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Math Help - Dice throwing problem

  1. #1
    Junior Member Greg98's Avatar
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    Dice throwing problem

    The problem is following:
    A and B takes turns throwing a dice. A begins. Who throws six first wins. What's the probability that A wins?

    I tried to examine probabilities that A wins on different rounds. Something like this:
    1. <br />
 \frac{1}{6}<br />

    2. <br />
 \frac{5}{6}*\frac{5}{6}*\frac{1}{6}<br />

    3. <br />
 \frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\f  rac{1}{6}<br />

    But I can't solve the general form (or something). I think it has something to do with binomial probability and geometric series... Any help is appreciated, thanks!
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  2. #2
    Senior Member
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    You're correct. Add up all the values you've gotten. This is a geometric series:
    (1/6) * [ 1 + x + x^2 +x ^3 + ...]
    where x = (5/6)(5/6).
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  3. #3
    MHF Contributor
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    Quote Originally Posted by Greg98 View Post
    The problem is following:
    A and B takes turns throwing a dice. A begins. Who throws six first wins. What's the probability that A wins?

    I tried to examine probabilities that A wins on different rounds. Something like this:
    1. <br />
 \frac{1}{6}<br />

    2. <br />
 \frac{5}{6}*\frac{5}{6}*\frac{1}{6}<br />

    3. <br />
 \frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\f  rac{1}{6}<br />

    But I can't solve the general form (or something). I think it has something to do with binomial probability and geometric series... Any help is appreciated, thanks!
    As the game can go on indefinately, you need to evaluate the infinite geometric series.

    The probability that A wins is

    \frac{1}{6}+\frac{5}{6}*\frac{5}{6}*\frac{1}{6}+\f  rac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\fra  c{1}{6}+........

    The first term is \frac{1}{6} and the common ratio is \left(\frac{5}{6}\right)^2

    For a geometric series,

    S_n=\frac{a\left(1-r^n\right)}{1-r}

    S_{\infty}=\frac{a}{1-r}
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