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Math Help - Combination and Permutations

  1. #1
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    Combination and Permutations

    What is the probability that a five-card poker hand has the following?
    (a) Two Pairs (not four of a kind or a full house).

    Choose a kind - 13
    Choose 2 of a kind - C(4,2)
    Choose another kind - 12
    Choose 2 of a kind - C(4,2)
    Choose last card - 44
    Multiply all number above for answer.

    Is that right?
    Another thing that came to mind is to replace the 13 and the 12 by C(13,2), I'm having trouble choosing one over the other. Can somebody confirm the right answer and explain why. Thank you.
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  2. #2
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    Lexington, MA (USA)
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    Hello, algorith!

    Your afterthought is correct . . .


    What is the probability that a five-card poker hand has the following?
    (a) Two Pairs (not four of a kind or a full house).

    Choose a kind - 13
    Choose 2 of a kind - C(4,2)
    Choose another kind - 12
    Choose 2 of a kind - C(4,2)
    Choose last card - 44
    Multiply all number above for answer.

    Is that right?

    Another thing that came to mind is to replace the 13 and the 12 by C(13,2),
    I'm having trouble choosing one over the other. .
    This one is correct.
    Can somebody confirm the right answer and explain why?

    We choose 2 values from the possible 13 values: . _{13}C_2\,=\,78 ways.
    . . (The order does not matter.)

    Choose a pair: . _4C_2\,=\,6 ways.

    Choose another pair: . _4C_2\,=\,6 ways.

    Choose the fifth card: .  44 ways.


    So there are: . 78\cdot6\cdot6\cdot44\:=\:123,\!552\,\text{ possible Two Pairs.}


    There are: . _{52}C_5 \:=\:2,598,960 possible 5-card hands.


    Therefore: . P(\text{Two Pairs}) \;=\;\dfrac{123,\!552}{2,\!598,\!960} \;=\;\dfrac{198}{4165} \;\approx\;4\frac{3}{4}\%
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  3. #3
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    Thanks a lot, that was brilliant.
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