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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?

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    $\begingroup$ 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. $\endgroup$ – Milo Brandt Jan 4 '16 at 18:40
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    $\begingroup$ 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. $\endgroup$ – A.S. Jan 4 '16 at 18:41
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    $\begingroup$ Because you need to condition on event "Presenter picked door 3" NOT on event "There is a goat behind door 3". $\endgroup$ – A.S. Jan 4 '16 at 18:56
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    $\begingroup$ 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. $\endgroup$ – A.S. Jan 4 '16 at 19:23
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    $\begingroup$ 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. $\endgroup$ – fleablood Jan 4 '16 at 19:37
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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.

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  • $\begingroup$ i.e. the "mistake" is in thinking that $P(A|B)$ is equal to the probability of "winning when you switch" in general. $\endgroup$ – T_M 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.

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  • $\begingroup$ You can make it more general by not assuming $P(B|A)=\frac 1 2$ $\endgroup$ – A.S. Jan 4 '16 at 19:36
  • $\begingroup$ 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. $\endgroup$ – David K Jan 4 '16 at 19:37
  • $\begingroup$ @David $P(B|A^C)=\frac 1 2$ doesn't require justification/modeling assumptioms, does it? $\endgroup$ – A.S. Jan 4 '16 at 19:50
  • $\begingroup$ @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. $\endgroup$ – David K Jan 4 '16 at 19:53
  • $\begingroup$ @David I don't follow. A host opens "another door" - that is NOT the door a contestant picked. This forces $P(B|A^c)=\frac 1 2$. $\endgroup$ – A.S. Jan 4 '16 at 19:57
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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.

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  • $\begingroup$ 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. $\endgroup$ – David K Jan 4 '16 at 19:59
  • $\begingroup$ @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. $\endgroup$ – fleablood Jan 4 '16 at 20:15
  • $\begingroup$ 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. $\endgroup$ – fleablood Jan 4 '16 at 20:20
  • $\begingroup$ 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. $\endgroup$ – fleablood Jan 4 '16 at 20:22

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