Let $S_A$ denote the probability of $A$ winning, given he has won the previous game (but not the game before that; he has not yet won). Let $S_B$ be the probability of $A$ winning, given $B$ won the previous game (but also $B$ has not yet been declared winner). Then we get a set of recurrence relations as
$S_A = \frac{2}{3} \cdot 1 + \frac{1}{3} \cdot S_B,$
and
$S_B = \frac{1}{3} \cdot 0 + \frac{2}{3} \cdot S_A.$
(Can you explain the above relations?)
These are two linear equations in two unknowns, so we can solve them for $S_A$ and $S_B$. Finally, the answer to the question is given by considering the possible outcomes of the first single game, and adding up the probabilities:
$S = \frac{2}{3} \cdot S_A + \frac{1}{3} \cdot S_B.$
Solving the first set of equations in $S_A$ and $S_B$ results in $S_A = \frac{18}{21}$ and $S_B = \frac{12}{21}$. The last equation thus gives us the solution $\boxed{S = \frac{16}{21}}$.