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) Find the expected number of points scored by the Experts team in 100 games.

b) Find the probability that in the next game, envelope No. 5 will be played.
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points, so in total there were

$6+k$ envelopes. In this case, the winning team definitely received a point for the question from the last envelope, while for the other $5+k$ questions, points could be received 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 $\quad \mathrm{E} X=0 \cdot C_{5}^{0} \cdot \frac{1}{2^{5}}+1 \cdot C_{6}^{1} \cdot \frac{1}{2^{6}}+\ldots+k C_{5+k}^{k} \frac{1}{2^{5+k}}+\ldots+5 C_{10}^{5} \frac{1}{2^{10}}$

Note that $k C_{n}^{k}=(n-k+1) C_{n}^{k-1}$, in particular, $k C_{5+k}^{k}=6 C_{5+k}^{k-1}$. Therefore,

$\mathrm{E} X=6 C_{6}^{0} \cdot \frac{1}{2^{6}}+6 C_{7}^{1} \cdot \frac{1}{2^{7}}+\ldots+6 C_{10}^{4} \cdot \frac{1}{2^{10}}=$

$6\left(C_{6}^{0} \cdot \frac{1}{2^{6}}+C_{7}^{1} \cdot \frac{1}{2^{7}}+\ldots+C_{10}^{4} \cdot \frac{1}{2^{10}}+C_{11}^{5} \cdot \frac{1}{2^{11}}+C_{12}^{6} \cdot \frac{1}{2^{12}}\right)-\left(C_{11}^{5} \cdot \frac{1}{2^{11}}+C_{12}^{6} \cdot \frac{1}{2^{12}}\right) .$.

In the first parenthesis, we have 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. Therefore,

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

We add 6 to get the total number of points (earned by both teams), and divide by two to get only the expected sum of points for the experts (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) Introduce indicators $\xi_{k}, 1 \leq k \leq 13: \xi_{k}=1$, if the envelope with number $k$ was played, and 0 otherwise. 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 from part a): $12 \cdot\left(1-\frac{C_{12}^{6}}{2^{12}}\right)$. 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 $\mathrm{E} \xi_{k}$ is also the probability of the $k$-th envelope being played.

## Answer

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