The Monty Hall problem
This problem has rapidly become part of the mathematical
folklore.
The American Mathematical Monthly, in its issue of January 1992,
explains this problem carefully. The following are excerpted from
that article.
Problem:
A TV host shows you three numbered doors (all three equally
likely), one hiding a car and the other two hiding goats. You get
to pick a door, winning whatever is behind it. Regardless of the
door you choose, the host, who knows where the car is, then opens
one of the other two doors to reveal a goat, and invites you to
switch your choice if you so wish. Does switching increases your
chances of winning the car?
If the host always opens one of the two other doors, you should
switch. Notice that 1/3 of the time you choose the right door (i.e.
the one with the car) and switching is wrong, while 2/3 of the time
you choose the wrong door and switching gets you the car.
Thus the expected return of switching is 2/3 which improves over
your original expected gain of 1/3.
Even if the hosts offers you to switch only part of the time, it
pays to switch. Only in the case where we assume a malicious host
(i.e. a host who entices you to switch based in the knowledge that
you have the right door) would it pay not to switch.
There are several ways to convince yourself about why it pays to
switch. Here's one. You select a door. At this time assume the host
asks you if you want to switch before he opens any doors. Even
though the odds that the door you selected is empty are high (2/3),
there is no advantage on switching as there are two doors, and you
don't know thich one to switch to. This means the 2/3 are evenly
distributed, which as good as you are doing already. However, once
Monty opens one of the two doors you selected, the chances that you
selected the right door are still 1/3 and now you only have one
door to choose from if you switch. So it pays to switch.
