Question 199: Two people, A and B, play a game with the following rules: In the odd-numbered rounds, A's probability of winning is $\frac{3}{4}$; in the even-numbered 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 person wins 2 more rounds than the other. Try to find: When the game ends, the expected number of rounds played by A and B.
Question 199, Solution: Let the required expectation be E.
The probability of the game ending in the first two rounds is: $\frac{3}{4} \times\left(1-\frac{3}{4}\right)+\frac{1}{4} \times\left(1-\frac{1}{4}\right)=\frac{3}{8}$.
If the game does not end in the first two rounds, then each of player A and B wins one round, returning to the initial state. From this point on, the mathematical expectation of the number of rounds played by A and B until the game ends is still E, which means that from the initial moment, the expectation for this scenario is $\left(1-\frac{3}{8}\right) \cdot(E+2)$. Therefore,
$E=2 \times \frac{3}{8}+\left(1-\frac{3}{8}\right) \cdot(E+2) \Rightarrow E=\frac{16}{3}$
In conclusion, the mathematical expectation of the number of rounds played by A and B is $\frac{16}{3}$.