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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)

#191

Post by rspeele »

kragen wrote: post #170
Chinjut wrote: post #165

Earlier quoted context omitted.

If you want the result where there is a lower conditional probability of your originally selected door having the prize versus the remaining unopened door, then it completely does matter to have the setup be "Monty always opens a door with a goat" instead of "Monty opens a door at random, and in this particular case, it happened to have a goat behind it".

nope, do the math edit: i ran a monty carlo simulation¹ and i was doing the math wrong. it really does matter if monty knows or not. here's the simulation where he knows: In [15]: non_censored_trials = got_car_trials = 0 In [16]: for trial in range(100_000): ...: car_door = random.randrange(3) # the other two doors have goats ...: your_door = random.randrange(3) ...: monty_door = random.choice(list({0, 1, 2} - {your_…

One way to think about is, suppose you switch -- why not switch back?

In the random-open case, you really know nothing new about either of the closed doors. If you can talk yourself into switching, you could make an equally good argument for switching back.

In the Monty-knows-and-always-shows-goat case, you have gained information about one of the closed doors. You haven't gained any information about your initial pick door. But the other remaining door, you know there's a 2/3rds chance that Monty was forced to avoid it so as not to reveal the car. Only in the 1/3rd case where you were already on the car does Monty have freedom to open either door willy nilly.

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

#192

Post by apetresc »

bazoom42 wrote: post #186
NoboruWataya wrote: post #161

Wow, I knew about the problem and was vaguely aware that it had generated some controversy amongst statisticians but I had no idea about the insanely arrogant and obnoxious (not to mention wrong) abuse Marilyn had received from the "intellectual elite". I have little sympathy for the "ambiguous question" defence. Not only is Marilyn's interpretation grammatically valid, it just wouldn't make sense in the context of t…

> it just wouldn't make sense in the context of the game show for the presenter (who knows where the goats are) to ever open a door to reveal the car and give you the option to switch. None of it makes sense because it is not a real game show, it is a thought experiment.

Apropos the topic of being confidently incorrect, it was indeed a real show: https://en.wikipedia.org/wiki/Let%27s_Make_a_Deal

Hosted by a real Monty Hall: https://en.wikipedia.org/wiki/Monty_Hall

It even had the gag prizes like goats (or, as in the photo from the wiki page, a llama)

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

#193

Post by KrazyButTrue »

jncfhnb wrote: post #115
KrazyButTrue wrote: post #100

Earlier quoted context omitted.

> Nowhere in the problem are Monty's knowledge and motivation stated. "and the host, who knows what’s behind the doors, opens another door [..] which has a goat." The question is clear: The host (1) knows what is behind each door and (2) always shows a goat. It's clearly a determinate problem.

Doesn’t matter. If you don’t have a guarantee that he will always open a goat door, his opening of the goat door doesn’t give you information.

> If you don’t have a guarantee that he will always open a goat door...

Well you do have that guarantee -- as it is stated in the problem. It's clear that no matter what door you open, you will be shown a goat.

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

#194

Post by XCabbage »

NoboruWataya wrote: post #161

Wow, I knew about the problem and was vaguely aware that it had generated some controversy amongst statisticians but I had no idea about the insanely arrogant and obnoxious (not to mention wrong) abuse Marilyn had received from the "intellectual elite". I have little sympathy for the "ambiguous question" defence. Not only is Marilyn's interpretation grammatically valid, it just wouldn't make sense in the context of t…

> it just wouldn't make sense in the context of the game show for the presenter (who knows where the goats are) to ever open a door to reveal the car and give you the option to switch

No, but it would make perfect sense for him to open a door at random and - if doing so reveals the car - tell you sadly that you've lost as a disappointed klaxon plays. Indeed this is almost the only possible way that the initial door opening could realistically make sense in a game show; if it works in any other way, there's no tension when the first door is opened since there's no risk!

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

#195

Post by topaz0 »

kragen wrote: post #170
Chinjut wrote: post #165

Earlier quoted context omitted.

If you want the result where there is a lower conditional probability of your originally selected door having the prize versus the remaining unopened door, then it completely does matter to have the setup be "Monty always opens a door with a goat" instead of "Monty opens a door at random, and in this particular case, it happened to have a goat behind it".

nope, do the math edit: i ran a monty carlo simulation¹ and i was doing the math wrong. it really does matter if monty knows or not. here's the simulation where he knows: In [15]: non_censored_trials = got_car_trials = 0 In [16]: for trial in range(100_000): ...: car_door = random.randrange(3) # the other two doors have goats ...: your_door = random.randrange(3) ...: monty_door = random.choice(list({0, 1, 2} - {your_…

You can tell because only trials where you chose a goat in the first round can be "censored". Your first pick is still twice as likely to be a goat as a car, but half of the times you do choose a goat on round 1 you won't get a chance to switch. Whereas if your first pick was a car, the game is guaranteed to complete (and switching is guaranteed to lose).

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

#196

Post by kragen »

dekhn wrote: post #190
kragen wrote: post #170

Earlier quoted context omitted.

nope, do the math edit: i ran a monty carlo simulation¹ and i was doing the math wrong. it really does matter if monty knows or not. here's the simulation where he knows: In [15]: non_censored_trials = got_car_trials = 0 In [16]: for trial in range(100_000): ...: car_door = random.randrange(3) # the other two doors have goats ...: your_door = random.randrange(3) ...: monty_door = random.choice(list({0, 1, 2} - {your_…

I dunno about anybody else, but to me teaching people how to convert word problems into code/simulations and how to interpret the results is one of the most important things a country that wants to be wealthy and powerful in the future should be doing. I've always admired folks who can do these sorts of things in their head, while it takes me a bunch of time to inspect the code and convince myself it's an accurate re…

in this case I couldn't verify it against an actual experiment, which i think is good practice for newly programmed simulations, but when I saw that removing car_door from monty's choices made the probability go from 50% to 67%, i was reasonably sure that i hadn't fucked up the code, just the stuff i did in my head

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

#197

Post by kragen »

topaz0 wrote: post #195
kragen wrote: post #170

Earlier quoted context omitted.

nope, do the math edit: i ran a monty carlo simulation¹ and i was doing the math wrong. it really does matter if monty knows or not. here's the simulation where he knows: In [15]: non_censored_trials = got_car_trials = 0 In [16]: for trial in range(100_000): ...: car_door = random.randrange(3) # the other two doors have goats ...: your_door = random.randrange(3) ...: monty_door = random.choice(list({0, 1, 2} - {your_…

You can tell because only trials where you chose a goat in the first round can be "censored". Your first pick is still twice as likely to be a goat as a car, but half of the times you do choose a goat on round 1 you won't get a chance to switch. Whereas if your first pick was a car, the game is guaranteed to complete (and switching is guaranteed to lose).

this is an excellent insight; thank you

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

#198

Post by jncfhnb »

KrazyButTrue wrote: post #193
jncfhnb wrote: post #115

Earlier quoted context omitted.

Doesn’t matter. If you don’t have a guarantee that he will always open a goat door, his opening of the goat door doesn’t give you information.

> If you don’t have a guarantee that he will always open a goat door... Well you do have that guarantee -- as it is stated in the problem. It's clear that no matter what door you open, you will be shown a goat.

Cite where it is stated then

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

#199

Post by calderknight »

lamontcg wrote: post #167

The asymmetry is that if you pick the one with the car first (only 1/3rd chance) then Monty Hall can show you either of the other doors. If you pick one without a car first (2/3rd chance) then Monty Hall must show you the door that doesn't have a car and the remaining door always has the car. So 2/3rd of the time you have a 100% chance of a car if you switch, 1/3rd of the time you have 0% chance of a car if you switc…

It's simple. To begin with it's 1/3 your pick is right. You know that a goat will be revealed. So it's absurd to suppose that the fact that a goat is revealed has increased the probability of your initial guess being correct.

Which door is opened to reveal a goat is totally irrelevant, as either would (under the absurd supposition) increase the probability of the initial guess to 50%. In other words, an increase to 50% would be guaranteed from the very beginning. So a contradiction is reached with the obvious fact that initial prob is 1/3.

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

#200

Post by tshaddox »

ordu wrote: post #137
niel wrote: post #86

Earlier quoted context omitted.

I'm trying to understand why the odds would be different if Monty opened empty doors by chance versus on purpose. Does it really change anything?

It changes. Monty's behaviour influenced by his knowledge were the car is. He is leaking information. It is a probabilistic leak: if you first picked by a chance the only door with a car, then Monty is free to open any of remaining doors. But if you had chosen a door with a goat, then Monty has no choice at all, he must open the only door with a goat that you didn't pick. It is a leak. From other hand if Monty picked…

> but he might open a door with the car accidentally.

Yes, but the fact that the problems never mention this possibility makes it pretty clear to me that, from the contestant's perspective, it is guaranteed that this will not happen. The original problem even mentions that Monty Hall knows what's behind the doors, which gives a clear idea of how this guarantee is implemented (versus, say, the contestant's memory being wiped and the game reset every time a car is revealed).

The language of the problem is still ambiguous, of course, because all human language is ambiguous. It could be that the car is a Hot Wheels car and the goat is actually a more valuable prize. We could quibble endlessly about the ambiguity of the problem statement, but I personally find the mathematical problem of the traditional intended interpretation more interesting.

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