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Math Help - Dice Probability

  1. #1
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    Dice Probability

    Hi. Is there a way to solve the following without actually listing out all the possibilities?

    What is the probability of rolling a total sum of 8 using three dice?

    Thank you!
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  2. #2
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    Yes, there is a way.

    There are 3 dice and you want to find all the sums of 8.

    \left(\sum_{k=1}^{6}x^{k}\right)^{3}

    Expand this out and check the coefficient of x^{8}

    That is the number of ways to sum to 8.

    The coefficent is 21x^{8}

    Therefore, there are 21 ways.

    There are 6^{3}=216 possible rolls with 3 dice.

    Probability is \frac{21}{216}
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  3. #3
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    Wow! Ingenious xD.
    Now can u explain why that is? I don't see it.

    Thanks again!
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  4. #4
    Eater of Worlds
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    A text on advanced counting methods and/or generating functions may prove useful if you are curious of how it's derived.
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  5. #5
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    Thank you. =]
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  6. #6
    Junior Member F.A.P's Avatar
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    Quote Originally Posted by galactus View Post
    A text on advanced counting methods and/or generating functions may prove useful if you are curious of how it's derived.
    This is a more direct approach for those not familiar with generating functions.
    Let Z be the sum of three dice. Let Y be the sum of the first two dice.

    P(Z=8)=\sum_{k=2}^{12}P(Z=8|Y=k)P(Y=k)

    Now

    P(Z = 8|Y = k) = 1/6, for k = 2, 3,...,7, and 0 elsewhere. Also it's easy to show that

    P(Y=k)=\frac{k-1}{6^{2}}, for k = 2, 3,...,7 (we won't consider higher k's)


    Hence we end up with

    P(Z=8)=\sum_{k=2}^{7}\frac{1}{6}\frac{k-1}{6^{2}}=\frac{21}{6^{3}}=\frac{21}{216}
    Last edited by F.A.P; December 20th 2006 at 02:59 PM.
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