Vega [@vega] showed that for the random tournament $H \sim \mathcal{T}(h,1/2)$ w.h.p. there is no ordering with $|F_{\sigma, H}|>\frac{1}{2}\binom{h}{2} + 2h^{3/2}.$ So it seems quite likely that the $\sqrt{\log h}$ term in the above lemma is not necessary. In fact for a different definition of quasirandomness we do obtain a result which is best possible up to a constant factor.

\[lem:simple-is-quasirandom2\] If $H$ is a non-transitive $h$-vertex [$\mathcal{T}_{cliq}$-forcing]{} tournament then for any disjoint subsets $U,W \subseteq V(T)$ we must have $|e(U,W)|\le \frac{1}{2}|U||W| + 2h^{3/2}$.

Assume towards a contradiction that there are disjoint sets $U,W\subseteq V(T)$ for which $|e(U,W)| > \frac{1}{2}|U||W| + 2h^{3/2}$. Using these sets we will find many orderings $\sigma$ such that $|F_{\sigma, H}|\ge h^2/4+2h^{3/2}$. Let $S= V(H) \setminus (U \cup W).$ We will always place all vertices of $U$ before all vertices of $W$ in the orderings $\sigma$. Also if $e(S,U \cup W) \ge e(U \cup W, S)$ we place all the vertices of $S$ before all vertices of $U \cup W$ and behind otherwise. Within the sets $U,W,S$ we take the orderings which have more forward edges than backward edges. Note that for any subset of $T$ exactly one out of each of its orderings and its 