A and B play a game of tennis. The situation of the game is as follows: if one scores two consecutive points after a deuce, he wins; if loss of a point is followed by win of a point, it is deuce. The chance of a server to win a point is 2/3. The game is at deuce and A is serving. Probability that A will win the match is (serves are changed after each game)

Let us assume that A wins after n deuces, n = 0, 1, 2, 3, …….

The probability of a deuce is (2/3) x (2/3) + (1/3) x (1/3) = (5/9). [A wins his serve, then B wins his serve or A loses his serve.] So, the probability that A wins game after n deuces is (5/9)ⁿ x (2/3) x (1/3). [After nth deuce, A serves and wins, then B serves and loses.] Therefore, the required probability of 'A' winning the game is

$\sum _{\mathrm{n}=0}^{\mathrm{\infty }} {\left(\frac{5}{9}\right)}^{\mathrm{n}}\times \frac{2}{3}\times \frac{1}{3}=\frac{1}{1-\frac{5}{9}}\times \frac{2}{9}=\frac{1}{2}$