Monty Redux

In the Monty Hall Problem a contestant is given a choice between one of three doors, with a fabulous prize behind only one door. After the initial door is selected the host, Monty Hall, opens one of the other doors that does not reveal a prize. Then the contestant is given the option to switch his or her choice to the remaining door, or stick with the original selection. The question is whether it is better to stick or switch.

The answer is that it is better to switch because the probability of winning after switching is two out of three, whereas sticking with the original selection leaves the contestant with the original winning probability of one out of three. Why?

The trick to understanding why this occurs is to view the situation not from the contestant’s viewpoint, but from Monty Hall’s. At the outset, from Monty’s point of view, the contestant has a one out of three chance of guessing the correct door. In the likely situation (two out of three) that the contestant chose wrongly, Monty then has to know where the prize is among the two remaining doors in order to open a door that does not reveal the prize. So Monty opens a door not revealing the prize and asks the contestant whether he or she would like to switch or not.

However, the contestant knows that in the likely (two out of three) situation that the initial choice was wrong, Monty had to know where the prize was in order to open the door that did not contain the prize. Since the contestant knows that Monty has to know where the prize is to make the correct choice, the contestant can (in this likely case) place him or herself in Monty’s shoes. At this point Monty knows that the remaining door is the one that contains the prize, and hence the contestant should switch.

If we consider the unlikely situation in which the contestant initially chose the door with the prize behind it, then this line of reasoning will not work. Imagine that Monty forgets the location of the prize every time the contestant guesses correctly. In this situation he can still open either of the remaining doors without ever ruining the game. From his perspective the location of the prize is unrelated to his actions; it played no part in his decision to open one door or another (he merely chose a door the contestant hadn’t).

So, in the one out of three case where the contestant initially selected the correct door, there is no way to deduce whether switching is beneficial based upon placing oneself in Monty’s shoes:  the situation where Monty has forgotten the prize’s location is indistinguishable from a situation in which he has not forgotten. Without any way to further analyze the situation and tilt the odds to over one out of three, the contestant should always assume that he or she is in the previous, more likely, situation and take the opportunity to switch.1


.

1Imagine that the contestant has a guardian angel that will let the game run its course if the contestant switches doors, but will change the location of the prize such that if the contestant sticks with the original door the angel will make sure that the contestant wins four out of five times. Then the probability of winning while switching will stay at 2/3 but the probability of winning while sticking will be 4/5. If the contestant had some way of divining that this was happening, this would be a case in which further analysis would be of benefit.


File translated from TEX by TTH, version 3.79.
On 13 Aug 2009, 13:48.

Posted in epistemology, game theory, logic, philosophy. Tagged with , , .

8 Responses