How to solve the Monty Hall problem using Bayes Theorem

Question

I've recently come across the Monty hall problem and while the reasoning behind switching doors makes sense intuitively to me I can't seem to understand the maths behind it.

I've seen many proofs online using Bayes Theorem and I manage to understand the majority of it aside from one thing.

In the classic setting of the Monty Hall problem let $A$ be the event that the car is behind the first door that I chose. Let $B$ the event that Monty reveals a goat behind door 2.

Then, using Bayes Theorem, we have

$$P(A \mid B)=\frac{P(B \mid A)P(A)}{P(B)}$$

Now, $P(B \mid A)=\frac{1}{2}$ because if the car is behind door 1 then Monty can choose either the second or third door. $P(A)=\frac{1}{3}$ because there's a one in three chance of the car being behind the first door.

This all makes sense to me - my struggle comes in finding $P(B)$.

I understand that there are 3 separate scenarios:

-$ \textbf{The car is behind door 1} $ As above in this case $P(B)=\frac{1}{2}.$

• $\textbf{The car is behind door 2} $ Clearly, $P(B)=0$ as Monty cannot reveal a goat behind door 2.

• $\textbf{The car is behind door 3} $ If the car is behind door 3, then Monty is forced to open door 2 and so in this case $P(B)=1$.

The way I see it, by combining these three scenarios $$P(B)=\frac{1}{2} + 0 + 1=\frac{3}{2}$$

But in every proof that I have seen they divide this by 3 for some unknown reason. I feel like I may be missing something blindingly obvious but I don't understand why this is true.

Could someone explain the reason that $P(B)=\frac{1}{2}$ as opposed to $\frac{3}{2}$?

Answer 1

Because each of the scenarios has probability $\frac13$. You're applying the law of total probability, and each term contains a factor $\frac13$.

Let's say it rains tomorrow with probability $\frac12$. If it rains, I brush my teeth. If it doesn't rain, I also brush my teeth. Is the probability that I brush my teeth tomorrow $1$ or $2$?

You also have an error in your calculation of $P(B\mid A)$. This is $\frac13$, not $\frac12$. This follows by symmetry – conditioning on an event that doesn't distinguish one of the three doors from the others can't make the probabilities for the doors being opened non-uniform.

Answer 2

In the classic setting of the Monty Hall problem let $A$ be the event that the car is behind the first door that I chose. Let $B$ the event that Monty reveals a goat behind door 2.

Ah! You want $B$ to be the event that Monty chooses door 2 (rather than door 3) since you always select door 1.

Let $A_n$ be the event that the car is behind door $n$.   $A_1$ is the event that it is behind the door you choose.

$$\mathsf P(A_1\mid B)~{=\dfrac{\mathsf P(B\mid A_1)\mathsf P(A_1)}{\mathsf P(B\mid A_1)\mathsf P(A_1)+\mathsf P(B\mid A_2)\mathsf P(A_2)+\mathsf P(B\mid A_3)\mathsf P(A_3)}\\=\dfrac{\tfrac 12\tfrac 13}{\tfrac 12\tfrac 13+0+\tfrac 11\tfrac 13}\\=\tfrac 13}$$

Answer 3

To answer your question:

The way I see it, by combining these three scenarios $P(B)=1/2+0+1=3/2$

Let us denote scenarios one through three $S_1, S_2,$ and $S_3$.

$P(B|S_1) = 1/2$

$P(B|S_2) = 0$

$P(B|S_3) = 1$

How to combine them:

$P(B) = P(B|S_1)*P(S_1) + P(B|S_2)*P(S_2) + P(B|S_3)*P(S_3)$

Given $P(S_1)=P(S_2)=P(S_3)=1/3$

$P(B) = P(B|S_1)*1/3 + P(B|S_2)*1/3 + P(B|S_3)*1/3$

$P(B) =$ {$P(B|S_1)*P(S_1) + P(B|S_2)*P(S_2) + P(B|S_3)*P(S_3)$}$*(1/3)$

Substituting in $P(B|S)$:

$P(B)=(1/2+0+1)*(1/3)=3/2*1/3=1/2$

(As for the proof, I am not familiar with it, and not certain that it is valid/relevant, only that it produces the correct answer.)