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0 pointsthe_af1y ago0 comments

That they are independent is the only reasonable assumption, everything else is going out of your way to complicate the problem.

If they are NOT independent because Monty only shows you a goat behind the door if you've picked the wrong (or right) door, this is giving the game away. You don't need to guess, you always know what to pick with 100% certainty based on Monty's algorithm.

(Also, the game show doesn't work like this. And the text doesn't mention Monty's motivations, which in standard logic puzzle formulation must mean they are irrelevant, just as the phase of the Moon is also irrelevant and you must not take it into consideration)

If Monty picks randomly instead of always a goat, and he shows a car, the game has ended and no probabilities are involved, because you don't get to switch anymore; you've lost.

If Monty opens a door and there's a goat, we're within the parameters of the problem as stated (and you should switch!).

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alexey-salmin1y ago· 2 in thread

> That they are independent is the only reasonable assumption

No, this is not true. From the mathematical viewpoint Monty can have any strategy as long as it satisfies the problem statement. Which is, he DID open the door for whatever reason, the rest is uncertain. This literally what Diaconis means when he says "the strict argument would be that the question cannot be answered without knowing the motivation of the host" -- yes, in the strict sense he is indeed correct. This thread started because ryao stated that Diaconis is wrong [1].

Now even if you try to play the card of "reasonable assumptions" and rule out "boring" strategies because they are "giving the game away" this still won't eliminate all "non-independent" cases. The space of possible probability distributions here is way bigger than your list above. I can come up with an infinite number of "reasonable non-independent" strategies for Monty.

For example:

1) He rolls a dice before the game in his dressing room, secretly from the audience.

2) If he gets 6: he will open a door if you guessed incorrectly. If you guessed correctly he won't open the door.

3) If he gets 1-5: he will open a door if you guessed correctly. If you guessed incorrectly he won't open the door.

The situation is still the same: you've made your guess, then Monty opened the door with a goat and now you need to figure out whether to switch or not. It matches the problem definition stated above: the door was opened but we don't know why.

Let's see your chances if we assume Monty follow the dice approach:

event A: you've guessed correctly from the first try

event B: Monty opened the door

P(A|B): probability that you've guessed correctly given that Monty opened the door -- if it's less than 50% you should switch

P(A) = 1/3

P(B) = (1/6)x(2/3) + (5/6)x(1/3) = 7/18

P(AB) = (5/6)x(1/3) = 5/18

P(A|B) = P(AB)/P(B) = 5/7

So, in this case Monty doesn't "give up the game" -- there's still a significant random aspect to it. However in this setup for you it's better for you to stay (5/7 of winning) rather than switch (2/7).

[1] https://news.ycombinator.com/item?id=43005371

UPD fixed a typo, it's 5/7 not 5/8

the_afOP1y ago

You're right: I was focused on Monty picking a door with a goat depending on whether you had picked the right door. That would certainly give the game away, but indeed is not the only option.

However,

> Now even if you try to play the card of "reasonable assumptions" and rule out "boring" strategies because they are "giving the game away" this still won't eliminate all "non-independent" cases. The space of possible probability distributions here is way bigger than your list above. I can come up with an infinite number of "reasonable non-independent" strategies for Monty.

None of the assumptions you proceed to list are "reasonable". They introduce enough to the puzzle that they ought to be stated as part of the problem. Since they aren't, it's safe to assume none of those are how Monty picks the door.

Your "dice rolling" formulation of the puzzle is nonstandard. If you want to go with it, you must make it clear in the presentation of the puzzle. There are infinite such considerations; maybe Monty observes the phase of the Moon, maybe Monty likes the contestant, and so on... it wouldn't work as a puzzle!

Given no additional information or context, all we're left with is assuming Monty always opens a door with a goat behind it.

If we want to introduce psychology: I bet you almost all of the naysayers to vos Savant's solution to the puzzle are a posteriori rationalizing their disbelief: they initially disbelieve the solution to the standard puzzle, then when shown it actually works, they stubbornly go "oh, but the problem is underspecified"... trying to salvage their initial skepticism. But that wasn't why they reacted so strongly against it -- it was because their intuition failed them! I cannot prove this, but... I'm almost certain of it. Alas! Unlike with probabilities, there can be no formal proofs of psychological phenomena!

alexey-salmin1y ago

> Given no additional information or context, all we're left with is assuming Monty always opens a door with a goat behind it.

If you're playing against an opponent and trying to devise a winning strategy against him you can't just say "given no additional information or context, all we're left with is assuming his strategy is to always do X" and viola: present a strategy Y that beats X.

In this case X is "always opens a door with a goat behind it" and Y is "always switch doors". This is fascinating but simply incorrect from the math standpoint.

> Your "dice rolling" formulation of the puzzle is nonstandard. If you want to go with it, you must make it clear in the presentation of the puzzle. There are infinite such considerations; maybe Monty observes the phase of the Moon, maybe Monty likes the contestant, and so on... it wouldn't work as a puzzle!

The "dice rolling" it's not a problem formulation, it's one of the solutions to that problem i.e. specific values of X and Y that satisfy all the requirements. I present it to prove that more than one solution exist and furthermore not all solutions have Y="always switch", so you can't establish Y independent of X.

They key difference here is that I don't consider it as a "puzzle", whatever that means. I consider it to be a math problem. Problems of this kind are often encountered in both Game Theory and Probability Theory. It's perfectly fine to reason about your opponents strategies and either try to beat them all or find an equilibrium: this is still math and not psychology.

You can argue that it's a puzzle instead and I don't mind. What I do mind however is saying that Diaconis was wrong. He specifically said "the strict argument would be..." meaning that his conclusions hold when you consider it as a math problem, not as a "puzzle". My whole point is to demonstrate that.

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