Bertrand's Box Paradox, which I wrote about in my own comment, applies to it. The upshot is that probability is not based on which prize placements _could_ lead the current game state, it is the set of all possible game states. Lets assume that the contestant starts off with door #3.

Case 1: The prize is behind door #1, and the host must open door #2. Probability 1/3.

Case 2: The prize is behind door #2, and the host must open door #1. Probability 1/3.

Case 3: The prize is behind door #3, and the host has a choice. Case 3A: The host opens door #1. Probability Q/3. Case 3B: The host opens door #2. Probability (1-Q)/3.

If the host actually opens door #1, the probability that door #2 has the prize is (Case 2)/(Case 2 + Case 3A) = (1/3)/(1/3+Q/3) = 1/(1+Q).

If the host actually opens door #2, the probability that door #1 has the prize is (Case 1)/(Case 1 + Case 3B) = (1/3)/(1/3+(1-Q)/3) = 1/(2-Q).

My point is that, since you get to see which door is opened, 2/3 is correct only if you assume Q=1/2. We aren't told what Q is, but we must assume it is 1/2 because otherwise the answer is different depending on which door is chosen.

Well if we frame the question as “what is the probability of winning by always switching” then this doesn’t play into it and the answer is indeed 2/3. Hence as a question about general strategy the standard answer is correct.

You’re right if we are asking about a specific case though.

This always bugged me. It isn't switching your choice that gives the 2/3 probability, it is exercising a choice once one of the doors has been eliminated. The odds are the same as flipping a fair coin to decide on which of the two doors remaining to pick.

Am I wrong?

Yes.

You pick a door. Probability 1/3 it has the car behind it. The host then picks another door, revealing the booby prize. That door now has probability 0 of having the car. But you still picked a door with probability 1/3 of having the car, which means there's a 2/3 chance the other door you didn't pick has the car. Write a program to run a Monte Carlo simulation... of the Monty Hall problem... and you will see this is the case.

The host has given you information about where the car is by revealing where it is not. Whether you can effectively use that information is another matter.