How to explain the Monty Hall problem to a disbeliever • Hacker News Board

Re: How to explain the Monty Hall problem to a disbeliever

#91

Post by firstlink » Thu, Mar 16, 2023, 6:59 PM UTC

mcv wrote: post #87

firstlink wrote: post #86

Earlier quoted context omitted.

This alternate problem (commonly known as "Monty Fall") has always infuriated me more than the original. I do not believe it is possible that it matters whether the host "knows" or doesn't. The problem asks for the probability of switching resulting in a win, conditioned on the host doesn't reveal the prize . It does not matter mathematically why the host doesn't reveal the prize (i.e., whether P(host doesn't reveal…

> 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 ran…

> 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.

Re: How to explain the Monty Hall problem to a disbeliever

#92

Post by EGPRC » Thu, Mar 16, 2023, 7:24 PM UTC

firstlink wrote: post #91

mcv wrote: post #87

Earlier quoted context omitted.

> 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 ran…

> 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…

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.

Re: How to explain the Monty Hall problem to a disbeliever

#93

Post by firstlink » Thu, Mar 16, 2023, 7:46 PM UTC

EGPRC wrote: post #92

firstlink wrote: post #91

Earlier quoted context omitted.

> 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…

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 th…

> 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.

Re: How to explain the Monty Hall problem to a disbeliever

#94

Post by mcv » Fri, Mar 17, 2023, 12:18 AM UTC

firstlink wrote: post #91

mcv wrote: post #87

Earlier quoted context omitted.

> 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 ran…

> 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…

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.

Re: How to explain the Monty Hall problem to a disbeliever

#95

Post by mcv » Fri, Mar 17, 2023, 12:27 AM UTC

firstlink wrote: post #93

EGPRC wrote: post #92

Earlier quoted context omitted.

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 th…

> 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…

> 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.

Knowledge and intentions don't matter? They most certainly do.

Imagine a third version of the problem (I'm afraid I don't have a good name for it) where the host actively tries to lure you away from the correct answer, and only opens an empty box if you've picked the correct box. Now, if you pick a box and the host opens an empty box, you should absolutely not switch, because the remaining box has 0% of being the right one.

Knowledge and intentions matter. The real problem happens when you don't know what strategy the host is using.