'A' and 'B' throws the dice alternatively and the sequence is ABABAB................
'A' wins sum of the dice is 8.
and probability of getting 8 is P(A)=5/36. [(2,6),(3,5),(4,4),(5,3),(6,2)]
same as 'B' wins sum of two dice is 7 and probability of getting sum 7 is P(B)=6/36.
Probability of 'A' wins the game is.
=P(A)+P($\bar{A}$)P($\bar{B}$)P(A)+P($\bar{A}$)P($\bar{B}$)P($\bar{A}$)P($\bar{B}$)P(A)+.........................
=$\frac{5}{36}$+$\frac{31}{36}*\frac{30}{36}*\frac{5}{6}$+$\frac{31}{36}*\frac{30}{36}*\frac{31}{36}*\frac{30}{36}*\frac{5}{6}$+.........
=$\frac{5}{6}$[1+$\frac{31}{36}*\frac{30}{36}$+$\frac{31}{36}*\frac{30}{36}*\frac{31}{36}*\frac{30}{36}$+.................]
=$\frac{5}{36}*(1-(\frac{31*30}{36*36}))^{-1}$
=$\frac{5}{36}*(\frac{36*36}{36*36-31*30})$
=0.49180
=$\frac{30}{61}$.
Hence option should be C
