## Task 19/76

In a dice game with 6 dice, only those throws should be evaluated where at least one 1 or at least one 5 or at least 3 of the same (any) numbers appear.

What is the probability that no evaluation takes place?
For each throw with 6 dice, each of which can show the numbers from 1 to 6, there are a total of $6^{6}=46656$ variations possible. The number of unfavorable cases can be found through the following consideration:

Unvalued are the cases where the numbers 2, 3, 4, or 6 occur at most twice, thus (initially without considering the order):

$$
\begin{aligned}
& 223344 \quad 223366 \quad 224466 \quad 334466 \\
& 223346 \quad 223446 \quad 233446 \quad 223466 \\
& 234466 \quad 233466
\end{aligned}
$$

The first 4 of these combinations can, due to the equality of two numbers each, occur in $\frac{6!}{2!\cdot 2!\cdot 2!}=90$ ways, and the remaining 6 in $\frac{6!}{2!\cdot 2!\cdot 1!\cdot 1!}=180$ ways. Thus, the number of unfavorable cases is $4 \cdot 90+6 \cdot 180=1440$, and the probability of an unvalued throw is

$$
\frac{1440}{46656}=\frac{5}{162} \approx 0.0309 \approx 3.1\%
$$