TITLE: Tennis related probability question
QUESTION [1 upvotes]: Found this question quite puzzling, was hoping to get some help with it :).

A tennis match is played as a series of POINTS, contained within
  GAMES, which in turn are contained within SETS.
If a player reaches 4 or more points won in a game, and also has won 2
  points more than their opponent, then they win that game.
If a player reaches 6 or more games won in a set, and also has won 2
  games more than their opponent, then they win that set. If both
  players reach 6 games, then a TIE-BREAK is played to determine the
  winner of the set.
Suppose Djokovic plays a set of tennis against Murray under these
  rules, and his probablity of winning a point is 0.54. Assuming the
  probability of winning a point remains constant throughout, what would
  be the probability of Djokovic losing the set 2-6?

I'm struggling to figure out the probability of Djokovic winning a game, let alone a set. My thinking so far is you could add the probability of the case where he wins 4 points in a row, with the probability that he wins four and Murray gets one as well, then add the infinite amount of scenarios where Djokovic wins by 2 points which meant adding this:
$$
\sum_{n=4}^{\infty}\binom{2n-2}{n}(0.45)^n(0.55)^{n-2}
\,
$$
That's when I realised I was way in over my head, grateful for any suggestions!

REPLY [2 votes]: If $p$ is the probability of winning a point, then the probability of winning a game can be found here or here and is $$P_g = p^4 + {4\choose 1}\cdot p^4(1-p) + {5\choose2}\cdot p^4(1-p)^2 + {6\choose 3}\cdot \frac{p^5(1-p)^3}{1-2p(1-p)}$$
To win $2$ games and lose $6$ games can be done in ${7\choose 2}$ ways (the last game is always a loss), so the probability for that set score is $$P={7 \choose 2}\cdot P_g^2(1-P_g)^6$$
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