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Math Help - Another dice question!

  1. #1
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    Another dice question!

    You have two dice What is the probability of rolling a "3" total before rolling a "7" total?
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  2. #2
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    Quote Originally Posted by syke2 View Post
    You have two dice What is the probability of rolling a "3" total before rolling a "7" total?
    Hi syke2,

    presuming you mean the probability of throwing the 2 dice, repeatedly if necessary,
    and you get a total of 3 before ever getting a total of 7, then

    there are 2 ways to get a total of 3

    1,2 and 2,1

    there are 6 ways to get 7

    1,6 and 6,1
    2,5 and 5,2
    3,4 and 4,3

    Hence the probability of getting neither a total of 3 or 7 is \frac{36-8}{36}=\frac{28}{36}

    Hence the probability of getting a total of 3 before a total of 7
    (not a total of 3 followed immediately by a total of 7) is

    a total of 3 on the first 2 throws
    or neither total on the first 2 throws and a total of 3 on the next 2
    or neither total on the first 4 throws and a total of 3 on the next 2
    etc...

    This probability is

    \frac{2}{36}+\left(\frac{28}{36}\right)\frac{2}{36  }+\left(\frac{28}{36}\right)^2\frac{2}{36}+\left(\  frac{28}{36}\right)^3\frac{2}{36}+........

    =\frac{2}{36}\left(1+\frac{28}{36}+\left(\frac{28}  {36}\right)^2+\left(\frac{28}{36}\right)^3+.....\r  ight)

    In brackets is an infinite geometric series... a=1,\ r=\frac{28}{36}

    for which the sum is \frac{a}{1-r}=\frac{1}{1-\frac{28}{36}}=\frac{1}{\frac{36-28}{36}}=\frac{1}{\left(\frac{8}{36}\right)}=\frac  {36}{8}

    hence the probability is

    \frac{2}{36}\left(\frac{36}{8}\right)=\frac{1}{4}

    Or...

    there are 8 chances of getting a total of 3 or 7,
    there are 2 chances of getting a total of 3.

    Hence if we got a total of 3 rather than a total of 7, that's 2 chances in 8.
    Last edited by Archie Meade; May 13th 2010 at 04:07 AM.
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  3. #3
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    Awesome! Thank you very much!
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