The movie "21" went over a statistics problem in an attempt to make the main character look smart. I'm not sure though that the Solution was correct, even though it does seem so. The problem is this: A game show host gives you a selection of 3 doors, two of which enclose goats, and the other encloses a car. Given that you selected the first door, the game show host reveals that door number 3 is a goat. He then offers you the option of changing your choice to door number 2. The proposed solution: You would want to switch because you're initial chance of being wrong was 66%, but when one of the doors was opened it placed a 66% probability on the 2nd door. It makes sense in theory, but I see a problem with it. In this situation, you can't place the 66% probability on door number 2, because a conflict arrises if you allow such an algorithm to determine probability. Please, tell me if I am wrong. Assume the movie is correct and when there is a 2/3 probability of one of two choices being accurate, the probability of either choice becomes 2/3 in the event that the other choice was eliminated as a possibility. That's essentially what the solutions suggests, but look what happens when you apply that concept to doors 1 and 3. Initially the probability of either door 1, or 3 being the door in front of the car is 66%. When Door three is eliminated, there is now a 66% chance that door 1 holds the car. but not both door 1 and 2 can have a 66%; that would be impossible, so the original assumption must not be true. Am I wrong?