Well, I think most of the confusion comes from the way that Rosenthal ignores a key part of the Monty Fall problem. In the monty fall problem, you first choose

Well, I think most of the confusion comes from the way that Rosenthal ignores a key part of the Monty Fall problem. In the monty fall problem, you first choose a random door. Assuming a random placement of the car and a random guess, you should have a 1/3 chance of choosing the correct door on your first guess, and therefore a 1/3 chance of winning no matter what else happens. I believe what's really happening in this example is that the Rosenthal is allowing for a possibility that he doesn't explicitly mention: the chance that the host opens the wrong door, nullifying the game and ruining the show. Since there is a 1/3 chance that you picked the right door, it's a probability of 2/3 that it is in one of the other two doors. The host accidentally opens one of the two remaining doors. Since the host randomly chooses one of the remaining doors, the chance that he picks the door with the car is 1/3. If he does that, the game is invalidated, so Rosenthal ignores that result. Therefore, the chance that you picked the correct door is 1/3, and the chance that the remaining door is the correct one is 1/3 (the remaining third is the chance that the host ruined the game). Since those are equal probabilities, you end up with an even chance whether or not you switch.