Let the probability that A gets a six be $\frac{1}{6}$

$\Rightarrow$ $p=\frac{1}{6}$ and hence $q=1-\frac{1}{6} =\frac{5}{6}$

Probability of A's winning $=P(A)+P(\bar {A} \bar {B} A)+P(\bar {A} \bar {B} \bar {A} \bar {B}A)+.....$

$=P+Pq^2+Pq^4+.....$

This is in geometric progression, where summation is $S_n =\frac{P}{1-q^2}$

Substituing for P and q, $\large\frac{1/6}{1-(5/6)^2}=\frac{1/6}{\frac{36-25}{36}}$

Therefore The required probability of winning is P(A's winning)=$\large\frac{6}{11}$

and P(B's winning)=$1-\large\frac{6}{11}=\frac{5}{11}$