Finding the max value of p

**lllll** · September 14th 2008, 08:49 PM

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.

**CaptainBlack** · September 14th 2008, 11:10 PM

Originally Posted by lllll

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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