We shuffle a deck of French cards, then draw the cards one by one. On which turn is it most likely to draw the second ace?
If $1 \leq n \leq 52$, let $P(n)$ denote the number of permutations of a 52-card deck where the second ace appears in the $n$-th position. Clearly, the most likely position for the second ace to appear is where $P(n)$ has the largest value.

If the second ace is drawn in the $n$-th position (so $n > 1$), then the first ace can be in any of the first $n-1$ positions with equal probability, and the other two aces can be in any two of the remaining $52-n$ positions, independent of the first two. Since the 48 other cards can be in any of the 48! permutations, which is irrelevant to the question of the maximum, we have

$$
\frac{P(n)}{48!}=(n-1)\binom{52-n}{2}
$$

which is maximized when the product $(n-1)(51-n)(52-n)$ is maximized.

The maximum of the above product can be determined by analyzing the continuous function $(x-1)(51-x)(52-x)$ using calculus, or by examining the possible values of $n$. Below, we present a relatively quick, elementary, though not entirely precise method.

We can write the product of interest as

$$
\frac{1}{2}(2 n-2)(51-n)(52-n)=\frac{1}{2} q(n)
$$

The sum of the factors in the product $q(n)$ is constant, so we can apply the arithmetic mean-geometric mean inequality:

$$
q(n) \leqq\left(\frac{101}{3}\right)^{3}
$$

Thus, since $n$ is an integer,

$$
(n-1)(51-n)(52-n) \leqq\left[\left(\frac{101}{3}\right)^{2} \cdot \frac{1}{2}\right]=19079
$$

If $n=18$, then

$$
(n-1)(51-n)(52-n)=19074
$$

and it can be seen that if the value of $n$ changes by 1, the value of the product changes by much more than $19079-19074=5$, so it can only decrease. Therefore, the function $(x-1)(51-x)(52-x)$ has a local maximum in the interval $(17,19)$. If the product were to take a value greater than 19073 anywhere else in the interval [1,52], then by the same reasoning, it would also have a local maximum there, which is not possible for a cubic function.