50.
Honestly, if someone responds correctly so fast to the MH problem it just means that he already heard it - not so strange in nerdy environments. You are not testing anything in particular.
True that. But this is a movie and the scene is showing that the young dude is a bit of a quick thinking genius. At the same time giving the audience the chance to recognize the question and feel good about it :-) Great script in my opinion, even if it's not super realistic.
That's true. Before hearing of this I would've never switched. Why? I didn't think about odds and believe in picking right the first time.
Kevin Spacey's great acting makes this scene so natural.
The simplest explanation is: You have 33,3% that the car will be behind the door You picked, 66,7% that the car will be behind the other 2 doors. If the host reveals one door that he knows there is no car, he only can chose the door, if You picked the correct door in the first place, otherwise he can't and can only open one door that he knows has got no car. Therefore when You are given the chance to pick again the propability for the doors doesn't change, it's still 33,3% for your door, or 66,7% the other doors (of which one was eliminated by the host). It's like the host asked You after your initial pick: do You want to open Your door, or all other doors.
If there were 100 doors, and You pick one door, you have 1% chance of finding the car, 99% that the car is in the other doors, if the host reveals 98 other doors he knows have no car and leaves one door. The question is: when there will be no car in the door the host left? Only when You picked the right one initialy, and the door You picked had 1% of being right, the other were 99%, that's why You pick the other.
This only applies if you know the host is always going to open a door after you make the first choice. If the decision to open a door or not is conditional, or arbitrary, this falls apart.
the problem takes as its premise an established game show that people were generally familiar with, so I think that's a little moot. Whatever the criteria are for whether or not to open a door (including complete randomness), the player would be able to leverage statistics to have a similar or greater chance of winning, provided he has access to the problem's history (e.g., previous episodes of the show).
For instance, If the decision whether to offer the switch is random, then the same logic applies: once the host opens a door and shows you a goat, you get a +33.3% boost by switching. If he doesn't open a door and offer you a switch then it's outside the bounds of the problem as there's no decision to be made, so those cases don't count.
Another example, the above reply about the host ONLY offering a switch if you picked the car means that upon being offered a switch, you'd have guaranteed 100% win chance by declining.
In fact I can't think of any criteria for how the host behaves that would leave you with worse than a 66.7% chance (either by staying or switching), once it's established that the player has been shown one of the doors and is offered a switch.
That's the whole point
...uhhh the host could definitely open a door if you chose a goat @@GregoireLamarche
Right. The screenwriters didn't understand the Monty Hall Dilemma. The host's free will changes the problem by introducing an unknown variable
@@3Torts You misunderstood GregoireLamarche's point. HYPOTHETICALLY, if the host only opens a goat-door when you've chosen the car, you should NEVER switch when he does so.
For those who are confused -
Initially the probability of winning was 33.3%(for the door 1) and probability of losing was 66.6% ( for other 2 doors). Now when its revealed that in door 3 theres a goat , the entire 66.6% of probability is shadowed on door 2.
U would think that its 50% but that would be incorrect as it doesn't follow the causality principal
Mythbusters did an incredible experiment on this, which concluded most people will stay with their first choice, yet should switch.
@@bullspun3594 Would you happen to have a link for that
@@p-opremont Actually of all the clips I do have from that show that one I don't have, I know it's from the episode Wheel of Mythfortune.
yep only 50/50 if the first choice is relinquished and whats behind the door is shuffled.
why does the percentage stay the same when you literally have two choices
When the host opened door 3 and revealed the goat, you are guaranteed to win if you switch and chose wrong originally; hence the probability of winning if you switch equals to the probability that you chose wrong originally, which equals 2/3.
The simplest way to understand is that say you have picked door A and say door C gets eliminated now the chances of door B being correct increase by some amount because it was not eliminated but the chances of door A didn't increase because it wasn't eliminated because we picked it
I prefer set theory. The doors you pick and didnt pick form 2 sets. The probability between those sets dont change. Your 33% vs the non-yours 66%.
Last time I heard, Leo Da Vinci actually has wrote a solution that got pretty close to equating gravity or something along those lines
For a more intuitive approach, consider instead of 3 doors there are 100.
You still pick 1 door initially, a 1% shot.
The host opens 98 doors, leaving your door and another door unopened. The prize is still not visible. Now one could say "well, now it's a 50/50 shot", but does that sound correct?
Do you really think there's a 50% chance that you chose correctly prior to all of them opening?
The fact is that there is still a 1% chance that you were correct and still a 99% chance that you are incorrect.
However, now, your actual OPTIONS have consolidated - the chance never changes, simply the option of representation did.
So you should switch your choice.
This is the best explanation and people just seem to ignore it.
I think they omitted something in this explanation. I totally DID NOT get it while watching the movie, but when Youtubing a few other explanations, it made perfect sense.
LOVE KEVIN 💕❤😅😅😅....Hope comes back to making movies 😀
Monty Hall never offered to switch. He would sometimes build tension by showing a door, but the contestant was locked into their choice. So the whole problem is much ado about nothing.
But - yeah - Ben's answer is correct, *under the following circumstances* :
1. The game show host _does_ know where the winning door is
2. The game show host _will_ always _choose_ to open a door where there is a goat.
3. The contestant does want a car and not a goat. (ref: xkcd #1284)
Haha.
A simple way to get it; when you choose to switch, you essentially pick the 2 other doors, then have a free pass to safely remove 1 goat room
I saw this in the theater with my middle school crush, her older sister drove us and watched it with us.
Had to weite it down to believe. It works.
There are 3 cases. W after door number means a win. Your first choise is always door 1.
Case 1: 1w,2,3. 3 is reveled as a loss. You switch to door 2 and you lose.
Case 2: 1,2w,3. 3 is revealed as a loss. You switch to door 2 and you win.
Case 3: 1,2,3w. 2 is revealed as a loss. You switch to door 3 and you win.
If you dont switch, you win only case 1.
If you switch, you win case 2 and case 3.
Your explanation is simple and enlightening for people like me.
I would say the simplest explanation is: there is a chance your host had no choice but to open door 3 because the car was in door 2. The fact this is a possible scenario makes door 2 more likely to be the right door statiscally. Before door 2 and door 1 were the same. Now door 2 is more special than door 1 because the host chose door 3 instead of door 2. Then consider all the scenarios of why they would choose door 3 to open. There's a 50% chance they choose it because the car is behind door 2. And there's a 50% chance they choose door 3 at random because a goat is behind both doors. You should add the likelihood that the host opened door 3 because they couldn't open door 2 to your overall probability that the car is behind door 2. That is why it's more likely to be in door 2.
Thanks for your concise explanation of this problem. You made the answer clear by stating that door 2 is more special. Much better than other explanations I have read!
To go further, is it fair to say that switching to door 2 doesn’t just improve the odds? Rather, it means a certain win, because the host obviously couldn’t open doors 1 (your door) or 2 (the car is there). As such, switching created a sure winner?
I was asked this game show problem in one of the interviews a year ago and I fked it up big time.
To put it more obviously, imagine there are 1000 doors, you pick one and the host opens 998 of the 999 doors you didn’t choose revealing nothing (or a goat if you want to stick with that analogy). Now, do you keep your answer or change it? Obviously you change it, the odds that you picked the right door are 0.1% and since only one other door could be the answer, the odds that it is correct is 99.9%
When I was at college (in the early 90s) it was called the Newton-Raphson method.
If the first door he chooses is really car then the host did all that and he changed his door, he be so mad at statistics after that 😂
You missed the point
@@gregai8456 no point is missed
@@Tiktokkaki you think so because you don't understand statistics.
@@gregai8456 then u also missed the point of my point
@@Tiktokkaki because your point is outcome based and irrelevant.
if you think the answer is 50/50 read this: its literally this easy: if you assume you always switch doors, in every scenario that you pick a goat at the beginning you win because you have a goat chosen, other one is revealed and the moment you swap the only thing you can land on is the prize. and the fact that there is a 2/3 chance to pick a goat at the beginning means you have a 2/3 chance to win. its this simple.
As an economics aspirant , I loved the way The Monty Hall problem is portrayed.
Best way to explain this:
Imagine you play the lottery.
Now imagine i make a bet with you. I bet you whether you won or lost the lottery. Its 50/50 right? You either won or lost the lottery.
By staying with the first door. You are affirming that your first choice was correct and that you won despite the odds being against you, it would be like making that additional bet that you won the lottery the first time.
Its the old Monty Hall problem, this scene is extremely exaggerated because most people studying statistics would of already heard of it and knew the answer. As for the "Inspired by real events" aspect, that was marketing Bulls@*$. The film was just a adaptation of the book "Bringing Down the House", most of the book was fictional.
The problem misses a component: the host is supposed to open a door with a goat not any door.
Otherwise, If the game host acts to maximize the chances of the player loosing, he would always open the door with the car if he has the opportunity to do so, resulting in the strategy being completely reversed (and the chances of winning being 33,33% in any case).
No it doesn’t… 1:32 Host knows whats behind the doors… F
@@iv4nGG read my comment. If the host knows what is behind the door and must make everything he can so that the player looses, he will always open the door where the car is (assuming that the player has chosen the wrong door) - thus preventing the player to win
True
@@jackroberts416 (small question- did you amend your answer? When I receive the e mail notification of your answer, it was a long one!)
@@burgerman1234567 well if he allows you to switch doors id just pick the door w the car then 😂
This looks like a good state university somewhere in the Midwest. I would like to have a patient professor like this😊
😂😂😂
When the lady that came up with this logic came up with this logic, she was laughed at professionally.
I wish someone - Marilyn Vos Savant, the makers of this movie, SOMEONE! - would finally get it right and explain that this only works when there is an a priori agreement (or sufficient previous observation) that the game show host is going to open a door after your first guess. Otherwise, the host could, for example, simply open your door immediately whenever you pick a goat - in which case it doesn't matter how frequently he exposes a goat otherwise, your best strategy is to stand pat.
Bridge players know of a variation on the Monty Hall problem called the Law of Restricted Choice, that has a similar issue with assumptions. If your opponent doesn't randomize when playing an honor out of king-queen, the simple formulation of Restricted Choice fails in a way similar to how the Monty Hall strategy argument fails.
You are incorrect.
@@aheroictaxidriver3180 explain how then!
@@soriba391 The solution is valid irrespective of any agreement or special knowledge involving the contestant. Since the contestant is WRONG 2/3 of the time with his first choice, switching gives him a 2/3 chance of being right. You're confused because you think the object is to find the car. Or maybe you think the object is to know what the host knows. Even if there is no actual car, and the contestant only believes there is one, switching is better.
@@soriba391 The remaining door is just the BEST GUESS at where the car is, if there is a car.
@@aheroictaxidriver3180 Oh damn, from a mathematical perspective you're absolutely right. But since the motivation of the contestant is still the car does it mean from that point on his decision, even if he doesn't switch and still get's the car in the end, is kinda illogical. Sorry, can't phrase it better (not my first language)
simply : if we say (goat-car-goat) and you choose door 1 and the host choose door 3 what happens exactly is that the host qualified one door between door 2 and 3 , but door 1 which you choose is a random choose , it is still really hard to think about this way so let me give you a huge and no way to doubt example >>>> lets say that we have 100 door instead of 3 and u choose door 1 for example , the host opened 98 doors from the remaining 99 doors and behind those 98 there are goats and give u the chance to change ur selection would u change ,,,, now it is clear that u should change to the other door because what happened is that (why would this only door which could be from door 2 to door 100) be chosen from the 99 doors , there is something special about this door that it has been qualified from a 99 doors from the host of coarse , if you changed the door it is 99% that this is the true door that has a car behind it ,,,, now if we go back to the first example it is now clear why the percentage is 66.6 to 33.3 not 50 to 50 and key word in the whole problem that changed the percentages I would call it (QUALIFICATION of doors)
So, can I see it like this:
Say I have two choices instead of one. I am my own host. I start by choosing a door of the three say A. I open it, see a goat,SWITCH to B or C. So here statistically I've got better odds by switching (i.e, the action of switching itself, 2/3 or 66%), then holding onto the door (1/3 or 33%).
When the game in question begins and I choose one of the three doors, chances are the car is in the OTHER one as SWITCHING itself gives me higher odds as aided by the host when a door with a goat is opened, as now switching has only one door left.
The problem statistically analizes the outcome by weighing between the choice to either switch (right 2/3 of the time) or hold (right 1/3 of the time). The car can be in the originally selected door then, just lower odds?
Very dexterous wielding of the index finger. It really paid off, all that practice with size YS fruit-of-the-looms.
Game is Door1Goat - Door2Car - Door3Goat
33% of game - If I choose Door 1, the host will open Door 3. If I change my choice to Door 2, I win.
33% of game - If I choose Door 2, the host will open Door 3 or Door 1, if I change my choice of Door 2, I lose.
33% of game - If I choose Door 3, the host will open Door 1, if I change my choice to Door 2, I win.
Looking at the game as a whole, if the host opens a door, I have a 66% (33% + 33%) chance of winning - only if I change my choice; if I don't change my choice, I have a 33% chance of winning.
I had to work it out for myself from the @Gonedop comment.
It's crazy smart - I work from choices, but the correct answer is from assessing each probability - that's why I'd never be able to work it out on my own.
But if the host already knows which door has the car, why wouldn’t he end the game right away since he knows Ben pick the wrong door? The only reason why he let Ben choose again is to give Ben a “choice”and hope that he switch the door and that only happens when Ben picked the correct door right from the beginning.
Then there would be no game show. The player loses every time? I'll pass.
I don't think you know how game shows work 😂
It's true, it's a key part of the problem to assume that the host MUST ALWAYS show you a goat behind one of the doors that you didn't pick, and then offer you the choice to switch.
If that is the consistent structure of the problem, then the math holds. Obviously if the host can do whatever they want, and didn't have to offer you any choice, or show you what was behind any doors, then you can't soundly make inferences anymore.
He ALWAYS gives the contestant a chance to change his mind. And he ALWAYS opens one of the bad doors. That's the way the show works. No matter which door Ben chose, there was a bad door to show.
@@kevinrosenberg4368 Incorrect. No matter what the host's motives or past behavior, in this specific sequence, you should change. That's the point.
I wrote a program to prove this and by swapping doors I found the car 66% of the time
Just saw a guy say "I'm a professional mathematician and I disagree with this." Like fr all you have to do is run the simulation yourself to get 2/3 and you don't even need a computer program to do it cuz it's not that complex lol. Some pro mathematician that guy is lol
@@BlaqEndeavor send me your email address. It's in python so you will need idle or similar to run it
Oh look its General Irons 😂
Look up the Monty hall problem if you want to see a quick simple explanation that makes sense.
Here’s the true problem…ben’s 97% paper lol 😂
I wonder what the director was thinking giving Ben 97% and not 100%?
yea. 97% is Asian Fail.
Because college professiors and graduate TAs are stingy assholes who never give 100%s
Because getting 100% on math papers at university is unrealistic..
The thing I don't get, is if instead of switching to door #2, you don't "stay" with door #1 but instead pick it again a fresh new time, wouldn't it also have a 50% chance?
I took some time to understand it too, what helped me was thinking like this:
There is 2 scenarios, the one that you switch and the one that you dont.
Now lets see what it would take to win in each one of this two scenarios:
If you are in the scenario that you dont switch: the only way to win would be if you picked the prize right away. Since there is one prize and two goats. The chances are 1/3.
Now in the scenario that you do switch: You win if you pick one of the goats in the first pick. Since you would be picking one goat, the host would eliminate the other, and since in this scenario you are garantee to switch, you would switch to the door with the prize. But the thing is, there is two goats, so your chances are 2/3.
Basically, if you switch you are aiming for the goats, if you dont you are aiming for the prize. And its easier to aim for the goats because there is more of them.
Hope this is clear enough, english is not my first language
1/3 of the time you pick the right door originally then you switch and pick the wrong door
2/3 of the time you pick the wrong door originally, now he will open the other wrong door meaning if you switch you will pick the correct door
So if you switch it's 2/3 that you get the correct door
How you can also see it is if u picked a door, the chance is 67% that it is a goat meaning the other 2 doors contain 1 goat and the car. When the host runs into that 67% that he has to close the goat, meaning the other door that you didnt pick 100% guarantees a car.
No because originally you had 33%, the guy let you know which one of the 2 remaining doors is bad. If he had taken away one of the incorrect doors BEFORE you picked an initial one as your best bet, then it would 50%, but since he took a false door away AFTER you already picked one, you instead have a 66% chance.
I find it funny, that the prof asks why Raphson doesn't get any credit for it, but Newton-Raphson method is just another name for this method. He does get credit.
But movie gotta movie. No one watching this will know because it's not the main plot device, it's just a cute little scene to get you interested in the movie.
Kevin Spacey is such a brilliant actor!