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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)

#111
post #86
post #66

Earlier quoted context omitted.

Your formulation is once again ambiguous. Instead of clearly stating that Monty had to open 998 empty doors, you state that he did that. Which could mean he happened to do that. Had to vs happened to makes a big difference. Had he opened them at random and by (extremely small) chance they happened to be empty, the odds are quite different.

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, it does.

You had a 1/1000 chance of picking the right door at once.

Monty had a 1/1000 chance of opening only empty doors.

He had a 998/1000 chance of opening a door to a car, but that didn't happen.

So it's either of the other two cases, with equal probabilities.

If he used his knowledge to never open a car, the 998/1000 case wouldn't exist, and there would be a 999/1000 chance that the door he left unopened had the car.

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

#112
post #101
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…

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, and it wasn't until I read some answers (not her explanation) that I understood I was wrong (and I went through a lot of anger, like many of the commenters, and thought she was very wrong before that).

If there had been exactly one more sentence saying "monty always opens a door showing a goat", I'm pretty sure I would have recognized that. BTW I was familiar with the game (I liked Jeopardy better, as Let's Make a Deal was a fairly dull show) and I don't think they had a game that was identical to the problem as described.

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

#113
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…

I assume that at the time of writing, most readers would have been familiar with the actual game show and the fact that the host always opens a door with a goat?

The show uses a different game.

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

#114
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.

I think I have a simple explanation. There's 1/3 chance of picking the car. 1/3 of the time you pick the car, switch and lose. 2/3 of the time you pick a goat, switch and win. Why? Because 2/3 of the time you picked a goat, Monty shows you the other goat, so if you switch you definitely get the car.

Yes but in fact Monty always shows you a goat. There's always a goat for him to choose regardless of your choice, he knows where it is, and he's not going show you the car or the goat you already picked.

Showing you the goat is the event that, as you say, guarantees that the other door has the opposite of your original choice behind it. Because nobody closed the curtain to shuffle the choices.

One of the things I think people struggle with -- and I struggle with -- is that probability isn't about hypothesising about a single event that happened and how it might have happened. It's about encapsulating all the possible ways a single specified scenario can play out in a single expression.

Monty shows you a goat this time, but this means Monty always shows you a goat. There's no scenario where he is unable to show you a goat. Just like you always only pick one door. And there's always only two goats and one car.

(Full marks for username choice)

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

#115
post #60

As stated, the answer is indeterminate. To demonstrate, consider these scenarios. - Savant's Monty: The host knows where the car is, and wants to prolong the game. Switching wins 2/3 of the time. - Ignorant Monty: The host has no idea where the car is. Revealing the goat was luck and your odds remain 50-50. - Malicious Monty: The host knows where the car is, and wants you to lose. The fact that the host didn't reveal…

> 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.

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

#116
post #39

The intuitive way for me to understand the Monte Hall problem is to pretended there are 1000 doors. You pick 1. There’s a 1 in 1000 chance you get it right. The host then opens 998 doors that don’t have the prize. Do you keep your original or do you switch? Are the odds 50:50?

I never understood why the 1000 door version was more intuitive. Here's how I understand it.

There's 1/3 chance of picking the car. 1/3 of the time you pick the car, switch, and lose. 2/3 of the time you pick a goat, switch, and win. Why? Because 2/3 of the time you picked a goat, Monty shows you the other goat, so if you switch you definitely get the car.

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

#117
post #60

As stated, the answer is indeterminate. To demonstrate, consider these scenarios. - Savant's Monty: The host knows where the car is, and wants to prolong the game. Switching wins 2/3 of the time. - Ignorant Monty: The host has no idea where the car is. Revealing the goat was luck and your odds remain 50-50. - Malicious Monty: The host knows where the car is, and wants you to lose. The fact that the host didn't reveal…

Ignorant Monty can't exist. Monty is the host of a game show, not a contestant. He's not trying to win a prize, and if he opens a prize door, it's a production failure. Whether he knows or not, somebody told him which door to open, and that door does not have the prize behind it.

Malicious Monty, however, can exist. He only ever opens a second door when you've picked correctly, but when you haven't simply accepts your choice.

The problem was stated correctly and clearly, though, as long as you assume that Monty is not a person, has no motivation at all, and as part of a math problem simply wants to reveal a goat after you've made a choice and has the means to do it. That's Savant's Monty: the Monty that doesn't add significant features to the question.

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

#118

Earlier quoted context omitted.

> it could be that the host always shows a goat That’s what the sentence says: “and the host, who knows what's behind the doors, opens another door […] which has a goat”. > or the host shows something at random and we just look at the cases where he shows a goat Nothing says that the host opens a door at random. Quite the contrary, we know that the host has a perfect knowledge of the situation.

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 guaranteed to be behind the other door? A car.

So what are the odds that behind the other door is a car? Two in three.

You should switch.

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

#119

Earlier quoted context omitted.

You don't need to know whether he does or does not do that all the time. Because he did it this time? Amazed this nitpicking is still going on; it's illustrative.

If he only opens a door when you picked the car, then you will lose 100% of the time by switching when he shows a goat. I instead love when people get uppity about others being wrong about Monty hall while still being wrong about Monty hall!

It doesn't matter what he only does, or always does.

It matters what he did in the problem as it is described. Because that is what you're solving.

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

#120
post #106
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…

There are probably a lot of people at this point who have never watched one of these types of game shows. Maybe if they thought about it, they'd realize that MH probably will always show a goat but it wasn't explicit in the initial question that Marilyn answered. Certainly, that constraint is important to the final solution.

I guess my point is that in the inevitable arguments, whether they are in advice columns or HN comment threads or around dinner tables, sometimes seem to their participants to be about how math works, when they're really about the problem description. Sometimes the disagreement amounts to "I agree that those constraints make sense and I wasn't properly accounting for them" and other times it amounts to "I was actually picturing a scenario where MH might have shown you a car". Either is ok, and there might be room for more discussion after getting there, but it feels like a big waste for either of these to get confused with a fruitless argument where both sides think the other is just doing the math wrong.
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