Game Theory Scene | 21(2008) | Now Playing 

I love how college classes in movies are only 4 minutes long.

imagine if he went to find Ben Campbell's exam and saw a "46% F"

The video is titled Game Theory, the class is named Nonlinear Equations, the question is asked is about Probability Theory.

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.

I love how the class on solving polynomials became a probability class.

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.

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.

This is not "Game Theory" - this is "The Monty Hall Problem".

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.

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

Say what you will about spacy. But man oh man the guy can act.

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.

i wish a nonlinear equations class was this easy in real life. one of my toughest undergrad classes right after topology

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

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.

Amo esse ator e esse jeito de falar. ❤

Some great acting by Spacey

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.

The comment section of only RU-vid video which helped me to learn this classical problem...

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%.

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?

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)

To audition Ben with the Monty Hall problem was simply genius

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

The method described initially is known as Newton-Raphson method. So Raphson did get credit for that

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.

His acting make the scene so much interesting ..

Kevin Spacy what an actor

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.

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.

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?”

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 💀😂

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.

It's called The Monty Hall Paradox this "paradox" was mentioned also in the series Better Call Saul Season 2 Episode 4

Спасибо за очередную полезную связку, все работает как в описании 🔥🔥🔥💯

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.

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).

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.

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: 👏👏

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.

imagine if Ben asked Fisher for extra help to take down Micky Rosa

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)

LOVE KEVIN 💕❤😅😅😅....Hope comes back to making movies 😀

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

This is called the Monty Hall problem

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.

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…

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.

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.

I'm very impressed by the fact that you can use non-linear equations to get out of groomer charges...

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.

This is not "Game Theory" - this is "The Monty Hall Problem".. Some great acting by Spacey.

Last time I heard, Leo Da Vinci actually has wrote a solution that got pretty close to equating gravity or something along those lines

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?

Kevin Spacey's great acting makes this scene so natural.

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 was asked this game show problem in one of the interviews a year ago and I fked it up big time.

Kevin Spacey is such a brilliant actor!

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

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)

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

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.

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.

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.

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.

When I was at college (in the early 90s) it was called the Newton-Raphson method.

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%

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 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.

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...

I love the fact how well the standard internet user is educated these days due to open source education and state education. Globalisation isn't bad as you'll see if you scroll down the comment section. Maybe I'm biased because it's a math related topic, but anyway, I'm proud to be human in moments like these.

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.

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.

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.

This entirely assumes the host would even consider opening an alternative door in the first place.... WHICH YOU DON'T KNOW.

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 😂

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.

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.

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.

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.

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.

At first, I did not understand the logic and theory explained in this scene. However, after going through several comments and explanations online, it did make sense. I will try to explain some important points to understand the theory. 1) After the door 3 is open, this is just the second part of the same problem. It is obvious that taken independently, there are then 2 doors and 50% chance of choosing right. However, it is important to see that as the second part of the same problem / equation and not an independent one. 2) It's statistics and probabilities. It doesn't mean the right door, in this scene the one with the new car, is door 1, 2 or 3.It's about understanding what choice / what door has the most chances of being the right one. If we keep doing this experiment thousands of times, what door will be correct the most often. 3) This is something which is not mentioned in the scene but which is implicit. This is after understanding this and it made sense to me. When you choose 1 door out of 3, let's say like Ben the door 1, you have 33.3% of chance choosing right (this would be the same if you chose door 2 or door 3).However, that implies that you have 66.7% of choosing wrong. Those 66.7% mean that the right door is elsewhere, either door 2 or door 3. We don't know which one, but statistically, it would be one of them. If it's either 2 or 3 and that the game show host indicates it's not door 3, then logically it should be door 2.Consequently, it is in our interest to switch from door 1 to door 2.Again, as explained in 2), it doesn't mean it's 100% correct. It means switching the door has the most chances of winning and will win the most often if we perform this experiment hundreds or thousands of times. It is similar to surveys...The larger the sample is, the more accurate / correct the outcome is. 4) To understand better, we can take the example of a deck of cards, 52 cards. Let's say you pick one without looking at it. What are the chances of you picking your favorite card, let's say for the example ace of spades?1 chance out of 52, about 2%.It is much more likely that the ace of spades is in the rest of the deck than the card in your hand. You don't know which card is the ace of spades, but you guess it's somewhere in the deck. Following probabilities, it is in your interest to switch your card with the rest of the deck. Now, if you reveal 50 cards out of the 51 cards left in the deck and the ace of spades isn't any of them, you will end up in a similar situation as the 3 doors and the movie scene. It may appear as a 50%/50%.However, we said earlier that even though we couldn't tell which card it was, it must be in the deck. In conclusion, it is likely the last card in the deck and it is in your interest to change your initial choice.

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.

This looks like a good state university somewhere in the Midwest. I would like to have a patient professor like this😊

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).

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)

Kevin Spacey is one hell of an actor

"Sorry, what's your name?" "I'm Peter. Peter Parker."

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.

Here’s the true problem…ben’s 97% paper lol 😂 I wonder what the director was thinking giving Ben 97% and not 100%?

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.

Movie is titled ‘21’ released in 2008

I wrote a program to prove this and by swapping doors I found the car 66% of the time

This was such an underrated film. The way Ben played Mickey at the end was absolutely priceless… and I mean that literally and figuratively 😂 Leave the bag… 🔫

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.

Very dexterous wielding of the index finger. It really paid off, all that practice with size YS fruit-of-the-looms.

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?

So, here's the problem with this and it was demonstrated by the late, great Monty Haul himself. If the game show host always opens one of the two remaining doors after you pick, then the correct move is to switch. But what if he doesn't? What if you select a door with a goat behind it and the host, knowing this, just opens it and doesn't give you a chance to switch? If the host has the option to allow you switch or not, then the question isn't one of math, it's "is the host messing with me?"

So for people who are confused, let me give you the reverse example. If the host gives you the opportunity to select 2 doors out of 3, you now have 66% chance of winning. Suppose you selected 1&2 and he shows you goat in 1 and ask if you want to switch from 2 to 3, you don't. Because when you selected your 2 doors, your probability of winning was 66%, and you should not return that higher probability back to host by switching it.

I saw this in the theater with my middle school crush, her older sister drove us and watched it with us.