Such that the contestant chooses 1 door, the host then opens 98, and they are given the opportunity to switch to the last remaining door?

That's actually a pretty brilliant way of explaining it. With numbers like that the answer becomes much more intuitive.

Why 100? Why not just 5? Some people would 'get' it at five, some people at 100, and some people at a million. If you have to choose out of a million doors, and no matter what the host opens all but one of them, so that your prize is either behind the door you picked, or behind the other one -- then should you switch your choice?

Well, obviously, you should - with a million doors, it becomes obvious that you have just a 1 in 1,000,000 chance of having picked it.

But thing is - that "obvious" thing 'should' be just as obvious with 1000 doors, 100, 20, 5, or...3....

It's a matter of degree - not kind.

So appealing to a way of intuiting it that is a lot more 'obvious' - while in fact having the exact same format of question, just goes to underscore how fickle intuition can be.

That said, taking individual variables to ridiculous extremes is a great way to thought experiment and an awesome way to get intuition to work better.

Yeah, I'm not wed to a particular number... I'm just observing that larger numbers seem to make it more intuitive than, say, 3 doors.

I never understood this assertion. Most people I explain it to in this way still think it's 50:50 because you only have two doors left.

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!

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.

Because it is near impossible to open 98 doors at random and not reveal the car.

To the people who have trouble with this, that's neither relevant nor helpful. They just say "two doors left, hence 50:50."