On the long running hit game show, "Let's Make a Deal," Monty Hall would ask the contestant which of three doors he or she wants. Behind each door is a prize, only one of which is valuable. After the contestant chooses a door, Hall opened one of the other doors and revealed a worthless prize. Then he would ask the contestant if he or she wanted to switch doors. To maximize chances of winning, one should always switch.

To analyze the problem, suppose I were to pick door number 1 and then was revealed door number 3 to be the worthless prize. Now I know the prize is either behind the door I chose or door number 2. It appears that the probability of winning is 50% and switching doesn't increase or decrease my chances.

Think of the problem this way. The probability that the prize is behind door number 1 given that the prize is not behind door number 3. The probability behind door number 1 equals 1/3 since there are 3 doors to choose from. The probability the prize is behind door number 1 or 2 knowing it is not behind door number 3 is 1/2.

The problem involves the use of conditional probability. Suppose you have events A and B, then the probability of A given B equals the probability of A and B divided by the probability of B. This is denoted as P(A/B) = P(A and B)/P(B).

Using the formula for conditional probability, you will find the probability of winning if switching is (1/3)/(1/2) = 2/3. Another way to think of this is that if you switch, the only way that you lose is if the prize was behind the door you initially picked, which has probability of winning 1/3. So the probability that you win is 1 - 1/3 = 2/3.

Although switching doesn't seem to give the best chance of winning, you can conduct an experiment. Play the game with someone assuming the role of Monty Hall and you'll find out as you increase the number of games you play, you'll win approximately 2 out 3 times.