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Math Help - probability: poker cards problem

  1. #1
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    Post probability: poker cards problem

    I don't understand how to solve this problem. It says "you select 5 cards without replacement from a standard 52 card deck, what is the probability of getting exactly 2 pairs?"
    I understand that my sample space is 52 choose 5. But I have trouble picking the values and kinds of the two pairs.
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  2. #2
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    Apr 2010
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    Two pairs [one pair of each two different face values and a card of
    a third face value]


    Prob(2 pairs) = (13C2 x 4C2 x 4C2 x 11 x 4)/(52C5) = ?


    The various terms in above expression arise as follows.

    13C2 = (13 choose 2) is the number of ways of choosing the two
    different face values for the two pairs.

    4C2 = (4 choose 2) is the number of ways of choosing the two suits for
    one pair, and the second 4C2 is the number of ways of choosing the two
    suits for the second pair.

    11 is the number of ways of choosing the face value for the 5th card.
    It must not be the same face value as either of the pairs.

    4 is the number of ways of choosing the suit for the 5th card.

    52C5 = (52 choose 5) is the number of unrestricted ways that 5 cards
    can be selected from the 52 in the pack.
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