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Probability: choice of boxes

  1. Mar 30, 2009 #1
    hi, I am writing a computer algorithm which descibes the change of choice.

    1.your friend puts $10 in a box among three (there are three boxes) but you don't know which.
    2.you choose one of them but do not open it.
    3.your friend opens (eliminates) one of empty boxes
    i.e. if you choose the lucky box, he eliminate either one of two empty boxes at equal probability
    and if you choose an unlucky one, he eliminates the empty remainder.
    4.then you decide, whether or not you change your choice between two remainings.
    5.repeat 1~4 many times and expect the maximum result(in $).

    Question: you'd better change your choice? or should not change? for the maximum outcome.

    I expected that the change of choice should not matter: the equal probabilities.

    but my computer algorithm tells that "if you change the choice, better"

    Can someone tell me whether I am wrong or my algorithm is wrong?
  2. jcsd
  3. Mar 30, 2009 #2


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    The algorithm is right. Search for Monty Hall problem (here or on the search engine of your choice). It's a popular topic. In short: picking the remaining box is really as good as picking both remaining boxes, and there's a better chance it's in one of the two remaining boxes than in your original 1.
  4. Mar 30, 2009 #3
    Thank you for such a quick reply!
    I will check it out. but your explanation helped a lot, CRGreathouse!
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