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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)

#281

Post by HarHarVeryFunny »

petesergeant wrote: post #273
HarHarVeryFunny wrote: post #239

Earlier quoted context omitted.

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…

> The simplest way to explain why switching is the correct strategy In my experience the most _intuitive_ explanation is to simply ramp it up to 100 doors, with Monty opening 98 of them, to make it clear that switching offers you the benefit of all unopened doors.

Maybe different explanations resonate with different people.

Another explanation:

After initial selection there is 1/3 chance car is behind your door, and 2/3 chance it is behind one of the other two doors.

When one of the other doors is opened revealing a goat, we NOW know (new information to be taken advantage of!) the 2/3 chance represented by those two other doors lies with the unopened one.

So, your choice is stick with original choice which has 1/3 chance of being correct, or switching to the other door which you now know has a 2/3 chance of being correct! Obviously you switch.

The big lie to "with two doors left it's 50/50" is that opening doors makes no difference. You're original choice had 1/3 chance of being correct, and you can't go back in time and change that.

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

#282

Post by [deleted] »

calderknight wrote: post #205
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…

I don't think your interpretation of the sentence is sensible. The sentence mentions that the host knows what's behind the doors. So, if he is allowed to open the door with the car, the problem would become insoluble and would just be about speculating on the host's personality. And it definitely doesn't support the conclusion that the probabilities become 50/50.

[deleted]

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

#283

Post by calderknight »

kragen wrote: post #276
calderknight wrote: post #220

Earlier quoted context omitted.

If the format of the game allows him to show the car to the player, how could his personality not enter into it? In every case where the player picks a goat-door, the host will be presented the option to either reveal the car or the goat. I mean, one can imagine various complicated scenarios in which the host might reveal the car exactly 50% of the time in such cases, but none seem like they can be reasonably arrived…

it turns out that there is in fact no way that his personality could not enter into it, but that was not obvious to me until i did the simulation. even if he chose to reveal the car exactly 50% of the time in such cases, that would be a result of his personality, wouldn't it?

You can imagine factors other than his personality, but they're all equally as speculative. Additional rules to the game, for example.

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

#284

Post by calderknight »

[deleted] wrote: post #282
calderknight wrote: post #205

Earlier quoted context omitted.

I don't think your interpretation of the sentence is sensible. The sentence mentions that the host knows what's behind the doors. So, if he is allowed to open the door with the car, the problem would become insoluble and would just be about speculating on the host's personality. And it definitely doesn't support the conclusion that the probabilities become 50/50.

[deleted]

Not logical. It doesn't follow from the assumption that the host would not pose the question "Hey, here's the car, wanna switch?" that the contestant would automatically lose. You could just as well speculate that if the host opened the door to the car the contestant would automatically win the car.

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

#285

Post by grogers »

harshreality wrote: post #249
lucas_membrane wrote: post #226

Earlier quoted context omitted.

>> The game would be reduced to "does Monty feel like gifting you a car or a goat today". But, in fact, that is the way the real show worked. He was more likely to give you a car when he hadn't given away any cars for a while. And Marilyn's description of the problem included no information about whether or not Monty offered to swap doors rarely, frequently, or always, or about the previously observed outcomes depend…

If you're playing the game as stated, and if the host opens a door revealing a goat, you must switch for the best odds. The offer to switch doors is part of the problem statement. You can't justify your decision based on the possibility in the real game that you might not get the offer to switch (which would mean that when you do it might be an intentional red herring). Even if there's a 50% chance there are all goat…

The person you are responding to is pointing out that if the host doesn't offer you to switch all the time, then it isn't the best strategy to always switch when you are given the option. For example, knowing that everyone switches when given the option, the host might only offer to switch when you've picked the correct door. In that case, not switching would be better. Hence the poker aspect.

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

#286

Post by asdfasvea »

Jeez, this thread should have it's own thread where we discuss it:

"Marilyn vos Savant and the Monty Hall Problem (2015) Problem (2024)"

That thread wouldn't be about the problem, or about the decades ago backlash to the correct answer about the problem, but about the backlash to how everyone in 2024 semantically picked apart the problem.

I lived though the initial backlash and it was annoyingly entertaining how high level mathematicians tried to show how she was wrong. Now everyone is trying to use semantics to show how they are right--or at least not wrong.

I don't know which is worse--pompous incorrectness (then) or pedantic semantics (now).

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

#287

Post by rspeele »

bemusedthrow75 wrote: post #260
rspeele wrote: post #241

Earlier quoted context omitted.

> It doesn't make a difference what causes Monty to reveal a goat. Oh, but it does! See the "Monty Fall" version of the problem, in which Monty accidentally trips and opens a door, which just happens to reveal a goat. In this variant there is no advantage gained by switching, because no more information was revealed about the remaining unopened door. The information gain only happens in the original game because we k…

Nope. But I am going to leave this to someone else to explain. I'm tired out now.

They are two different problems. Map out all the scenarios exhaustively and you'll find the difference.

In both cases we were originally 1/3 chance of being right. That is not in dispute.

In the original (fully defined) "Monty Hall", Monty was going to show us a goat no matter what. It's part of the rules, he has to show a goat. So the fact that we see a goat behind the revealed door is no surprise, and no new information. But which of the two unchosen doors was the goat, is valuable information because in 2/3s of the scenarios Monty's hand was tied and he HAD to show that door to avoid revealing the remaining car.

In the "Monty Fall" problem, the fact that we see a goat at all is interesting information. This becomes more likely when we picked the car in the first place, because if we had initially picked the car, and a random other door is opened, it's 100% going to be a goat, whereas if we had picked the goat in the first place, we are only 50% likely to see a goat when a random other door is opened. Let's call the goats Alice and Bob to illustrate this point. We know we DID see a goat, but we don't know which of these equally probable scenarios led to that:

1. We picked the car and saw Alice

2. We picked the car and saw Bob

3. We picked Alice and saw Bob

4. We picked Bob and saw Alice

Notice how "we picked the car" originally had 1/3 odds but represents half the scenarios that remain possible, because there are two ways to see a goat with that start, while only one way to see a goat with the others.

This kind of brings the problem back around to similar territory as Bertrand's Box[0] where the fact that you drew a gold coin is already hinting to you that you're more likely on the "both gold" box than on the "half gold" box.

[0]https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox

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

#288

Post by hulium »

dekhn wrote: post #112
ghaff wrote: post #101

Earlier quoted context omitted.

I think this is the original "Ask Marilyn": Suppose you’re on a game show, and you’re given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors? Craig F. Whitaker Columbia…

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 cannot always assume uniform probability only because there are 2 options.

(I wanted to point out the flaws in your thinking because this has also been a confusing problem for me)

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

#289

Post by jncfhnb »

zachm0 wrote: post #280
jncfhnb wrote: post #267

Earlier quoted context omitted.

It’s still wrong. He needs to declare (or at least he needs to consistently) open an empty box; not any random box or a box of his choosing at his whim. If he opens a random box; you have not learned anything about the keys GIVEN he opened an empty box This paper adds that in as an assumption after the prompt, which I’m pretty sure is not the original prompt

If I understand correctly, you’re saying Monte’s intention (randomly picking an empty box vs purposely picking an empty box) is effecting the odds that the box in hand has keys? Also, do you have any evidence that this isn’t the original?

> If I understand correctly, you’re saying Monte’s intention (randomly picking an empty box vs purposely picking an empty box) is effecting the odds that the box in hand has keys?

Sort of. If you’re saying the next box is chosen at random then there are 2 of 6 possible end games in which the key is chosen; 2 of 6 in which you pick an empty box and can switch for the keys; and 2 of 6 that both of the remaining are empty.

Since the prompt says that you did in fact open an empty box, that removes the 2 where you open the keys. So it’s 50/50.

When you know for a fact that the keys will never be chosen, the probability of picking an empty box when you chose an empty box goes from 50% to 100%. Meaning it now occupies twice the probability space. That’s now it’s 2/3 chance of winning.

You truly learn nothing if Monte randomly opens one of the doors and it is not the keys.

> Also, do you have any evidence that this isn’t the original?

The quote in the article?

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

#290

Post by hulium »

digitalscribe wrote: post #233
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…

> 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.

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