Petya and 9 other people are playing a game: each of them rolls a die. A player wins a prize if they roll a number that no one else rolls.

a) What is the probability that Petya will win a prize?

b) What is the probability that at least someone will win a prize?
a) Suppose (for illustrative purposes only) that Petya throws first. He scores some number of points. The probability that each subsequent player will score a different number of points is 5/6.

Since the throws are independent, the probability that the nine remaining players will all score differently from Petya is

$(5 / 6)^{9} \approx 0.194$.

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

$\mathrm{P}(A)=(2 / 6)^{9}$.

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

$\mathrm{P}\left(A_{1} \cap A_{2}\right)=1 \cdot \frac{5}{6} \cdot\left(\frac{4}{6}\right)^{8}=\frac{5 \cdot 4^{8}}{6^{9}}$

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

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

We are interested in the probability of the event $A=\{$ 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 (the inclusion-exclusion principle), we find:

$\mathrm{P}(A)=\mathrm{P}\left(A_{1}\right)+\mathrm{P}\left(A_{2}\right)+\ldots+\mathrm{P}\left(A_{10}\right)-\left(\mathrm{P}\left(A_{1} \cap A_{2}\right)+\mathrm{P}\left(A_{1} \cap A_{3}\right)+\ldots+\mathrm{P}\left(A_{9} \cap A_{10}\right)\right)+$

$+\left(\mathrm{P}\left(A_{1} \cap A_{2} \cap A_{3}\right)+\mathrm{P}\left(A_{1} \cap A_{2} \cap A_{4}\right)+\ldots+\mathrm{P}\left(A_{8} \cap A_{9} \cap A_{10}\right)\right)-\ldots-\mathrm{P}\left(A_{1} \cap A_{2} \cap \ldots \cap A_{10}\right)$.

The last term (with a minus sign) is the probability that all ten players win simultaneously. In reality, more than five players cannot win, so all terms involving the probability of six or more players 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.

Thus, $10 \cdot \frac{5^{9}}{6^{9}}-45 \cdot \frac{5 \cdot 4^{8}}{6^{9}}+120 \cdot \frac{5 \cdot 4 \cdot 3^{7}}{6^{9}}-210 \cdot \frac{5 \cdot 4 \cdot 3 \cdot 2^{6}}{6^{9}}+252 \cdot \frac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1^{5}}{6^{9}} \approx 0.919$.

## Answer

a) $(5 / 6)^{9} \approx 0.194 ;$ b) $\approx 0.919$.

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