\big )
                = 
                (n)_h \times p_H(\alpha ).$$ This gives *(i)*.

To see *(ii)*, note that $p_H(1/2) = 2^{-\binom {h}{2}}$ from the definition of $p_H(x)$ since $|F_{\sigma ,H}| + |B_{\sigma ,H}| = \binom h 2$ for every $\sigma \in S_h$. Also note that if $T$ is non-transitive then $B_{\sigma, H}>0$ for all $\sigma$. So, each term of $p_H(x)$ is zero for $x=1$ implying $p_H(1) = 0$.

Lastly, we show *(iii)*. For each $\sigma \in S_h$ let $\overline {\sigma }$ denote the ‘reversal of $\sigma $’, given by $\overline {\sigma } (i) = {\sigma }(h+1-i)$ for all $i\in [h]$. As this gives a bijection from $S_h$ to itself and $F_{\sigma ,H}=B_{\overline{\sigma} ,H}$, for all $\sigma \in S_h$, we have $$\begin{aligned}
            p_H(x)  & = \frac {1}{2h!} \sum _{\sigma  \in S_h} \Big ( x^{|F_{\sigma ,H}|} (1-x)^{|B_{\sigma ,H}|}+x^{|F_{\overline{\sigma} ,H}|} (1-x)^{|B_{\overline{\sigma} ,H}|} \Big )\\
                    & = \frac {1}{2h!} \sum _{\sigma  \in S_h} \Big ( x^{|F_{\sigma ,H}|} (1-x)^{|B_{\sigma ,H}|}+x^{|B_{\sigma ,H}|} (1-x)^{|F_{\sigma ,H}|} \Big )
        \end{aligned}$$ As $x^a(1-x)^b+x^b(1-x)^a$ is symmetric around $1/2$ for any $a,b\in[0,1/2]$ so is $p_H(x)$.

We are now ready to give the formal definition of [$\mathcal{T}_{cliq}$-forcing]{} which plays a central role in our arguments.

A tournament $H$ is said to be *[$\mathcal{T}_{cliq}$-forcing]{}* if $p_H(x) \neq 