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Math Help - conditional probability

  1. #1
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    conditional probability

    Would you help me with this, please!
    I have to solve it but it doesen't work in the standard way.

    You play with 2 opponents with whom you alternately play with probability p_A and p_B.
    p_B >= p_A.
    We want to minimize the number of games you need to play to win 2 in a row.
    Should we start with A or B?

    Thanks!
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  2. #2
    MHF Contributor matheagle's Avatar
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    I don't understand this game.
    Is PA the probability you beat person A?
    And you play A and B back to back?

    I would say that, if you have a better chance of beating person B, play him/her.
    But I'm not sure what these rules are.
    Last edited by matheagle; August 25th 2009 at 04:27 PM.
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  3. #3
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    But I have to know the probability of playing with them. It is not 1/2 for sure because it is conditional.
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  4. #4
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    Quote Originally Posted by matheagle View Post
    I don't understand this game.
    Is PA the probability you beat person A?
    And you play A and B back to back?

    I would say that, if you have a better chance of being person B, play him/her.
    I don't think so.

    The probability of winning twice in the first two games is pA*pB no matter what opponent you choose.

    The probability of winning twice first in the 2nd and 3rd games is:

    pA*pB*(1-pA) if you choose person A and

    pA*pB*(1-pB) if you choose person B,

    clearly, the second is smaller, so you have higher probability of winning in the 2nd and 3rd games first, if you choose A.

    To win the 3rd and 4th games first the probability is:

    pA*pB*(pA*(1-pB) + (1-pA)*(1-pB)) if you choose A

    and

    pA*pB*(pB*(1-pA) + (1-pA)*(1-pB)) if you choose B

    In this case the probability is bigger if you had chosen B.

    I think that it will alternate between the two but it will be best if you choose A.
    Last edited by pedrosorio; August 25th 2009 at 10:35 AM.
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