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You can distill it all to this:

If your initial guess is correct, then switching always loses.

If your initial guess is wrong, then switching always wins.

Everyone who understands the game can agree on those facts without controversy. If they don’t then there’s no point proceeding because they do not understand the rules of the game.

Therefore, the odds of winning when you switch are the inverse of the odds of winning from your initial guess.

If your initial guess wins 1/3 of the time, that means always switching loses 1/3 of the time and conversely always switching wins 2/3 of the time.

A slightly different reframing of this idea makes it even more clear to me:

When you make a choice, you divide the doors into two sets: the set containing your choice, and the set containing the other doors. Obviously your set has a 1/3 chance of containing the prize, and the other set has a 2/3 chance.

When the host reveals a door in the second set, it has no effect on those initial probabilities. The set as a whole still has a 2/3 chance of containing the door, and now you have the choice of selecting the only item in that set which you know isn't wrong.

This reasoning is just as obvious when you scale up the doors too. With 100 doors, your set has a 1% chance of containing the prize, and the set of unchosen doors a 99% chance. All but one door in that set are revealed to be wrong, then you get a "50/50" choice of swapping to that last remaining item... of course that's what you want to do!

Consider an alternative game

1. Pick a door

2. Monty asks if you want to switch to both of the other two doors

3. If you switched, Monty will remove a losing door from one of your two doors

I think it is easy to see that this game has the same odds as the original, and also that switching increases your odds from 1/3 to 2/3.

The one which made it clear for me was: imagine you have 4 doors instead. You choose one at random. The host then chooses two of the remaining doors he knows has goats. Is it in your best interest to switch to the one he didn't open?

Suppose you have 100 doors. You choose one at random. The host then chooses 98 of the remaining doors he knows has goats. Is it in your best interest to switch?

In all of these situations, the basic question is: what is more likely - that you chose the one with the car, or that you didn't, and the door remaining is only remaining because the host knew it was the one with the car?

Yeah, simply imagining a much larger number of doors makes the logic obvious.

I used an entire deck of cards (52 of them) to demonstrate this to a friend, and he still wouldn't believe it. I gave up explaining it after that.

It's really easy to get people arguing about this one, even though the correct answer is so well documented.

I remember reading someone who couldn’t understand it - so he build a computer program (in python I seem to recall) to simulate it over millions of goes. And the numbers worked out as expected.

It's an easy python program to write, because there's nothing complex about the process. Pick a language here: https://rosettacode.org/wiki/Monty_Hall_problem

Every one of these prove the effect beyond a doubt.

To be precise they demonstrate it; a proof (mathematically) comes from axioms.

Yeah, this is exactly the easiest way to think about the problem. In summary, by switching you win when you initially picked a goat, which happens two thirds of the time.

This is the first thing to get, the second is that Monty won't pick the car. If you do not understand these two, then you will say 1/2.

I've heard the analogy with 1 000 doors and opening 998 of them before but that doesn't explain anything at all. If you think that makes you understand then you're mistaken; that made you understand something else.

The key is that you have 1/3 of being right, 2/3 of being wrong. So in 2/3 of the cases, you will be offered to pick the car. So 2/3 switching gives you the car and 1/3 loses you the car. So always switch, it has better odds.

Any explanation that doesn't involve scaling up the number of doors to a arbitrarily large amount where the probability becomes obvious is really lacking if you ask me. The only reason this is so confusing in the first place is because it's only 3 doors.

With an infinite amount of doors, the chance of picking the right one at first is zero. When the host opens all doors except one other, it's almost certain that it's the door with the car.

It’s good for the aha intuition moment for some people. It’s the thing that made me get it as a kid. There’s something unsatisfying about it though. You can have 100 doors, and open 98 to leave the initial guess and 1 more. But this is scaling up the rule as “all but one”. I don’t see an obvious reason that the rule shouldn’t scale up as “open one more door”. Then you have 99 closed and 1 open and the result isn’t nearly as obvious, even though switching is the better move.

Yeah I thought about that, both +1 door and infinity-1 door options are both equally unfaithful to the original I suppose. Keeping the 50% door opening ratio would make more direct sense if you wanted to scale it up for other reasons, but isn't nearly as clear in intuitively showing the concept.

I agree that it feels weird at first, but thinking about it this way fixed that for me:

The whole reason for the apparent paradox is having the "50/50" choice at the end. You either keep the door you picked, or swap with the other door, so it feels like no advantage.

If they only revealed a single door, then you would have multiple doors to swap to, so you wouldn't have the "50/50" choice. The only way to keep the "50/50" choice at the end is to reveal all but one door and give you the option to switch to it.