rvern
Joined 12 years ago
Joined 12 years ago
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.
That's where you make a mistake. Whatever the reason, the Wikipedia articles on the g factor, the intelligence quotient and Cattell–Horn–Carroll theory represent better what is widely accepted in the field than any other source. This somehow also turns out to be true for other academic fields.
Anyway, when source A says explicitly that source B is wrong but source B makes no such claim about source A, you should usually believe source A. But in this case you can know that the study is right by just learning for yourself what does happen to be widely accepted in the field and why it is widely accepted.