I'm no way an expert, so even if Ms. Savant beautifully explained it; my puny brain thinks... "how is it different from starting the game with 2 doors?". If you choose again, how is the probability not 1/2
Because you aren't starting from scratch - that you had a 1/3 chance to begin with is important. Say there are three doors, A B and C. You pick A:
P(A wins) = 1/3
P(A loses) = P(B _or_ C wins) = 2/3
The host then reveals that there was a goat behind door B. This doesn't change the state of anything (there is always a losing door in the two you didn't pick, and he is always choosing to show that one, and none of the items move). This means the probabilities remain as they were above. However, we know that B didn't win, so we can simplify it to:
P(A wins) = 1/3
P(A loses) = P(C wins) = 2/3
Therefore, if you switch to door C, you have a 2/3 chance of winning, rather than 1/3.
The only way the probabilities would go back to 1/2 for the second choice is if the prize and the goat were shuffled after B is revealed. However, they are not, so the chance that you picked the right door initially is fixed when you picked it.
To think about it another way, your initial choice of A means there's a 1/3 chance it is in A, and a 2/3 chance it is not and is behind one of the other doors. By ruling out B, we don't change the 1/3 chance that it was initially behind A. That means when we are asked again, there is still a 1/3 chance it was behind A, and a 2/3 chance that it was not. However, there's now only one thing that is not A, so there is a 2/3 chance it is behind door C.
P(A wins) = 1/3
P(A loses) = P(B _or_ C wins) = 2/3
The host then reveals that there was a goat behind door B. This doesn't change the state of anything (there is always a losing door in the two you didn't pick, and he is always choosing to show that one, and none of the items move). This means the probabilities remain as they were above. However, we know that B didn't win, so we can simplify it to:
P(A wins) = 1/3
P(A loses) = P(C wins) = 2/3
Therefore, if you switch to door C, you have a 2/3 chance of winning, rather than 1/3.
The only way the probabilities would go back to 1/2 for the second choice is if the prize and the goat were shuffled after B is revealed. However, they are not, so the chance that you picked the right door initially is fixed when you picked it.
To think about it another way, your initial choice of A means there's a 1/3 chance it is in A, and a 2/3 chance it is not and is behind one of the other doors. By ruling out B, we don't change the 1/3 chance that it was initially behind A. That means when we are asked again, there is still a 1/3 chance it was behind A, and a 2/3 chance that it was not. However, there's now only one thing that is not A, so there is a 2/3 chance it is behind door C.