Adam and Eve play a series of games of tennis, stopping as soon as
one as them has won $n$ games. Suppose that they are evenly matched
and that Adam wins each game with probability $1/2$, independently
of the other games. What is the probability that the loser has won
exactly $k$ games when the match is over?
The answer I get is $$\dbinom{k+n-1}{k}(1/2)^{k+n}$$
[here1]. Is it correct?
Best Answer
The loser has won $k$ games if the ultimate winner wins $n-1$ of the first $n+k-1$ games, and then wins the $(n+k)$-th game.
Thus the probability that Adam has won $k$ games when Eve wins the match is $$\binom{n+k-1}{n-1}(1/2)^{n+k-1}(1/2).$$ The probability Eve has won $k$ games before Adam wins the match is the same. So the required probability is $$2\binom{n+k-1}{n-1}(1/2)^{n+k-1}(1/2),$$ for $k=0$ to $n-1$, twice the answer in the OP.