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Math Help - Dice Toss Problem

  1. #1
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    Dice Toss Problem

    Hi everyone,

    I have the solution to a problem, but I am nore sure how the answer was arrived at, or whether it is 100% correct. The problem involves tossing 4 dice, and finding the probability of the sum being 10.

    The solution to this problem is ((9 choose 3)-4) / 6^4.

    Any ideas as to how this solution was derived? Any insight is appreciated.
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  2. #2
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    Re: Dice Toss Problem

    Quote Originally Posted by KelvinScale View Post
    The problem involves tossing 4 dice, and finding the probability of the sum being 10.
    There several ways to get a ten.
    \begin{gathered}  1234 \hfill \\  1126 \hfill \\  1225 \hfill \\  2224 \hfill \\  1333 \hfill \\  1144 \hfill \\  1135  \hfill \\  2233 \hfill \\ \end{gathered}
    Now you have to count all of those.
    See if I missed any.

    POST SCRIPT.
    Here is a neat trick.
    Look at this webpage.
    Look for the coefficient of x^{10}.
    That tells us how many ways the the four dice add to 10.
    Last edited by Plato; September 23rd 2011 at 12:53 PM.
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  3. #3
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    Re: Dice Toss Problem

    Hello, KelvinScale!

    I have the solution to a problem, but I am not sure how the answer was arrived at,
    or whether it is 100% correct.
    The problem involves tossing 4 dice, and finding the probability of the sum being 10.

    The solution to this problem is: . \frac{{9\choose3} - 4}{6^4}

    Any ideas as to how this solution was derived?
    Any insight is appreciated.

    I have a theory . . . see if it makes sense.


    Place ten objects in a row, leaving a space between each pair.
    . . \circ\;\circ\;\circ\;\circ\;\circ\;\circ\;\circ- \circ\;\circ\;\circ

    Select 3 of the 9 spaces and insert "dividers".

    So that: . \circ\;\circ\;\circ\:|\:\circ\:\circ\:|\:\circ\:| \:\circ\;\circ\;\circ\;\circ . represents the roll [3,2,1,4]

    There are: . {9\choose3} ways to place the dividers.

    But these include the division: . \circ\:|\:\circ\:|\:\circ\:|\:\circ\;\circ-  \circ\;\circ\;\circ\;\circ\;\circ
    . . which represents [1,1,1,7] which is not permitted.

    There are four such divisions: [1,1,1,7], [1,1,7,1], [1,7,1,1], and [7,1,1,1].

    Hence, there are: . {9\choose3} - 4 .ways to get a sum of ten on four dice.

    And there are: . 6^4 possible outcomes.

    Therefore: . P(\text{sum of 10}) \:=\:\frac{{9\choose3} - 4}{6^4}

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  4. #4
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    Re: Dice Toss Problem


    And Plato's approach is absolutely correct!


    . . \begin{array}{cc} \text{Combination} & \text{Permutations} \\ \hline \\[-4mm]1234 & 4! \:=\:24 \\ 1126 & \frac{4!}{2!}\:=\:12 \\ \\[-4mm] 1225 & \frac{4!}{2!} \:=\:12 \\ \\[-4mm] 2224 & \frac{4!}{3!} \:=\:\;4 \\ \\[-4mm] 1333 & \frac{4!}{3!} \:=\:\;4 \\ \\[-4mm] 1144 & \frac{4!}{2!2!} \,=\:6 \\ \\[-4mm] 1135 & \frac{4!}{2!} \:=\:12 \\ \\[-4mm] 2233 & \frac{4!}{2!2!} \,=\:\:6 \\ \\[-4mm] \hline & \qquad\quad 80\end{array}


    Therefore: . P(\text{sum of 10}) \:=\:\frac{80}{6^4}

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