In a tennis match, player A wins a point with probability p, and winning 4 or more points with a lead of two wins a game. A set of tennis is won by the first player to win at least six games, with a lead of two. (Ignoring tiebreaks for this question.)
What is the probability that player A wins a set by $a$ games, where $a>2$?
Following Probability of winning a game in tennis?, the probability of winning a game is given by:
$$P(Win\ a\ game) = p^4 + {4\choose 1}\cdot p^4(1-p) + {5\choose2}\cdot p^4(1-p)^2 + {6\choose 3}\cdot \frac{p^5(1-p)^3}{1-2p(1-p)}$$
Let $$G = P(Win\ a\ game)$$
then to win a set a games to b: $$ P(Score\ a:b) = {a + b-1\choose b}\cdot G^a \cdot (1-G)^b$$