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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)

#291

Post by dekhn »

hulium wrote: post #288
dekhn wrote: post #112

Earlier quoted context omitted.

Yes, I'm literally talking about this sentence: "host, who knows what’s behind the doors, opens another door, say #3, which has a goat." I parsed that as "50% of the time, monty opens another door and it has a car and you win immediately, and 50% of the time, monty opens another door and it has a goat". In retrospect I think my brain just sort of pictured that and proceeded to assume there was no reason to switch, an…

> opens another door, say #3, which has a goat There is no randomness in this sentence. The door has a goat, not a car. Then again, you might read it as an example outcome. But even if the door was chosen randomly, it would not justify the 50%/50% answer. Independently of that, assuming that the host chooses the door with the car and the goat with 50% probability each is the same mistake that confused so many, you ca…

I think you are correct about my thinking: at the time, I assumed it was an example outcome and my brain turned off at that point and simply assumed the probs were still 50-50. In some sense I was both intellectually wrong (in my assumption about it being an example) and intellectually lazy (in my failure to work through the implications of my assumption).

Personally the whole thing taught me is one true hallmark of intelligence is the ability to eliminate unnecessary ambiguity and find the "right" answer.

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

#292

Post by heleninboodler »

HarHarVeryFunny wrote: post #239
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…

I think the problem statement is clear, and is independent of how the TV show actually operated. The problem posed is that you have three closed doors, behind one of which is a car, and behind the other two are goats. You get to pick one of the closed doors, and will win whatever is behind it (you want the car). One of the doors you did not pick is now opened, revealing a goat. You therefore now know the car is eithe…

It's a lot easier to see the 2/3 nature if you imagine that instead of revealing the goat and then asking if you want to switch, Monty said "you've chosen this door that has a 1/3 chance. I'm offering you to switch to choosing both of these two remaining doors, at least one of which definitely has a goat by definition." This is functionally the same offer. It's 2/3 to switch because you're effectively choosing two doors and having Monty automatically grant you the car if it's behind one of them.

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

#293

Post by kragen »

rspeele wrote: post #191
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_…

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 p…

that's all true, but i had concocted explanations that sounded equally convincing to me of why it was wrong

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

#294

Post by jlarocco »

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…

I think the phrase ", say #3," confuses people. Taking it out removes the ambiguity:

"[The] host, who knows what’s behind the doors, opens another door which [...] has a goat."

So he always shows a goat. If he opened the door with the car then the game would be ruined.

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

#295

Post by perrygeo »

Let's build a realistic model of a game show host. What is the causal process at work?

The game show host isn't going to reveal YOUR chosen door - that would ruin the whole game.

Likewise, the host isn't going to reveal the CAR and end the game immediately without a winner - it's bad television.

So the host opens a third class of "irrelevant goat door" in order to offer you the suspenseful proposition that is sure to get millions on the edge of their seat - switch or stick?

In the case of 3 doors there is only 1 irrelevant goat door. In the case of 100 doors, there are 98 irrelevant goat doors, driving the point home further.

We have to assume the host knows which are irrelevant. All irrelevant goats are eliminated. Which means the probability of the car behind your door is 1/n and the probability of car behind the remaining door is always (n-1)/n. Always. Given this assumed behavior of the host.

Knowing the above gives you a clear empirical advantage for modeling and inference. Why would you reject knowledge that may be directly relevant to the question at hand?

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

#296

Post by tshaddox »

jncfhnb wrote: post #215
tshaddox wrote: post #203

Earlier quoted context omitted.

The statement of the problem is the guarantee, just like the statement of the problem is the guarantee that there is always 1 car and 2 goats, that you always get to initially choose 1 door, etc.

Nope! It states you opened a door and then Monty opened a door. It does not state that Monty would have opened a door if you had picked a different one. If this is not communicated you don’t gain any knowledge from his reveal. The typical proper Monty hall formula states this assumption clearly that he will always open a goat door. The original one does not state this.

I don’t disagree that adding “always” to every clause of the problem statement technically makes it less ambiguous.

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

#297

Post by topaz0 »

SeanLuke wrote: post #275
topaz0 wrote: post #95

The explanation in the article is fine as far as it goes, but I think it's much more helpful to emphasize one key fact, which is that MH has to show a goat, and can't open your first choice door, even if it's a goat. That means that if you did pick a goat the first time around, he can only have revealed the other goat, in which case switching will necessarily give you the car. That happens with probability 2/3. This…

> The explanation in the article is fine as far as it goes, but I think it's much more helpful to emphasize one key fact, which is that MH has to show a goat, and can't open your first choice door, even if it's a goat. The second part I agree with, but not the first part; because it sounds like you're saying MH is required by the rules to reveal a goat. But the problem works fine if he always opens a remaining door a…

This has been addressed several times in the comments on this post, and the answer is no, it does matter that he has to select a goat. If he might have selected a car and just got lucky in selecting the goat, then all of the following had equal probability:

1. you picked car first, he picked the first goat

2. you picked car first, he picked the other goat

3. you picked first goat first, he picked the car

4. you picked first goat first, he picked the other goat

5. you picked other goat first, he picked the car

6. you picked other goat first, he picked the first goat

You happen to know that he picked a goat, so you can eliminate 3 and 5, but the remaining four possibilities are still equally likely. Half of them let you win by switching. This is also the "MH stumbles into a door on accident" variant that others have mentioned.

Another way of getting intuition about your version is to realize that, if he might have shown you a car, then the fact that he has opened a goat is evidence against the hypothesis that your first pick was a goat, because half of the times you choose a goat he would open a car, whereas if you chose a car he would have to show you a goat every time. It eliminates some of the ways things could have turned out if you started on a goat without eliminating any of the ways things could have turned out if you started on a car, so the former is relatively less likely in retrospect than it was before MH opened the goat. More obvious in the 100-door case: if your first pick was a goat, then the fact that he hasn't shown you a car after opening 98 doors at random (or equivalently leaving one door closed at random) either means that you were very lucky in the first pick and got the can (in which case it's easy to open 98 doors without a car) or he got very lucky in his random choices. Most games played in this way (98%) would end with MH revealing the car, and the fact that you don't happen to be in one of them only tells you that one or the other of you got very lucky. The probabilities work out that the two options are equally likely.

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

#298

Post by listenallyall »

HarHarVeryFunny wrote: post #239
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…

I think the problem statement is clear, and is independent of how the TV show actually operated. The problem posed is that you have three closed doors, behind one of which is a car, and behind the other two are goats. You get to pick one of the closed doors, and will win whatever is behind it (you want the car). One of the doors you did not pick is now opened, revealing a goat. You therefore now know the car is eithe…

It absolutely is not "independent of how the show operated." The host knowing in advance where the car is, (thereby ensuring the revealed door is always a zonk) is essential. If the host just picked a random door every day, and sometimes did reveal the car prematurely, then it would in fact be a 50-50 decision to switch.

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

#299

Post by tga_d »

hulium wrote: post #290
digitalscribe wrote: post #233

Earlier quoted context omitted.

> Not only is Marilyn's interpretation grammatically valid I would consider both interpretations equally valid. If anything, in law we have the "rule of the last antecedent" (phrase/clause modifies nearest antecedent noun) so that the hater's interpretation could be more correct if based only on this rule. This language is ambiguous: "host ... opens another door, say #3, which has a goat." The two options are Marilyn…

If the the door contains a car, then it doesn't matter whether you switch or not. If it is a goat you get the problem as intended. You still have the intended problem as a subproblem, so this is just a distraction and not a serious objection.

If a goat was revealed randomly instead of intentionally, there is no advantage to switching. See the "Monty Fall Problem" in this paper already linked elsewhere in the discussion: https://web.archive.org/web/20230706235720/https://probabili...

To use the 100 door extension some people find helpful: If Monty always reveals the 98 doors that don't contain goats, then 99% of the time (every time you picked a non-car door), the other door will have the car. If Monty is opening a random set of 98 doors, then 98% of the time Monty will reveal the car, 1% of the time you'll have picked the car on the first guess, and 1% of the time the car is in the other door. When you're in those 2% of cases where no car is revealed, you have a 50/50 shot of being in either of the 1% states where that happens.

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

#300

Post by krackers »

full_monty wrote: post #127
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?

Yes. If Monty opens the door by chance there are 3 equally likely cases: There's a 1/3 chance you picked the car and Monty shows you a goat, 1/3 chance you picked a goat and Monty shows you the other goat, 1/3 chance you picked a goat and Monty shows you a car. So if Monty shows you a goat you have equal probability of being in one of the first two cases. If Monty doesn't open the door by chance then he never shows y…

Yup, the fact that you only have an advantage in switching when Monty "leaks information" in explicitly choosing _not_ to open a certain door as [1] pointed out, is likely the crux of what makes this unintuitive, since it is a very unusual prior.

[1] https://news.ycombinator.com/item?id=39514463

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