I find the Monte Hall problem is simple to grasp once you explain it properly. But you need math for that.

I think it can be explained just using intuition too.

Instead of 3 doors, imagine there are 1e9 doors. Then you choose one door (at random), and the host reveals all but one other door.

I think then it's clear why you would switch doors.

The problem is that it is not obvious why the generalisation mus be that the host opens n-2 doors. Many people view it as the host opening just one door, and doing so also in the 1e9 case.

What complicates things is that these people are right, too. The generalisation of the problem does not fix the number of doors opened by the host. It could be that the host opens just one door. Or 5e8 doors, splitting the set in the middle. Or any other number in between. In fact, the number of doors opened by the host could be decided by dice before each round starts.

So now the person you talk to goes, "I'm still not convinced, but let's say for the sake of the argument that switching is good in the situation you mention. Now tell me why it is better to switch in all other cases."

You're ending up in formal proof country a lot faster than you'd want for an intuitive explanation.

(Odd observation: if k is the number of doors opened by the host, I find any case where n=2k+1 a lot harder to explain than when k=1 and n is some large number. Maybe that can give some insight into why it's a tough problem.)

It's fairly obvious to me (after careful elaboration, that is): if you don't switch, you'll win iff you got the right door the first time (1/3). If you do switch you'll win iff you picked one of the wrong doors (2/3).

My intuitive explanation when I was tutoring statistics was the following. (This works better while drawing I find.)

Assume you pick one door. It's the right door with a chance (1/3) and the wrong door with a chance (2/3).

Now if you could pick both other doors (2/3) you would choose those. And this is exactly what you're doing when you switch after the reveal.

The other two doors always have one wrong one, which can be revealed. So switching after the reveal is the same as having the chance to pick two doors before the reveal.

Sure, there are many possible generalizations, some of which are more intuitive than others. But it should be clear that n=3 is a specific case of this particular generalization, for which I think the intuition is clear, which should help illuminate why the n=3 case holds.

I don’t think you do. Imagine the Monty hall problem had 1,000 doors. You choose one, then 998 others are opened to reveal goats. It’s abundantly clear that switching in this case is better odds, without even formalizing the mathematics.

Look at kqr's comment, which shows why this is not an equivalent problem, and also points out that questioning it is quite right in this case.

It is an equivalent problem. kqr's comment is wrong in its implications. Whether you open one door or 998 doors, you are revealing information about the likelihood of what is behind the last door. It's just more obvious that this is the case when N-2 doors are opened than just one.