Suppose a series of games is played between two teams and that in any individual game, one of the teams wins and the other team loses. Let the probability that team A wins an individual game be p and hence the probability that team B wins is q = 1 � p. Assume that the outcomes of individual games are independent. The series continues until one team wins N games in which case they are declared the overall winner.
(a) If N = 2, what is the probability that A is the overall winner? (Your answer will depend on p).
(b) If N = 2 and X is the number of games needed to determine the overall winner, what are the possible values of X? What are the expected value, variance, and standard deviation of X? (Your answer should be a function of p and no other variables).
(c) What value of p maximizes the expected value of X when N = 2?
Ok, I have the answers and a somewhat loose way to get the answers, but I do not understand the reasoning for b and c. I understand a pretty well, but can you explain the logic behind b and c.
Best Answer
$\mathsf P(A) = p$ is the probability that team $A$ will win a game against team $B$. Thus $0\leq p \leq 1$, since that's the range of any probability; from impossible to certain.
Since the games are independent, the probability that $A$ will win two games in a row is: $\mathsf P(AA) = \mathsf P(A)\mathsf P(A) = p^2$.
Similarly the probability that $B$ will win two games in a row is: $\mathsf P(BB)=\mathsf P(B)\mathsf P(B)=(1-p)^2$.
Then, since these are mutually exclusive $\mathsf P(AA\cup BB) = p^2+(1-p)^2$.
Likewise: $\mathsf P(AB\cup BA) = \mathsf P(A)\mathsf P(B) + \mathsf P(B)\mathsf P(A) = 2 p(1-p)$
Notice that: $1 = (p+q)^2 = p^2 + 2pq + q^2$