0
$\begingroup$

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 $a>2$?

$\endgroup$
0
$\begingroup$

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: $$ P(Score\ a:b) = {a + b-1\choose b}\cdot G^a \cdot (1-G)^b$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.