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Probability of obtaining 7 before 8 with pair of fair dice

  1. Sep 23, 2008 #1
    1. The problem statement, all variables and given/known data

    A pair of fair dice is rolled until the first 8 appears. What is the probability that a sum of 7 does not precede a sum of 8.

    2. Relevant equations

    Geometric series


    3. The attempt at a solution

    P(sum of 7 does not appear before sum of 8) =

    5/36 + 5/36 * 25/36 + 5/36 *(25/36)^2 + ...

    = (5/36)/(1-25/36) = 5/11

    do I have to use geometric series for this or just find the prob of 7 before 8?

    Thanks!
     
    Last edited: Sep 23, 2008
  2. jcsd
  3. Sep 23, 2008 #2

    Dick

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    Re: Probability

    Yes, I think you need to use a geometric series. Either 7 or 8 appearing ends the game, right? So I think you need to find the probability that neither 7 nor 8 appears in k-1 rolls times the probability that 8 appears on the kth roll. Then sum over all k.
     
  4. Sep 23, 2008 #3

    Dick

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    Re: Probability

    I think that's correct 5/36 is the probability of 8 and 25/36 is the probability of neither 7 nor 8. You did do a geometric series. How can you compute the probability of 7 before 8 without it?
     
    Last edited: Sep 23, 2008
  5. Sep 24, 2008 #4
    Re: Probability

    I don't think it is possible without using a geometric series because it's a problem of infinite sum.
     
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