Yeah, it's starting more and more to make sense.

Question: As soon as you pick a door, your chance of finding the car is 1 in 3. And the chance of the car being behind your door is also 1 in 3. Suppose Monty then reveals nothing but just gives you the option of switching. You switch to, say, door No. 2.

The chance of the car being behind door No. 2 is still 1 in 3, exactly the same as being behind either remaining door. Because the car doesn't move. But the chance that *you'll have found it* is 2 in 3 because *your choice* has "moved." You've been given two swings at the same pitch, if I understand correctly. And it doesn't matter if you "hit" the first time and never find out: your chances once you switch remain 2 in 3, but you can still lose.

But I stipulated that Monty doesn't have to do anything but offer you a second chance. Does it still work without seeing a goat? (Right now I think it does.)

A second contestant, facing two doors, has a different situation. There's a 1 in 2 chance that the car is behind either door. His odds of finding it are also 1 in 2. Switching won't increase them because with only two doors and only four possible outcomes there are no available odds between 1 in 2, and 2 in 2 (certainty). Certainty is impossible, so (as anyone would expect)the odds stay at 1 in 2 no matter how many times he switches between closed doors. Contestant 1 has the odds advantage because he started with 3 doors and nine possible outcomes; he could build from odds of 1 in 3 to odds of 2 in 3. Certainty was not the sole alternative to odds of 1 in 3.

Switching works with 3 doors but not with 2.

This made sense as I was writing it, but now I'm not so sure.

The accepted solution so defies (my) common sense that it's hard to transfer the principle to situations that are similar but not identical.