Curt points to:
> http://www.straightdope.com/classics/a3_189.html

Ah, that's provided a clear explanation.

There are three doors.  You pick one, at which point your odds are 1 in 3 of
having the willing door.

Therefore, the odds that one of the other two doors has the prize *must* be
2 in 3.

Given the choice of keeping your original door, or being able to have *both*
the other doors, you would obviously switch, because now you get two of the
three instead of just one.  And that's what's going on here.

The fact that one of the other two doors is opened to reveal a non-winning
solution has no effect other than to compress the 2 in 3 chance of the
remaining doors down to a single door, and by being given the chance to
switch to that door you get both it's original 1/3 chance of wining plus the
1/3 chance from the door that was opened :-)

The opening of one of the two doors you didn't pick is irrelevant, since a
door can always be opened to reveal a non-wining prize, no matter how you
choose.

So the problem is equivalent to the question of whether you would rather
choose one of three possibilities or two of three possibilities.

For those who still think it's 50:50, I'm going to bring three walnut shells
and a dried pea to HP World. :-)

G.