this gives $p_H\big ( \frac {1+\varepsilon }{2} \big ) = 2^{-\binom {h}{2}}$ and so $H$ is not [$\mathcal{T}_{cliq}$-forcing]{}.

The following lemma shows that, in addition to being almost regular, any [$\mathcal{T}_{cliq}$-forcing]{} $H$ must be strongly quasirandom.

\[lem:simple-is-quasirandom1\] If $H$ is a non-transitive $h$-vertex [$\mathcal{T}_{cliq}$-forcing]{} tournament then for every ordering $\sigma$ of $H$ $$|F_{\sigma, H}|=\frac{1}{2}\binom{h}{2} \pm h^{3/2}\sqrt{\log h}.$$

Assume that there is an ordering $\sigma$ such that $f=|F_{\sigma, H}|\ge h^2/4+h^{3/2}\sqrt{\log h}$. If we let $b=|B_{\sigma,H}|$ since $f+b=\binom{h}{2}$ we get $b \le h^2/4- h^{3/2}\sqrt{\log h}$ and $f-b \ge 2h^{3/2}\sqrt{\log h}.$

We will show that a single term of $q_H(x)$, $(1+x)^f(1-x)^b>h!$ if we choose $x=\frac34 h^{-1/2}\sqrt{\log h}$. This shows that $q_H(x)>1$ and we can complete the argument as in the previous lemma. To show the above inequality note that $$\begin{aligned}
    (1+x)^f(1-x)^b & > e^{(x-x^2)f+(-x-x^2)b}\\
                   & > e^{x(f-b)-x^2h^2/2}\\
                   & > e^{h \log h} =h^h >h!
    \end{aligned}$$ where in the first inequality we used $1+t>e^{t-t^2}$ for $|t| \le 1/\sqrt{3}$.

By a result of Spencer [@spencer1] it is known that for any tournament $H$ there is an ordering of its vertices $\sigma$ for which $|F_{\sigma, H}| = \frac{1}{2}\binom{h}{2} + \Omega(h^{3/2}).$ So the above result is best possible up to the $\sqrt{\log h}$ factor. Fernandez de la 