It would seem a paradox, but it is better to change the initial choice: doing so doubles the probability of winning. How is this possible?

At the beginning there are three possible scenarios, each having probability 1/3:
(1) The player chooses the first non-winning door (number 1), The host shows the other non-winning door, (number 2): by changing the player wins. By not changing he loses.
(2) The player chooses the second non-winning door (number 2). The handler chooses the other non-winning door (number 1): by changing, the player wins. By not changing, he loses.
(3) The player chooses the winning door. The handler shows one of the two losing doors. By changing, the player loses. By not changing, he wins.
In the first two scenarios the player wins only by changing; in the third scenario the changing player does not win: the “changing” strategy leads to victory two out of three times (2/3), while the “not changing” strategy leads to victory in only one out of three cases (1/3).

The resolution of this problem is so counter-intuitive that several academics did not recognize it until it was explained to them in detail. A simple way to understand it is to increase the number of doors, and to simplify you can use playing cards, the figure represents the winning card, the cards represent the doors.

Let’s put this time 4 doors, 3 losers and one winner. The player, as before, chooses one door, the host, however, shows two – losers, as before – and later asks the player if he wants to change or keep his initial choice. Thus there will be 3 out of 4 eventualities in which the player wins only if he changes and 1 out of 4 in which he wins only if he does not change. The result is valid and, indeed, is even clearer the more you increase the number of doors, and with cards it is very easy to increase iul number of doors at will.