Large-order reverse-strategy conjecture for the permutation avoidance game

Let $k\geq 3$ be the pattern length, let $S_n$ denote the symmetric group on $n$ elements, and let $\operatorname{sg}(S_n,k)$ denote the Sprague–Grundy value of the permutation avoidance game PAP on $S_n$ with pattern length $k$. The reverse strategy is the strategy in which Player II responds by choosing the reverse of Player I's chosen pattern.

Large-order reverse-strategy conjecture. For every $k\geq 3$, there exists $n_0(k)$ such that the reverse strategy is a winning strategy for Player II on $S_n$ for all $n\geq n_0(k)$. In particular,

$$\operatorname{sg}(S_n,k)=0$$

for all such $n$.

Computations suggest that small values of $n$ can obstruct the reverse strategy, while the paper proves that it succeeds for $k=4$ whenever $n\geq 10$. The conjecture asserts that an analogous eventual result holds for every pattern length $k\geq 3$.