Monty-Hall is a simple problem that’s attracted a disproportionate amount of controversy. It’s originally formulated as follows:

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?

The question is posed in a way that makes you think the answer is no, since you have a ^{1}⁄_{2} chance of winning whether you switch or not.

The key to thinking about it properly, which isn’t really clear in the original formulation, is that the host won’t open door 2 or 3 if he knows the car is behind it, but chooses uniformly at random between 2 and 3 if the car is in door 1. Working out the probabilities, it’s then clear that switching to door 2 will give you a ^{2}⁄_{3} chance of winning. Different interpretations may lead to different solutions.

In the equations below, $\Pr(C_i)$ denotes the probability that the car is in door $i$, while $\Pr(D_i)$ denotes the probability that the host opens door $i$. Here’s the correct interpretation:

By contrast, if $P( C_1 \mid D_1) = \frac{1}{3}$ – that is, the host always chooses door 3 when the car is in door 1, chooses door 3 when the car is in door 2, and chooses door 2 if the car is in door 3 – the probability of winning is now ^{1}⁄_{2} whether you switch or not.

Now consider if the host chooses between doors 2 and 3 uniformly at random. The probability of winning outright is now ^{1}⁄_{3} - the car is in the host’s chosen door, so you can just choose that one. Then $P(C_2 \mid D_3 \wedge \neg C_3)$ is ^{1}⁄_{2} – either it’s in door 1 and he chooses the wrong door or in door 2 and he chooses the wrong door, making the total prob. of winning still ^{2}⁄_{3}.

$$ P(Winning) = P(picks correctly) + P(picks incorrectly \wedge guess correctly) $$ $$ = P( C_2 \wedge D_2) + P( C_3 \wedge D_3) + P(C_1 \wedge D_3 \wedge \neg C_3) + P(C_1 \wedge D_2 \wedge \neg C_2) $$ $$ = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{2}{3} $$