Suppose that two teams play a series of games that ends when one of them has won $i$games. Suppose that each game played is, independently, won by team $A$with probability $p$. Find the expected number of games that are played when

(a) $i=2$and

(b) $i=3$. Also, show in both cases that this number is maximized when $p=\frac{1}{2}$.

Two teams play a series of games that ends when one of them has won $i$games. And that each game played is, independently, won by team $A$with probability $p$.

Define $X$as the random variable that marks the number of games that are played. Observe that there will be maximally played $3$games and minimally two games. Hence $X\in \{2,3\}$. If $X=2$, that means that some of the team has won both of the first two games. Thus

$P(X=2)=p^2+(1-p{)}^2$

If $X=3$, that means that two games have been won by one of them, but the remaining team had to won some of the first two games. Hence

$P(X=3)=2p^2(1-p)+2p(1-p{)}^2$

The expected number of games is

$E\left(X\right)=2·\left(p^2+(1-p{)}^2\right)+3·\left(2p^2(1-p)+2p(1-p{)}^2\right)$

$=-2p^2+2p+2$

This is quadratic polynomial which reaches maximum at $-\frac{b}{2a}=\frac{1}{2}$.

Two teams play a series of games that ends when one of them has won $i$games. And that each game played is, independently, won by team $A$with probability $p$.

Define $X$as the random variable that marks the number of games that are played. Observe that there will be maximally played $5$games and minimally three games. Hence $X\in \{3,4,5\}$. means that some of the team has won all three first games. Thus

$P(X=3)=p^3+(1-p{)}^3$

If $X=4$, that means that three games have been won by one of them, but the remaining team had to won some of the first three games. Hence

$P(X=4)=3p^3(1-p)+3p(1-p{)}^3$

If $X=5$, that means that three games have been won by one of them, but the remaining team had to won two games out of the four three games. Hence

$P(X=5)=6p^3(1-p{)}^2+6p^2(1-p{)}^3$

The expected number of games is

$E\left(X\right)=3·\left(p^3+(1-p{)}^3\right)+4·\left(3p^3(1-p)+3p(1-p{)}^3\right)$

$+5·\left(6p^3(1-p{)}^2+6p^2(1-p{)}^3\right)$

$=3+3\left(p-p^2\right)+6{\left(p-p^2\right)}^2$

On interval $(0,1)$function $p\mapsto p-p^2$is strictly positive, so finding the maximum of $E\left(X\right)$is equivalent of finding maximum of $p-p^2$. But, similarly as in (a), we have that the maximum is reached in$p=\frac{1}{2}$.