13. The game "Unicum". Petya and 9 other people are playing such a game: each throws a die. A player wins a prize if they roll a number of points that no one else has rolled.

a) (6th - 11th grades, 1 point) What is the probability that Petya will win a prize?

b) (8th - 11th grades, 3 points) What is the probability that someone will win a prize?
Solution. a) Suppose (for clarity only) that Petya throws before everyone else. He scores some number of points. The probability that each subsequent player will score a different number of points is $\frac{5}{6}$. Since the throws are independent, the probability that the nine remaining players will all score differently from Petya is $\left(\frac{5}{6}\right)^{9} \approx 0.194$.

b) Let's number the players and consider the events $A_{j}=\{j-\text{th player wins a prize}\}$, where $j=1,2, \ldots, 10$. From the solution in part a), we see that

$$
\mathrm{P}\left(A_{1}\right)=\left(\frac{5}{6}\right)^{9}
$$

The probability that the first two players win prizes is found similarly:

$$
\mathrm{P}\left(A_{1} \cap A_{2}\right)=1 \cdot \frac{5}{6} \cdot\left(\frac{4}{6}\right)^{8}=\frac{5 \cdot 4^{8}}{6^{8}}
$$

Similarly, the probability that the first $k$ players win prizes is

$$
\mathrm{P}\left(A_{1} \cap A_{2} \cap \ldots \cap A_{k}\right)=1 \cdot \frac{5}{6} \cdot \frac{4}{6} \cdot \ldots \cdot \frac{7-k}{6} \cdot\left(\frac{6-k}{6}\right)^{10-k}, \text { where } k=1, \ldots, 6
$$

We are interested in the probability of the event $A=\{\text{at least one player wins a prize}\}$. This event is the union of the events $A_{j}: A=A_{1} \cup A_{2} \cup \ldots \cup A_{10}$. By the formula for the probability of the union of events (sometimes called the inclusion-exclusion principle), we find:

$$
\begin{aligned}
\mathrm{P}(A)= & \mathrm{P}\left(A_{1}\right)+\mathrm{P}\left(A_{2}\right)+\ldots+\mathrm{P}\left(A_{10}\right)- \\
& -\left(\mathrm{P}\left(A_{1} \cap A_{2}\right)+\mathrm{P}\left(A_{1} \cap A_{3}\right)+\ldots+\mathrm{P}\left(A_{9} \cap A_{10}\right)\right)+ \\
& +\left(\mathrm{P}\left(A_{1} \cap A_{2} \cap A_{3}\right)+\mathrm{P}\left(A_{1} \cap A_{2} \cap A_{4}\right)+\ldots+\mathrm{P}\left(A_{8} \cap A_{9} \cap A_{10}\right)\right)- \\
& -\ldots \\
& \ldots-\mathrm{P}\left(A_{1} \cap A_{2} \cap \ldots \cap A_{10}\right)
\end{aligned}
$$

The formula looks daunting, but its essence is simple - the probabilities of each winning are added, then the probabilities of each pair winning are subtracted, then the probabilities of each trio winning are added, and so on. The last term (with a minus sign) is the probability of all ten winning simultaneously. In reality, more than five players cannot win, so all terms involving the probability of six or more winning simultaneously are zero. Note also that in each row, the terms are the same, and their number is $C_{10}^{k}$, where $k=1, \ldots, 5$ is the row number. For example, there are $C_{10}^{2}=45$ terms of the form $\mathrm{P}\left(A_{i} \cap A_{j}\right)$.

Taking these two observations into account, we get:

$$
\begin{aligned}
& \mathrm{P}(A)=C_{10}^{1} \mathrm{P}\left(A_{1}\right)-C_{10}^{2} \mathrm{P}\left(A_{1} \cap A_{2}\right)+C_{10}^{3} \mathrm{P}\left(A_{1} \cap A_{2} \cap A_{3}\right)-C_{10}^{4} \mathrm{P}\left(A_{1} \cap A_{2} \cap A_{3} \cap A_{4}\right)+ \\
& \quad+C_{10}^{5} \mathrm{P}\left(A_{1} \cap A_{2} \cap A_{3} \cap A_{4} \cap A_{5}\right)= \\
& =10 \cdot \frac{5^{9}}{6^{9}}-45 \cdot \frac{5 \cdot 4^{8}}{6^{9}}+120 \cdot \frac{5 \cdot 4 \cdot 3^{7}}{6^{9}}-210 \cdot \frac{5 \cdot 4 \cdot 3 \cdot 2^{6}}{6^{9}}+252 \cdot \frac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1^{5}}{6^{9}} \approx 0.919
\end{aligned}
$$

Surprisingly, the probability that at least one player will win is quite high.

Answer: a) $\left(\frac{5}{6}\right)^{9} \approx 0.194 ;$ b) approx. 0.919.