6. In a match without ties, the game ends when one person wins 2 more games than the other, and the one with more wins is the winner. It is known that in the odd-numbered games, the probability of A winning is $\frac{3}{5}$; in the even-numbered games, the probability of B winning is $\frac{3}{5}$. Then the expected number of games played until the match ends is $\qquad$
6. $\frac{25}{6}$.

Let the expected number of games played until the end of the match be $E$.
The probability that player A wins the first game and player B wins the second, or player B wins the first and player A wins the second, is
$$
\frac{3}{5} \times \frac{3}{5}+\frac{2}{5} \times \frac{2}{5}=\frac{13}{25} .
$$

In this case, the expected number of additional games to be played is $E$, and the total expected number of games is $E+2$.

If player A wins both of the first two games or player B wins both, the probability is $1-\frac{13}{25}=\frac{12}{25}$, then the match ends exactly after two games. Therefore, $E=\frac{12}{25} \times 2+\frac{13}{25}(E+2) \Rightarrow E=\frac{25}{6}$.