[\min ( i-1, n/4)]$, this contradicts the quasirandom property $\mathcal{P}_5$ from [@CG] which requires that $$\sum_{v \in V(T_i)}|d^+_{T_i}(v)-d^-_{T_i}(v)|=o(n^2).$$ Thus $T_i$ is not quasirandom and so $H$ cannot be globally forcing.

**Remark:** Note that our argument shows that in order for $H$ to be globally forcing $N^*_T(H)$ must be asymptotically maximised when $T$ is the random tournament. In other words if we can find a sequence of $n$-vertex tournaments $T_n$ such that $N^*_{T_n}(H) \ge (2^{-\binom{h}{2}}+\delta)n^h$ our argument above (switching edges incident to one vertex at a time) shows that $H$ is not globally forcing. In fact it is enough to find a single such $T_m$ for some $m\ge h$ since if we let $T'_n$ be the blow up of $T_m$ with parts of size $\frac{n}{m}$ then $N^*_{T'_n}(H) \ge (2^{-\binom{h}{2}}+\delta)m^h\left(\frac{n}{m}\right)^h=(2^{-\binom{h}{2}}+\delta)n^h$ so $T'_n$ provides us with our sequence.

Local forcing and the counting polynomial {#sec: counting polynomial}
=========================================

To study locally forcing tournaments we introduce the following polynomial.

\[dfn: counting poly\] The *counting polynomial* of an $h$-vertex tournament $H$ is given by $$p_H(x) 
                := 
            {\mathbb E}_{\sigma \sim S_h} 
             \Big ( x^{|F_{\sigma ,H}|} (1-x)^{|B_{\sigma ,H}|}\Big ) = 
             \frac {1}{h!} \sum _{\sigma  \in S_h} \Big ( x^{|F_{\sigma ,H}|} (1-x)^{|B_{\sigma ,H}|} \Big ).$$

The following lemma 