The Monty Hall paradox is a famous problem in the field of game theory and probability for which the best strategy of the player is counter intuitive. The problem statement is as follows:

You are on a game show and in front of you are three doors. You know that there are goats behind two of the doors and there is a new car behind the third door (the assumption is that you would prefer the car over a goat). You are asked to pick a door and you will win whatever is behind it. After you pick your door, the host, who knows how the prizes are positioned, will always open one of the doors that you did not pick to reveal one of the goats. He then offers you a choice to either keep your current door or to switch to the remaining door that is unopened. What do you do?

If this is your first time hearing this problem, you might convince yourself that it does not matter whether a switch is made. Clearly, there is 1/3 probability that I pick the door with the car behind it. After the host opens a door with a goat behind it, there are only two doors left unopened. Since we have no further information about the two unopened doors, it follows that the probability of the door I picked having a car behind it has been conditioned to 1/2. Thus it would not make a difference if I switched doors.

However, the best strategy is actually to switch doors when given the opportunity. The formal derivation of this solution involves the application of probability theory which you can read about at http://en.wikipedia.org/wiki/Monty_Hall_problem#Bayes.27_theorem if you are interested. However, we can simply demonstrate by exhaustion that switching doors is indeed the best strategy to follow.

G = goat, C = car

|C|G|G|

|1|2|3|

Assume that the positioning of the prizes is as shown above (the numbers on the doors do not matter so this case is equivalent to any other starting position if we just change the door numbers) and lets say that you always switch doors when given the opportunity. Then the possible outcomes of the game are as follows.

pick door 1 and then switch = win goat

pick door 2 and then switch = win car

pick door 3 and then switch = win car

Notice that the cases in which you win the car are the ones where you pick the door with the goat behind it to start out with. Clearly there is a 2/3 probability of your starting door having a goat behind it. Thus we conclude that you will win with 2/3 probability if you follow the strategy of always switching when given the opportunity.

This is the same result that you will get if you work out the formal probability theory. Thus, we have demonstrated that the best strategy is actually to always switch doors.