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Math Help - Rolling a die

  1. #1
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    Rolling a die

    There is a game with 2 players (A&B) who take turns to roll a die and have to roll a six to win. What is the probability of person A winning?


    I got (5/6)^(2n) x (1/6)

    Is this correct?
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  2. #2
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    This question is similiar to a thread I created before... see the second post, change homer and marge to A and B, and the probability from \frac{2}{6} of winning to  \frac{1}{6}

    So assuming that A goes first;


    P(A)=\frac{1}{6} \cdot \frac{1}{1-\frac{25}{36} } =\frac{1}{6}\cdot \frac{36}{11}=\frac{6}{11}<br />
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  3. #3
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    Quote Originally Posted by Robb View Post
    This question is similiar to a thread I created before... see the second post, change homer and marge to A and B, and the probability from \frac{2}{6} of winning to  \frac{1}{6}

    So assuming that A goes first;


    P(A)=\frac{1}{6} \cdot \frac{1}{1-\frac{25}{36} } =\frac{1}{6}\cdot \frac{36}{11}=\frac{6}{11}<br />
    How do you get a geometric sum? The variable n is not defined, so Homer/A could win the game in the first game or the 301st game.. etc?
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  4. #4
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    Yup, exactly.. he can win on his first roll, his second roll, his third roll etc..

    So, if we let W be the event that a 6 is rolled, and N be the event any other number, and A rolls first, then A will win if the following happens;
    W
    NNW
    NNNNW
    NNNNNNW
    NNNNNNNNW
    etc.
    So the probability of these happening;
    1/6
    \frac{5}{6}\cdot\frac{5}{6}\cdot\frac{1}{6}=\frac{  25}{36}\cdot\frac{1}{6}
    \frac{5}{6}\cdot\frac{5}{6}\cdot\frac{5}{6}\cdot\f  rac{5}{6}\cdot\frac{1}{6}=\left(\frac{25}{36}\righ  t)^2 \cdot\frac{1}{6}
    etc.

    So the Probability of A winning is the sum of the combinations of those events happening
    P(A winning)=\frac{1}{6}+\frac{25}{36}\cdot\frac{1}{6}  +\left(\frac{25}{36}\right)^2 \cdot\frac{1}{6}+....
    Which is a geometric series
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