Probability of winning first n games.

Question

Two players A and B play tennis ,where they serve alternately each point.

Probability of player A winning a point on his serve is p.

Probability of player A winning a point on player B's serve is q.

Player who first wins n points will win the match.What is the probability of player A winning the match?

Player A starts serving first.n is even.

Total number of game to be played is n+k;$0\leq k \leq n-1$.

So n wins of A can divided into two parts

1.m wins when A is serving.

2.n-m wins when B is serving.

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Formally stated question:

Two players A and B play tennis for several games. Player A has different probability to win at different games. At $(2k-1)$-th game, the probability that A wins is $p$, while at $2k$-th game, the probability that A wins is $q$, where $k = 1,2,3,\cdots,$. The player whose number of winning games firstly achieves $n$ will be the final winner ($n\geq 1$ is a given fixed number). The question is, what's the probability that A become the final winner.

Answer 1

Since my answer is a complicated sum of items, I am not sure whether it's the best answer, but I am sure it's correct (even if it's almost of no use since it could be too tedious to compute that sum!).

Let's begin from a simpler setting, where $p = q$. Suppose that they keep playing even if one player wins $n$ points first, until the total number of games achieves $2n-1$. Then

{$A$ first wins $n$ points} = {$A$ wins $k$ points in the first $2n-1$ games, for $k = n,\cdots,2n-1$}.

The probability of the right hand side can be easily computed since $k$ has a binomial distribution, thus

$$P(LHS) = P(RHS) = \sum_{k = n}^{2n-1}C_{2n-1}^k p^k(1-p)^{2n-1-k},$$ where $C_b^a$ is the binomial coefficient, and is sometimes denoted by $_b C_a$ or $(^b_a)$. Now we come to deal with the original problem. The equation

{$A$ first wins $n$ points} = {$A$ wins $k$ points in the first $2n-1$ games, for $k = n,\cdots,2n-1$}

is still correct, but now the hard point is, $P(RHS)$ is harder to compute. But we can figure it out by analysing more details. Suppose that $A$ wins $k$ points in the first $2n-1$ games, and among those $k$ points, $m$ points are from odd-ordered games (i.e., whose serial number is an odd number), and $k-m$ points are from even-ordered games, then

$$P(RHS) = \sum_{k = n}^{2n-1}\left[ \sum_{m = 1}^{n} \left(C_{n}^m p^m(1-p)^{n-m}\right)\cdot \left(C_{n-1}^{k-m} q^{k-m}(1-q)^{n-1-k + m}\right) \right],$$ where one should notice that there are only $n$ odd numbers and $n-1$ even numbers in $1,\cdots,2n-1$.