**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: $1,4,2,5,3,6$. In 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. (*EDIT.* user14111 showed in the comment section that this relation is not transitive; the situation reminds me a little of the [intransitive dice](https://en.wikipedia.org/wiki/Intransitive_dice).)

**Questions.**  

1) 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:

2) 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 $f_0$?