Ok so the inequality on this problem is throwing me off
Suppose that two teams (team A and team B) play a series of games to determine a winner. In a best-of-three series, the games end as soon as one team has won two games. In a best-of-five series, the games end as soon as one team has won three games, and so on. Assume that team A’s probability of winning any one game is p, where .5 < p < 1. Also assume that the outcomes are independent from game to game.
What is the exact probability team A will win a three game series, as a function of p? I know this should be straightforward, I just can’t seem to grasp it
P(A win series) = P(WW)+P(LWW)+P(WLW) $$P(A)=p^2+p^2(q)+p^2(q)=p^2+2p^2q$$ P(B win series) = P(LL)+P(LWL)+P(WLL) $$P(B)= q^2+2q^2p$$
$$P(A)+P(B)=p^2+q^2+2p^2q+2q^2p=1$$
In a best of $2n+1$ case $$P_{2n+1}(A)=p^{n+1}+\binom{n+1}{1}p^{n+1}q+\binom{n+2}{2}p^{n+1}q^2....+\binom{2n}{n}p^{n+1}q^n \\ =p^{n+1}\sum_{k=0}^n\binom{n+k}{k}q^k$$