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Math Help - Combination problem

  1. #1
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    Combination problem

    \text{Poker Dice is played by simultaneously rolling 5 dice.  Show that P(Two Pair) = .2315}\\\text{I'm having a hard time finding the solution}
    Last edited by downthesun01; August 24th 2013 at 07:48 AM.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Combination problem

    We have 6 choices for the first pair, 5 choices for the second pair, and 4 choices for the odd die. Multiply this by the number of ways to choose 2 from 6, and divide by the 6^5 different outcomes, hence:

    P(\text{two pair})={6 \choose 2}\frac{6\cdot5\cdot4}{6^5}=\frac{25}{108}
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  3. #3
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    Re: Combination problem

    Please explain the 6 choose 2

    Also,

    The solution I was given is

    \frac{4 {6 \choose2} {5 \choose2} {3 \choose2}}{6^5}

    It gives the same value as your solution, but I was hoping that you could offer some reasoning behind it?
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Combination problem

    The 6 choose 2 represents the number of ways to choose the two pairs from the 6 different values of a die.
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  5. #5
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    Re: Combination problem

    Quote Originally Posted by downthesun01 View Post
    The solution I was given is
    \frac{4 {6 \choose2} {5 \choose2} {3 \choose2}}{6^5}
    It gives the same value as your solution, but I was hoping that you could offer some reasoning behind it?
    Do you understand that there is no one correct way of doing this?

    I would have done as \frac{\dbinom{6}{2}(4)\dfrac{5!}{(2!)^2}}{6^5}

    Now I can explain reasoning. The six choose 2 is the number of ways to have two pair.
    The four is the number of ways to have the odd number out.
    The numbers of ways to arrange AABBC is \dfrac{5!}{(2!)^2}.
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