The answer is (b). He should change his choice. The point is that there are two boxes containing nothing, and the one which is removed may be either of them.
Suppose the contestant chooses Box A, If the prize is in fact in Box A, and he switches, he will get nothing.
If the prize is in Box B, Box C will be taken away, and if he switches, he'll get the prize. If he sticks with A, he'll lose.
If the prize is in Box C, then Box B will be taken away. If he switches, he'll win. If he sticks with A, he'll lose.
So the chance of his being wrong is not reduced by taking one of the boxes away. If you don't believe this, try a practrical experiment (using, for example, the traditional pea and thimbles).
This seems to be a recent variant of an old poser about three cards. One card is white on both sides, one is black on both sides, one is black on one side and white on the other. One of the cards is laid on a table, and the side facing up is black. What are the chances that the other side is also black? Again the answer is 2 chances in 3, because the side you see may be the black side of the black-and-white card, or EITHER side of the all-black card.
Jul 04 2002, 10:05 AM