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.

* L. Gillman* ** The Car and the Goats** * American Mathematical Monthly,* January 1992, pp. 3-7.

Fri Feb 20 21:45:30 EST 1998