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 $\frac{3}{2}$?
$\endgroup$ 23 Answers
$\begingroup$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.)
$\endgroup$ $\begingroup$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.
$\endgroup$ 1 $\begingroup$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.
$$\mathsf P(A_1\mid B)~{=\dfrac{\mathsf P(B\mid A_1)\mathsf P(A_1)}{\mathsf P(B\mid A_1)\mathsf P(A_1)+\mathsf P(B\mid A_2)\mathsf P(A_2)+\mathsf P(B\mid A_3)\mathsf P(A_3)}\\=\dfrac{\tfrac 12\tfrac 13}{\tfrac 12\tfrac 13+0+\tfrac 11\tfrac 13}\\=\tfrac 13}$$
$\endgroup$
