3 ms·

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 helpfu

by 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.
        asdfologist 13y ago
        You're wrong and I wrote a simulation to prove it. To repeat my original assertion, the probability of winning if you switch is (N-1)/(N(N-2)). For N=4, this equals 3/8. Run the following simulation and you'll get roughly 0.375, as my formula predicts. http://pastebin.com/i89M4GZU http://pastebin.com/i89M4GZU
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.

by gruseom 13y ago

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.
      - asdfologist 13y ago
        You're wrong and I wrote a simulation to prove it. To repeat my original assertion, the probability of winning if you switch is (N-1)/(N(N-2)). For N=4, this equals 3/8. Run the following simulation and you'll get roughly 0.375, as my formula predicts. http://pastebin.com/i89M4GZU http://pastebin.com/i89M4GZU

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.