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Marilyn vos Savant and the Monty Hall Problem (2015)

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Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#171

Post by bemusedthrow75 »

jncfhnb wrote: post #155
bemusedthrow75 wrote: post #147

Earlier quoted context omitted.

> You’ve incorrectly assumed that I would be showing you a goat in every case. But that is not included in the prompt. Yes. It is. It is a fixed part of the scenario. Monty opens a door and shows you a goat. He knows it is going to be a goat (he is "well-aware of what is going on behind the scenes"). He's showing you a goat as part of the problem which is: should you switch? Again: think through what the problem actu…

No! Read the prompt very carefully. > Imagine that you’re on a television game show and the host presents you with three closed doors. Behind one of them, sits a sparkling, brand-new Lincoln Continental; behind the other two, are smelly old goats. The host implores you to pick a door, and you select door #1. Then, the host, who is well-aware of what’s going on behind the scenes, opens door #3, revealing one of the go…

> You do not know if he would have shown door 3 had you picked door 2! This is NOT in the prompt. You cannot assume that.

Monty isn't a contestant. Monty's action and outcome is part of the fixed description of the problem. He opens one of the doors and reveals a goat. Full stop.

If you think revealing one of the two goats (the other of which might or might not be behind your current selection) isn't information of value in assessing whether your chances in that scenario are improved by switching, I'd encourage you to consider why.

And now I really must leave it to someone else to help you, if they will. But I very much appreciate your politeness.

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#172

Post by bazoom42 »

> host, who knows what’s behind the doors, opens another door, say #3, which has a goat.

There is quite a lot of ambiguitiy in how this is phrased. So the host knows what is behind the doors, but how does this affect his choice to open another door?

Does he always open another door, or does it depend on what is behind the door the contestant selected?

Is the second door selected by random or will it always be one with a goat?

The implicit rule is that the host: 1) always opens one of the other two doors, regardless of what is behind the selected door. 2) deliberately opens one of the two doors which doest have a car behind it.

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#173

Post by juped »

dekhn wrote: post #93

There's the Monty Hall Problem and then there's the Second Order Monty Hall Problem Problem. That is, did the wording as it was originally printed unambiguously define the problem to be solved, or was it ambiguous enough that some people who got it wrong got it wrong for the right reason? Like many others, I completely missed that when Monty opens a door to show you a goat, he always shows you a goat. I think under c…

No, there's not.

There's the Monty Hall problem and then there's the "can I get away with falling back on claiming the question was somehow wrong after embarrassing myself by getting the Monty Hall problem incorrect" problem. No one who claimed it was unclear or ambiguous did so before getting it wrong.

It's really tiresome to have to always pretend to assume good faith when it's really clear what's going on.

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#174

Post by NoboruWataya »

bilsbie wrote: post #73

Has anyone else spent hours on understanding this and just accepted they’ll never accept it? I completely get all the explanations but it just feels too weird.

In addition to the "100 door" formulation that helped me a lot, the way I rationalise it is that:

* When you choose your first door, the chance the car is behind that door is 1/3, for obvious reasons.

* When Monty opens the goat door, nothing actually changes that is relevant to your odds. He is always going to open a goat door, whether your first choice is a car or a goat, so it's basically irrelevant. He hasn't move the car, he hasn't given you any information you didn't already know when you made your first choice. So the odds of your first choice being correct can't have changed. They are still 1/3.

* The only possible outcomes are that your first-chosen door has the car, or the other remaining door has the car. The probability if your first-chosen door having the car is (still) 1/3, and the probabilities of all possible outcomes must add up to 1, so the probability of the other door having the car are 1 - 1/3 = 2/3.

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#175

Post by topaz0 »

TotempaaltJ wrote: post #158
dekhn wrote: post #93

There's the Monty Hall Problem and then there's the Second Order Monty Hall Problem Problem. That is, did the wording as it was originally printed unambiguously define the problem to be solved, or was it ambiguous enough that some people who got it wrong got it wrong for the right reason? Like many others, I completely missed that when Monty opens a door to show you a goat, he always shows you a goat. I think under c…

Vos Savant mentions this, and she's kept track of who did and didn't understand the conditions: > And a very small percentage of readers feel convinced that the furor is resulting from people not realizing that the host is opening a losing door on purpose. (But they haven’t read my mail! The great majority of people understand the conditions perfectly.) https://web.archive.org/web/20181118225305/http://marilynvos...

Nice find! I have wondered this. It's easy to say after you have come around that you had first misunderstood the scenario and done the math right in that context, even if in truth you (implicitly) understood the scenario but just neglected to account for the constraints.

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#176

Post by klyrs »

Swizec wrote: post #162
empyrrhicist wrote: post #154

Earlier quoted context omitted.

Your re-framing is actually not quite right either (or at least, not complete). Key point: if the host picks the car, the game essentially re-sets. Imagine the extreme scenario with 999 goats and 1 car. You select the first door, the host opens 998 doors at random, leaving your selected door and one other. Two scenarios are now possible: 1. There was a car in the 998 doors that got opened. Tough luck, you lose the ga…

Also you're picking between Reward and Bigger Reward. Positive EV no matter your choice so eh don't worry about it too much :) A nice goat fetches up to $1000 to according to a quick google.

A nice goat, you say? How much for a mean goat, jaded from years of being the spoiler prize that nobody wants?

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#177

Post by dunk010 »

bazoom42 wrote: post #172

> host, who knows what’s behind the doors, opens another door, say #3, which has a goat. There is quite a lot of ambiguitiy in how this is phrased. So the host knows what is behind the doors, but how does this affect his choice to open another door? Does he always open another door, or does it depend on what is behind the door the contestant selected? Is the second door selected by random or will it always be one wit…

The key is that if the contestant chose a goat then Monty is FORCED to reveal the other goat. It’s this which skews the odds. He has (often) added information to the game.

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#178

Post by jncfhnb »

bemusedthrow75 wrote: post #171
jncfhnb wrote: post #155

Earlier quoted context omitted.

No! Read the prompt very carefully. > Imagine that you’re on a television game show and the host presents you with three closed doors. Behind one of them, sits a sparkling, brand-new Lincoln Continental; behind the other two, are smelly old goats. The host implores you to pick a door, and you select door #1. Then, the host, who is well-aware of what’s going on behind the scenes, opens door #3, revealing one of the go…

> You do not know if he would have shown door 3 had you picked door 2! This is NOT in the prompt. You cannot assume that. Monty isn't a contestant. Monty's action and outcome is part of the fixed description of the problem. He opens one of the doors and reveals a goat. Full stop. If you think revealing one of the two goats (the other of which might or might not be behind your current selection) isn't information of v…

> Monty isn't a contestant. Monty's action and outcome is part of the fixed description of the problem. He opens one of the doors and reveals a goat. Full stop.

Incorrect and I invite you to cite the line of the prompt that says otherwise. This is the critical missing piece of information that makes it the commonly understood “Monty hall problem” and not the improperly stated prompt that we actually have.

“Monty did this” != “Monty would have done this regardless of your previous choices”.

Unless this is stated, you just have an incomplete problem. If you don’t know the mechanics by which Monty decides to share information, it does not give you probabilistic info.

Probability questions frankly often require you to make judgements about the probable starting conditions based on information given so far. You frequently need to recognize that your current situation is the result of prior processes that are or are not defined.

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#179

Post by kristjansson »

topaz0 wrote: post #168
dekhn wrote: post #93

There's the Monty Hall Problem and then there's the Second Order Monty Hall Problem Problem. That is, did the wording as it was originally printed unambiguously define the problem to be solved, or was it ambiguous enough that some people who got it wrong got it wrong for the right reason? Like many others, I completely missed that when Monty opens a door to show you a goat, he always shows you a goat. I think under c…

What astonishes me is not that people interpret the scenario differently, but that they are so quick to conclude that others are doing the math wrong and feel superior or even angry, rather than thinking about whether you might be talking at cross purposes about different scenarios. So I see the second order problem a bit differently: it's not "was the statement unambiguous", it's about what you do with the (possibil…

Humans don't backtrack well :)

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#180

Post by LegibleCrimson »

bemusedthrow75 wrote: post #118
jncfhnb wrote: post #102

Earlier quoted context omitted.

Incorrect. The host doesn’t need to choose randomly. His purposeful behavior could also be random. The only way you know the switching to be optimal is if it specifically says he picks a goat door to open. If it says he opens a door, and it has a goat, then you’ve learned nothing about the remaining two.

> His purposeful behavior could also be random. It doesn't matter. Because the problem explains what happens: he opens a door and reveals a goat. That is crucial information. There's now only one goat and one car left. But the choices have not been shuffled: you're definitely still pointing at your original choice. What were the odds that your original choice is a goat? Two in three. If you picked a goat, what is gua…

It's actually counter-intuitive. I was going to argue on your side, and then I wrote up a quick program that proved me wrong[0]. Let's go through the scenarios, with Goat A, Goat B, and Car C. In the scenario where Monty picks a door purposefully, always selecting a goat, the scenarios are:

    You picked A, Monty showed you B, and you switch to get C.
    You picked B, Monty showed you A, and you switch to get C.
    You picked C, Monty showed you either A or B, and you switch to get the other goat.
So a 2 / 3 chance of getting the car if you always switch.

If Monty is choosing randomly, we have the following scenarios:

    Initial Choice | Monty's choice | Remaining Door
    A | B | C
    A | C | B
    B | A | C
    B | C | A
    C | A | B
    C | B | A
But we know in the problem statement that Monty hall showed us a goat, so we can eliminate possibilities 2 and 4 to get:

    Initial Choice | Monty's choice | Remaining Door
    A | B | C
    B | A | C
    C | A | B
    C | B | A
Whether you switch or not, you have a 50/50 chance.

I'm not great with probabilities, but the major difference I can see is that in the first scenario, if you pick the car, Monty will either show you the goat A or B with equal probability as a part of the same 1/3 scenario. So you have really:

    1/3: You picked A, Monty showed you B, and you switch to get C.
    1/3: You picked B, Monty showed you A, and you switch to get C.
    1/6: You picked C, Monty showed you A, and you switch to get B.
    1/6: You picked C, Monty showed you B, and you switch to get A.
But in the second scenario, each of those options is actually 1/4, because he was choosing randomly. Most importantly, each option was a 1/6, but two options where you selected a goat were eliminated because those were ones where Monty selected the car.

[0]: https://gist.github.com/Taywee/2ba202b1bf7af40293ecffb01c2ab...

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