Solution :

Probability of getting a 6 = $$\frac{1}{6}$$

Probability of A Winning is = $$\frac{1}{6}$$

Probability of A losing is = 1 – $$\frac{1}{6}$$ = $$\frac{5}{6}$$

Probability of B winning is = $$\frac{1}{6}$$

Probability of B losing is = $$\frac{1}{6}$$

If A starts the game , the probability of A winning ,

P(A) + $$P(\bar{A})P(\bar{B})P(A)+P(\bar{A})P(\bar{B})P(\bar{A})P(\bar{B})P(A)$$

⇒$$\frac{1}{6} + \frac{5}{6}* \frac{5}{6} *\frac{1}{6} +\frac{5}{6}* \frac{5}{6}* \frac{5}{6}* \frac{5}{6}* \frac{1}{6}* $$+ ……………….

⇒$$\frac{1}{6}\left [ 1+ \frac{5}{6}* \frac{5}{6}+\frac{5}{6}* \frac{5}{6}*\frac{5}{6}* \frac{5}{6}+….\right ]$$

⇒ $$\frac{1}{6}\left [1 + (\frac{5}{6}) ^{2}+(\frac{5}{6}) ^{4}+ …. \right ] $$

⇒ $$\left [ 1 + a^{2} +a^{4}+….\right ]$$

$$\Rightarrow $$ $$\left [ \frac{1}{1-a^{2}} \right ]$$

a = $$\frac{1}{6} $$, r =$$ (\frac{5}{6})^{2}$$ =$$ \frac{25 }{36}$$

a = $$\frac{1}{6} $$ , r = a = $$\frac{25}{36} $$

$$S_{\infty } $$= $$\frac{\frac{1}{6}}{1-\left ( \frac{25}{36} \right )}$$

$$\frac{\frac{1}{6}}{(\frac{36-25}{36})} $$ = $$ \frac{6 }{11}$$