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 $$\mathtt{E}_A^{>s}p\iff (\exists A_0\subset A:|A_0|\geq s|A| \text{ and }a\in A_0\implies \mathtt{B}_ap)$$

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 of $s=1$ 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.