11. (3b,9-11) In the conditions of a chess match, the winner is declared as the one who outperforms the opponent by two wins. Draws do not count. The probabilities of winning for the opponents are equal. The number of decisive games in such a match is a random variable. Find its mathematical expectation.
# Solution.

Let $\mathrm{X}$ be the number of successful games. At the beginning of the match, the difference in the number of wins between the two participants is zero. Let's list the possible cases of two successful games, denoting a win by the first participant as 1 and a win by the second participant as 2: $11, 12, 21, 22$. Two out of the four cases mean a win in the match, specifically, 11 and 22. The other two cases bring us back to the initial state with a zero difference in the number of wins. Thus, with a probability of $\frac{2}{4}=\frac{1}{2}$, the number of wins $X=2$ and with the same probability $X=2+\tilde{X}$ where $\tilde{X}$ is distributed the same as $X$. From this, we derive the equation for the expectation

$$
\begin{gathered}
\mathrm{M} X=2 \cdot \frac{1}{2}+(2+\mathrm{M} \tilde{X}) \cdot \frac{1}{2}=2 \cdot \frac{1}{2}+(2+\mathrm{M} X) \cdot \frac{1}{2} \\
\frac{1}{2} \mathrm{M} X=2 ; \mathrm{M} X=4
\end{gathered}
$$

Answer: 4.