A and B alternately throw a pair of dice. A wins if he throws 6 before B throws 7; and B wins if he throws 7 before Athrows 6. What are their respective chances of winning, if A throws the dice first?

The probability that A throws '6' = $$\frac{5}{36}$$ and 

The probability of A not getting '6' = $$1 - \frac{5}{36} = \frac{31}{36}$$ 

The probability that B throws '7' = $$\frac{6}{36}$$

The probability of A winning the game is given by,

= (In the first throw if A wins) or (In the 1st throw A looses, B also looses and in the 2nd throw A wins) or ( A and B looses in the 1st and 2nd throw's and A wins in the 3rd throw .....it goes on like that)

= $$\frac{5}{36} + \frac{31}{36}.\frac{30}{36}.\frac{5}{36} + \frac{31}{36}.\frac{30}{36}.\frac{31}{36}.\frac{30}{36}\frac{5}{36} + $$.......

It is in arithmetic progression,

= $$(\frac{5}{36})(\frac{1}{1 - \frac{31}{36}.\frac{30}{36}})$$

= $$(\frac{5}{36})(\frac{1296}{1296 - 930})$$

= $$(\frac{5}{36})(\frac{1296}{366})$$

= $$\frac{5 \times 6}{61} = \frac{30}{61}$$

Probability of B winning the game is given by,

= $$1 - \frac{30}{61}$$

= $$\frac{31}{61}$$

Hence, option B is the correct answer.