Monty Hall Problem (extended math version)
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- Опубликовано: 22 май 2014
- Shorter version at: • Monty Hall Problem - N...
And more: bit.ly/MontyHallProb
This video features Lisa Goldberg, an adjunct professor in the Department of Statistics at University of California, Berkeley.
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Numberphile is supported by the Mathematical Sciences Research Institute (MSRI): bit.ly/MSRINumberphile
Videos by Brady Haran
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A run-down of Brady's channels: bit.ly/bradychannels
It's actually quite simple and intuitive: If your strategy is to
1. *always switch,* then you win whenever your *first choice was wrong.* (probability = 2/3)
2. *always stick to your first choice,* then you win whenever your *first choice was correct.* (probability = 1/3)
Best explanation I've ever read about this problem, thank you !
Brilliant explanation! Thank you.
Yes, it really is that simple. Amazing that anyone can think it's 50/50.
@@woufff_ I second that. This comment is the explanation I find easiest to understand.
@@rwb966 If the goat was revealed accidentally, it actually would be 50.50.
I remember getting this question when I was like 12 or something in Maths class. I quickly submitted the answer 50-50. When the teacher insist that I was wrong and get me to look into it again. I tried to prove her wrong with a tree diagram back at my desk, and had my mind blown instead.
I was shown how it worked with a tree diagram and kind of understood, but now I get it
Shame on your math teacher, you were right the first time
@@martinmulligan4327 But they weren’t…did you even watch the video?
@@martinmulligan4327 they weren't right the first time. It's not 50-50
@@martinmulligan4327 I suspect you are right, the 'first' game in the typical 3 game table has two permutations. If you play the game over and over you dont get 3 permutations with a double switch win but you have to play the 'first' game twice to get a true relationship. To clarify, car is behind door 1, Monty can choose door 2 OR 3. This requires two games to be played. If the car is behind 2, assuming you pick door 1 he can only pick 3. Car is on 3 he can only pick 2. 2 stand and win plays and 2 switch and win plays. 50/50.
Just in anticipation --- these different versions remain unlisted for the first day or two here on the "second channel", then become public later.
This is mainly to increase the probability that people who subscribe the main channel and this "extras" channel come across the main video first!
However making this unlisted ensures this video can be accessible via links and annotations right from the start.
You could also give people the choice to subscribe to one of 3 channels. One of them being the main channel, but the viewer would not know that beforehand. Then you show one of the channels that is not the main channel and give them the option to choose between the two not-shown channels. They will be more likely to choose for the main channel!
Thanks for this, Brady and Friends...but probability still utterly eludes my intuition. I'd like to make a request (with no expectation of it being requited): please do more videos on probability to help us all develop some intuition for it. :)
Thade Look up some extra vids on youtube!
Go over permutations, combinations, Bayes', conditional probability, covariance, correlation, probability mass functions, probability density functions.
Numberphile2, you should make your channel profile picture of 'Tau' because your numberphile channel has Pi, and 2 x Pi = Tau, because it's Numberphile2.
Brady, do you remember when you did a video about educational videos and your interview subject said that you have to address the misconceptions first? I think the comment section on the other video illustrates why.
Let's say the prize is behind door A
If you pick door A and you don't switch, you win.
If you pick door B and you don't switch, you lose.
If you pick door C and you don't switch, you lose.
If you don't switch, your chance of winning is 1/3
The prize is still behind door A.
If you pick door A, Monty shows you either B or C. If you switch, you lose.
If you pick door B, Monty shows you door C. If you switch, you win.
If you pick door C, Monty shows you door B. If you switch, youl win.
If you switch, your chance of winning is 2/3.
Wow, finally after all these videos a simple comment makes me understand it. thanks!
musicalcolin this is the best explanation i've seen
It took me years! Thank you so so much.
You made a huge mistake in counting odds.
In the second paragraph, you said, "The prize is still behind door A. If you pick door A, Monty shows you either B or C. If you switch, you lose."
You should have said the following rather than the way above,
「The prize is still behind door A. If you pick door A, Monty shows you B. If you switch, you lose.
The prize is still behind door A. If you pick door A, Monty shows you C. If you switch, you lose.」
If you pick door B, Monty shows you door C. If you switch, you win.
If you pick door C, Monty shows you door B. If you switch, you win.
===========================================================
Since you calculated the odds by adding up the number of possible incidents, you should not "pretend" that (Monty shows you either B or C) is one incident. For two incidents , you must count twice.
黃正邦
In the first selection we are suppose to pick each of the contents 1/3 of the time, that is, if we did multiple trials, we should pick the car in about 1/3 of them, the goat1 in 1/3 of them, and the goat2 in 1/3 of them. So, those two cases you mention in which Monty can show door B or C (because both have goats) must fall into the 1/3 when you picked the car door, therefore each will have 1/2 * 1/3 = 1/6. They cannot be counted as if they had the the same probability as the other two cases, because they are half as likely as each of them. In order to get probabilities by counting cases you have to make sure that everything you count as a case is equally likely.
If you don't see it with fractions, it could be clearer with trials. Let's say you played 900 times and let's put the numbers, assuming that the sample proportions exactly match the probability.
1) In 300 games your door has the car.
1.1) In 150 of them the host reveals the goat1.
1.2) In 150 of them the host reveals the goat2.
2) In 300 games your door has the goat1. In all those 300 the host reveals the goat2.
3) In 300 games your door has the goat2. In all those 300 the host reveals the goat1.
- If goat 1 is showed, you can only be in case 1.1) or in case 3), so you are under a subset of 450 games.
You win by staying in 150 of those 450 (case 1.1) and by switching in 300 of those 450 (case 3).
• 150 represents 1/3 of 450.
• 300 represents 2/3 of 450.
- If goat 2 is showed, you can only be in case 1.2) or in case 2), so you are under a subset of 450 games.
You win by staying in 150 of those 450 (case 1.2) and by switching in 300 of those 450 (case 2).
• 150 represents 1/3 of 450.
• 300 represents 2/3 of 450.
This is because the host does not it randomly; he knows the positions and is forced by the rules to avoid revealing the car. He must reveal a goat always.
There is a small error in Dr. Goldberg's reasoning at 6:08, though it does not affect her (correct) conclusion. She explains that P(y) = 1/2 because Monty will always know that Door #2 and another door contain a zonk. But that's not the whole story. Sometimes the contestant will pick the door with the car, in which case Monty does indeed open Door #2 or the other zonk with 50% probability each. But sometimes the contestant will pick Door #2 (in which case the probability of Monty opening Door #2 is zero), and sometimes the contestant will pick the other zonk (in which case the probability of Monty opening Door #2 is 100%). P(y) does wash out to 1/2, because (1/3)(.5) + (1/3)(0) + (1/3)(1) = 1/2, but not for the stated reason.
I was looking for an explanation of P(y)=0.5, but I think this one is still not correct? You differentiate between 3 things which supposedly all happen with 1/3 odds to make 1 total, but isn't there overlap between "choosing door 2" and "choosing the door with prize/zonk"? I think it goes a bit further than this, but I don't know how.
I got as far as this now. What if you take as P(y) "the chance that the host opens door 2 given that you choose door 1". Note that this differs from P(y|x) because the car does not have to be behind door one, we only need to have chosen door 1.
I then get:
* You choose door 1 (as a given), and the prize is behind door 1 (1/3 chance) -> odds of opening door 2 by the host is 50% (door 2 and 3 are both zonks)
* You choose door 1 (as a given), and the prize is behind door 2 (1/3 chance) -> odds of opening door 2 by the host is 0% (door 2 has the prize)
* You choose door 1 (as a given), and the prize is behind door 3 (1/3 chance) -> odds of opening door 2 by the host is 100% (door 3 has the prize)
I then also get 1/3*50%+1/3*0%+1/3*100% = 50% but then with three chances that actually are 100% together (prize behind door 1,2 or 3.
The only problem I have is that this pre-assumes that we have chosen door 1. Since there was no precondition for P(y), it seems odd that we have to do it like this?
We don't have to assume that you chose any particular door. Instead of taking as a given that you choose door 1, take as a given that the prize is behind door A, and the zonks are behind doors B and C. And assume that you're equally likely to pick A, B, or C.
* If you pick A, Monty opens either B or C with 50% probability each.
* If you pick B, Monty opens door C with 100% probability.
* If you pick C, Monty opens door B with 100% probability.
Then for either of the two "zonk doors" B or C, (1/3)(50%)+(1/3)(100%)+(1/3)(0%) = 1/2. And this is P(y).
Yes, but this assumes that the prize is behind door 1. This is equivalent to P(y|x) = "Odds that door 2 is opened when the prize is behind door 1". I thought that we wanted P(y), so the chance that door 2 is opened, without any further conditions whatsoever. When you do that, you get P(y)= 1/3 which is obviously not what we want. If you either assume that you pick door 1 (as I did), or that the prize is behind door 1 (as you did), then it works, but then they should have explained that P(y) is not entirely unconditional.
I'm not at all assuming that the prize is behind door 1, which is why I renamed them doors A, B, and C. I'm merely assuming that the prize is behind one of the three doors, which I am arbitrarily calling door A. Door A might be Door #1, Door #2, or Door #3. No matter which door is the prize door, the analysis holds: you are equally likely to pick Door A (50-50 for whether Monte opens B or C), Door B (0% for whether Monte opens B), or Door C (100% for whether Monte opens B). If you know nothing about which door the contestant picked, only that Door #2 is not the prize door, then the probability that Door #2 will be opened is 50%.
It seems incredibly counterintuitive that the probability doesn't change with the sample size. I understand the mathematics she went over, but it's still a crazy thought. Math is beautiful.
Chaosdude341 Maybe I'm missing your point, but that's how all probability works. If using a strategy yields a 30% good outcome in one game, then when applied to 10 million games we'd expect the same strategy to yield about 3 million good results. There's nothing unique to the Monty Hall problem which demonstrates this fact specifically, we could show the same principle with 10 million coin flips. Or are you talking about the number of doors? In that case the probabilities do change; the probability of winning by swapping is given by (n-1)/n, where n is the number of doors we start with.
TedManney i think he's probably referring to the fact that once one of the doors is revealed to be empty, you're left with just 2 doors and a lot of people think this means your odds are 1/2 no matter what....but they are not understanding and using the information we are given when Monty reveals a blank door which is invaluable and changes the "straight odds".
chaosdude did you not see the example with 100 doors?? when Monty goes through the 99 other doors and reveals that 98 of them are worthless....you have to ask yourself, what are the chances that I somehow randomly picked the correct door to begin with(1/100)....then remembering that Monty KNOWS where the prize is, you ask yourself what are the chances that the ONE friggin door Monty didn't open is the correct choice(1/2)....so in the 100-door example....IF you switch doors, your odds are 50% of getting the prize....IF you don't switch, your odds are 1% of getting the prize.
I understood the point more once Brady put up that new Numberphile video. It's the use of the knowledge that Monty has that was throwing me for a loop.
Love the sarcasm .
And 8 years ago I see.
I think the easiest way to understand it is, that you pretend Monty doesnt open any doors, but you get the chance to open ALL the remaining doors or stay with your first pick.
Jichael Mackson but that's not the same problem...
@@FirstNameLastName-tc2ok it is exactly the same
No, that isn't necessarily the same. If you just get to trade your door for both of the other doors, naturally, you win 2/3 of the time. But that is not the same as Monty revealing a goat, and you getting to then take in essence both of the other doors, his and the other. Monty has to be avoiding the car. If Monty opened the goat door randomly, the odds would just be 50.50. Any understanding of the puzzle must involve that Monty's intentional avoidance of the car is what matters.
I like the man behind the camera..... He asks interesting questions and spontaneously cracks jokes
Does the fluorescent green leotard increase or decrease my odds of winning?
Definitely increases.
I would appreciate more Bayes Theorem videos, if that's possible
In case anyone else was a bit confused by the choice of 1/2 for P(y), note that *all* of the probabilities are given that you first *chose* door 1, which at first I was confusing with x, which is the event that the *car* is behind door 1. The fact that you first *chose* door 1 just happens not to affect P(y|x) or P(x). Given that you did choose door 1, there is a 1/2 probability that Monty will open door 2, because if the car is behind door 1, the assumption is that he opens door 2 with probability 1/2 and otherwise, there's a 1/2 probability of the car being behind door 3, which is the other case where he'd open door 2.
Thanks, I had that confusion
I think the biggest reason why this was never really brought up when the show was actually around is very simple: It assumes Monty is playing by rules that he's never actually been obligated to follow.
Specifically, the assumption that he's obligated to open a zonk and give the contestant the option to switch was never true. If the contestant's first pick was a zonk, he was certainly able (and, I believe, often would) open the door they picked and show that they lost, without giving them a choice to stay or switch.
Naturally, that changes the equation significantly, as now we know that if our first pick is a zonk, we don't get a choice, while if our first pick is correct, we do get a choice... and since we only get a choice to switch if our first pick is correct, then staying with the first pick is the better choice. (Of course, if people started to catch on to the trick, he'd start offering the choice to switch if they pick the wrong door and show people that they would have won if they switched some of the time... then go back to forcing them to open their door without a choice once people start switching again. So in the end, there's really no strategy beyond out-psyching Monty.)
+Sethala It all comes down to the information we are given to solve the question. If we have no information that the host has only offered a switch because we chose the car, we are unable to assume that he has done so. Equally, we are unable to assume that he has done the opposite and we chose the goat. The effect is therefore either absent or random. The information we do need is that the host is not acting randomly, and we are given the information that he knows what is behind the doors so we are able to infer this. There is no need to out-psych the host; it is our own intuition that we need to manage.
I saw greedy people switch many times they got zonked.
The reason it was never brought up at the time is simply because the actual game was never played that way.
Thanks a lot Brady for these great videos. I too would like to see more on probability as applied to game systems, this being a perfect example.
A way I like to think about it is this:
Pick a door. Monty then turns to you and says: "I'm going to give you two choices. You can either stick with that door, or you can pick the other two".
What do you do? You surely go for the other two! But this is precisely the same problem, because when you decide to switch, you are basically just opening all of the other doors. If you switch and the car was in one of those doors, you keep it, if it was in your original door, then you don't, just as in the original problem. The same reasoning works for any number of doors (bigger than 2).
People get confused, since they feel that the probability should be 50-50 at the final stage. But, the thing is, it isn't just a choice between two doors. The door that you didn't pick is SPECIAL: it is enriched with some extra information: assuming that you didn't pick the car, it would already have been opened if there was a goat behind it. So it's not a choice of two doors, it's a choice of two doors with some extra information.
Just as a quick note, you don't even really need Bayesian analysis for this problem. What probability do we have of winning if our strategy is always to stick? Well, it's 1/3, since we win precisely when we pick the car first (which is probability 1/3). That means that the probability of winning when we DON'T stick (that is, when we switch) is 2/3.
Brilliant way to think about it
But it isn't embeued with information. Why should it be? If you chose right switching always looses, but if you chose wrong switching only looses 50/50.
The critical flaw here is that the two events (picking 1st choice and picking 2nd choice) do not happen at the same time.
The concept commonly spoke of is that the 2nd choice door if you switch (be it door 47 in the 1/100 example, or whichever in the 3 door) is embued with some special properties, namely the acquiring of the given doors 1/3 chance- but that's not how reality works. Rather, that 3rd doors 1/3 probability is granted to both choices evenly, since your initial choice is rendered irrelevant from the hosts actions.. The result is a 50/50.
Ben Leza I'll try explaining in a more esotheric way
Let's say Monty just doesn't open a door. You pick one and that's final, he opens your door and reveals your prize? What are your odds? 1/3, right?
Now, considering he does open a door you didn't pick, containing a zonk, what is the probability that he opens a door? 100%, right? Whatever you choose, he will always open a door.
But then, why does your door suddently get 1/2 of chance of having the car when he opens that door? Where does that increased probability come from? It suddenly became more probable that your door is correct because he did something he always could?
If you choose a door, he can ALWAYS open one of the other doors. So if you chose a zonk (which is the more likely scenario), he can ONLY open one door, and whenever you choose a zonk (2/3 of the time) it will be on the door he didn't open. 1/3 of the time you will choose a car, and it will be on the door you chose.
So the door Monty didn't open is embeued with the increased probability that you initially chose a zonk, considering that if you did, switching ALWAYS gives you the car.
+Ben Leza
*"But it isn't embeued [sic] with information"*
It is. It is imbued with the information that you didn't pick it.
When Monty opens another door, you are left with two doors. One is the first you picked, the other is the one you didn't. The one you didn't pick is more likely to have the prize. The original door that you picked had a 1/3 chance of being correct, and that probability remains even after Monty opens the door (indeed, how would the probability increase? He was never allowed to open that door so you have learned nothing about the likelihood of this door being correct).
I really don't understand why people argue about this problem. Just do an experiment for yourself. After a few trials, you should see that the probability of winning is 2/3 if you always switch.
jamma246
I think
Why people argued about this display is because of the illogical thought.
Monty Holl solution was the sample that you uploaded and it is absolutely right.
But this vedio just shrinks the possibility with very simple and basic method. three became two, then the car should be either one or the other. So the proportion of possibilities would 1/2 - 0 - 1/2. As I type, it can't be solved in a way in this video.
I'm not a professional mathematician, just a student, so could you explain me why this video's solution is logical?
razorborne couldn't agree more. The number of times that I've watched an interesting video and wanted more and just had to sit through the same stuff all of over again. Please just put the additional stuff up.
I have always loved probability and statistics, and this is another excellent example of the outcome of mathematical analysis in these fields being counter intuitive. I hope we see more of Prof. Lisa Goldberg, she is great and this video was entertaining and informative.
Hear about that in the past, but never really got, why you should switch. The example with the hundred door really made it a lot easier for me to understand. Thanks for the video!
I've got a different modification that helps to illustrate the right answer.
Say you're taking a multiple-choice quiz with three answers: A, B, and C. You think it's A. C is obviously wrong. Your teacher is Monty Hall, and so, after you write down an answer, Monty will tell you one wrong answer that you didn't pick, and you can switch if you want.
One strategy is to pick A. Then your teacher says it's not C (or says it's not B), you stick with your answer, and you get points if A was actually right.
Another strategy is to pick C. Then your teacher might tell you it's not B, in which case you switch to A, or your teacher tells you it's not A, in which case you switch to B. You get points unless it really was C.
The second strategy is better; you get the points unless the obviously wrong answer was actually right. Clearly, you want to start with the least likely answer and switch to whichever of the more likely answers doesn't get eliminated; you win unless the answer you liked least was correct.
Now, if you didn't study at all and all the answers are equally attractive, it's the traditional Monty Hall problem, but the same strategy applies, aside from the fact that the least likely answer is as likely as either other answer. You still do better trying to guess one of the two wrong answers at the start than trying to guess the one right answer.
So, imagine a multi-choice quiz with lots of questions each with three possible answers A, B and C, where one possible answer of each is obviously wrong and the other two are plausible (this actually tends to reflect reality!). You go through the quiz without reading the answers, selecting A every time. You then read the answers. If the obviously wrong answer is B, then you switch to C, if the obviously wrong answer is C then you switch to B. The Monty Hall principle would suggest that your chance of getting these correct is now 2/3. If the obviously wrong answer is A, then choose at random between B and C - a 50-50 chance.
For a large enough quiz, you should get correct answers (2/3 + 2/3 + 1/2)/3 i.e. 11/18 of the time.
On the other hand, if you read through the quiz answers first and discount the obviously wrong answer and then pick at random from the two that are left, you should get correct answers 50% or 9/18 of the time.
So where is the logic flaw in the above?
You're misunderstanding when the Monty Hall answer is applicable. It applies to the situation as a whole, not to individual cases like that. It says that, if you randomly pick one of three options, there's a 2/3 chance that you're currently wrong; if you have a trick for definitely getting full credit by starting with a wrong answer, you'll win 2/3 of the time.
In your situation, it's more complicated because if A is wrong, there's a 50/50 chance that it's obvious; on the other hand, if A is correct, it's definitely not the obviously wrong answer. So you get credit for the situation where A is wrong and not obvious, which is 2/3 * 1/2 = 1/3; you also get credit for the situation where A is wrong, is obvious, and you guess correctly, which is 2/3 * 1/2 * 1/2 = 1/6. That adds up to 1/2, like guessing between the two plausible answers.
Charles Gaskell I introduced the idea of an "obviously wrong" answer, to try and get rid of the all-knowing Monty Hall. It begs the question "obvious to whom?"
On "Who Want to be a Millionaire", you have a choice of 4 answers, 3 wrong and 1 right. In the situation where you haven't a clue which is right, all are equally possible, so choosing any one has a 1 in 4 chance of being right, and 3 in 4 of being wrong. One of the lifelines is to go 50:50, where the computer removes two wrong answers.
In the scenario where you choose your answer at random (but not "final answer"), then go 50:50, if the computer doesn't remove your initial choice, should you stick with your original choice, switch to the other remaining choice, or doesn't it matter? What is the maths for this?
One other observation is that Monty Hall is not forced to open the door and make his offer (and apparently he did not always do this), so knowing that the contestant knows about the Monty Hall problem, (and given that he knows where the car is hidden) he could choose mainly to make the offer where the contestant's choice was originally correct...
In all of these cases, the regular intuition that weird schemes for picking an answer at random don't affect anything is correct. If you do out the maths, the fact that the computer might have removed your original guess if it was wrong but couldn't have removed it if it was right precisely balances out the benefit you get from switching away from a 1 in 4 guess. (Of course, assuming that probability theory works right, variations on a technique that don't affect what happens also don't affect the probabilities, so it's not a surprise that it balances perfectly.)
The usual intuition is only wrong in the (mathematically ideal) Monty Hall case because you've got an extra source of information (Monty) who is forced by the script for the show to act in a way that can be helpful. If Monty is an active participant with a meaningful choice playing against you, it's generally best to ignore him. (Which may mean that the solution to what has become known as "The Monty Hall Problem" wouldn't have been a good technique for Let's Make a Deal; or maybe you should also rush the early rounds so he needs to fill up time at the end.)
Charles Gaskell
The logic flaw is that the obv wrong answer can be A. In which case you don't get a zonk revealed to you. It is equal to Monty having the option of not giving you the option to switch if you choose the predetermined zonk. So if the Monty Hall problem is changed so that if you choose zonk 1 or the car, you get the switch option, if you choose zonk 2, you are stuck. Then the probability is 50-50 (same as your example above).
Your problem isn't the same as the Monty Hall problem. You have assumed that the obv wrong answer is never A.
The 100 doors example is a great example to understand your thesis, thanks!
Thank you so much, math is beautiful
I thought I'd never be able to understand this... until today. Thank you, numberphile 4 life, yo!
Great explanation. Would be curious to know if some contestant actually did the maths during the lifetime of this game show.
DEAL OR NO DEAL EXPLANATION
For me the most intuitive way of explaining it is as follows.
If your strategy is to switch:
There is a 2/3 chance that you picked a door with a goat behind it. Monty Hall then shows you the other goat. If you switch you are guaranteed to get the car (i.e. 1/1 probability).
There is a 1/3 chance that you picked the door with a car behind it. Monty Hall then shows you a goat. If you switch then you are guaranteed to get the remaining goat.
Thus the probability that you get the car if you switch is 2/3 * 1/1 = 2/3 and the probability of getting a goat is 1/3 * 1/1 = 1/3.
Excellent explanation! :)
Just start by trying to pick a losing door, because your probability of picking a losing door is 2/3. So if you are right and picked a losing door on your first pick then by switching you will have to win. So your chance of winning are 2/3.
did someone count the results of the show?
was it near 66-33%?
I wish you would include the full version on the main channel, and the less mathematical version on n
Numberphile 2.
i think that maybe it´s better to only show the "extras" as before, so the ones like me that have already watched the main channel video, don´t need to watch again a lot of the same content.
that is and will be the normal course of action... there were a few reasons this was a different case. it is not something that will happen often.
Well, that is exactly what I hoped to read somewhere down here in the comments. :)
hmm, I would imagine people not switching more than switching on the gameshow.
Given ignorance of the probability (or wrongly thinking it's 50/50) I would hypothesise people would regret switching and then losing over not switching and losing. Therefore choosing not to switch.
This has been demonstrated in other psychological tests. People are averse to risk and to change. The see switching and losing as somehow worse than staying put and losing.
***** Veritasium did a video about this, can't remember what it's called
6:42 I can't believe all that was necessary to get to a 1/3 probability - for one correct answer out of three doors
I'm not saying that adjunct professor is wrong, I just have a question.
if the situation was to happen like this:
1) Monty from the beginning would open a door, let us say that it's door 2
2) Monty asks the contestant to choose a door ( 1 and 3 are the ones left)
3) after the contestant has chosen a door, Monty asks the contestant if he would like to switch to the other unopened door (keep in mind that Monty has already opened one of the doors before the contestant has chosen).
isn't this situation is the same as having only two doors and the probability would be fifty to fifty?, is the order of opening doors very important?, if it is(and I think it is from the formulas used) would you post a video explaining the formula used here and how it was derived. I think it would be an exciting video.
If Monty opens a zonk from the beginning, he's picking from two doors that have zonks. While in the normal way, 2/3 of the time he is not picking anything, he's only opening the door that has a zonk and the contestant didn't pick. Only 1/3 of the time, when the contestant picked the door leading to the car, does Monty actually make a choice of which door to open.
Otherwise, yes, I too would like some videos about Bayes' Theorem.
It isn't the same situation. Picking a door before Monty opens one blocks him from choosing your door. Then either you've picked the car and he can choose one of the other two to open, or you've missed the car and he's forced to open the other zonk door.
If Monty picks first he always has a choice of two zonk doors to open, and afterwards it's 50 50 to you which remaining door has the car.
your situation is just like this: there are two doors, one winning, one not. you choose one. the host ask if you switch. that's it. another opened door added or not, doesn't matter.
You're telling me that I have a choice between a rusty gas-guzzling internal combustion engine sitting on a few clunky wheels, or an animal that will provide a lifetime supply of delicious feta cheese!? I think I'll stick with my initial choice.
I really like Lisa's way of explaining this, very entertaining.
I still love this video. She was the only one who made this solution make sense.
You can understand it via pigeonhole principle three layouts two outcomes, at least two layouts lead to the same outcome.
htmiguy88, you said: "You can understand it via pigeonhole principle three layouts two outcomes, at least two layouts lead to the same outcome."
You keep mentioning the pigeonhole principle. I'm not familiar with the term. Is it a phrase used in probability courses?
Combinatorics and cardinality of sets.en.m.wikipedia.org/wiki/Pigeonhole_principle
Thanks for the link.
The comment sections on these 2 videos are an incredible read, it's like professor dave's flat earth debunk comment section.
Bayesian problem, awesome explanation!
an important way to clarify the problem and why Monty disrupts the probabilities is to examine a somehow "different" problem. Suppose that Monty open the the other door _only_ when you choose the car (otherwise he opens the car and tells you that you lost). If that was what Monty does then the probability if you keep the door is 100% and if you change, the probability is 0%. So simple.
If Monty opens an other door randomly (he does not know or does not care where the car is) then the probabilities are even. It is 50% and 50%.
Only if Monty only opens a zonk the strategy to change is right and the 1/3 and 2/3 probabilities are right.
I'm sorry this is a big fail in describing the set up.
But I see this has been corrected in a subsequent video.
There are 4 important things to emphasise:
- the host knows what is behind all three doors.
- the host asks the contestant to randomly select one of the three doors.
- the host must then open one of the two doors you have not selected.
- the door the host opens must have a goat behind it.
- the host must then offer you a switch to the other unopened door.
Would it be possible to think about it this way, which I found more intuitive, but I could be terribly wrong
1. There are 2 possible outcomes of your first pick, either it is C, car, or M, mule.
2. If C, then the leftover door after the host has opened the other one, must be M
3. If M, then the leftover door after the host has opened the other one, must be C (since from the 2 remaining doors, the host must open the one that contains a mule)
4. The probability that your first pick leads to M is 2/3
5. Therefore, probability of leftover door being C is 2/3
There is nothing wrong in your method of understanding the problem.
Regards.....
I just realized this is at my school. Brady when were you at Berkeley!?
This results completely counter-intuitive for me, and this problem used to drive me crazy. Thanks for actually running the math!!
I think that the most amazing thing about this problem is that so many people fail to grasp its simplicity. There's no need for Bayes' Theorem, nor tricky psychology, nor any sort of convoluted reasoning involving changing probabilities or whatever. There's no need even to number the doors.
Sticking with the basic 3-door problem, the probability of choosing correctly initially is one in three, or 1/3. Correspondingly, the probability of choosing wrongly is two in three, or 2/3. There is thus a probability of 2/3 that the car lies behind one of the non-chosen doors. But Monty knows where the car is, and always opens a door with a goat behind it. If you have chosen correctly, he can simply open either of the remaining doors. If you have chosen wrongly (which is twice as likely), then he must open the one remaining goat-hiding door. Either way, when he gives you the opportunity to change, you should take it without hesitation, as you will thereby double your chance of winning (1/3 if you stay; 2/3 if you change).
The 100-door problem makes it even clearer, and vastly increases your chance of winning, provided you accept the offer to swap.
I found that if I actually simulated the game by making rows of three boxes on a piece of paper and trying the game over and over I got the results described in this video. I first heard of the Monty Hall Game Show problem on an NPR program called Car Talk
This is the video that should have been on the mail channel, as it includes the critical math necessary for the solution, and the math is really not very difficult at all, and Bayes' theorem can in fact be used to solve many similar probability brain-teasers, such as Bertrand's Box.
Ok so how does the 2/3 concentrate?
is she saying "flash car"?
this is cool!
A simpler explanation.
Let's say you've decided on a strategy to always make the switch.
Now all you have to do is initially pick a door with a zonk, and you'll win the car after the switch. Chances of picking a zonk is 2/3 => it's a good strategy.
In other words, with this strategy you only lose if you pick a door with a car in the first place. The probability for that to happen is just 1/3.
It's not about what is the best strategy. It's about answering the question: Should you switch or does it matter? And that can only be determined by assessing what affect the revealing if a goat by the host have?
Not true
thank you I really wanted to understand this and you made it sooooo easy. new sup
I'm not one of those who do not get it; I understand the correctness of switching, which increases the chance of winning. This could be demonstrated by using a model to repeat the scenario hundreds or thousands of times. But one side point has not been made by anyone I've seen so far. In the actual TV show, we don't really know without going back and reviewing every episode whether there was any bias at all in Monty's offer to switch doors. He didn't always do it, and there is a possibility that he offered the choice more (or less) often when the contestant originally chose the winning door. This is just a note of interest and has no bearing on the outcome of the given problem, where a goat door is always revealed, and switching is always the best option.
I looked at the conditional probability logic behind this earlier this year in A-Level Maths - seemed a bit weird to think of it like this.
So.... How many of their contestants switched?
See this is the way to go!
It is the pre-knowledge of Monty knowing what is behind the doors that is the key i.e. P(y)! Correct me if I'm wrong! Cheers
What happens when you try to directly calculate the probability of door 3?
If we call the probability of a car behind door 3 "P(z)" would the maths look like this?
P(z I y)*P(y) = P(y I z)*P(z)
P(z I y) = P(y I z)*P(z) / P(y)
And P(y I z) is 1; P(z) is 1/3; P(y) is 1/2
Therefore P(z I y) = 1*(1/3)*(1/2) = 2/3
Right?
It's amazing how much math is required to make sense of intuition!
My only issue with this problem is that ... if you think about it you aren't actually choosing between three doors at all are you? There is always going to be one door removed from the dataset, and the door removed is always going to be one which has a zonk behind it. Thus, in actual fact you are always choosing between 2 doors, it just doesn't appear that way.
I still can't get it... To me, at the moment you have chosen Door #1, and Monty chose Door #2, you remain with Doors #1 and #3. So you can imagine that, instead of chosing between Staying or Changing your choice, you have to choose again a door, between the two remaining. So now, the probability of getting the car is 1/2, cuz you have one car that could equally be behind both the doors...
Of course, I agree with the maths explained in the video, but to me, it's only one way of considering the problem between others...
Assume the contestant chooses door 1, we know that if Monty chooses door 3, then combination #3 (GGC) cannot occur and should be eliminated when calculating the probability (mutually exclusive).
This leaves only two combinations which is 50/50 for keeping or changing doors. If Monty chooses door 2, then combination #2 cannot occur and should be eliminated. This is again 50/50 for keeping or changing doors.
Google "Monty Hall problem test". Play around with some of the computer simulations, or try the game yourself with three playing cards. The solution to this problem is not up for debate, it's empirically understood, so it's a waste of time to continue to try and come up with arguments that invalidate the prediction. My previous comment should already have illustrated why it's a mistake to think the player has a 50% chance of having the car behind their door *at any point in the game*.
G Hamilton Actually, taking into account which door specifically was opened, here are the possibilities, given the contestant chooses door 1:
1a) CGG (1/3 chance) AND Door 2 opened (1/2 chance since host had 2 options to choose from) = 1/6 chance
1b) CGG (1/3 chance) AND Door 3 opened (1/2 chance) = 1/6 chance
2) GCG (1/3 chance) AND Door 3 opened (1/1 chance since host had only 1 option) = 1/3 chance
3) GGC (1/3 chance) AND Door 2 opened (1/1 chance) = 1/3 chance
If the host chooses door 3, then possibilities 3 AND 1a could not have occurred (mutually exclusive). You are left with comparing possibilities 1b (1/6 of initial probability space) and 2 (1/3 of initial probability space) meaning you are twice as like to be in possibility 2 and twice as likely to win by switching.
Right off the bat you're more likely to pick a zonk door (2/3 more likely). Once one of the other doors is revealed as a zonk, you will do better switching to the "untouched" door, since you probably chose the wrong door to begin with. Revealing one of the zonk doors minimizes (or zones in on) the options for what's actually the "right" door.
Same if there were 5 doors. Pick door number 2 (you're 4/5 likely to have chosen the wrong door). Doors 1 and 3 are revealed as zonks. Your options: stay with door number 2 (where you're 1/5 likely to have chosen the right door), switch to door 4 (1/3 likely to be the correct door) or to door 5 (again, 1/3 likely to be the correct door). I would switch to door 4 or 5.
Why is the cover or image not tau?
Ms. Goldberg's analysis is wrong. She has taken p(y|x)=p(y)=1/2. But p(y) per se does not presuppose anything on behind which door the car actually is. The should be left to the conditional probabilities, p(x|y) and p(y|x). But what is p(y)? The following diagram on the pick and open sequences based on where the car is each time might enlighten. The first column is the contestant's initial pick, the second column is the door Monty opens, and the third column the contestant's final decision.
1) The car is behind door 1
1 2 3
1 2 1
1 3 2
1 3 1
2 3 1
2 3 2
3 2 1
3 2 3
2) The car is behind door 2
2 1 3
2 1 2
2 3 1
2 3 2
1 3 2
1 3 1
3 1 2
3 1 3
3) The car is behind door 3
3 2 1
3 2 3
3 1 2
3 1 3
2 1 3
2 1 2
1 2 3
1 2 1
So the probability the car is behind door 2 is the number of sequences with 2 in the second column divided by the total number of sequences, i.e. 8/24 = 1/3. Therefore p(x|y)=1/2.
"Given that the player has selected door #1, the probability that the host will eliminate door #2 is 1/2, so the probability of the scenario on the first line is (1/3 x 1/2) = 1/6." But the two events are not independent in order for us to calculate their probability occurring together by multiplying. The "player's" (what happened to contestant? come to think of it what happened to the principle of indeterminacy?) choosing the door with the car unties the hands of Monty Hall, who now can open any of two doors. One thing I learned from studying probability is that to calculate the probability of dependent events, it's best to draw up a table and take stock. On the whole in my table the car could be anywhere and the probability of the contestant picking a particular door first is 8/24=1/3 as you require. It's a two-person game and certain openings provide more degrees of freedom than others. You don't weight each different game in chess. You start with the assumption that all are equally likely but some prolong the gameplay, say, up to fifty moves. These give more freedom to the opponent's development and are more reasonable and likely than others.
Dimitri Boscainos No matter what your argument, you are demonstrably incorrect. Please Google "Monty Hall problem test" - you'll find dozens of explanations (the Wikipedia article is actually quite good), videos of people testing the strategies, and even online simulators you can try yourself. Or, you can set up an experiment using nothing more than three playing cards. Do a bunch of trial runs and record the results. As you do, it will probably occur to you where your thinking has been errant. You're probably the billionth person to argue against the Monty Hall problem, and the billionth person to be wrong about it - that's what makes it a good puzzle.
Dimitri Boscainos There's no "probability" to which door the host opens. The host is not acting randomly, but with knowledge and according to rules; the host is not a random factor. The host follows the following rule in a not random way: "open a door the player did not pick which does not reveal the prize." Also, it's important not to get hung up on the numbering of the doors; it is arbitrary because the player picks any door as a random chance (1/3 of the time the correct door) and the host acts with knowledge and according to rules and not randomly at all to reveal a door that is not the prize.
*****
Wouldn't his assertion be Argumentum ad Populum?
Your tables seem to assume that the player is twice as likely to choose the door with the car than they are to choose either other door.
Yes, if you somehow are doubly likely to choose correct, then the odds of winning are 1/2. Because your odds of choosing correctly was 1/2 to begin with.
But it's not.
It's easier to understand if you begin with the assumption that you've first picked the wrong door. If there are 5 doors and you choose #2 you have a 4/5 chance of being wrong. Say Monty reveals doors 4 and 5 to be zonks. This does not increase your chances of having picked the right door, there could still be a zonk behind door #2. What it does do, however, is reduce the possibility of you choosing the wrong door again. What are the probabilities of again choosing another wrong door (by switching) ? They've been reduced since now we have 3 doors to select from instead of 5.
Now I am convinced :)
The reason this happens is because Monty KNOWS which door not to open. At the begining, there's a 2/3 chance for the car to be in one of the doors that you didn't choose. Assuming that this is the case, the guys that run the show tell Monty what door to open and he opens it. When it happens, he basically tells you that this is not the door with the car, and that the car is behind the other door.
You could make a youtube experiment for this main video + two goat videos and a car video. Then count the views.
but why does the probability concentrate behind door no. 3 and does not separate between door 1 and 3?
Limboman1988 There's a 1/3 probability the car is behind the door you initially pick, 2/3 probability that it's behind one of the two doors you didn't pick. The host will always deliberately choose a goat door from among the ones you didn't pick, and open it. Doing so does not magically change the probability that we chose the door with the car, that probability stays at 1/3. But provided the car was among one of the two doors we didn't pick (still 2/3), we now know which of the two doors it was behind. Swapping gives us the benefit of always winning the car as long as it was among the doors we didn't initially select. There's simply no reason why the probability would be redistributed after the host's elimination.
Moreover, we can take the formula given at 4:57 to determine what is the probability that the car is behind #3 given that you choose #1 and that Monty opens #2 (considering that you just opened #1). According to the equation: this equals to ((Probability that you choose #1 x Probability that Monty opens number 2 (considering that you just opened #1)) | given that the car is behind number #3) x [(Probability that the car is behind #3) / (Probability that you choose #1 x Probability that Monty opens #2 (considering that you just opened #1))]. So, we have: (1/3 x 1/1) (1/3 / (1/3 x 1/2)) = 2/3. Thus, the probability that the car is behind #3 given that you choose #1 and that Monty opens #2 (considering that you just opened #1) is 2/3.
(Don't know if this has been mentioned - I'm not going to go through all 250 comments to check)
Actually, Monty Hall *NEVER* asked the contestants if they wanted to switch. That was just added for this problem. See Monty's own comments on the problem in "Haircut in Horse Town: & Other Great Car Talk Puzzlers" by Tom and Ray Magliozzi.
What is a zonk?
After you choose Door #1, Monty has his boys move the car behind the other door you didn't choose initially, minus the one with the zonk, under the premise that you will stay with Door #1, giving you the lower probability choice by sticking to your initial door.
Therefore behaviorally, a person is most likely to stick with their "original" door, raising the probability that if Monty has his boys move the car from Door #1 to Door #3, this will favor the "house"...
This is a pure mathematics problem, psychology and behavior have nothing to do with it and only confuse the situation. There is also no logic in your shady proposal because we're not talking about a "house" (i.e. casino), we're talking about a game show. Game shows *want* players to win.
Its really simple.
when he reveals the goat behind one of the other two doors. After this, two scenarios can happen.
scenario 1:
If the door you chose had a goat, the car is behind the other door.
scenario 2:
if the door you chose had the car, the other door has a goat.
The probability that "scenario 1" happens is 2/3.
The end.
Pranshu Upadhaya you can't say it's "really simple" by just stating the conclusion without an explanation. Everyone's heard the answer, they just don't understand how it was concluded.
Using Bayes's Formula was certainly the right thing to do, and is logically equivalent to using the "total formula for probability" derived from it. But I wish she had done a more convincing job of explaining why P(y/x) is 1/2 at 5:56. Same for her "base probability".
probability is not concentrating, its just the initial choice that makes all the difference...
If you pick a door, you'd be 33% sure of picking the car behind that chosen door. that means you'd be 67% sure that the prize sits behind one of the other two doors. now if Monty opens one of the doors and out comes a zonk then you become 67% sure that the car sits behind the only door left closed. so simply go for the switch and park your ass in your brand new car!!!
i like the way you have to have a whole other video for the "extended math version" on a maths channel‽
also you know maths has an 's' on the end... what are you playing at?
+JP05CPSN Europeans and other English speakers who insist that "math" must have an 's' on the end and Americans who insist that the 's' in "maths" should be removed are pretty much on the same low intellectual level. It's amusing that you imagine your comment puts you above the American intellect, when in fact it shows that if you were to move here, you'd fit in best in the arrogant, undereducated Deep South.
+TedManney Sorry dude, I forget that Americans' lack the ability to understand sarcasm.
+JP05CPSN When it's on the internet, unfunny, not accompanied by a smiley, and is immediately preceded by a serious comment, it doesn't register as sarcasm to anyone from any country. What, were you born yesterday or something?
+JP05CPSN " Americans' "
While we're on spelling, nowhere apostrophizes non-possessive plurals. Now, back to the math ...
+TedManney ;)
But when Monty picks a door with a goat, one of the 3 scenarios goes away, leaving one of two chances available
G Hamilton Just because there are two doors left does not mean that each one is equally likely to contain a prize. Say we're playing Monopoly, I'm stuck in jail and need to roll doubles to get out. After I roll, there are 2 possible results - I either get out of jail, or I stay in for another round. However, clearly the fact that there are 2 possible results doesn't necessarily mean that each one is equally likely to occur. So it is with the Monty Hall problem. As long as the player's first selection is a zonk, the host has no choice but to eliminate all other zonks from the game, and in that case swapping will always win. Swapping therefore wins as long as the player's initial selection was incorrect. Clearly, the player will have picked the prize *less than half* the time when selecting blindly from among three doors. There is no opportunity for a 50/50 chance to even enter in to the problem.
Actually I've just thought about this a little more after reading some comments (im also on the toilet, where I do my best thinking). Another way to look at this problem is that you have a 2/3 chance of getting a zonk when you choose door 1. Monty then opens door 2 and reveals a zonk. There is now a 50/50 chance that door 3 has a zonk, which is better than 2/3 if you want to avoid zonks.
I couldn't get the explanation of the 3 doors, I just agreed in the case of 100 doors. So I tried to do a blind test. Took Tree cups, two white and one blue, put a little ball in a white cup, then I always knew that the ball wasnt in the blue (like to open one empty door). So I closed my eyes and changed ramdonly the order of the cups and always picked the left white cup. I did 12 times switching the choice and 12 times not switching. I was shocked at the result: I won exactly 8 times by switching the cup, and just 4 times by not switching the cup. it seems like that calculation I didnt understand is right.
that is not truly a blind test if you know where the ball isn't.
Plus you knew exactly where the ball was because you put it in the cup and moved the cup that had the ball in it, just did the experiment, I always knew where the ball was.
So what would happen if two people participated in the game show, they each chose a different door and the door Mr. Hall opens does not contain the car (so we ignore the 1/3 of the times that it would be a car). Would not both contestants technically have a 2/3 chance of winning if they switched? Or does the added factor of a decision being made without the game show host reduce it to 1/2 per door?
Person A chooses door 1, person B chooses door 2. Both have a chance of 1/3 of winning. door 3 is opened and does not contain the car. The chances of this happening are 1/3 , which is split among the other doors (1/6 each), leaving person A and B with a chance of (1/3+1/6) = 3/6 = 1/2, or 50% of winning.
It would be different. With one player Hall never reveals what's behind your door, would he be able to do that with two players? Then the rules would be different. If not he has only one choice, but that choice could be the car. With one player he never reveals the car. So it would be different either way.
The change in the rules would make the chance 50/50, either way I think with your scenario.
You've changed the game if there's a chance for Monty to get the car!
Did each contestant see the other's choice? If yes, and they chose different doors then they both have a 1/2 chance of winning with either door at that point, and a 2/3 chance of having got that far without Monty getting the car, so a 1/2 x 2/3 = 1/3 chance of winning with that strat from the start. They'd have done better playing as one.
If they don't see each other's choices (so they don't know whether the other has picked the same door or a different one), things are still different. There was a 2/9 chance that Monty would have to get the car when he opened the door. If that hasn't happened then both players have a 3/7 chance to win if they stick and 4/7 if they switch. That's true whether they actually picked the same door or different doors.
The audience who can see both choices but not the car will know whether they picked the same door and will judge the probabilities to be 1/3 2/3 if they did or 1/2 1/2 if they didn't (and they'll see the first case come up 1/3 of the time, the second 4/9ths of the time and Monty get the car in 2/9ths of games.) To Monty who knows where the car is the probabilities are 1 for the door with the car and 0 for the other one. Probability is all about knowledge. If you have different knowledge about something it has different probability for you.
We are assuming that Hall can't open a door if you have chosen it and he can't open a door if the car is behind it. If two people are playing and both pick goats, then Hall can't open a door. Therefore, both choose to switch to the remaining door.
If one person picks a car and the other picks a goat, then we run into a problem. In the one player game, if a person picks a car, then Hall can open any door he wants; if the person picks a goat, then Hall only has one door left to open. Here, we have someone picking a goat, yet Hall still only has one door he can open. If Hall opens a door, then we know that someone picked the car initially, but we don't know who. We know that if both people choose the same strategy, then there is a 100% chance someone wins; however, there is an equal 50-50 chance that either person wins.
We are all counting that the two men can't choose the same door, that mean that the first one that choose have a 1/3 to take the door with the price, and the other choose only between two doors. If the first one has more possibilities to fail the door (2/3), the other one, inside the 2/3 has a 1/2 to choose the right one, meaning that when Monty open the last door with the goat, the second who choose has the 2/3 to win while the first has the 1/3 to win.
That is not valid if the two men choose the door randomly at the same time and can choose the same door.
For clarifing if three persons choose between 3 doors, and can't choose the same, the first one can fail in 2/3, it's more probably that the other select the right one, and making that the first who select will win lesser times.
What about the Bayes update?
P(X = 1) = 1/3 for any X1, X2, X3
so P(X1 = 1 | X2 = 0) =
P(X1 = 1, X2 = 0) / P(X2 = 0) =
1/3 / (1- 1/3) =
1/2
That is not Bayes' theorem, that is the conditional probability formula.
Bayes says this:
P(A|B) = P(B|A)*P(A) / P(B)
The problem with your calculus is that you are not taking into consideration which door you selected, so you are missing a very important fact. Notice that your calculus doesn't imply that you have selected X1.
Having the X2 opened, the probability of X1=1 is not the same if you selected X1 or you selected X3, because supposing X1=1, if you selected X1, the host could have opened X2 or X3 as well, so having X2 opened covers half of this case, but if you selected X3, the host was forced to open the X2, so having the X2 opened completely covers this case.
We can use Bayes' theorem to check this. Instead of "X2 = 0", I will say "X2 opened" to emphasize the fact that the host opened the X2. Suppose you select X1.
P(X1=1 | X2 opened) = P(X2 opened | X1=1) * P(X1 = 1) / P(X2 opened)
Now,
• P(X2 opened | X1=1) = 1/2, because the host had two possible doors to choose, X2 or X3.
• P(X1=1) = 1/3
To calculate P(X2 opened), we have to use the total probability formula:
P(B) = P(B|A1)* P(A1) + P(B|A2)* P(A2) + ... + P(B|An)* P(An)
This only if A1, A2, ..., An form a partition of the sample space.
In our case:
P(X2 opened)
= P(X2 opened| X1=1)* P(X1=1) + P(X2 opened | X3=1)* P(X3=1)
= 1/2 * 1/3 + 1 * 1/3
= 1/6 + 1/3
= 3/6
= 1/2
I didn't put the case of X2=1 because since the host cannot open the car door, P(X2 opened | X2=1) is 0. The P(X2 opened | X3=1) is 1 because the host cannot open the X3 since it's the car door, and he cannot open X1 since it's the door you selected.
So, the final result is:
P(X1=1 | X2 opened) = (1/2) * (1/3)/ (1/2) = 1/3
Omg Ronald, and I thought just giving the 'thumbs up' was complicated.
Anytime you pick a bad door, Monty is forced to open the other bad door. Since both bad doors are picked, switching is advised.
You have a 2/3 chance of picking a bad door, and that's what you WANT to do so Monty will open the other bad door. When both bad doors are chosen by you and Monty, you'll switch to the good door. The only one remaining.
Sorry for my poor language, but this problem took me about an hour to figure out.
I just realized that this applies to another popular more recent game show: Deal or No Deal... only it's closer to the 1/100 example than the 1/3
That one is a little more interesting because of it's structure, but is also closer to the Monty Fall Problem than the Monty Hall Problem because the elimination is at random and can take out the six high value prizes just as easily as the 20 low value ones.
Nope. It doesn't apply to the Deal or No Deal, as Noel doesn't eliminate all the low numbers for you. You eliminate the prizes randomly with no information gain from Noel, so it is impossible to have anything more than a 1/20 chance of getting the £250,000.
In fact, that game can be constructed using traditional probability easily. You have a 2/20 chance of having £250,000 in your final two boxes, and then the symmetry tells us there's a 1/2 chance. So we get 1/20. Why is this problem symmetric where Monty Hall isn't? Well here it is clear that you have effectively just chosen 2 boxes randomly (choosing 1, and then eliminating 18 is the same as choosing 2).
With Month Hall the asymmetry comes from the fact that as he must reveal a Zonks to you, he must leave the car in the "switch" position provided you chose a Zonk to start with.
gotcha
It doesn't because the Dead or No Deal is random. Each time a case is taken out the probability has to be recalculated based on the new information. In the monty hall problem you have no new information, you already knew he was going to open one of the goat doors.
Deal or no deal is actually a lot more complicated due to the fact that it's the contestant who choose the briefcases, NOT the host. You have to take into consideration the probabilities of each successive pick the contestant makes.
Suppose you were playing deal or no deal, and you manage to open all of the briefcases except the one you picked at the beginning, and one more remaining case. The remaining amounts are one really high value and one really low value. Which is more likely? 1) You picked the high value at the very beginning. OR 2) You did NOT pick the high value for each of the cases you picked besides the first.
I actually don't know the answer. =P just saying it's more complicated.
the logic is actually really simple if you think about it in a certain way... if you pick a door and stay with your choice, you have a 66% chance of losing. if you lose while staying with your choice, you would have won by switching your choice. therefore the chance of winning while switching is 66%.
Derek Lance your explanation already assume we know you have a 1/3 chance of losing by keeping the door
If I pick door 1 and Monty picks door 3, then we know a goat is behind door 3. There are two possible scenarios where a goat is behind 3 - one scenario with a car behind 2 and one with a goat behind 2, therefore 50-50 chance of getting a car if you switch to door 2 (or don't switch).
G Hamilton This is the wrong way to break down the possibilities. The probability that the car was behind one of the doors the player didn't initially select is 2/3, not 1/2. Therefore, there is a 2/3 probability that Monty was *forced* to reveal the only goat left among those doors. In that case, swapping wins.
While it is correct to say that there are two reasons the host could have chosen door #3 given that the player has chosen door #1, it is incorrect to assume that those two reasons are equally likely to occur. If the car was behind door #2, then Monty had no choice but to open door #3 (in other words, he will open door #3 100% of the time given that the car is behind door #2). If one of the goats was behind door #2, then Monty did have a free choice and either flipped a coin or had some other way of randomly determining which door to reveal given the option (in other words, he will open door #3 only 50% of the time given that the goat is behind door #2). The requirements for the latter scenario are more difficult to meet and will occur with less frequency.
G Hamilton "There are two possible scenarios where a goat is behind 3 - "
Yes, but they are:
Scenario (1) You picked a goat (2/3 chance) and Monty opened 3 because 2 has the car, and
Scenario (2) You picked a car (1/3 chance) and Monty could open 2 or 3. In other words:
" - one scenario with a car behind 2 and one with a goat behind 2"
the first scenario is 2/3 likely and the second is 1/3 likely. The proportion of outcomes is not the same as their likelihood; Scenario (1) is 1/2 the number of possible outcomes but 2/3 of the likelihood, while Scenario (2) is 1/2 the number of outcomes but 1/3 the likelihood.
I remember hearing about this problem in school and I never understood why it should be a 2/3 chance until today. In retrospect, I must have been such a fool.
Maybe you could teach us one or two more things about probability using a card trick or some kind of dice game? At least those two were the first things coming to mind... I am sure you can think of something as astonishing as the "Monty Hall Problem".
Lets try another thing. What if instead of picking a door you didn't choose Monty picked a zonk at random regardless of if you picked it or not. Now I might be doing the math completely wrong, but if your door gets picked (say #2) you would have a 50% chance of picking the right door that has the car (#3). Now lets say door #1 gets picked even though I picked #2. Would I have a higher chance if I switched or would it still be 50/50.
+OrinjFlames +OrinjFlames Did nobody ever answer this? It deserves a response, so here's my attempt. This is different to the 50/50 obtained by Monty revealing one of the non-chosen doors at random and it happening to be a goat. So two scenarios: (a) he reveals any door at random and on this occasion it happens to be a goat, or (b) he identifies the two goat doors and flips a coin. And, according to your question, both happen to be one you didn't choose first (although it could have been). I think (a) is the equivalent of Monty making a first choice in the same way you did, with the same odds - if it happens to be a goat the car is 50/50 between the other two. And in (b), he does not deliberately avoid the chosen door but does so by chance. So that's all down to chance too and still 50/50 where the car is.
Just like James Grime's explanation (the 100 door thing)
If you are stuck on this problem, it is probably because it is about as clear as mud, here's a thought that really helped me, please read it carefully and think about it...
There is a bag of marbles, 1 red and 6 blue. The objective is to pick the red marble. You pick one at random but you can't see what you have chosen. Then 5 blue marbles are removed from the bag, the marble you have in your hand has a probability of 6/7 of being blue (since you have just selected it from the 1-red-6-blue mixture inside the bag). If the marble in your hand is blue then the marble in the bag is red and you should swap. Thus in 6/7 cases you will win if you swap.
If you are thinking that once the 5 blue marbles are removed, there is obviously a 50/50 chance of your marble being red/blue, you need to remember that the marble you have taken out of the bag was selected by you from a bag containing 1 red marble and 6 blue marbles, and thus it had a probability of 6/7 of being blue, and the act of swapping will give you the opposite of what you originally had. thus choosing and swapping will give you the red marble 6/7 times (because you swap the probabilities when you choose to swap).
Alternatively think about it this way, you have decided that you will never swap and you want to trust your instincts, ok, so you choose a marble randomly from the bag, it has a 6/7 chance of being blue (remember you want to pick red), and since you will never swap then the bag is discarded and your choice is revealed... this means that if you always stay with your first choice then you will lose 6/7 of the time.
If this did not help then I'm sorry...
Another thing I just thought of is this...
Imagine the car is like orange squash molecules, and you're diluting it in zonk molecules with the ratio 1/2 - car/zonk... this means that its harder (less likely) to pick the car as your first choice than if it were an equal measure of car and zonk molecules, then Monty removes half of the zonk molecules (this is the tricksy confusing bit that misleads people). Whatever you originally picked has more of a chance of being a zonk because you picked it from a dataset diluted in zonks. so if it is more likely that you picked a zonk to start with, then swapping will lead to a car so it's better to swap.
hope this is at least vaguely helpful, if not sorry for wasting internets :)
Thomas Dennis
Mmm....... orange and goat juice, delicious
Does this theory match reality? I mean, has anyone analysed the actual show results in a matter of whether the car was behind the door of the original choice or was it behind the other one and then compared results to this strategy which states, that your original choice is only 1/3 of the time worth sticking with?
Dovile Arm Despite what Prof. Goldberg may claim, Let's Make a Deal didn't work exactly like this. It's pretty easy to simulate the general idea of the Monty Hall problem and then compare the results of swapping / not swapping strategies to verify that switching is indeed the better move.
Dovile Arm Yep it does match reality. If you look at the short version of the video, many people in the comments have run simulations and experiments that prove this.
They're like Monty can you be my baby daddy; I'm like yeah
Yet another way to think of it is this: regardless if the remaining two doors both have zonks behind them, which there is a 1 / 3 chance of, or if the dream car is behind the remaining two doors, which there is a 2 / 3 chance of, Monty will open a door with a Zonk from those remaining two doors. Given that, does that mean that there is no longer a 2 / 3 chance of the dream car being behind the two doors that you didn't pick just because Monty happens to open one of those doors? No... It does not mean that. There is still a 2 / 3 chance that when Monty is opening a door from the two that you didn't pick that the dream car is behind one of those two doors. That is just wild, lol. (This is my first time understanding this, heh.)
Think of it this way
3 doors, only 1 has car, the rest has nothing, forget about zonks.
You choose door number 1 (1/3 chance you win, 2/3 chances the car is in the other doors)
You are given the opportunity to whatever is behind of number 2 AND number 3 (2/3 chances of winning). I think we all agree you should always switch
The problem here is exactly the same, except the host opens one of the empty doors for show, IRRELEVANT DISTRACTION, he still offers you to switch to 2/3 chances of winning, you should always switch.
I just wonder one thing. If the same game is played twice then 1st time you pick door 1 and door 2 opens then there is 2/3 chance on door 3. Nvm the answer you play again, and then you pick number 3 and then open door 2 and switch to door 1 which means that door 1 and door 3 have 2/3 chance.So if you think door one has also 2/3 chance in the first game,becouse the game is set up in the same way, so if you think a little you might notice that the chance of both is equal.Anyway is it 1/2 or 2/3 Hmmmm.....
I made simple program checking winnings in 1,000,000 games and the avarge
winning games are about 66-67%
The one thing that looked odd to me, even though I know the professor is right, is her application of Bayes' theorem. For P(B|A) she puts in 1/2, which is clear, because if the car is behind A, then there is a 1/2 chance Monty picks door two since both doors are goat doors. The important part of that seems to be that it is assuming the car is behind door number one though, as A has to be true. But then, she also puts in 1/2 for P(B), which is where I get a little confused. P(B) is just the probability he picks door two, not the probability he picks door two assuming door one has the car, which would lead to a different value for P(B) wouldn't it, because then there is also the chance that the car isn't behind door one? If we just assume P(B) is the same as P(B/A), then why doesn't Bayes Theorem just simplify to P(A/B) = P(A) (which it shouldn't, which is where I get confused)? I feel like I'm missing something important/blatantly obvious, but I'd appreciate if someone could point it out.
I've always used the Law of Total Probability to calculate P(B), so
P(B)=P(B!CarA)*P(CarA) + P(B!CarB)*P(CarB) + P(B!CarC)*P(CarC)
= 1/2*1/3 + 0*1/3 + 1*1/3 = 1/2 (which gives the same answer as the video)
where P(B!CarA) is Prob Monty opens DoorB given the car is behind DoorA
Would this strategy still be valid if, for example, there were 5 doors, you pick one, then only one door is revealed, and you can switch to any of the other doors. This continues until you stick with a choice or you end up with 2 doors left?
Edit: Hmm, cool, so even if the process is repeated, you should keep switching right to the very end
Yes, because you would have a 1/5 (20%) chance initially, leaving 4/5 (80%) for the rest of the doors. When one of those doors is removed, that 4/5 is then "concentrated" onto 3 doors, providing a 4/15 (26%) chance that the prize is behind a different door. And then onto 2 doors providing a 4/10 (40%) chance, then onto one door providing a 4/5 (80%) chance.
Yes, but only in a minor %. If you choose 1 of 5 doors you will get a 20% to get the price. If they show you 1 of the other 4, (that represent a 80%, 20 per door), that mean that the other 3 doors has the 80%, and that mean that if you change you will get a 26,6666% to win the price. Isn't much, but is something.
Hmm, this is interesting.
Well let's think about this, without the repeating part. The first door has a 1/5 chance of being right. He then removes one door, so the remaining 3 collectively have a 4/5 chance of being correct. 4/5/3 = a 4/15 chance that any of the other doors has the car, which is slightly better than your chance from the first door (1/5 = .2 and 4/15 ~ .26). So yes, switching would help, but only by a little.
The key thing to remember is that you're distributing a higher probability over LESS choices when you expose one of the wrong answers. Where it used to be even, it now favors the other side, even if only slightly. It's a lot better in the original problem, though, since you only have one choice to move to, so the probability is concentrated into just that choice.
Yes.
There would be 4/5 chance of the car being behind one of the other doors, and even if you distribute this probability evenly amongst the other 3 remaining doors (would yield 4/15) it would still be larger probablity than the 1/5 chance when sticking with the initial door.
Initially, you have three options. A third of a chance of a reward, two thirds chance of a fail.You removed one option, that was identified as a dud. Now, you just have two options. The door you have now has 1/2 probability, just as much as the other. There are two options. Now you have to choose between two options, forget that you originally chose from three. Now start over and choose from two. You are halfway likely to get what you hope. Doesn't this seem balanced? Or does concentration hold a prominent alteration to these plans? Must you multiply to find the new chances?
If I give you cards, you have to choose which one to choose as a reward, now I give you two boxes as your only options. I can say your card corresponds with one of the boxes. The boxes were not apart of the original problem, but none of the options of the original are there. only the same rules apply.
It's very difficult for me to get over the fact that it's not 50/50. It just seems like that would be the case because it goes from 3 doors to 2 doors.
It's a 1/3 chance you would get it right with your first guess, right? Then it's a 2/3 chance the car is behind one of the other doors. One of the other doors must have zork behind it, since there are 2 zorks and 1 car, right? So him revealing that there is a zork behind one of the other doors give you no new information. Since there is no new information the odds don't change, there is still only a 1/3 chance you got it right.
If you initially pick the door that has the goat, the host will open the door with the other goat. So, switching will give you the car.
You have 2 out of 3 chances to pick the goat. However, in both of those chances, if you switch, you get the car. So, by switching, you get a 2 out of 3 chances to get the car.
And yes, a lot of people get confused here. But just because you go from 3 doors to 2 does not make it 50/50
Just because you have 2 door reamining doesn't mean there is a 50/50 chance. When you chose the initial door you "eliminated" one choice but the door is still there so 3 doors = 1/3 vs. 2/3 chance.
wariolandgoldpiramid Oh, I understand now. You put it very beautifully. I didn't quite pay attention to fact that there are two zonks and one car, or 2 losses and 1 win. I was thinking of it more or less as 3 potential wins. If I'm understanding it correctly, there's a 2/3 chance to pick a door with a zonk. When he opens the other zonk door keeping in mind that 2/3 of the time you will be on zonk, it eliminates the chance of switching to a zonk, but doesn't entirely change the scenario, so it remains 1/3 by staying on your own door.
benblue3
You're on the right track.
Another alternate explanation that might help some people still struggling with the problem.
Let's play a different game. There are just two doors, one with a car behind it, and one with a goat. Monty opens the door with the goat, and asks if you would like to leave, or open the door and take what is behind it.
You have 100% of getting the car if you open the door.
Now lets change the game a little, there is a 1/3 chance that there is no car and a 2/3 chance that there is a car/
If there is a car, you will win it.(2/3 times there will be)
If there isn't a car you can't win it.(1/3 times there wont be)
Now we can go back to the original game, now your first choice is determining if there is a car, and your second choice is like our first game.
You could have posted a textbox at the beginning of the short video to this video. I found out about it at the end of your other video and now have to watch through this video where I've already seen 50%.
Thanks
I find it a bit odd that Marilyn vos Savant was never mentioned. Otherwise it was a great explanation! Thanks.