Item — HN Reader

Here's how it helps me to look at it. As others have said, you can play with the numbers, but the mechanics are basically this:

First step: you are choosing one option from all available choices. Second step: you are now betting whether your original choice was the correct one.

By consciously narrowing down all the remaining options to one, the host is essentially asking you: "Do you think you guessed right?" As long as there are more than two available options at the start, the odds are greater than you guessed wrong.

As others have said, the important dynamic here is that the host _knows_ where the prize is and proceeds to open all the other doors, leaving only yours and one remaining closed. That remaining door then symbolizes an aggregate of all the doors you didn't select.

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Consider the version where Monty does not know what door the car is behind. He randomly opens one of the other two doors, and (per the rules) if he reveals a car, game over too bad. This time, he happened to reveal a goat. Why is the question not "did you guess right originally?"?

I'm quite convinced that it is different if the host knows versus if he does not, but I've never quite been able to put my finger on why in an intuitive way.

Imagine the host doesn't know. If you guess right, he opens a door with a goat behind it and you find yourself in the classic situation.

But if you guess wrong, there's a 50% chance that the host reveals a car by accident.

So the probablities look like:

33%: you guess right, the host reveals a goat 33%: you guess wrong, the host reveals a goat 33%: you guess wrong, the host reveals a car

So, of the situations where the host reveals a goat, the car will be behind your door 50% of the time.

If the host knows and avoids the goat, the probabilities look like this:

33% you were right; the host reveals a goat 33% you were wrong; the host reveals a goat behind door B 33% you were wrong; the host reveals a goat behind door C

So in the situations where the host knows about and avoids the goat, 66% of them have the car behind the other door.

In the case where the host doesn't know about the car, picking the car on your first choice makes it more likely for the host to reveal a goat. That's the difference.

That's a great question! I think the person who can articulate that intuitively should win the car!

My stab at it: If he doesn't know, then he's essentially another contestant, and his odds are the same 1 in 3 as yours. You've made selections one after the other, but since your selection has yet to be revealed as either right or wrong, you are essentially just picking different options from the same scenario. There are exactly equal chances that you won the car, he flubs and reveals the car, or he reveals a goat.

I like frankc's intuitive explanation of the original game given in another comment:

The way I explain is to expand it but also change the frame of reference to so its clear that the host is an adversary. Imagine we play a game called "who has the Ace of Diamonds"? I deal you one card face down and I deal me 51 cards. I look at my cards and then choose 50 of them to show you, none of which are the Ace of Diamonds. Do you want to keep your card or take the one I have not turned over?

Now imagine the same set up, but instead, after dealing the cards, I don't look at them. I then proceed to turn 50 of mine over one after the other and, though unlikely, I happen to not reveal the Ace of Diamonds. Now, is it any more likely the last card I haven't turned over yet is the card, than it is that you have it?

The game is no different in this case to one person lining up all 52 cards and turning them over one by one along the line. If you get to 50 cards and you still haven't found it, that obviously doesn't mean it's more likely to be the end card than the penultimate one.

Whether we arrive there by chance or by deliberate action, once we know that the open door has a goat, the odds for winning by switching remain the same. The odds only change for the overall game (where there's now a 1/3 chance we lose/win when the host reveals the car). If the goat is revealed, we're still in the original game: 2/3 for switching, 1/3 for staying.

EDIT: Leaving my wrong answer, but bvk is correct.