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The Monty Hall problem is described this way:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

I am interested in finding the probability of winning when you switch. I already know it's $2/3$ but I want to show it with Bayes Rule.

I tried this:

$A$ = car behind door $1$

$B$ = goat is behind door $3$

$$P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{1 \cdot 1/3}{1-1/3} = \frac{1}{2}$$

$P(B|A)$ = the probability that a goat is behind door $3$ given that the car is behind door $1$. This is equal to $1$ because if we know where the car is, then any other door must have a goat.

$P(A)$ = the probability of the car being behind door $1$. Assuming any door is equally likely to contain a car before we open any doors, this is $1/3$.

$P(B)$ = the probability of a goat behind behind door $3$. This is equal to $1$ minus the probability that the car is behind door $3$, so $1-1/3$.

Where is my mistake?

Jam
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AJJ
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    It's notable that this is exactly the calculation you'd do if the host revealed that there was a goat behind door 3, and you wanted to know if the car was behind door 1. However, it is, apparently, of some significance that you already picked door 1 before the host revealed door 3. That is, I'm not sure $P(A|B)$ is what you want to calculate. I'm not quite sure how to fix that though. – Milo Brandt Jan 04 '16 at 18:40
  • It sounds reasonable enough, doesn't it? You are calculating $P(A|B)$, in which it is given that the goat is behind door $3$. Not just any door other than the one with the car behind it, but door $3$. Then the car can only be behind door $1$ or $2$, making it 50/50 of it being behind door $1$. – Mankind Jan 04 '16 at 18:40
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    Because you are not conditioning on $B$ but on the fact that the presenter chose door 3. Suppose instead of door 1 you would have picked door 3. Then presenter couldn't reveal a goat behind door 3 - even if that was the case. If you were told upfront that there is a goat behind door 3, of course, the car is equally distributed behind the two other doors. – A.S. Jan 04 '16 at 18:41
  • What he said. The host chose his door based on your previous choice of door 1. – GEdgar Jan 04 '16 at 18:41
  • But why should any of that matter? I am computing this probability as if I were actually on the show. I choose door 1. Presenter opens door 3 to show a goat. Do I switch? Bayes Rule computes conditional probability and takes conditioning into account through the prior probability (or whatever they're called), so I don't understand why this is wrong. – AJJ Jan 04 '16 at 18:45
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    Because you need to condition on event "Presenter picked door 3" NOT on event "There is a goat behind door 3". – A.S. Jan 04 '16 at 18:56
  • Think about this in these terms. You had a 1/3 chance to be correct on the first pick. Anything presenter does afterwards - apart from showing you the car - doesn't affect your chances of being right. – A.S. Jan 04 '16 at 19:11
  • @A.S. I still get the same answer using that approach. $P(A|B) = ((1/2)(1/3)) / (2(1/3)(1/2)) = 1/2$ – AJJ Jan 04 '16 at 19:13
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    Let $C$ be the car-door and $PD$ - door that presenter opened. Then $P(PD=3)=\frac 1 3+\frac 1 3\cdot c$ where $c$ is the probability presenter would pick door 3 if both 2 and 3 had goats (that is $C=1$). Then

    $$P(C=1|PD=3)=\frac{P(PD=3|C=1)P(C=1)}{P(PD=3)}=\frac c {1+c}\in[0,\frac 1 2]$$

    You get the classical $\frac 1 3$ if $c=\frac 1 2$. Unless $c=1$ always, it makes sense to switch.

     – A.S. Jan 04 '16 at 19:23
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    It does depend on the rules that host goes by. If the rules are: The host will randomly show a door you didn't pick. Then 1 out of 3 times the host will show you a door with the car (too bad). If that is the case and he shows you a door with a goat then it is even 1/2 to switch and your calculations are correct. But the assumption is that the host isn't randomly showing you a door you didn't pick but specifically showing you a door with a goat. You aren't calculating P(A|B) but P(A|host showed you door 2) which is different. – fleablood Jan 04 '16 at 19:37
  • There are yet other interpretations of the rules that are consistent with the problem statement. Just read the problem statement as a story: it tells what happens on the one occasion when you play the game, not what happens every time someone plays the game. Then the answer depends on the (unstated) strategy according to which Monty decides whether to open a door and give you another chance. (Better presentations of this problem are explicit about Monty's strategy; the "standard model" is that he will always show a goat and give you a chance to switch no matter which door you choose.) – David K Jan 04 '16 at 19:49
  • The standard telling of the problem is that the host will always show you a goat behind a door you did not pick. If so, and as @DavidK points out that is a BIG if, then you should always switch. The problem in the calculation the OP made is P(B) shouldn't be interpreted as prob goat behind door 3 but host showed door 3. On the other hand if we interpret the story as "this happened; we don't know why" then B does equal goat behind door 3 but switching and staying have equal expectations and the calculations yielding 1/2 are correct, after all. – fleablood Jan 04 '16 at 20:30
  • @flea This is not correct. In alternative reading ("this is what happened") it's possible that the host reveals an extra door only if your initial choice was correct - in which case $P(C=1|PD=3)=1$. We have to somehow restrict set of presenter's actions to make any conclusions. – A.S. Jan 04 '16 at 20:37
  • @A.S. exactly my point. The standard telling is that "these are the rules" and that is fair but then the OP didn't set up the correct revealed events. If the interpretation "this is what happened" all we know is one door has a car and it is not door 3. – fleablood Jan 04 '16 at 21:05
  • @flea Again, no. We know that the "host opened door 3" - not that "the car is not behind door 3". The doors weren't opened by wind or RNG, but by the, supposedly, free-willed host, who could have based his decisions of which door to open (and whether to open them at all) on correctness of your initial pick. – A.S. Jan 04 '16 at 21:11
  • @A.S. We are on the same page. To calculate the probability we need to know the conditions that cause the host to do what he does. The "standard telling" of the problem spells out what they are. To get back the the OP's question "why didn't Bayle's formula work". Well, because it was set up for wind or RNG or circumstances. For wind or RNG or circumstances it did work. For host deliberately showing doors under certain guidelines it simply wasn't set up right. To find P(B) "host shows door 3 is a goat" requires calculating P(B) with further Bayle's theorem – fleablood Jan 04 '16 at 23:37

3 Answers3

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If you define event $B$ simply as 'there is a goat behind door 3', then of course $P(A|B)=\frac{1}{2}$, for there are two options left for the car. And your use of Bayes' theorem to show $P(A|B)=\frac{1}{2}$ is also correct, for indeed with the $B$ defined this way, you have $P(A)=\frac{1}{3}$, $P(B)=\frac{2}{3}$, and $P(B|A)=1$

Put differently: asking what the chance is that door 1 has a car given that door 3 has a goat is effectively ignoring the whole 'game play' behind this problem. That is, you are not taking into account that Monty is revealing a door as a result of your choice, and whatever other assumptions are in force (such as: Monty knows where the prize is; Monty is certain to open a door with a goat; if you initially pick a door with the car, Monty will randomly pick one of the remaining two). Instead, event $B$ simply says: "there is a goat behind door $3$'. Indeed, as such, the problem statement might as well be:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. Oh, and another thing: door $3$ contains a goat. You pick door $1$. Is it to your advantage to switch your choice?

OK, what we need to do is take into account Monty's actions: it is indeed Monty's-act-of-picking-and-revealing-door-$3$-to-have-a-goat that is all important here.

So, instead, define $B$ as: "Monty Hall shows door $3$ to have a goat"

Notice how this is crucially different: for example, if door $1$ has the car, then door $3$ is certain to have a goat, but Monty is not certain to open door $3$ and reveal that: he might also open door $2$.

Now, let's use the standard assumptions that Monty is always sure to reveal a door with a goat and that, if both remaining doors have a goat, Monty chooses randomly between them to open.

OK, now $P(A)$ is still $\frac{1}{3}$, but otherwise things change radically with this new definition of $B$:

First, $P(B|A)$: Well, as pointed out above, this is no longer $1$, but becomes $\frac{1}{2}$, since Monty randomly chooses between doors 2 and 3 to open up.

Next, $P(B)$: what is the probability Monty opens door 3 to reveal a goat? There are two cases to consider in which Monty opens door 3: door 1 has the car, or door 2 has the car, each having a probability of $\frac{1}{3}$ Now, if door 1 has the car then, as we saw, there is a probability of $\frac{1}{2}$ of Monty revealing door 3 to have a goat. If door 2 has the car, then Monty is certain to reveal door 3 to have a goat. So: $P(B)=\frac{1}{3}\cdot \frac{1}{2}+\frac{1}{3}\cdot 1 = \frac{1}{2}$

Plugging this into Bayes' rule:

$$P(A|B)=\frac{P(B|A)\cdot P(A)}{P(B)}=\frac{\frac{1}{2}\cdot \frac{1}{3}}{\frac{1}{2}}=\frac{1}{3}$$

I believe the difference between these two $B$'s is actually at the heart of the Monty Hall Paradox. Most people will treat Monty opening door 3 and revealing a goat simply as the information that "door 3 has a goat" (i.e. as your initial $B$), in which case switching makes no difference, whereas using the information that "Monty Hall opens door 3 and reveals a goat" (i.e. as the newly defined $B$), switching does turn out to make a difference (again, within the context of the Standard Assumptions regarding this puzzle). And this is hard to grasp.

Bram28
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  • i.e. the "mistake" is in thinking that $P(A|B)$ is equal to the probability of "winning when you switch" in general. – SBK Dec 14 '18 at 20:54
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Answering my own question but this appeared to work.

Define:

$A$ = car is behind door $1$, my chosen door

$B$ = presenter opened door $3$ to show a goat

$P(B) = P(B|A)P(A) + P(B|\neg A)P(\neg A)= (1/2)(1/3) + (1/2)(2/3)$

$$P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{(1/2)(1/3)}{(1/2)(1/3) + (1/2)(2/3)} = \frac{1}{3}$$

I hope I did not just get lucky, though.

AJJ
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  • You can make it more general by not assuming $P(B|A)=\frac 1 2$ – A.S. Jan 04 '16 at 19:36
  • It appears that you decided that $P(B \mid A) = 1/2$ and that $P(B\mid\neg A) = 1/2$. Both statements are correct (in the "standard model" for Monty Hall) and can easily be justified, but you did not mention how you came to those conclusions. Perhaps the conclusions are "obvious", but in a paradoxical problem like this I would prefer to justify everything. Still, I think this is actually quite a neat way of reaching the correct conclusion. – David K Jan 04 '16 at 19:37
  • @David $P(B|A^C)=\frac 1 2$ doesn't require justification/modeling assumptioms, does it? – A.S. Jan 04 '16 at 19:50
  • @A.S. It depends on how you read the problem. Read the problem again: does it rule out the possibility that the previous contestant choose door 1, after which Monty opened door 1, revealing a goat, and said, "Sorry, no car for you"? In that case it may be that $P(B\mid \lnot A) = 0$. Part of the "standard model" I mentioned before is we assume Monty does not play tricks like that. – David K Jan 04 '16 at 19:53
  • @David I don't follow. A host opens "another door" - that is NOT the door a contestant picked. This forces $P(B|A^c)=\frac 1 2$. – A.S. Jan 04 '16 at 19:57
  • @A.S. Where exactly did the problem statement specify that the host had to open another door? Why is it not possible that the action "open another door" was itself conditioned on which door the contestant chose? – David K Jan 04 '16 at 20:01
  • @David "You pick a door [...] and the host, who knows what's behind the doors, opens another door [...] which has a goat." – A.S. Jan 04 '16 at 20:02
  • Let us continue this discussion in chat. – David K Jan 04 '16 at 20:06
  • You didn't just get lucky: this is exactly right! – Bram28 Dec 14 '18 at 20:33
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What are the hosts rules?

1) Host will randomly pick a door you didn't choose.

You chose door 1.

Case A: Car behind 3. Host shows you 3. Should you switch? Doesn't matter; neither door has the car. too bad. But this didn't happen.

Case B: Car behind 3. Host shows you 2. Should you switch? Yes.

Case C: Car behind 2. Host shows you 3. Should you switch? Yes.

Case D: Car behind 2. Host shows you 2. Should you switch? Doesn't matter. But this didn't happen.

Case E & F: Car behind 1. Host shows you 3 and 2. Should you switch. No.

So P(A|B) = 1/2 is just fine. It is a 1/2 chance. This is what you calculated.

Rule 2: The host shows a random door with a goat. Here B isn't that Door 3 had a goat. It is that the host showed you door 3.

So... P(A) = 1/3. Okay. P(B|A) = 1/2. (If car is behind door 1 the probability that the host shows you door 3 is 1/2). P(B) is... lessee. If car is door 1 it's 1/2. If car is door 2 it is 1. If car is door 3 is is 0. That (1 + 1/2 + 0)/3 = 1/2.

So P(A|B) = P(B|A)P(A)/P(B) = (1/2)*(1/3)/(1/2) = 1/3. You should switch.

fleablood
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  • Rule 3: if your first choice is the car, the host opens another door and offers you the chance to switch; otherwise the host opens the door you chose and you lose instantly. Rule 4: if your first choice is the car, the host shows it and you win; otherwise the host opens another door and offers a switch. – David K Jan 04 '16 at 19:59
  • @DavidK and so on... Rule 3-> switch good: 0, Rule 4--> switch good: 1. Supposedly thirty years ago Monty Hall talked about the program and claimed he could coax anyone to always switch or stay by psychology and strongly implied his rules were not necessarily to show a random loser door. If my memory serves me right you had 3 doors with unknown items of unknown and he'd show one (not nesc. random) and give the option to switch. You'd have no idea if the prize shown was a bad prize or a good prize nor why he chose to show you that one. – fleablood Jan 04 '16 at 20:15
  • and the rules would be switched. Sometimes you knew there'd be a car and you picked a door and he'd show you what you picked was a washer dryer set and give you the option to switch. One would be the car and the other would be unknown prize which may be more the the washer dryer or less. – fleablood Jan 04 '16 at 20:20
  • Rule 6: Host randomly shows you a goat which may or may not have been the door you picked. If it is the door you picked you can switch to one or the other door. If so, and you picked 1 and he showed 3, it's even odds you should switch. – fleablood Jan 04 '16 at 20:22