There is an uncountable set of $q\in (1/2,1)$ for which there are only two sets $M$ (which are complements of each other) such that $\sum_{m\in M}q^m=\frac{1}2\cdot \frac{1}{1-q}$ - and uncountably many of these forced sets are not of asymptotic density $1/2$.

To prove this, let us define, for any subset $M$ of the natural numbers, the summing function
$$f_M(q)=\frac{\sum_{m\in M}q^m}{\sum_{n\in\mathbb N}q^n}=(1-q)\cdot \sum_{m\in M}q^m$$
Note that for any $M$, the function $f_M$ is continuous where $|q|<1$ - and, in fact, the family of such functions is equicontinuous on this range.

Let's also define the left shift operator
$$L(M)=\{m' \in \mathbb N : m'+1\in M\}$$
and note that
$$f_{M}(q) = (1-q)\cdot \chi_{M}(0) + qf_{L(M)}(q)$$
where $\chi_M$ is the indicator function of $M$.

Let's say a set $M$ has $n$-bounded runs if there are no values $k$ such that $k\in M$ but $k+1,k+2,\ldots,k+n$ are all not in $M$ or such that $k\not\in M$ and the following $n$ integers are all in $M$.

We can now state a lemma:

**Lemma:** For any $n$, there is some $\varepsilon > 0$ such that if $M$ has $n$-bounded runs, then for all $q\in[1/2,1/2+\varepsilon)$ and all $M'\subseteq \mathbb N$ we have $f_M(q)=f_{M'}(q)$ implies $M=M'$.

**Proof.** First, we need to choose $\varepsilon$ so that $\left|f_{M}(q) - \frac{1}2\right| > \varepsilon$ for all sets $M$ with $n$-bounded runs and all $q\in [1/2,1/2+\varepsilon)$ - this is possible since $|f_M(1/2)-\frac{1}2|$ can never be less than $\frac{1}{2^{n+1}}$ for such a set, considering binary representations, and the family of functions $f$ is equicontinuous, so the values $|f_M(q)-\frac{1}2|$ must remain bounded away from zero for all $q$ close enough to $\frac{1}2$ - which allows us to choose a $\varepsilon$ satisfying our condition. Note that this, more helpfully, ensures that $f_{M}(q)$ cannot ever be in $[1-q, q]$.

One can now fix any $M$ with $n$-bounded runs and any $M'$ with $f_M(q)=f_{M'}(q)$ for some $q\in [1/2,1/2+\varepsilon)$ and prove inductively that $M=M'$. Consider the equation:
$$f_M(q)=f_{M'}(q)$$
$$(1-q)\chi_{M}(0)+qf_{L(M)}(q) = (1-q)\chi_{M'}(0)+qf_{L(M')}(q)$$
Note, however, that this expression is either less than $1-q$ (in which case it must be that $\chi_{M}(0)=\chi_{M'}(0)=0$) or greater than $q$ (in which case it must be that $\chi_{M}(0)=\chi_{M'}(0)=1$) by choice of $\varepsilon$. Thus, $0$ is either in both or neither of the sets. However, this also immediately then simplifies as:
$$f_{L(M)}(q)=f_{L(M')}(q)$$
which then allows us to repeat this argument inductively to see that $M=M'$, proving the lemma.

Then we finish with another lemma:

**Lemma 2:** Let $M$ be a non-empty set with $\ell$-bounded runs such that $0\not\in M$. Define $M_n = \{0\} \cup (M + n + 1)$ to be a right shift of $M$ with an initial element appended. For all large enough $n$, there exists some $q\in [1/2,1]$ such that if $f_{M'}(q)=1/2$ and $0\in M'$, then $M'=M_n$.

**Proof.** Note that $L(M_n)$ is always a set with $\ell$-bounded runs since the initial large run of elements not in the set is not preceded by an element that is in our set, as our definition would require - and all other runs must be in $M$ itself. Moreover, note
$$f_{M_n}(q)=(1-q) + q^{n+1}f_{M}(q).$$
Note, in particular, that $f_{M_n}(1/2)\geq \frac{1}2$ and that if $q^{n+1} \leq q - \frac{1}2$, then $f_{M_n}(q)\leq \frac{1}2$ - and note that, for any $\varepsilon>0$ there is some large enough $n$ so that this is true of some $q$ in $(1/2,1/2+\varepsilon)$. In particular, suppose we choose some $\varepsilon$ so that Lemma 1 applies to sets with $\ell$-bounded runs with that $\varepsilon$ and choose $n$ big enough so that $q_0^{n+1}\leq q_0 - \frac{1}2$ for some $q_0\in (1/2,1/2+\varepsilon)$. Then, by the intermediate value theorem, there has to be some $q\in [1/2,q_0]$ such that $f_{M_n}(q) = \frac{1}2$.

However, if we let $M'$ be such that $f_{M'}(q)=\frac{1}2$ and $0\in M'$, we can then realize that $$f_{L(M')}(q) = f_{L(M_n)}(q)$$ and apply Lemma 1 to see that $L(M')=L(M_n)$, which then, since both $M'$ and $M_n$ contain zero, implies that $M'=M_n$, completing Lemma 2.

To then reach our end goal: fix any pair $(q,M)$ with the property from Lemma 2 that if for any $M'$ with $0\in M'$ we have $f_{M'}(q)=\frac{1}2$ implies $M=M'$. Note that if we had some arbitrary set $M''$ with $f_{M''}(q)=\frac{1}2$, we would also have that $f(\mathbb N\setminus M'')=\frac{1}2$ - and either $M''$ or its complement must contain $0$ - thus must equal $M$.

Together, this gives us a method to construct a $q$ for which fairly splitting soup is impossible from any set with bounded runs but not with asymptotic density $1/2$. This gives an uncountable set of $q$ for which fairly splitting soup is not possible. Note that if you choose the set $M$ to be have a periodic pattern, the produced counterexamples will be algebraic.