Many people believe that it makes no difference whether the contestant switches. Given that there are now two doors, one with a prize and one without, many individuals (including for a while the great mathematician Paul Erdös) think that each door has a 50-50 chance of having the prize. For more information about the Monty Hall problem, its history and its solution, see the Wikipedia article on it.

The best course of action is for the contestant to switch. By doing so, the contestant doubles his or her chances of winning from 1/3 to 2/3. The easiest way to see this is that the contest originally had 1/3 of a chance of being correct and the opening of the door by the host has not changed this. Since the probability of the selected door having the prize plus the probability of the remaining door having the prize must sum to one, the latter must necessarily be 2/3. Below, we demonstrate this result through the enumeration of cases.

Of course, some people may think that the revealing of an empty door has altered the probability that the originally selected door is correct. Perhaps the easiest way to see that the two doors do not have the same probability of having the prize is to consider the following variation. Suppose there are 1000 doors, the contestant picks one and the host opens 998 other doors, all of whom have no prize. Originally, the contestant had one chance in a thousand of being correct, and this probability actually does not change after the host opens doors. The two remaining doors do not each have a 50-50 chance of having the prize. Think about it: by opening 998 doors is not the host leading the contestant in part to the correct door among the 999 if the prize resides behind one of them? Surely all this information about the 999 doors suggests that the one remaining unselected door is very likely to have the prize. Indeed, is it not suspicious that the host suddenly skips over one of the 999 doors?

Because the host cannot open a door with a prize behind it, there is a constraint on the value of r. Of the d-1 doors that the contestant has not picked, there might be p prizes among them thereby leaving at least d-1-p doors without prizes. Therefore, the number r of doors the host can reveal must be less than or equal to d-1-p otherwise the host might be forced to open a door with a prize behind it. In summary,

In the simple example above, d=3, p=1, and r must be less than or equal to d-1-p=3-1-1=1 and, in fact, is 1.

Given that there are d doors, p of which have prizes behind them, the chance that a contestant picks a door with a prize is

Now if one sums the probabilities of winning for all d doors, then one gets d × (p/d) = p. This sum must remain unchanged if the host reveals r doors without prizes. The originally selected door of the contestant has a chance of p/d of winning, while the d-1-r remaining doors must all have the same probability p

Solving for p

Hence, if the contestant switches to one of the other doors, he or she will have a chance of p

For the single-prize (p=1), three-door (d=3), one-door-revealed (r=1) case, p

As an example of equation (3), consider 10 doors, 2 prizes and 6 empty doors revealed by the host. The original chance of the contestant winning is 2/10 = 1/5. This is increased by the enhancement factor (10-1)/(10-1-6) = 3. So by switching, the contestant triples his or her chances to 3/5.

Without loss of generality, we can assume that the contestant selects door A since by symmetry the other two possibilities are equivalent. Each of the three cases in figure (a)-(c) is equally probable. If the contestant sticks with his or her choice, then the contestant wins a pot of gold in figure (a) and loses in figures (b) and (c), and hence the chance of winning is one out of three or 1/3. If the contestant switches, then he or she loses in figure (a) and but wins in figures (b) and (c) because after the host has revealed the door without the gold, the contestant will automatically pick the one with the gold. By switching the chances of winning are two out of three or 2/3.

For the mathematically inclined, one can do the enumeration-of-cases analysis for the general problem. There are two situations: (i) a prize is behind the initially selected door and (ii) a prize is not behind that door. In switching, one needs to compute the contributions from these two situations. For (i), there are C

where n! means n factorial. For (ii), there are C

which gives the same result as in Eq.(2) after some algebra is performed. Clearly, the method used above is simpler than the enumeration-of-cases approach.

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