Three friends A, B and C are playing with a pair of dice. They throw two dice alternately. Coming of a doublet on two dice leads to a success and the game stops. If A starts the game, then the probability of his winning, is :

$\frac{125}{216}$

According to problem three friends A, B and C are playing with a pair of dice. They throw two dice alternately. Coming of a doublet on two dice leads to a success and the game stops. 

The probability of getting a doublet is a success 

Possible Case= (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) = 6

Total no. of cases in throwing two dice = (6)2

                                                        = 36

So, the probability of getting a doublet in tossing two dice = 6/36 = 1/6

If A start the game then he will win if he gets a doublet so the possible case for this = A  gets a doublet and B & C will not get a doublet

             = (1/6)(5/6)(5/6)

             = 25/216

  But there are total three players 

So, the required probability = 3*(25/216)

                                         = 125/216