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

    so far I have:

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

    now taking the derivative to find to max I get:

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

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

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

    now it states that p is maximized at $\displaystyle \frac{1}{2}$, 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 $\displaystyle p$ and the probability for team B to win is $\displaystyle (1-p)$. The winner of the series must win $\displaystyle i$ games. If i=4, find the probability that a total of 7 games are played. Also show that this probability is maximized when $\displaystyle p=\frac{1}{2}$

    so far I have:

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

    now taking the derivative to find to max I get:

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

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

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

    now it states that p is maximized at $\displaystyle \frac{1}{2}$, 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:

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

    where probability of success in a single trial of $\displaystyle p$.

    RonL
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