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Math Help - Probability in a game of Bridge

  1. #1
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    Probability in a game of Bridge

    I've researched this quite a bit. First off I'll post the problem then I'll show you some of what I think the work is to prove that I've attempted it.

    The question:
    In a bridge game each player has 13 cards so you and your partner have a total of 26 cards. What is the probability that you and your partner have exactly x Aces when x equals 0,1,2,3,4.

    MY WORK (so far):
    4/52 * 3/52 * 2/52 * 1/52 (this would be the equation IF you were only given 4 cards). Right??? Or is it something like (4!/52!). Totally confused.

    We now have to add on the odds considering that you are getting 22 more cards which would be something like this...
    48/52 * 47/52 * 46/52 * 45/52 * 44/52 * 43/52 * ......... * 28/52 * 27/52. Can you help me finalize this please? I've researched all over gooooogle and everywhere else I can think of.

    I tried Probabilities Miscellaneous: Bridge Odds to get some insight but that didn't help me either.
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  2. #2
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    Hello, cunoodle2!

    Your fractions are wrong.
    You can't keep the denominator (52) constant.


    In a bridge game each player has 13 cards, so you and your partner have a total of 26 cards.
    What is the probability that you and your partner have exactly x Aces for x = 0,1,2,3,4.

    There are: . {52\choose26} possible outcomes.


    There are 4 Aces in the deck, and 48 Others.

    To get exactly x Aces, there are: . {4\choose x} ways.
    We want 26-x cards from the 48 Others: . {48\choose 26-x} ways.
    . . Hence, there are: . {4\choose x}{48\choose 26-x} ways to get exactly x Aces.


    Therefore: . P(x\text{ Aces}) \;=\;\frac{{4\choose x}{48\choose 26-x}}{{52\choose26}}

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