## Task A-4.4.

Players $A, B$ and $C$ take turns rolling a die. What is the probability that $C$ gets a higher number than the other two players?
## Solution.

Let $\mathcal{D}$ be the event "player $C$ has a higher number than the other two players," and let $p_{k}$ be the probability of event $\mathcal{D}$ given that player $C$ has rolled the number $k$ (for $k=1,2,3,4,5,6$).

If player $C$ rolls the number $k$, then his number is the highest if players $A$ and $B$ roll numbers less than $k$, i.e., some of the numbers $\{1, \ldots, k-1\}$.

The probability of this is $p_{k}=\frac{(k-1)^{2}}{36}$.

Finally, the desired probability is

$$
\frac{1}{6} p_{1}+\frac{1}{6} p_{2}+\frac{1}{6} p_{3}+\frac{1}{6} p_{4}+\frac{1}{6} p_{5}+\frac{1}{6} p_{6}
$$

so by substitution we get

$$
\begin{aligned}
\frac{1}{6}\left(p_{1}+p_{2}+p_{3}+p_{4}+p_{5}+p_{6}\right) & =\frac{1}{6}\left(\frac{0}{36}+\frac{1}{36}+\frac{4}{36}+\frac{9}{36}+\frac{16}{36}+\frac{25}{36}\right) \\
& =\frac{0+1+4+9+16+25}{6 \cdot 36}=\frac{55}{216}
\end{aligned}
$$

Note: A student who determines only some of the probabilities $p_{k}$ should receive:

3 points if they determined all except one probability;

2 points if they determined at least two probabilities and at least one of $p_{3}, p_{4}, p_{5}$;

1 point if they determined at least one of the probabilities $p_{3}, p_{4}, p_{5}$ or both $p_{2}$ and $p_{6}$.