33. Two players $A$ and $B$ play rock-paper-scissors continuously until player $A$ wins 2 consecutive games. Suppose each player is equally likely to use each hand-sign in every game. What is the expected number of games they will play?
33. Answer: 12 .
Let $E$ be the expectation. If $A$ does not win, the probability is $2 / 3$ and the game restarts. If $A$ wins and then does not win, the probability is $(1 / 3)(2 / 3)$ and the game restarts. The probability that $A$ wins two consecutive games is $(1 / 3)(1 / 3)$. Then
$$
E=\frac{2}{3} \times(E+1)+\frac{2}{9} \times(E+2)+\frac{1}{9} \times 2 \text {. }
$$

Solving the equation, we get $E=12$.