$A$ and $B$ are playing a tournament consisting of some matches; the first one to win $(n+1)$ matches wins the tournament. If $A$ has probability $p$ of winning a match, find the probability that $A $ wins the tournament.
$A$ can win in the following ways; wins $(n+1)$ matches in a row, wins $n$ matches, loses $1$ match and wins the next, and so on till $A$ wins $n$ matches, loses $n$ matches and wins the $(2n+1)$th match.
Hence, this probability becomes $$P_A=\sum_{k=0}^{n}\binom{n+k}{n}p^{n+1}(1-p)^{k}$$ Or, $$P_A=(1-P_B)=1-\sum_{k=0}^{n}\binom{n+k}{n}p^{k}(1-p)^{n+1}=1-(1-p)^{n+1}\sum_{k=0}^{n}\binom{n+k}{n}p^{k}$$
However, even in the case of the slightly simplified sum in $P_B$ (which represents the probability B of winning the tournament), I don't know how to evaluate it. Is there any clever probabilistic reasoning I can use to arrive at an answer for this summation or a way to evaluate this summation to obtain a closed-form expression for the probability of $A$ winning the tournament?