How to explain the Monty Hall problem to a disbeliever

> the remaining unopened box is different from the box your picked originally.

Because you now know that every other box is empty. So by process of elimination you know that your box and the remaining one are different.

There are a couple approaches I like:

"Which of the three boxes do you choose?"

"Box A."

"Is the prize probably in Box A?"

"No, it's probably in one of the two other boxes."

"So it's probably not in Box A."

"Correct."

"Look: Box C doesn't contain the prize. Do you want to stay with Box A which probably doesn't have the prize, or switch to one of Box B or C, which probably does have the prize?"

"Well, since Box A probably doesn't have the prize, I should switch to Box B or C. But I know Box C doesn't contain the prize, so I'll switch to Box B, which probably does."

Another approach is to make a lot of boxes.

"Here are 100 boxes. Choose one that has the prize."

"I choose box 17, which probably doesn't have the prize."

"Now I'm going to open 98 more boxes and show you that none of them contain the prize. As you see, only box 68 remains closed (as well as your box 17). Do you wish to switch to box 68?"

"Hell yes!"

I think the 100 box explanation misses a subtlety in the problem. Conditional on those 98 boxes being empty, box 17 and box 68 are equally likely to have the prize. The key is that you have a large probability of backing the host into a corner as it were: forcing him to choose 98 of 98 openable boxes vs 98 of 99. This is where your large probably of winning comes from, but it's no clearer in the large (100-box) example than the small. Therefore I posit that the large example exploits a misunderstanding about the solution rather than elucidating it.

> I do not believe it is possible that it matters whether the host "knows" or doesn't.

I think it matters a lot, and I suspect this is the issue that Monty Hall nonbelievers (who think the chance is 50% even if the host knows) are struggling with, but from the opposite direction of you.

Let's try to work out the probabilities:

You randomly pick box (let's call it A), 1/3 chance of being correct. The host randomly randomly opens one of the other two (let's call it B). 1/3 chance he opens the box with the prize, invalidating the game. 2/3 chance he opens an empty box. If he picks an empty box (2/3 chance), each of the other boxes (A and the remaining box, let's call it C) still has 1/2 chance of holding the prize, so 1/2 * 2/3 = 1/3 chance each a priori, but 1/2 after eliminating the 1/3 chance of invalidating the game.

So now we've got 1/3 chance of A having the prize, 1/3 of the game being invalid, 1/3 of the third box holding the prize. After eliminating the invalid game, there's still an equal chance of A and C holding the prize.

When the host knows which box holds the prize and uses that information to always pick an empty box (which he always can, because there will always be at least one unpicked empty box), there's no chance of invalidating the game, so the remaining box has a 2/3 chance of holding the prize.

> The problem states that the host does not reveal the prize, not that the game is void if he does.

Fair point, but the host can only reliable pick an empty box if he knows which boxes are empty. If he doesn't, there's always a chance of him accidentally opening the box with the prize. If he opens an empty box by luck, then you dodged that 1/3 chance of invalidating the game. You're now in the changed probability space given that the host didn't accidentally invalidate the game. Each box had an a 1/3 priori chance of being right, but given the 2/3 chance that the host didn't invalidate the game, they now each have a (1/3)/(2/3)=1/2 chance of being the correct box.

It is the knowledge of the host that eliminates the chance of an invalid game and puts the 2/3 chance on the remaining box.

> If the host "does not" reveal the prize, then it really doesn't matter why.

It does. You dodged a bullet, and that has an influence on the remaining probabilities. In the original Monty Hall problem, the host provides extra information, in Monty Fall, he doesn't, but you dodge a bullet. That difference matters.

You can get that the answer must be 50% once the host just reveals a goat by chance using reductio ad absurdum. Notice that since the host is doing it randomly, it would be the same if you (the contestant) were who made the revelation instead of the host. By the end both are doing it without knowledge so the results shouldn't tend to be different. For example, you could pick door 1 and then decide to reveal door 2. But in this way what you are doing is basically selecting which two doors will remain closed for the second part. I mean, selecting door 1 and then revealing door 2 is the same as picking both door 1 and door 3 at once, and then discarding the other. So, if door 2 results to have a goat, which one do you think is which should have 2/3 probabilities of having the car, door 1 or door 3? Consider that the doors don't "know" if you picked them at once or one after the other. The result would have been the same if you had first declared door 1 as your staying option and number 3 as your switching one, or viceversa. So, neither of them can be more likely than the other.