Why is my Monty Hall answer wrong using Bayes Rule?

Question

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?

Answer 1

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.

Answer 2

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.