Probability to win a dice game A and B alternately throw a pair of die until one of them wins. A is considered winner
if he gets a total 7 and B is considered winner, if he gets a total of 5. If A starts the
game. then what is the probability that one of the die shows face 3 on the last throw.
I am not getting any start . Can anybody help me in this . 
 A: First off: The chance that A wins in any single attempt is 6/36, i.e. 1/6 - if you write out all possible results for a 2-dice-throw, this gets obvious:
(1,6), (6,1), (2,5), (5,2), (3,4), (4,3) are possible wins for A.
Chance to win for B is 4/36, i.e. 1/9:
(1,4), (4,1), (2,3), (3,2)
The chance that there is a three in the result is 1/3 for A and 1/2 for B.
With this knowledge you can now build an infinite sum:
Possibility 1: A wins and throws 3: $\frac16*\frac13$
Possibility 2: A loses, B wins and throws 3: $\frac56*\frac19*\frac12$
Possibility 3: A loses, B loses, A wins and throws 3: $\frac 56*\frac89*\frac16*\frac13$
Possibility 4: ... : $\frac56*\frac89*\frac56*\frac19*\frac12$
...
So the total sum is:
$$\frac1{18}+\frac5{108}+\sum_{i=1}^\infty\left(\frac56*\frac89\right)^i*\frac 1{18}+\sum_{j=1}^\infty\left(\frac56*\frac89\right)^j*\frac56*\frac1{18}=$$
$$\frac1{18}+\frac5{108}+\frac{20}7*\frac1{18}+\frac{20}7*\frac56*\frac1{18}=\frac{11}{28}$$
Are you sure that the result is 9/28?
EDIT: (I can't answer your comment directly, as I'm a new member). The sums should be correct, write the first few summands out, if you have doubts.
