14. What? Where? When? In the game "What? Where? When?", 13 envelopes with questions from the audience are played. The envelopes are chosen in a random order using a spinner. If the Experts answer correctly, they earn a point; if incorrectly, one point goes to the audience. The game ends as soon as one of the teams scores 6 points.

Assume that the teams of Experts and Audience are equally matched.
a) (8th - 11th grades, 2 points) Find the expected value of the number of points scored by the Experts team in 100 games.

b) (8th - 11th grades, 1 point) Find the probability that in the next game, envelope number 5 will be played.
Solution. a) Suppose that the losing team earned $k$ points in one game. The winning team earned 6 points, so a total of $6+k$ envelopes were played. The point for the question from the last envelope was definitely earned by the winning team, while for the remaining $5+k$ questions, points could be earned by either team with a probability of $1 / 2$. Therefore, the expected value of the random variable $X$ "the number of points earned by the losing team in one game" is

$$
\mathrm{E} X=0 \cdot C_{5}^{0} \cdot \frac{1}{2^{5}}+1 \cdot C_{6}^{1} \cdot \frac{1}{2^{6}}+\ldots+k C_{5+k}^{k} \frac{1}{2^{5+k}}+\ldots+5 C_{10}^{5} \frac{1}{2^{10}}
$$

We use the fact that $k C_{n}^{k}=(n-k+1) C_{n}^{k-1}$. In our case, this gives: $k C_{5+k}^{k}=6 C_{5+k}^{k-1}$;

$$
\begin{aligned}
& \mathrm{E} X=6 \cdot C_{6}^{0} \cdot \frac{1}{2^{6}}+\ldots+6 C_{5+k}^{k-1} \frac{1}{2^{5+k}}+\ldots+6 C_{10}^{4} \frac{1}{2^{10}}=6\left(C_{6}^{0} \cdot \frac{1}{2^{6}}+\ldots+C_{5+k}^{k-1} \frac{1}{2^{5+k}}+\ldots+C_{10}^{4} \frac{1}{2^{10}}\right)= \\
& =6\left(C_{6}^{0} \cdot \frac{1}{2^{6}}+\ldots+C_{5+k}^{k-1} \frac{1}{2^{5+k}}+\ldots+C_{10}^{4} \frac{1}{2^{10}}+C_{11}^{5} \frac{1}{2^{11}}+C_{12}^{6} \frac{1}{2^{12}}\right)-6\left(C_{11}^{5} \frac{1}{2^{11}}+C_{12}^{6} \frac{1}{2^{12}}\right)
\end{aligned}
$$

In the first parenthesis, the sum of probabilities that the losing team earns $k-1$ points ( $1 \leq k \leq 7$ ) when playing to 7 wins; it equals 1.

$$
\mathrm{E} X=6-\frac{6}{2^{12}}\left(2 C_{11}^{5}+C_{12}^{6}\right)=6-\frac{6}{2^{12}}\left(C_{11}^{5}+C_{11}^{6}+C_{12}^{6}\right)=6-\frac{6}{2^{12}} \cdot 2 C_{12}^{6}=6-\frac{6 C_{12}^{6}}{2^{11}}
$$

We add 6 to get the total sum of points (earned by both teams), divide by two to get only the expected sum of points for the losing team (since the teams are equally strong): $6-\frac{6 C_{12}^{6}}{2^{12}}$. Now we multiply the result by 100 since 100 games were played.

b) Again, we apply the method of indicators. Introduce random variables $\xi_{k}$, $1 \leq k \leq 13:$

$$
\xi_{k}=\left\{\begin{array}{l}
1, \text { if the envelope with number } k \text { was played, } \\
0, \text { if the envelope with number } k \text { was not played. }
\end{array}\right.
$$

Note that the sum of points earned by both teams in one game is equal to the sum of all $\xi_{k}$. The expected sum of points earned by both teams is known: $12 \cdot\left(1-\frac{C_{12}^{6}}{2^{12}}\right)$. All envelopes are equal, so we divide the found number by 13: $\mathrm{E} \xi_{k}=\frac{12}{13} \cdot\left(1-\frac{C_{12}^{6}}{2^{12}}\right)$. But the expectation of $\xi_{k}$ is also the probability of the envelope number $k$ being played.

Answer: a) $600 \cdot\left(1-\frac{C_{12}^{6}}{2^{12}}\right) \approx 465$; b) $\frac{12}{13} \cdot\left(1-\frac{C_{12}^{6}}{2^{12}}\right) \approx 0.715$.