For how many dice is the probability the greatest that exactly one six will appear when the dice are rolled simultaneously?
With a die, we can roll six different numbers, and with $n$ dice, this results in $6^{n}$ possibilities. Among these, there are $n \cdot 5^{n-1}$ cases where exactly one six is rolled, as this six can appear on any of the $n$ dice, and the other $n-1$ dice can independently show any of the other 5 numbers. The probability of rolling exactly one six is therefore

$$
V_{n}=\frac{n \cdot 5^{n-1}}{6^{n}}
$$

The question is, for which $n$ will this probability be the largest. It can be seen that $V_{1}=\frac{1}{6}, V_{2}=\frac{10}{36}, V_{3}=\frac{3 \cdot 25}{6 \cdot 36}=\frac{25}{72}$, i.e., $V_{1}<V_{2}<V_{3}$. If $n>5$, then $V_{n}>V_{n+1}$. (After this, the $V_{n}$ probability decreases). Therefore, the probability in question will be the largest for 5 and 6 dice, and it will be the same, namely $\left(\frac{5}{6}\right)^{5} \approx 0.4$ (which is surprisingly large).

Based on the work of Ambrus Gergely (Szeged, Radnóti M. Kísérl. Gimn., 10th grade)