6、A and B play a game with the following rules: In the odd rounds, A's probability of winning is $\frac{3}{4}$, and in the even rounds, B's probability of winning is $\frac{3}{4}$. There are no ties in any round. The rule is: The game ends when one of them wins 2 more rounds than the other. What is the expected number of rounds played by A and B when the game ends? $\qquad$
6. $\frac{16}{3}$

Analysis: Let the expected number of games played by A and B be $x$. It is easy to see that in the first two games, the probability of a winner being determined is $\frac{3}{8}$, and the probability of no winner being determined is $\frac{5}{8}$. When no winner is determined, the expected number of additional games to be played is $x$, so we have $\frac{3}{8} \times 2+\frac{5}{8} \cdot(x+2)=x$, which gives $x=\frac{16}{3}$