Motivation. Suppose we are given $6$ boxes, arranged in the following manner:
$$\left[\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}\right]$$
Two of these boxes contain a present, and the remaining $4$ are empty. From the outside, no-one can tell which boxes contain a present. There are ${6 \choose 2} =15$ ways to put the presents in the two boxes. The goal is to find one box containing a present as quickly as possible.
Anna opens the boxes in row-wise, that is, in the order: $1,2,3,4,5,6$. Bert opens the boxes column-wise: $1,4,2,5,3,6$. The $15$ ways to distribute the two presents into the $6$ boxes, Anna finds the first present quicker than Bert $5$ times and Bert beats Anna only in $4$ scenarios (and there are $6$ ties). I think it is crazy that one arbitrary method (Anna's) should be better than another (Bert's) to uncover the first presents even if the two gift locations are picked at random!
Formalization and generalization. For $n\in\mathbb{N}$ we write $[n] := \{1,\ldots,n\}$. If $X$ is a set, then let $[X]^2 = \big\{\{x,y\}:x\neq y \in X\big\}$. By slight abuse of notation, we write $[n]^2$ instead of $[[n]]^2$. The collection of bijections $f:[n]\to[n]$ is denoted by $S_n$.
For the remainder of this post, let $n\geq 3$ be an arbitrary, but fixed integer.Let $a, b\in S_n$. We can think of $a$ as being the method that Anna uses to open the boxes $\{1,\ldots,n\}$, and $b$ is Bert's way of trying to find one present. The locations of the $2$ presents is encoded by an element $P\in [n]^2$. The number of $P\in [n]^2$ such that Anna finds one present quicker than Bert, and therefore wins, is
$$W(a,b) = \Big|\{P\in [n]^2: \min \big(a^{-1}(P)\big) < \min \big(b^{-1}(P)\big)\}\Big|.$$
We say $a\in S_n$ is better than $b\in S_n$ if $W(a,b) > W(b,a)$.
To me, it is not clear, whether "betterness" as defined above is a transitive relation, but this is not my main focus for this question. (It could well be that this relation is not transitive, see the intransitive dice.)
Questions.
Are there infinitely many integers $n\geq 3$ such that for every $f\in S_n$ there is $f' \in S_n$ such that $f'$ is better than $f$? And related to this:
Are there infinitely many integers $n\geq 3$ such that there is $f_0\in S_n$ such that no element of $S_n$ is better than $f_0$?