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0 pointsmcv3y ago0 comments

> 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.

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4 comments · 1 top-level

firstlink3y ago· 3 in thread

> Fair point, but the host can only reliable pick an empty box if he knows which boxes are empty

You're ignoring the difference by posing a non mathematical objection to the statement of the problem. This is like having a physics problem about frictionless spherical cows and objecting that cows aren't really frictionless and spherical, and giving an answer for a different problem. Sure you might come up with a valid answer for your different problem, but it isn't a valid answer for your original problem. Likewise 1/2 is a valid answer to the invalidated trials version of the problem here, but not a valid answer to the problem as stated, which remains 2/3.

mcvOP3y ago

Keep in mind we're talking about two different problems here. Monty Hall and Monty Fall (I hadn't heard of that name before, but let's stick to it). The difference between these two problems is subtle and yet dramatic, because whether or not the host knows which box is empty and is therefore guaranteed to open an empty box, or doesn't know and is therefore lucky to open an empty box, makes the difference between whether the remaining box has a chance of 2/3 or 1/2.

These are the valid answers to the two problems. It's not clear to me what you think I'm ignoring, but if you can make that clear, perhaps I can explain my meaning a bit better.

EGPRC3y ago

The Monty Fall problem asks you about the case when the host has managed to reveal a goat. That only includes a subset of the total attempts that you would start selecting a door, not all. Once a goat is revealed, you could only be in the 1/3 case of when the player's choice is correct, or in the 1/3 case of when the other door randomly left closed is correct, so each represents 1/2 of this subset, not one 1/3 and the other 2/3.

firstlink3y ago

> That only includes a subset of the total attempts that you would start selecting a door, not all.

In the trial invalidation version of the problem, but not in the problem as stated. The problem as stated provides, the host does not reveal a car (or your original door).

Let me put it another way. If the question had asked, "Conditional on the host not revealing a car (or your original door), what is the probability of winning if you switch?", then that too would be 1/2. My problem is that this is not what is asked. We don't get to answer an easier or different version of the problem. And the reason it matters is because people start spouting woo about how the host's knowledge or intentions are what mattered, when that isn't true at all. What matters, as you have demonstrated and I have tried to clarify (we aren't really disagreeing), is whether we are exploring the conditional probability in the uniform distribution over a sample space where the host might open another door (1/2), or the probability in the uniform distribution over a sample space where he cannot (2/3). If one denies the difference between these versions of the problem, one ends up in woo-space where the host's knowledge or intentions matter. They don't.

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