“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 A be the event that the car is behind door 1, B that the car is behind door 2, and C that the car is behind door 3. Let E be the event that the host opens door 3. We assume that the car is initially equally likely to be behind each door and that the host opens a door with a goat at random, never opening the door we picked.

Since A, B, and C are exhaustive and mutually exhaustive propositions, we can calculate the marginal probability of E by using the law of total probability:

P(E) = P(E ∧ A) + P(E ∧ B) + P(E ∧ C).

Bayesians like to define joint probability from conditional probability instead of the reverse; that is, define P(A ∧ B) as P(A | B) P(B) instead of P(A | B) as P(A ∧ B) / P(B).

So P(E ∧ A) = P(E | A) P(A). P(E | A) is 1/2 because we picked door 1, the car is behind door 1, and the host chooses at random a door that has a goat, of which there are two: 2 and 3. P(A) is 1/3. Therefore P(E ∧ A) is 1/2 × 1/3 = 1/6.

Similarly, P(E ∧ B) = P(E | B) P(B). P(E | B) is 1 because we picked door 1 so the host will not open door 1 and we assume the car is behind door 2 so the host will not open door 2, leaving only door 3 to be opened. P(B) is 1/3. Therefore P(E ∧ B) is 1 × 1/3 = 1/3.

P(E ∧ C) = P(E | C) P(C). P(E | C) is 0 because the host will never open the door the car is behind. P(C) is 1/3. Therefore P(E ∧ C) is 0 × 1/3 = 0.

So P(E) = 1/6 + 1/3 + 0 = 1/2. We know that the host opened door 3 (this is E), so the car cannot be behind door 3. How likely is it to be behind door 1? By Bayes’ theorem,

P(A | E) = P(E | A) P(A) / P(E).

We said earlier that P(E | A) is 1/2, P(A) is 1/3, and P(E) is 1/2. So P(A | E) = (1/2 × 1/3) / (1/2) = 1/3.

Given E, the car must be behind door 1 or door 2 since the host opened door 3. Therefore the sum of P(A | E) and P(B | E) must be 1. P(A | E) is 1/3, so P(B | E) is 2/3. The car is more likely to be behind door 2 than door 1. We initially picked door 1, so, if we want the car, we should switch.

Thanks, this looks quite a bit more succinct than other examples of the math. I think I will have to make an app that renders out results using this. :)