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Math Help - Finding the max value of p

  1. #1
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    Finding the max value of p

    During a series there are two teams, team A and team B. The probability for team A to win is p and the probability for team B to win is (1-p). The winner of the series must win i games. If i=4, find the probability that a total of 7 games are played. Also show that this probability is maximized when p=\frac{1}{2}

    so far I have:

    \binom{7}{4}p^4(1-p)^3= 35p^4(1-p)^3

    now taking the derivative to find to max I get:

    35(4p^3)(1-p)^3+3(1-p)^2(-1)p^4 = 0

    35p^3(1-p)^2(4-7p) = 0

    so my values would be 0, \ \ \frac{4}{7}, \ \ 1

    now it states that p is maximized at \frac{1}{2}, can anyone help me figure out what I'm doing wrong.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by lllll View Post
    During a series there are two teams, team A and team B. The probability for team A to win is p and the probability for team B to win is (1-p). The winner of the series must win i games. If i=4, find the probability that a total of 7 games are played. Also show that this probability is maximized when p=\frac{1}{2}

    so far I have:

    \binom{7}{4}p^4(1-p)^3= 35p^4(1-p)^3

    now taking the derivative to find to max I get:

    35(4p^3)(1-p)^3+3(1-p)^2(-1)p^4 = 0

    35p^3(1-p)^2(4-7p) = 0

    so my values would be 0, \ \ \frac{4}{7}, \ \ 1

    now it states that p is maximized at \frac{1}{2}, can anyone help me figure out what I'm doing wrong.

    The probability that the game ends on the 7-th round is the probability that the score stands at 3,3 on the 6-th round:

    p(3,3|p)=\frac{6!}{3!3!}p^3(1-p)^3

    where probability of success in a single trial of p.

    RonL
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