Let $A$ be a finite set of agents and $\mathtt{B}_a$ a modal operator where $\mathtt{B}_ap$ means agent $a$ believes proposition $p$. For now I don't assume any properties of $\mathtt{B}_a$, though preserving conjunctions seems reasonable.

Given $0<s<1$ define an "$>sA$ believes" modal operator $\mathtt{E}^{>s}_A$ as follows $$\mathtt{E}_A^{>s}p\iff (\exists A_0\subset A:|A_0|> s|A| \text{ and }a\in A_0\implies \mathtt{B}_ap)$$

When is there a proposition $q$ that satisfies the following property? $$q\iff \mathtt{E}_A^{>s}(p\wedge q)$$

For instance take $s=\tfrac 12$. In this case the property is the following equivalence. $$q\iff \text{over half the agents believe }(p\wedge q)$$

I am struggling to write down such a $q$, so I thought maybe something is going on.

In the usual case (see below) one can use finite conjunctions to prove the usual infinitary definition of common belief has this fixed point property. However, the proof uses preservation of conjunctions, which $\mathtt E_A^{>s}$ does not satisfy.

Added.

Define $\mathtt{E}_A^k=\bigwedge _{(a_1,\dots ,a_k)\in A{^\times k}}\mathtt B_{a_1} \cdots \mathtt B_{a_k}$. Morally, $\mathtt E^k_Ap$ can be read like a Horner scheme: $$\text{everyone believes ($\cdots$ everyone believes (everyone believes $p$)$\cdots$)}$$ Define "common belief" as $\mathtt C_A=\bigwedge _{k\geq 1}\mathtt E^k_A$.