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The Time Everyone “Corrected” the World’s Smartest Woman (2015)
161–170 of 331 posts
Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)
#162So 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…
When you choose the first door, you are partitioning the "board" into two parts: the door you chose, and "everything else". By opening a door, Monty lets you cover 100% of the "everything else" partition using only one guess. So now you get to choose between partition 1, which covers 1/3 of the board, and partition 2, which covers 2/3 of the board.
I think the confusing part for many people is the unstated fact that the host takes all prizes into account when revealing a door, and that the host is forbidden from revealing the car. If the host chooses randomly then his pick adds no information (other than the extra revealed prize)
Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)
#163I still don't get it. 1. Three doors. 2. Two doors have a goat, and one has a car. 3. The contestant has a 1/3 chance of winning the car? 4. The contestant loses. 5. New game, and odds? 6. There's a 50/50 chance of winning? (I'm assuming Monte Hall has no clue to where the car is. He is just opening doors.) 7. Could someone explain it to me, and thanks in advance. (Off topic but a fawn had two babes in my back yard.…
You have three doors. As you note, you have a 1/3 chance of winning.
Monty now reveals a losing door. At this point, your door has a 1/3 chance of still winning - the probability of that choice can’t change. However, as we now know one door has a 0/3 chance of winning (it’s been revealed) the remaining door must have (1-1/3) chance of winning. Thus, the remaining door has a 2/3 chance of winning.
Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)
#164Earlier quoted context omitted.
It does change your odds -- you're now living in a world in which the car didn't get revealed, and by Bayesian reasoning that means it's more likely that you live in a world where you picked the correct door.
I didn't state that it doesn't change your odds. Obviously going from 33% to 0% because the car isn't possible to win anymore will be a change of your odds. I stated it doesn't change anything, because in one situation you go from 33% to 66% by changing and in the other you go from 0% to 0% by changing. So there is no difference to the logic of always switching because it either improves your odds or keeps them the s…
If the host reveals a goat at random rather than by special knowledge, your odds don't go from 33% to 66% by switching. The case where the host reveals a goat is only two out of three, not three out of three because one third of the time, the host will reveal the car. In one of those two cases, you picked a goat and in the other you picked a car. So you're at 50% odds whether you switch or not, if the host doesn't know where the car is.
If the host does know, then there are three out of three cases where the host reveals a goat. In one out of three cases you picked the car but in the other two cases you picked the goat. So that's why your odds go up if you switch.
Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)
#165There 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…
* Most of all, it's the calculation vs intuition that I like. You can do the calculation, or run simulations and prove correctness.* It puzzles me that so many smart people got it wrong. It’s so easy to check your working. When I first heard of this problem, I got it wrong too. Then I was told the answer, and to prove it to myself, it’s trivial to list the scenarios and simulate on paper. It’s still uncomfortable to…
I think people didn't check because the answer of 50% is so strikingly obvious (despite being wrong) that it seems like there's no work to check.
Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)
#166Earlier quoted context omitted.
Indeed, that is a crucial bit of information that is not made explicit in the original framing of the problem.
I think it's made pretty explicit. In this version: "Then, the host, who is well-aware of what’s going on behind the scenes, opens door #3, revealing one of the goats." The host has to open a door that doesn't show the car.
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.
Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)
#167Earlier 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…
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 also irrelevant since the problem is backwards and the supposed aim is to get the car).
Even if that last step repeats 1000 times with 1000 doors and 1 car, each removing a goat-door, the only relevant choice is still the first one as the host appears to be adding new information, but it's always irrelevant information as a new choice is always made at the end.
Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)
#168Earlier quoted context omitted.
The presentation of the problem included the phrase "the host, who knows what's behind the doors, opens another door", so I think in that case the interpretation that Monty deliberately reveals a goat is more appropriate (in addition, obviously, to being what was intended), although I agree there is wiggle room enough to admit both possibilities and clarity about which case you are discussing is important as it chang…
> as it changes the answer. It doesn't. If the host randomly reveals a car, then you have 0% chance to win. If he doesn't you have 66% chance to win by switching.
I once thought that had to be the case. I wrote the simulation to demonstrate that I was correct. I learned that I wasn't. I encourage you to do likewise.
I think I now have an understanding of why the right answer is the right answer, but I thought I had that before; I am more confident in the answer than my reasoning.
Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)
#169There 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…
Maybe someone can help me with this -- what if for round two, you flipped a coin: heads you pick the door you picked previously, tails you pick the other door. Does that change things at all, since you're "switching" each time?
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Edited -- Actually @haberman's comment that "By opening a door, Monty lets you cover 100% of the "everything else" makes the most sense to me as I think about this. Imagine there is no Monty hall, just three doors, and I say "you can choose one door and you win if there's a car behind it, or choose two doors and see if there's a car behind it." Clearly you're better off choosing two doors. And that's in fact, functionally what happens by switching after a door is eliminated.
Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)
#170Earlier quoted context omitted.
> as it changes the answer. It doesn't. If the host randomly reveals a car, then you have 0% chance to win. If he doesn't you have 66% chance to win by switching.
I once thought that had to be the case. I wrote the simulation to demonstrate that I was correct. I learned that I wasn't. I encourage you to do likewise. I think I now have an understanding of why the right answer is the right answer, but I thought I had that before; I am more confident in the answer than my reasoning.
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