[Shamanmuni]

That's odd, I found the 100 doors example is the most efective way of explaining it.

It's easy, pick a door, then the host discards 98 doors in which the car isn't. Do they still think that the probability of the other door left is the same than the one they picked? I want to play gambling games with them!

[ColinWright]

So people I ask say this:

    Pick a door, the host discards 98 doors
    where the car isn't.  There are now two
    doors left, so it's 50:50.

Actually, I'm with them (except for the 50:50 bit). I don't find the 100 door version any more convincing than the 3 door version. Under the usual assumptions the reveal of the other doors gives no information about the one you picked, so that will always remain 1/N. The remaining door will therefore be (N-1)/N, which is bigger if N>2. So switch.