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What a shame that the title doesn't reveal what this is -- it's the original appearance of the Monty Hall problem, when Marilyn Vos Savant answered the problem

by lutusp 13y ago

What a shame that the title doesn't reveal what this is -- it's the original appearance of the Monty Hall problem, when Marilyn Vos Savant answered the problem correctly, only to be criticized (wrongly) by any number of well-educate academics, like this one: "Since you seem to enjoy coming straight to the point, I’ll do the same. You blew it! Let me explain. If one door is shown to be a loser, that information changes the probability of either remaining choice, neither of which has any reason to be more likely, to 1/2. As a professional mathematician, I’m very concerned with the general public’s lack of mathematical skills. Please help by confessing your error and in the future being more careful. -- Robert Sachs, Ph.D." Doctor Sachs was entirely wrong, as were about ten thousand other academics who ganged up on Vos Savant in the months following the appearance of this column. The real outcome is that, if the contestant doesn't change doors, his probability of winning is 1/3, but if he does change doors, his chance goes up to 2/3. Details: http://en.wikipedia.org/wiki/Monty_Hall_problem http://en.wikipedia.org/wiki/Monty_Hall_problem

gruseom 13y ago
Thanks for the comment; I wouldn't have looked at the article otherwise, and it's great, both historically and as an explanation of the problem. The most helpful part is where she simply lists all six possibilities in a table. That makes it clear that since the odds are 2/3 that you originally chose a goat, the odds must be 2/3 that "all the remaining doors" contain the car. I also like how she suggested that people just run the experiment themselves.
- StavrosK 13y ago
  What crystallized it for me was this: Suppose you have a million doors, you pick one, and the host opens the other 999,998 doors. Should you switch or not? Clearly, clearly the prize is behind the unopened door, merely because the other door you picked had a one-in-a-million chance. Now it becomes a "so, do you think the prize is behind the door you picked, or the one I have conspicuously left unopened?".
  - gruseom 13y ago
    I agree, that is a clever way to shed light on the problem. It's easier first to see that the host is giving you 999,999 choices for the price of one, and then see that the same logic applies when there are only 3 doors.
  - dnautics 13y ago
    completely agree - that's what made me finally happy with the reasoning, even though I'd done the math by hand. It does get more subtle though if you have a million doors, and the host opens one, should you switch? The answer is still yes, but the margin is tiny.
    - StavrosK 13y ago
      Yep, exactly 10^-6.
      - asdfologist 13y ago
        I think it's actually about 1^-12. Given that you have N doors, you (N-1)/N chance of picking a door with a goat. If the host then shows you another door with a goat, and you decide to switch, then you have 1/(N-2) odds of picking the door with the car. So your overall probability of picking the door with the car if you switch is (N-1)/N * 1/(N-2) = (N-1)/(N(N-2)). For N=3, this is 2/3. For N=10^6, this equals roughly 10^-6 + 10^-12. So you have only a 10^-12 margin over the original 10^-6 shot of picking the car.
        StavrosK 13y ago
        I don't think that's right. Why is the probability of winning after switching the product of the initial probability times the one after the switch (which I think is also wrong)? You pick one door, the probability of winning is 1/N. The probability of losing is (N-1)/N, which stays fixed no matter how many doors the host opens. If you switch, the probability of winning is (1/(N-2))-(1/(1/N)). If he opens all doors except two, and you switch, the probability of either of the two having the car is 50% minus the original 10^-6/2. Each door decreases the probability by one in a million, not one in a trillion. If that were the case, it would be highly unlikely that you won even if you switched after all but one of the doors were opened.
- madlee 13y ago
  i've seen this problem several times, and I still think its a case of the statistics not representing the reality. For example, if you consider the case where you have originally chosen the correct door, there are two possibilities to consider (host reveals door A vs door B) in which switching/staying results in the same outcome, but considering results in 8 possibilities and puts the odds for both actions at 1/2
  - narcissus 13y ago
    One of the core 'assumptions' is that the host would never reveal the door with the goat behind it. Therefore, in your scenario where you choose the winning door in the first place, there is only one option for the host: reveal the door without the goat / car / whatever.
  - humbledrone 13y ago
    Read the article; your theory has been disproven by hundreds of independent experiments. Better yet, go write your own simulation and find for yourself that you are incorrect.
TallboyOne 13y ago
Is there a particular reason that so many people were wrong... that seems like a glitch in the matrix almost. I'm hungover, and suck at math, and understand after she explained it... Even after she explained it she still received so many replies saying she was wrong, I just am wondering what's going on here that's causing this. That's just so bizarre.
- MPSimmons 13y ago
  Human brains didn't evolve to deal with these kinds of problems, so when we encounter them, we take mental shortcuts, like "There are three doors. There is one good prize. The odds are one in three". It's actually a really big cognitive leap to get from there to the idea that, because the gameshow host knows the answer, that the visible answer differs from the mental shortcut. I've got a good friend who speaks at a lot of risk management seminars, and a good deal of people who deal with financial risk every day get this question wrong. He's got a really good blog at http://behavioralrisk.net/ http://behavioralrisk.net/
- asdfologist 13y ago
  From the wikipedia article: "Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation confirming the predicted result."
  - lutusp 13y ago
    Now I don't feel so bad for not getting it at first. :)
- lutusp 13y ago
  > Is there a particular reason that so many people were wrong ... Yes. The difficulty with the Monty Hall problem has to do with grasping all the probabilities represented by all the outcomes at once. I personally didn't fully understand it until I wrote a list of all the possible outcomes and summed the probabilities.
- Nzen 13y ago
  It just seems so easy to discount the first part of the exercise. It's hard to distinguish between contestants & the 'alien' who comes in after. I found her matrix rather messy & convinced myself with the following: | chosen | revealed | hidden | | car __ | empty __ | goat__ | | empty_ | goat ___ | car __ | | goat _ | empty __ | car __ |

by lutusp 13y ago

gruseom 13y ago

Thanks for the comment; I wouldn't have looked at the article otherwise, and it's great, both historically and as an explanation of the problem. The most helpful part is where she simply lists all six possibilities in a table. That makes it clear that since the odds are 2/3 that you originally chose a goat, the odds must be 2/3 that "all the remaining doors" contain the car. I also like how she suggested that people just run the experiment themselves.

StavrosK 13y ago
What crystallized it for me was this: Suppose you have a million doors, you pick one, and the host opens the other 999,998 doors. Should you switch or not? Clearly, clearly the prize is behind the unopened door, merely because the other door you picked had a one-in-a-million chance. Now it becomes a "so, do you think the prize is behind the door you picked, or the one I have conspicuously left unopened?".
- gruseom 13y ago
  I agree, that is a clever way to shed light on the problem. It's easier first to see that the host is giving you 999,999 choices for the price of one, and then see that the same logic applies when there are only 3 doors.
- dnautics 13y ago
  completely agree - that's what made me finally happy with the reasoning, even though I'd done the math by hand. It does get more subtle though if you have a million doors, and the host opens one, should you switch? The answer is still yes, but the margin is tiny.
  - StavrosK 13y ago
    Yep, exactly 10^-6.
    - asdfologist 13y ago
      I think it's actually about 1^-12. Given that you have N doors, you (N-1)/N chance of picking a door with a goat. If the host then shows you another door with a goat, and you decide to switch, then you have 1/(N-2) odds of picking the door with the car. So your overall probability of picking the door with the car if you switch is (N-1)/N * 1/(N-2) = (N-1)/(N(N-2)). For N=3, this is 2/3. For N=10^6, this equals roughly 10^-6 + 10^-12. So you have only a 10^-12 margin over the original 10^-6 shot of picking the car.
      - StavrosK 13y ago
        I don't think that's right. Why is the probability of winning after switching the product of the initial probability times the one after the switch (which I think is also wrong)? You pick one door, the probability of winning is 1/N. The probability of losing is (N-1)/N, which stays fixed no matter how many doors the host opens. If you switch, the probability of winning is (1/(N-2))-(1/(1/N)). If he opens all doors except two, and you switch, the probability of either of the two having the car is 50% minus the original 10^-6/2. Each door decreases the probability by one in a million, not one in a trillion. If that were the case, it would be highly unlikely that you won even if you switched after all but one of the doors were opened.
madlee 13y ago
i've seen this problem several times, and I still think its a case of the statistics not representing the reality. For example, if you consider the case where you have originally chosen the correct door, there are two possibilities to consider (host reveals door A vs door B) in which switching/staying results in the same outcome, but considering results in 8 possibilities and puts the odds for both actions at 1/2
- narcissus 13y ago
  One of the core 'assumptions' is that the host would never reveal the door with the goat behind it. Therefore, in your scenario where you choose the winning door in the first place, there is only one option for the host: reveal the door without the goat / car / whatever.
- humbledrone 13y ago
  Read the article; your theory has been disproven by hundreds of independent experiments. Better yet, go write your own simulation and find for yourself that you are incorrect.

TallboyOne 13y ago

Is there a particular reason that so many people were wrong... that seems like a glitch in the matrix almost. I'm hungover, and suck at math, and understand after she explained it... Even after she explained it she still received so many replies saying she was wrong, I just am wondering what's going on here that's causing this. That's just so bizarre.

MPSimmons 13y ago
Human brains didn't evolve to deal with these kinds of problems, so when we encounter them, we take mental shortcuts, like "There are three doors. There is one good prize. The odds are one in three". It's actually a really big cognitive leap to get from there to the idea that, because the gameshow host knows the answer, that the visible answer differs from the mental shortcut. I've got a good friend who speaks at a lot of risk management seminars, and a good deal of people who deal with financial risk every day get this question wrong. He's got a really good blog at http://behavioralrisk.net/ http://behavioralrisk.net/
asdfologist 13y ago
From the wikipedia article: "Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation confirming the predicted result."
- lutusp 13y ago
  Now I don't feel so bad for not getting it at first. :)
lutusp 13y ago
> Is there a particular reason that so many people were wrong ... Yes. The difficulty with the Monty Hall problem has to do with grasping all the probabilities represented by all the outcomes at once. I personally didn't fully understand it until I wrote a list of all the possible outcomes and summed the probabilities.
Nzen 13y ago
It just seems so easy to discount the first part of the exercise. It's hard to distinguish between contestants & the 'alien' who comes in after. I found her matrix rather messy & convinced myself with the following: | chosen | revealed | hidden | | car __ | empty __ | goat__ | | empty_ | goat ___ | car __ | | goat _ | empty __ | car __ |