(1) MH asks P to pick a door from a row of three that are on the stage. Behind one of the doors is a prize; behind each of the other two, a brick. P obligingly picks a door (say, door 1).

(2) MH now opens one of the doors not picked by P (say, door 3) and purposely reveals a brick.

(3) Finally MH gives P the opportunity to switch the original pick to the other door (which is now door 2). Before the right door is opened, P must choose either to stick with the original pick or to switch. (The crowd goes wild giving advice at this point.)

The problem is to decide which, if either, strategy is most likely to get P the prize.

The following is a more "common sense" approach to convince the skeptical that P is actually better advised to switch than to stick with the door originally picked. A more formal solution, using Bayes' theorem, may be found by looking here.

Let's suppose P decides to stick with the original pick. Then the probability of success is 1/3; i.e., if the "stick" game were to be repeated 300 times, P would make the right pick approximately 100 of those times. Now MH, knowing where the prize is, always reveals one of the doors hiding a brick. This means that the prize is behind the remaining door approximately 200 out of the 300 trials. Now suppose there is another player, a kibbitzer Q playing over P's shoulder, who follows P's first pick but who then plays the "switch" strategy. Then Q picks the prize exactly when P doesn't; i.e., Q wins the prize approximately 200 out of the 300 trials. Thus Q's chance of success is a whopping 2/3.

The moral of the tale is that a person playing the "switch" strategy has twice the chance of winning the prize as a person playing the "stick" strategy.