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 game is, (serves are changed after each point)
(a) 3/5
(b) 2/5
(c) 1/2
(d) 4/5
The answer is c)
but i am getting D)
This is how i am getting D)
The probability of getting a deuce is 4/9
so probability that A wins is $$\frac{2}{3})^2[1+4/9.....]$$
 A: *

*$\Pr\left(A\text{ wins in }2\text{ scores}\right)+\Pr\left(A\text{ looses in }2\text{ scores}\right)=1-\Pr\left(\text{deuce after }2\text{ scores}\right)$

*$\Pr\left(A\text{ wins in }2\text{ scores}\right)=\Pr\left(A\text{ looses in }2\text{ scores}\right)$
The first is evident.
The second because in both cases one of them must win a serve point and win
a non-serve point.
Also we have:
$\Pr\left(A\text{ wins}\right)=\Pr\left(A\text{ wins in }2\text{ scores}\right)+\Pr\left(\text{deuce after }2\text{ scores}\right)\Pr\left(A\text{ wins}\right)$
Combining this leads to:
$$\Pr\left(A\text{ wins}\right)=\frac{1}{2}$$
A: The answer is $\frac 12$, because of the symmetry. To note this find a sequence of points in which A wins and then you can easily find a sequence of points with the same probability that wins for B and vice versa.
A: My answer is:
case 1: if A wins the first point 
   probability of winning of A = {(2/3)(1/3)}+{(2/3)(2/3)(2/3)(1/3)}+{(2/3)(2/3)(2/3)(2/3)(2/3)*(1/3)}...... and so on.
   =2/9[1+(4/9)+(16/81).....]
the above term is a G.P.
   =2/9[1/(1-4/9)]
   =2/9[9/5]
   =2/5
case 2: if B wins the first point
   probability of winning of = {(1/3)(1/3)(2/3)}+{(1/3)(1/3)(1/3)(1/3)(2/3)}....and so on
the above term is also a G.P.
   =2/(3^3)[1+(1/9)+(1/81)....]
   =2/27[1/(1-(1/9)]
   =(2/27)*(9/8)
   =2/24
probability= case 1 + case 2
           = (2/5)+(2/24)
           = (48+10)/120
           = 29/60.
