For how many dice is the probability the greatest that, when the dice are rolled simultaneously, exactly one of the numbers rolled is a six?
When $n$ dice are rolled simultaneously, the result can be represented by an ordered sequence of $n$ numbers, each of which is one of the first six natural numbers. (We imagine the dice are numbered, and we write down the numbers in the order of the dice.) Considering all such outcomes equally likely, we can reason as follows: the total number of cases is $6^n$, since we are dealing with the $n$-th class permutations with repetition of 6 elements. The number of favorable cases is $n \cdot 5^{n-1}$, because if the first element of a favorable sequence is 6, then the others are one of the numbers $1,2,3,4,5$, i.e., a permutation with repetition of 5 elements of the $(n-1)$-th class, and the same is true if the second, ..., $n$-th element of the sequence is the only 6. Therefore, the probability we are looking for as a function of $n$ is

$$
P(n)=\frac{n \cdot 5^{n-1}}{6^{n}}=\frac{n}{6} \cdot\left(\frac{5}{6}\right)^{n-1}
$$

We will show that the function $P(n)$ defined on the set of natural numbers has a maximum at $n=5$ and $n=6$. Consider the ratio of $P(n+1)$ and $P(n)$:

$$
q(n)=\frac{P(n+1)}{P(n)}=\frac{n+1}{n} \cdot \frac{5}{6}=\frac{5 n+5}{5 n+n}
$$

We see that for $n=5$, $q(5)=1$, meaning that the probabilities $P(5)$ and $P(6)$ are equal. Furthermore, if $n<5$, then $q(n)>1$, so $P(n+1)>P(n)$. In other words, moving backward from 5 and forward from 6, we get smaller and smaller values of $P(n)$. This proves our statement. (It is also worth mentioning that the common value of $P(5)$ and $P(6)$ is $(5 / 6)^{5}=0.40187 \ldots, P(3)=0.35, P(10)=0.32$.)

Katona Klára (Budapest, Fazekas M. Gyak. Gimn., II. o. t.) Tarjányi Zsolt (Szeged, Radnóti M. Gimn., III. o. t.)

Remarks. 1. If $P_{k}(n)$ is the probability that exactly $k$ out of $n$ rolls are 6, then

$$
\begin{gathered}
P_{k}(n)=\binom{n}{k} \frac{5^{n-k}}{6^{n}} \\
q_{k}(n)=\frac{P_{k}(n+1)}{P_{k}(n)}=\frac{5 n+5}{5 n+5+n-6 k+1}
\end{gathered}
$$

from which, similarly to the given solution, it follows that the probability is maximal for $n=6 k-1$ and $n=6 k$.

2. Many solutions used a "big gun": they sought the maximum of the function $6 \cdot P(x) = x \cdot (5 / 6)^{x-1}$ extended to all positive $x$ by differentiation: let $5 / 6 = t$ (a constant), then

$$
6 P^{\prime}(x)=t^{x-1}+x t^{x-1}(\log t)=t^{x-1}(1+x \log t)
$$

where "log" denotes the natural (base $e$) logarithm. The derivative vanishes only at the following point:

$$
x=-\frac{1}{\log t}=\frac{1}{\log 6-\log 5}=\frac{\lg e}{\lg 6-\lg 5}=\frac{0.4343}{0.7782-0.6990}=5.5
$$

This is the only place where a maximum can occur.

However, it is still necessary to examine the function at the points 5 and 6 that bracket this point, and to prove that it is indeed a maximum. The second derivative is even more complicated, and the numerical calculations that were performed in the simpler solution cannot be avoided.

Moral: it can be useful—and not shameful—to occasionally include simple numerical trials in our examination.