Probability of winning a tennis match with a two point difference? Let $A$ and $B$ be two tennis players, playing against each other.
A win of the game is considered when one of the two players is leading by two points (e.g. $A$ scores $5$ points vs. $B$ scoring $3$ points).
The probability of a winning a point when $A$ is serving is $\frac{3}{5}$ the probability of a winning a point when $B$ is serving is $\frac{2}{3}$
$A$ serves first and they take turns after each serve. What is the probability of $A$ winning the game.
My progress so far:
I realise that A wins the whole game when the following sequence occurs: $WLW$ (where $W$ denotes a win and $L$ denotes a loss). However, also a game win can occur when we have $WLLLW$ or $WLLLLLW$ etc. If $A$ wins a point and $B$ wins a point, the score is essentially re-set. From here on, I am unsure how to continue...
 A: P(wins first point) = 3/5
P(wins second point) = 1/3
so he can win those two points with probability 1/5 and win in first two points - but if he wins the first point, then loses the second point, with probability 2/5 - he is back to square one - if he loses the first point and wins the second point (prob 2/15) - he is back to square one - if he loses two points - he lost (prob 4/15)
so after two points, he either wins, loses, or is back to square one with the same probability of winning - therefore 
P(winning) = M = P(wins in two points) + M x P(back to a draw after two points)
P(winning) = M = (3/5)(1/3) + (3/5)(2/3)M + (2/5)(1/3)M + 0 x(2/5)(2/3)
M = 1/5 + (2/5)M + (2/15)M
M = 3/15 + (6/15)M+ (2/15)M
M = 3/15 + (8/15)M
(7/15)M = 3/15 
M = 3/7
Answer - 3/7
A: Let E be the event sania winning the match, then $P(E)
                                        =0.63$ (given)
Since $P(\text{Sania loosing the match } )= P(\text{not } E)=P(E)
                                 =1-P(E)=1-0.63
                                 =0.37$
