Players A and B roll a pair of dice in turn, with player A rolling first. A's objective is to obtain a sum of 6 (on one roll), while B's ojective is to obtain a sum of 7. The winner is the first player to reach his or her objective. What's the probability that player A wins?

Does my approach make any sense? I'm probably making this way too hard.

I let $\displaystyle A_{W} $ be the event that A wins, and $\displaystyle A_{i} $ be the event that player A rolls a sum of i (i=2,3,...,12) on his first roll.

so $\displaystyle P(A_{W}) = P(A_{W}|A_{2})P(A_{2})+ P(A_{W}|A_{3})P(A_{3}) + P(A_{W}|A_{4})P(A_{4}) $ $\displaystyle + P(A_{W}|A_{5})P(A_{5}) + P(A_{W}|A_{6})P(A_{6}) + P(A_{W}|A_{7})P(A_{7}) + P(A_{W}|A_{8})P(A_{8}) $ $\displaystyle + P(A_{W}|A_{9})P(A_{9}) + P(A_{W}|A_{10})P(A_{10}) + P(A_{W}|A_{11})P(A_{11}) + P(A_{W}|A_{12})P(A_{12}) $

when $\displaystyle i \not= 6$, $\displaystyle P(A_{W}|A_{i}) = \frac {30}{36}P(A_{W}) $, which is the probability that player B does not win on his first roll times the probability that A wins (assuming independence of course)

then $\displaystyle P(A_{W}) = \frac {30}{36}P(A_{W})\frac{1}{36} + \frac {30}{36}P(A_{W})\frac{2}{36} $ $\displaystyle + \frac {30}{36}P(A_{W})\frac{3}{36} + \frac {30}{36}P(A_{W})\frac{4}{36} $ $\displaystyle + \frac{5}{36} + \frac {30}{36}P(A_{W})\frac{6}{36} + \frac {30}{36}P(A_{W})\frac{5}{36} $ $\displaystyle + \frac {30}{36}P(A_{W})\frac{4}{36} + \frac {30}{36}P(A_{W})\frac{3}{36} $ $\displaystyle + \frac {30}{36}P(A_{W})\frac{2}{36} + \frac {30}{36}P(A_{W})\frac{1}{36}$

and then solve for $\displaystyle P(A_{W}) $

I get 30/61, or about 49%