5 ms·

Since a lot of university mathematicians, engineers, and physicists famously got the answer wrong, I've classified The Monty Hall Problem as more of an "English

by jasode 9y ago

Since a lot of university mathematicians, engineers, and physicists famously got the answer wrong, I've classified The Monty Hall Problem as more of an "English parsing" problem instead of a math problem. From wikipedia[1], the problem as stated: >"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 No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?" The key phrase that everyone glosses over is: ", and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat." 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. 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. [1] https://en.wikipedia.org/wiki/Monty_Hall_problem https://en.wikipedia.org/wiki/Monty_Hall_problem

OscarCunningham 9y ago
I agree that the wording of the problem is important, but I'm not sure if that's what was causing the mathematicians to get the wrong answer. I think I would have naturally assumed that the host was opening a door without the car in order to raise tension, and that he would therefore never open the door with the car. It would be interesting to survey mathematicians, but I don't think there's enough of them who haven't already heard the problem.
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.
chki 9y ago
I think that is a bit far-fetched. In the context of a game show it would be strange to think that the host just makes a random choice. Also, why would there be the additional information that the host knows what's behind the doors if he just chooses randomly.
- jasode 9y ago
  >In the context of a game show it would be strange to think that the host just makes a random choice. A lot of people reading the MH problem were unfamiliar with the actual game show and therefore parsed it as an isolated word problem.
- lancebeet 9y ago
  Since the game show stands to lose by you getting the car, it's not an unreasonable assumption that you would only be given the choice to switch had you picked the car in the first place, especially from someone who may know about game theory but may not be an expert on American TV shows.
  - OscarCunningham 9y ago
    Optimising excitement and therefore viewership is probably more important to them than the value of individual prizes.
    - lancebeet 9y ago
      Of course, but deducing this on the spot if you have little or no interest in television goes far beyond the domain of the original problem.
dalbasal 9y ago
This might be a small difference, but is this an english parsing or reality parsing issue? I mean, shouldn't these questions come up as you're mentally interrogating the scenario, and trying to win the car? What happens if the host opens a car door, do I win? If the host opens doors randomly, I have a 2/3 chance of winning anyway. But, I didn't have enough information to know that when I picked a door first. By the time I do, I'm already back to 50-50 odds. I don't think people make either 2/3 game their mental model of the game. IDK about non-mathematicians, but I taught a parlour-games-ish parent day class at a high school once. We solved this problem until the kids got it. Then they had to explain it to other kids. My hint was emphasising "every week he shows a goat." Then basically this program, "playing" over and over on blackboard and recording results. Either way, it takes effort before people think of the entire game (including the 2nd choice) as a single game mechanic. Almost everyone gets it if you make them play the host. Maybe mathematicians are less prone to suspicion... Kids and adult non-mathematicians usually suspect the host of being up to something, so they are animating him in their mental model. Anyway, I think there's something interesting about this problem, and how people have difficulty thinking about it. I wonder if the same mechanic, but with a differently decorated story would be easier for people? A robot host, for instance.
- loup-vaillant 9y ago
  > What happens if the host opens a car door, do I win? I've always assumed that you lose in this case.
  - foo101 9y ago
    The host never opens a car door because the host knows where the car is. The host always opens a non-car door and gives the player a choice to make the situation interesting.
    - loup-vaillant 9y ago
      > The host never opens a car door How do you know that? Depending on the exact wording of the problem, or even the actual television show, the host can totally open the car door. He just didn't, this time.
pvg 9y ago
Since a lot of university mathematicians, engineers, and physicists famously got the answer wrong Did they? The wikipedia page you link says "After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them claiming vos Savant was wrong (Tierney 1991)." And there's a story about Erdős and the number of stories about Erdős far exceed the number of stories that would be plausible about anyone, even Erdős.
trevelyan 9y ago
If the host could open a door with the money behind it the correct response would be to switch doors anyway.
svara 9y ago
I don't think it's the wording - for me the problem is that I get the intuition that, at the time just after the game show host opened a door with a goat behind it, the system can be in one of four states of equal probability: * Closed door 1 chosen (incorrect) * Closed door 2 chosen (incorrect) * Closed door 1 chosen (correct) * Closed door 2 chosen (correct) ... each with a probability of 1/4. This is probably simply because that's the kind of situation we're familiar with. And it's completely wrong in this case. The actual state space at that point in time is (you can easily verify this by drawing the state transitions down on a piece of paper) * Incorrect door 1 chosen, incorrect door 2 open (1/3, losing) * Incorrect door 2 chosen, incorrect door 1 open (1/3, losing) * Correct door chosen, incorrect door 1 open (1/6, winning) * Correct door chosen, incorrect door 2 open (1/6, winning) ... and I think the reason that's not intuitive at first is that we're unfamiliar with thinking about a system transitioning through states with different probabilities: If you chose the incorrect door at first, the system state after the game show host intervenes is 100% determined (he will open the only one other incorrect door, and these states have 1/3 probability), but if you chose the correct door at first, he has a 50% choice and you will end up in one of two different states (1/3 * 1/2 = 1/6 probability). So, in effect, you have four states of unequal probability after the game show host opens the door, and by switching doors you reverse the probabilities to 1/3 losing and 2/3 winning.
sethammons 9y ago
I disagree that it is an English parsing problem. It is a logic parsing problem. Though, it would be great to poll people in other languages to try to verify if language or culture affect how people approach the problem. It is a logic problem because people tend to look at one part of the problem instead of the whole. They see a door taken from the pool of choices, now seeing two doors, they jump to the idea that it is 50/50. If they approach it as a simulation, it becomes clearer. A related question: if I have two children, and one is a boy, what is the probability the other is a boy? Got your answer? Most people, just like in the Monty Hall problem, get it wrong. If you run the simulation of options, it is clear. The options are BB, GB, BG (the option GG does not exist). So the answer is 1/3. Even seeing this, many folks have a mental block and want to say that the independent act of the other child is a given gender is still 50/50, which, ignoring the information already given in the scenario, would be true. Back to Monty Hall. Your choices are between C, G1, and G2. Your starting position can be any of those. C -> G1 or G2 is shown G1 -> G2 is shown G2 -> G1 is shown In two of those situations, the car has not yet been picked, so you have a 2/3 chance of picking the car if you switch your door choice after the goat reveal. People tend to think of the last door choice as independent of their earlier information. Again, I think it is a human logic flaw, but I would love a study that showed different languages or cultures approaching this problem differently!
- deleted 9y ago
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- xorcist 9y ago
  > if I have two children, and (number) one is a boy, what is the probability the other is a boy? The people who parse the sentence as the gender of the first child is given, will insist the odds are 50/50. Once you see the sample space, the sentence is unambigous.
mannykannot 9y ago
I would have thought that most times someone starts going down that road - of thinking that the host's choice had been a random selection from the two remaining doors - she would immediately see that it would be the wrong way to think about the problem because the prize would sometimes be revealed, a situation that does not occur in the question as posed, and which, when it does occur, renders the question of switching moot. In other words, if one starts by assuming that the question implies a random choice, it quickly becomes apparent that the question would not be well-formed under that assumption. To be a well-formed question under that assumption, it would have to say something like "the host flips a coin to pick one of the closed boxes which, when opened, reveals a goat", and that is a different question, not the original one rephrased.
mcphage 9y ago
> 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. If the host opened a door at random, then sometimes that door would contain the car. At which point, the entire exercise is meaningless—the host would in effect be asking "do you want to switch your door (which you know contains a goat) for another door (which you know contains a goat)?" Asking the question at all when they just lost—especially given that the host just cost them the game is clearly not a desirable scenario for any game show. So any interpretation of the problem which would allow for this scenario can be easily discarded as incorrect.
mannykannot 9y ago
Your rephrasing actually changes the question. It says that the scenario is run repeatedly, while the question is about a one-off event. By phrasing it as one of a series of events, you have told the subject how to go about answering the problem. In an earlier reply, I suggested that the only way to interpret the question as posed is that the host is not making a random selection, and the question would have to be phrased differently to allow an interpretation in which the host's choice is random - e.g."the host flips a coin to pick one of the closed boxes which, when opened, reveals a goat".

by jasode 9y ago

OscarCunningham 9y ago

I agree that the wording of the problem is important, but I'm not sure if that's what was causing the mathematicians to get the wrong answer. I think I would have naturally assumed that the host was opening a door without the car in order to raise tension, and that he would therefore never open the door with the car. It would be interesting to survey mathematicians, but I don't think there's enough of them who haven't already heard the problem.

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.

chki 9y ago

I think that is a bit far-fetched. In the context of a game show it would be strange to think that the host just makes a random choice. Also, why would there be the additional information that the host knows what's behind the doors if he just chooses randomly.

jasode 9y ago
>In the context of a game show it would be strange to think that the host just makes a random choice. A lot of people reading the MH problem were unfamiliar with the actual game show and therefore parsed it as an isolated word problem.
lancebeet 9y ago
Since the game show stands to lose by you getting the car, it's not an unreasonable assumption that you would only be given the choice to switch had you picked the car in the first place, especially from someone who may know about game theory but may not be an expert on American TV shows.
- OscarCunningham 9y ago
  Optimising excitement and therefore viewership is probably more important to them than the value of individual prizes.
  - lancebeet 9y ago
    Of course, but deducing this on the spot if you have little or no interest in television goes far beyond the domain of the original problem.

dalbasal 9y ago

This might be a small difference, but is this an english parsing or reality parsing issue? I mean, shouldn't these questions come up as you're mentally interrogating the scenario, and trying to win the car? What happens if the host opens a car door, do I win? If the host opens doors randomly, I have a 2/3 chance of winning anyway. But, I didn't have enough information to know that when I picked a door first. By the time I do, I'm already back to 50-50 odds. I don't think people make either 2/3 game their mental model of the game. IDK about non-mathematicians, but I taught a parlour-games-ish parent day class at a high school once. We solved this problem until the kids got it. Then they had to explain it to other kids. My hint was emphasising "every week he shows a goat." Then basically this program, "playing" over and over on blackboard and recording results. Either way, it takes effort before people think of the entire game (including the 2nd choice) as a single game mechanic. Almost everyone gets it if you make them play the host. Maybe mathematicians are less prone to suspicion... Kids and adult non-mathematicians usually suspect the host of being up to something, so they are animating him in their mental model. Anyway, I think there's something interesting about this problem, and how people have difficulty thinking about it. I wonder if the same mechanic, but with a differently decorated story would be easier for people? A robot host, for instance.

loup-vaillant 9y ago
> What happens if the host opens a car door, do I win? I've always assumed that you lose in this case.
- foo101 9y ago
  The host never opens a car door because the host knows where the car is. The host always opens a non-car door and gives the player a choice to make the situation interesting.
  - loup-vaillant 9y ago
    > The host never opens a car door How do you know that? Depending on the exact wording of the problem, or even the actual television show, the host can totally open the car door. He just didn't, this time.

pvg 9y ago

Since a lot of university mathematicians, engineers, and physicists famously got the answer wrong Did they? The wikipedia page you link says "After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them claiming vos Savant was wrong (Tierney 1991)." And there's a story about Erdős and the number of stories about Erdős far exceed the number of stories that would be plausible about anyone, even Erdős.

trevelyan 9y ago

If the host could open a door with the money behind it the correct response would be to switch doors anyway.

svara 9y ago

I don't think it's the wording - for me the problem is that I get the intuition that, at the time just after the game show host opened a door with a goat behind it, the system can be in one of four states of equal probability: * Closed door 1 chosen (incorrect) * Closed door 2 chosen (incorrect) * Closed door 1 chosen (correct) * Closed door 2 chosen (correct) ... each with a probability of 1/4. This is probably simply because that's the kind of situation we're familiar with. And it's completely wrong in this case. The actual state space at that point in time is (you can easily verify this by drawing the state transitions down on a piece of paper) * Incorrect door 1 chosen, incorrect door 2 open (1/3, losing) * Incorrect door 2 chosen, incorrect door 1 open (1/3, losing) * Correct door chosen, incorrect door 1 open (1/6, winning) * Correct door chosen, incorrect door 2 open (1/6, winning) ... and I think the reason that's not intuitive at first is that we're unfamiliar with thinking about a system transitioning through states with different probabilities: If you chose the incorrect door at first, the system state after the game show host intervenes is 100% determined (he will open the only one other incorrect door, and these states have 1/3 probability), but if you chose the correct door at first, he has a 50% choice and you will end up in one of two different states (1/3 * 1/2 = 1/6 probability). So, in effect, you have four states of unequal probability after the game show host opens the door, and by switching doors you reverse the probabilities to 1/3 losing and 2/3 winning.

sethammons 9y ago

I disagree that it is an English parsing problem. It is a logic parsing problem. Though, it would be great to poll people in other languages to try to verify if language or culture affect how people approach the problem. It is a logic problem because people tend to look at one part of the problem instead of the whole. They see a door taken from the pool of choices, now seeing two doors, they jump to the idea that it is 50/50. If they approach it as a simulation, it becomes clearer. A related question: if I have two children, and one is a boy, what is the probability the other is a boy? Got your answer? Most people, just like in the Monty Hall problem, get it wrong. If you run the simulation of options, it is clear. The options are BB, GB, BG (the option GG does not exist). So the answer is 1/3. Even seeing this, many folks have a mental block and want to say that the independent act of the other child is a given gender is still 50/50, which, ignoring the information already given in the scenario, would be true. Back to Monty Hall. Your choices are between C, G1, and G2. Your starting position can be any of those. C -> G1 or G2 is shown G1 -> G2 is shown G2 -> G1 is shown In two of those situations, the car has not yet been picked, so you have a 2/3 chance of picking the car if you switch your door choice after the goat reveal. People tend to think of the last door choice as independent of their earlier information. Again, I think it is a human logic flaw, but I would love a study that showed different languages or cultures approaching this problem differently!

deleted 9y ago
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xorcist 9y ago
> if I have two children, and (number) one is a boy, what is the probability the other is a boy? The people who parse the sentence as the gender of the first child is given, will insist the odds are 50/50. Once you see the sample space, the sentence is unambigous.

mannykannot 9y ago

I would have thought that most times someone starts going down that road - of thinking that the host's choice had been a random selection from the two remaining doors - she would immediately see that it would be the wrong way to think about the problem because the prize would sometimes be revealed, a situation that does not occur in the question as posed, and which, when it does occur, renders the question of switching moot. In other words, if one starts by assuming that the question implies a random choice, it quickly becomes apparent that the question would not be well-formed under that assumption. To be a well-formed question under that assumption, it would have to say something like "the host flips a coin to pick one of the closed boxes which, when opened, reveals a goat", and that is a different question, not the original one rephrased.

mcphage 9y ago

> 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. If the host opened a door at random, then sometimes that door would contain the car. At which point, the entire exercise is meaningless—the host would in effect be asking "do you want to switch your door (which you know contains a goat) for another door (which you know contains a goat)?" Asking the question at all when they just lost—especially given that the host just cost them the game is clearly not a desirable scenario for any game show. So any interpretation of the problem which would allow for this scenario can be easily discarded as incorrect.

mannykannot 9y ago

Your rephrasing actually changes the question. It says that the scenario is run repeatedly, while the question is about a one-off event. By phrasing it as one of a series of events, you have told the subject how to go about answering the problem. In an earlier reply, I suggested that the only way to interpret the question as posed is that the host is not making a random selection, and the question would have to be phrased differently to allow an interpretation in which the host's choice is random - e.g."the host flips a coin to pick one of the closed boxes which, when opened, reveals a goat".