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Math Help - Help with deck of cards probability

  1. #1
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    Help with deck of cards probability

    I have been working on these problems for a several hours and am stuck. Can anyone help me at all? Here are the problems:


    A poker hand is defined as drawing five cards at random without replacement from a deck of 52 playing cards. Find the probability of each of the following poker hands:

    A. Two pairs (two pairs of equal face value plus one card of a different value)
    B. One pair (one pair of equal face value plus three cards of different values)
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  2. #2
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    Quote Originally Posted by cb22hawk View Post
    I have been working on these problems for a several hours and am stuck. Can anyone help me at all? Here are the problems:


    A poker hand is defined as drawing five cards at random without replacement from a deck of 52 playing cards. Find the probability of each of the following poker hands:

    A. Two pairs (two pairs of equal face value plus one card of a different value)
    B. One pair (one pair of equal face value plus three cards of different values)
    Google:

    poker two pair probability

    You will get everything you want and more.
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  3. #3
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    Hello, cb22hawk!

    A poker hand is defined as drawing five cards at random
    without replacement from a deck of 52 playing cards.
    There are: . {52\choose5} \:=\:2,598,960 possible poker hands.


    Find the probability of each of the following poker hands:

    A. Two pairs (two pairs of equal face value plus one card of a different value)

    There are: . {13\choose2} = 78 choices for the values of the pairs.

    There are: . {4\choose2}{4\choose2} \:=\:6^2 ways to get the two pairs.

    And there are 44 choices for the fifth card.

    Hence, there are: . 78 \times 6^2 \times 44 \:=\:123,552 ways to get two pairs.


    Therefore: . P(\text{two pairs}) \:=\:\frac{123,552}{2,598,960} \:=\:\frac{198}{4165}




    B. One pair (one pair of equal face value plus three cards of different values)

    There are: . 13 choices for the value of the pair.

    There are: . {4\choose2} = 6 ways to get the pair.

    There are: . \frac{48\cdot44\cdot40}{3!} = 14,080 choices for the other three cards.
    . .
    (They must not contain another pair.)

    Hence, there are: . 13 \times 6 \times 14,080 \:=\:1,098,240 ways to get one pair.


    Therefore: . P(\text{one pair}) \;=\;\frac{1.098,240}{2.598,960} \;=\;\frac{352}{833}

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