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The Time Everyone “Corrected” the World’s Smartest Woman (2015)

priceonomics.com

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Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#191

Post by rfrey »

lachstar-x wrote: post #102
6gvONxR4sf7o wrote: post #13

There are a number of things I like about the monty hall problem. There's the history, the unintuitiveness, the subtle easy-to-screw-up nature of probability problems, the calculation, the sociology, and the overconfidence of wrong experts. Most of all, it's the calculation vs intuition that I like. You can do the calculation, or run simulations and prove correctness. In fact, it would be much harder to be so widely…

Can you help me understand - in my view the choice to switch doors or keep the same door is irrelevant, because even if you keep the same door you're making a choice that is now 2/3 of the right answer. If you switch or keep, you're still choosing from two doors that contain a car and a goat. The other door is no longer relevant and doesn't affect the new state at all. It's your perspective (narrowing the choice down…

The trick is that the host does not pick a door to open at random... If they did there would be a 1/3 probability they opened your door. If they didn't open yours under that circumstance, there would be a 50/50 chance you already picked the good door.

But it's not random... Monty will never open the door you chose. That's what messes up the intuitive probabilities.

Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#192

Post by ledauphin »

stordoff wrote: post #172
lachstar-x wrote: post #102

Earlier quoted context omitted.

Can you help me understand - in my view the choice to switch doors or keep the same door is irrelevant, because even if you keep the same door you're making a choice that is now 2/3 of the right answer. If you switch or keep, you're still choosing from two doors that contain a car and a goat. The other door is no longer relevant and doesn't affect the new state at all. It's your perspective (narrowing the choice down…

When you first pick a door, you have a 1/3 chance of it being the right door. There's a 2/3 chance of it being behind a door you didn't pick. When the host then opens a door, there's still a 2/3 chance that it is behind one of the doors you didn't pick. However, there's now only one door in this set, so there's 2/3 chance that it's behind _that_ door. To look at another way, imagine if the host didn't reveal the cont…

I've seen this explained ten different ways but this is the way I'm going to explain it from now on. thank you.

Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#193

Post by stordoff »

bnkamalesh wrote: post #155

I'm no way an expert, so even if Ms. Savant beautifully explained it; my puny brain thinks... "how is it different from starting the game with 2 doors?". If you choose again, how is the probability not 1/2

Because you aren't starting from scratch - that you had a 1/3 chance to begin with is important. Say there are three doors, A B and C. You pick A:

P(A wins) = 1/3

P(A loses) = P(B _or_ C wins) = 2/3

The host then reveals that there was a goat behind door B. This doesn't change the state of anything (there is always a losing door in the two you didn't pick, and he is always choosing to show that one, and none of the items move). This means the probabilities remain as they were above. However, we know that B didn't win, so we can simplify it to:

P(A wins) = 1/3

P(A loses) = P(C wins) = 2/3

Therefore, if you switch to door C, you have a 2/3 chance of winning, rather than 1/3.

The only way the probabilities would go back to 1/2 for the second choice is if the prize and the goat were shuffled after B is revealed. However, they are not, so the chance that you picked the right door initially is fixed when you picked it.

To think about it another way, your initial choice of A means there's a 1/3 chance it is in A, and a 2/3 chance it is not and is behind one of the other doors. By ruling out B, we don't change the 1/3 chance that it was initially behind A. That means when we are asked again, there is still a 1/3 chance it was behind A, and a 2/3 chance that it was not. However, there's now only one thing that is not A, so there is a 2/3 chance it is behind door C.

Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#195

Post by FabHK »

mkl wrote: post #185
listenallyall wrote: post #166

Earlier quoted context omitted.

You're quoting Priceonomic's 2021 description of the problem. This is NOT the 1990 Parade Magazine description of the problem that generated all the responses.

The 1990 Parade Magazine description is almost identical (and actually more explicit, since "say #3" is a removable parenthetical): "the host, who knows what’s behind the doors, opens another door, say #3, which has a goat" -- https://web.archive.org/web/20130121183432/http://marilynvos... The host's knowledge is explicitly mentioned, and the only purpose this could have is that so he can use it to avoid giving the g…

That is literally not explicit (where by "literally" I mean literally, not figuratively, and by "explicit", I mean explicit, not implicit). It is sort of hinted at, but it is not explicitly said that the host will mechanically reveal a door that has a goat. It is quite conceivable that the host picks a door randomly, or in fact that he picks the door with the car with a certain probability (saving the show quite some money).

Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#196

Post by admax88q »

lachstar-x wrote: post #102
6gvONxR4sf7o wrote: post #13

There are a number of things I like about the monty hall problem. There's the history, the unintuitiveness, the subtle easy-to-screw-up nature of probability problems, the calculation, the sociology, and the overconfidence of wrong experts. Most of all, it's the calculation vs intuition that I like. You can do the calculation, or run simulations and prove correctness. In fact, it would be much harder to be so widely…

Can you help me understand - in my view the choice to switch doors or keep the same door is irrelevant, because even if you keep the same door you're making a choice that is now 2/3 of the right answer. If you switch or keep, you're still choosing from two doors that contain a car and a goat. The other door is no longer relevant and doesn't affect the new state at all. It's your perspective (narrowing the choice down…

Here's another thought experiment.

Forget the opening of the door.

Start with picking a door at random, you have a 1/3 chance of having picked the car. On that I think we all agree.

Now let's say that the host offers to let you switch from the door you picked, to the other _two_ doors combined. He hasn't opened any doors, they are all closed, you're allowed to stick with your initial guess of one door, or switch to a combined guess of the other two doors.

If that's the case it should be fairly obvious that you have a 2/3 chance of getting the car by switching to the combined 2 doors.

Now that you've switched, would it really make a difference to your odds if the host opens one of your doors to reveal a goat? Would that lower your probably to 1/2 or would it remain 2/3?

Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#197

Post by TheDong »

tmule wrote: post #141

There’s a bit of historical revisionism at play in the article. Here’s an accurate representation by Aaron Brown of what transpired: https://www.quora.com/Why-do-some-PhDs-argue-against-Marilyn... Quoting for those who don’t want to go off-site: “You have to have lived through this ancient travesty to care about it. It’s probably best forgotten. Marilyn Vos Savant published an incorrect answer to the Monte Hall probl…

That quora answer does not quote the actual question she received. This archive claims to have the actual question: https://web.archive.org/web/20130121183432/http://marilynvos...

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

The question, if that was indeed the formulation, very explicitly does state that the host both knows what's behind the doors, and uses that knowledge to show a goat.

Unless someone provides evidence of more ambiguity in the original question, I'm going to have to trust Marilyn's website to be quoting the question correctly... And the question seems to make the assumption very clear.

Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#198

Post by jarpadat »

lachstar-x wrote: post #167
zeroimpl wrote: post #122

Earlier quoted context omitted.

The other door is not irrelevant. When you first picked a door, there was a 33% chance it was right, and a 67% chance one of the other two doors was right. Once the other door got opened, it is still a 67% chance that the other two doors is right, but you now know which one of those two it would be - the one which wasn’t opened.

Thats what I'm stuck on, I suppose. Imagine the entire situation is reversed. The Reverse Monty Hall problem. I'm given a choice between two doors, once of which contains a car and one contains a goat. I have to choose, 1 or 2. Then the game show host reveals that there was also a third door, which contained a goat, which is no longer relevant and never was relevant. I'm then also asked to choose a door (which is als…

I think it boils down to the fact that time isn’t reversible, so the “reverse problem” isn’t like the “forward” version.

Let’s take some new problems. Suppose you have 100 doors and no switching. Your probability is 1/100, even if the host later opens a goat door, so in that sense the new information is irrelevant. But if we “reverse it” and the host opens the door first and you guess second, your probability improves to 1/99. So now suddenly the same information is relevant. Two things to observe here, one is that the forward and reverse problems are different, the other is that the relevance or irrelevance of the information depends on the direction of time. If you learn the information before you act it is relevant, afterward it is irrelevant.

One way to think about Monty Hall is you’re deciding which of these games to play. If you will stick with your first decision, you are sorta turning it into the toy problem above, where you decide the door first and then the goat information is irrelevant. Vs if you will switch, the goat door is opened before you decide, which is relevant.

Another way to think about it is with two contestants. Let’s say I pick the door initially, then someone opens the goat door, and finally you decide whether to switch. In this scenario, you don’t have self-preference bias to stick with “my” original door. In fact, my decision was the irrelevant information. It doesn’t matter at all what door I picked, what matters is whether you pick the right door, and involving me at all is a kind of misdirection to anchor you to the 1/3 probability.

Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#200

Post by nonpolitical »

javajosh wrote: post #23

So this is at least in part about the "Monty Hall Problem" and why it's solution not intuitive. The article missed an important angle: when the host opens a door, he's giving you more information , which explains why it's better to switch. If you're the host, you need to know which door the car is behind to do your job 2/3 of the time, to avoid revealing it. It's this quality of unexpected information exchange that I…

This was one of the explanations that I find elucidating as well, but just now I realize that the change from 1/3 as an estimate to 1/2 is exactly the result of absorbing new information, which makes this explanation unsatisfactory.
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