Monty Hall problem with Bayesian inference

One simple way to think about the Monty Hall problem is to use Baye’s rule. For the sake of completeness, here is the problem statement:

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?

Let $X_2$ be the event that the car is behind door 2, $Y_1$ be the event that you pick door 1, and $H_3$ be the event that the host opens door 3. The mental gymnastics is now the compare $P(X_2 \mid Y_1)$ with $P(X_2|Y_1,H_3)$. Well, we can use Baye’s rule to compute this:

\[\begin{align} P(X_2\mid Y_1,H_3) = \frac{P(H_3\mid X_2,Y_1)P(X_2\mid Y_1)}{P(H_3\mid Y_1)} \end{align}\]

Now $P(X_2\mid Y_1) = P(X_2) = \frac{1}{3}$, since the car is equally likely to be behind any of the doors and is independent of your choice. Second, $P(H_3 \mid X_2, Y_1) = 1$ because if the car is behind door 2 and door 1 is chosen by you, the host must open door 3. Finally, $P(H_3 \mid Y_1) = \frac{1}{2}$ because the host can open either door 2 or door 3 with equal probability if the knowledge of the car’s location is not taken into account. So it is clear that

\[P(X_2\mid Y_1,H_3) = \frac{1\cdot 1/3}{1/2} = \frac{2}{3},\]

which implies that switching is the better option.