Recently, while we were in the car to Québec city with some PhD students, someone mentioned the Monty Hall paradox, and we discussed possible extensions… The Monty Hall paradox is usually presented from tv show,

Craig F. Whitaker wrote the problem as follows, in Parade Magazine, September 1990, « Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice? ». Actually, Bertrand proposed the same problem, but with boxes instead of doors…. but the problem was the same.

Assume that the candidate chooses door 1 (without loss of generality since the problem is clearly symmetric). The probability that the car is behind door 2 is . The animator can either open door 2 or door 3:

- if the car is behind door 3, he has to open door 2,
- if the car is behind door 2, he has to open door 3,
- if the car is behind door 1, he can open door 2 or 3, and we assume that the opening is equiprobable,

Assume that the animator says “the second box is empty”, what should the candidate do ?

To formalize the problem let denote the event that the car is behind door , and the event that the animator opens door . So, if he opens door 2, the probability that the car is behind door 3 is

where

from the previous discussion, since he cannot open door 1 (the candidate chose it) and the cannot open door 3 (since the car is behind). Further

from equiprobability. And for we get, similarly

Thus,

while

So the optimal strategy is to open the third door (even if I chose the first one)… It is usually seen as a paradox, but if you consider a much larger number of doors (say 4),

and that the animator opens 2 doors, then should we still change, and open the door that is still closed ? The higher the number of doors, the higher the probability to have something behind the other door…

For instance, with 4 doors, or boxes, if the candidate still chose the first door, and that the animator opens doors 2 and 3, then, the probability that the car is behind the fourth one is

i.e.

( that appears at the denominator since we focus on the pair ) while

Once again, we have that

i.e. the opening of a doors bring us no information about our choice, but it will after conditional probability for the remaining door.

More generally, with n doors, if the animator opens doors,

then

while

And here the result is even more intuitive: we have to open the door that was left closed. Actually, it is possible to see that it can be extend to the case where the are doors (or boxes), the candidate chooses doors, and the animator opens (out of the remaining doors). Then, behind each door chosen by the candidate, the probability does not change

where i goes from 1 to , while

where i goes from to .

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