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I'm not sure I agree. In particular: > If it was reworded as "the host who knows what's behind the doors and _always_ opens the door with a goat and _never_ o

by Lazare 9y ago

I'm not sure I agree. In particular: > If it was reworded as "the host who knows what's behind the doors and _always_ opens the door with a goat and _never_ opens the door with the car", the mathematicians would have seen the 2/3rds probability. There seems to be extensive examples of mathematicians encountering very clearly worded descriptions of the problem, but still failing to jump to the right answer. > The original wording with the nonchalant "say" in the middle of ", opens another door, say No. 3, which has a goat." makes it seems like there was no link between the host's knowledge of what's behind the door and the opening of door #3 and therefore random. Therefore the 50/50 probability seemed to be the correct answer. Yeah, but it's not 50/50. I think you're focusing on the wrong part there. The issue isn't the host's knowledge, it's the question of whether he always opens a door. If he always opens a door at random, and if in this particular run through he happens to open a door with a goat, then you will still double your odds if you switch. The host doesn't need to know what's behind the door for this to work! What does need to be the case is that the host always opens a door (or chooses whether or not to open a door for reasons other than his knowledge of what's behind them). In a hypothetical variant where Monty only opens a door and offers you the chance to switch if he knows you picked a car, you should never switch. In a variant where he only does so if he knows you didn't, you always should. :)

jasode 9y ago
>, and if in this particular run through he happens to open a door with a goat, then you will still double your odds if you switch. Yes, I agree. >The host doesn't need to know what's behind the door for this to work! The host has to know it and always open the goat door to prevent mathematicians from making a repeated simulation in their minds where the host sometimes opens a door with a car behind it. That would be over-reading and misinterpreting the MH wording but that's why the problem is so enduring -- it uses clever text to misdirect a lot of smart people away from the correct 2/3rds answer.
- deleted 9y ago
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- yen223 9y ago
  Weirdly enough, in the scenario where the host doesn't know which door has the goat and just happens to open the right door, your odds of winning the car if you switch actually does go back to 50-50. Edit: I made a quick IPython notebook to demonstrate this: https://github.com/yen223/monty_fall/blob/master/Monty%20Hall%20vs%20Monty%20Fall.ipynb https://github.com/yen223/monty_fall/blob/master/Monty%20Hal...
  - mannykannot 9y ago
    In your program, you have to classify all cases where the host picks the prize on his second selection as 'passed', and put them aside. This affects the result, as in all those cases the subject had picked one of the wrong boxes, so you have biased the cases you actually consider towards those where the subject originally picked the right box.
- Scarblac 9y ago
  In the case where he just opens random doors, it doesn't matter that you switch, your door and the other unopened have the same chance. After all, it'd be the same if you chose the door to be opened. If that mattered, you should always mentally choose a door, choose another to randomly open, switch and make that your choice. That's absurd. The whole thing is that be cause the host has to open a door with a goat, switching means you basically get to choose both other doors.
- dalbasal 9y ago
  I think if the mathematician (or anyone) gets to that point, he's solved the problem even if the answer is in the form of a probing question. There might be confusion about whether or not the host "cheats" and always shows a goat, but if you're asking this question you get to the answer. The issue is that people don't get to this question. I think that's not a wording-misdirection thing, it's something about how people think.
- deleted 9y ago
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- mannykannot 9y ago
  To be clear, as we are all in agreement that the correct answer is the same whether the question states the host has knowledge or not, there is no doubt here that the question, exactly as stated, is unambiguous and has one correct answer. If someone chooses to use faulty reasoning in solving it, they have made a mistake as surely as if they had made an arithmetical error in calculating the odds incorrectly. Mathematicians and philosophers, in particular, are supposed to know how to be rigorous. I think this has wandered across the threshold into "if they were asked a different question, they would have given a different answer", which might be true, and, if tested experimentally, might throw some light on why this problem seems unreasonably baffling, though it would be hard to find sufficient suitable candidates for this experiment (maybe among schoolchildren who have shown they have a good grasp of the relevant statistics). Personally, I would guess that people are led down the wrong path by causal intuition - i.e. "opening the second box makes no difference to where the prize is", or perhaps by an intuition that the choice to switch or not is as random a choice (i.e. over uniform odds) as their initial pick. I realize that at this point, these guesses have no more evidence for them than does the hypothesis that people are being misled by language clues.
Lazare 9y ago
To late to edit, but: I was incorrect to assert that you should still switch if the host opens the door at random. My bad. (A good reminder of how hard it is to reason about this type of stats I think...)
kmm 9y ago
> Yeah, but it's not 50/50. I think you're focusing on the wrong part there. The issue isn't the host's knowledge, it's the question of whether he always opens a door. If he always opens a door at random, and if in this particular run through he happens to open a door with a goat, then you will still double your odds if you switch. The host doesn't need to know what's behind the door for this to work! No, if the host always opens a door, but has no clue what's behind them, you chances will not go up by switching. That's the central clue of the Monty Hall problem, which really helped me to grok why the odds go up by switching. By opening a door, a Monty Hall who knows where everything is, is giving you information in a way that a random choice cannot do. Let's assume you always pick door 1. Then there are 6 equiprobable possible universes: 1. Prize behind door 1, Monty opens door 2 2. Prize behind door 1, Monty opens door 3 3. Prize behind door 2, Monty opens door 2 4. Prize behind door 2, Monty opens door 3 5. Prize behind door 3, Monty opens door 2 6. Prize behind door 3, Monty opens door 3 If no prize appears behind the door he opens, we're sure we're not in universes 3 and 6. That leaves behind 4 universes, in 2 of which we'll win the prize if we switch, so a 50/50 chance. What's the difference between this version and the normal Monty Hall? In the usual version, universes 3 and 6 never existed in the first place, and universes 1 and 2 are the same. Whether he opens the 2nd or the 3rd door depends on the protocol he uses, but is irrelevant.
- deleted 9y ago
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- marvelous 9y ago
  The 6 universes are not equiprobable. 1+2, 3+4 and 5+6 are equiprobable (1/3 each). 1 and 2 are equiprobable (1/6 each), 3 and 6 are impossible (0 each), 4 and 5 are equiprobable (1/3 each). You should switch in 2/3 of the cases.

by Lazare 9y ago

jasode 9y ago

>, and if in this particular run through he happens to open a door with a goat, then you will still double your odds if you switch. Yes, I agree. >The host doesn't need to know what's behind the door for this to work! The host has to know it and always open the goat door to prevent mathematicians from making a repeated simulation in their minds where the host sometimes opens a door with a car behind it. That would be over-reading and misinterpreting the MH wording but that's why the problem is so enduring -- it uses clever text to misdirect a lot of smart people away from the correct 2/3rds answer.

deleted 9y ago
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yen223 9y ago
Weirdly enough, in the scenario where the host doesn't know which door has the goat and just happens to open the right door, your odds of winning the car if you switch actually does go back to 50-50. Edit: I made a quick IPython notebook to demonstrate this: https://github.com/yen223/monty_fall/blob/master/Monty%20Hall%20vs%20Monty%20Fall.ipynb https://github.com/yen223/monty_fall/blob/master/Monty%20Hal...
- mannykannot 9y ago
  In your program, you have to classify all cases where the host picks the prize on his second selection as 'passed', and put them aside. This affects the result, as in all those cases the subject had picked one of the wrong boxes, so you have biased the cases you actually consider towards those where the subject originally picked the right box.
Scarblac 9y ago
In the case where he just opens random doors, it doesn't matter that you switch, your door and the other unopened have the same chance. After all, it'd be the same if you chose the door to be opened. If that mattered, you should always mentally choose a door, choose another to randomly open, switch and make that your choice. That's absurd. The whole thing is that be cause the host has to open a door with a goat, switching means you basically get to choose both other doors.
dalbasal 9y ago
I think if the mathematician (or anyone) gets to that point, he's solved the problem even if the answer is in the form of a probing question. There might be confusion about whether or not the host "cheats" and always shows a goat, but if you're asking this question you get to the answer. The issue is that people don't get to this question. I think that's not a wording-misdirection thing, it's something about how people think.
deleted 9y ago
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mannykannot 9y ago
To be clear, as we are all in agreement that the correct answer is the same whether the question states the host has knowledge or not, there is no doubt here that the question, exactly as stated, is unambiguous and has one correct answer. If someone chooses to use faulty reasoning in solving it, they have made a mistake as surely as if they had made an arithmetical error in calculating the odds incorrectly. Mathematicians and philosophers, in particular, are supposed to know how to be rigorous. I think this has wandered across the threshold into "if they were asked a different question, they would have given a different answer", which might be true, and, if tested experimentally, might throw some light on why this problem seems unreasonably baffling, though it would be hard to find sufficient suitable candidates for this experiment (maybe among schoolchildren who have shown they have a good grasp of the relevant statistics). Personally, I would guess that people are led down the wrong path by causal intuition - i.e. "opening the second box makes no difference to where the prize is", or perhaps by an intuition that the choice to switch or not is as random a choice (i.e. over uniform odds) as their initial pick. I realize that at this point, these guesses have no more evidence for them than does the hypothesis that people are being misled by language clues.

Lazare 9y ago

To late to edit, but: I was incorrect to assert that you should still switch if the host opens the door at random. My bad. (A good reminder of how hard it is to reason about this type of stats I think...)

kmm 9y ago

> Yeah, but it's not 50/50. I think you're focusing on the wrong part there. The issue isn't the host's knowledge, it's the question of whether he always opens a door. If he always opens a door at random, and if in this particular run through he happens to open a door with a goat, then you will still double your odds if you switch. The host doesn't need to know what's behind the door for this to work! No, if the host always opens a door, but has no clue what's behind them, you chances will not go up by switching. That's the central clue of the Monty Hall problem, which really helped me to grok why the odds go up by switching. By opening a door, a Monty Hall who knows where everything is, is giving you information in a way that a random choice cannot do. Let's assume you always pick door 1. Then there are 6 equiprobable possible universes: 1. Prize behind door 1, Monty opens door 2 2. Prize behind door 1, Monty opens door 3 3. Prize behind door 2, Monty opens door 2 4. Prize behind door 2, Monty opens door 3 5. Prize behind door 3, Monty opens door 2 6. Prize behind door 3, Monty opens door 3 If no prize appears behind the door he opens, we're sure we're not in universes 3 and 6. That leaves behind 4 universes, in 2 of which we'll win the prize if we switch, so a 50/50 chance. What's the difference between this version and the normal Monty Hall? In the usual version, universes 3 and 6 never existed in the first place, and universes 1 and 2 are the same. Whether he opens the 2nd or the 3rd door depends on the protocol he uses, but is irrelevant.

deleted 9y ago
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marvelous 9y ago
The 6 universes are not equiprobable. 1+2, 3+4 and 5+6 are equiprobable (1/3 each). 1 and 2 are equiprobable (1/6 each), 3 and 6 are impossible (0 each), 4 and 5 are equiprobable (1/3 each). You should switch in 2/3 of the cases.