# Thread: Finding the max value of p

1. ## 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.

2. Originally Posted by lllll
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