Let $\omega$ be a word in the free group on generators $x\_1,x\_2,\ldots,x\_n,g\_1,g\_2,\ldots,g\_k$, where $n>0$ and $k\geq 0$. For any finite group $G$ and elements $g\_1,g\_2,\ldots g\_k$ in $G$ we obtain an equation:
$$\omega(x\_1,x\_2,\ldots,x\_n,g\_1,g\_2,\ldots,g\_k)=e$$
Let $S=S\_{\omega,g\_1,g\_2,\ldots,g\_k}$ be the set of its solutions in $G^n$.


**Question:**
Is it true that for every $\omega$ there exists a constant $c<1$ such that for every finite group $G$ and its elements $g\_1,g\_2,\ldots,g\_k$ if $|S|>c|G|^n$ then $S=G^n$? 




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Comments: There exist many papers on problems related to the number of solutions of such equations (look [here](https://mathoverflow.net/a/198313/16678) for an example) but I couldn't find the above anywhere...


Some very easy cases: for $\omega(x,g)=xgx^{-1}g^{-1}$ we have $c={1\over 2}$, for $\omega(x\_1,x\_2)=x\_1x\_2x\_1^{-1}x\_2^{-1}$ we may take $c={3\over 4}$.


Why do I ask: if the answer is *yes* then for some very nice subsets of a profinite group positive Haar measure would imply nonempty interior.



Finite groups: equations with many solutions
In the case $k=0$ this is a conjecture of Dixon. See <http://people.math.carleton.ca/~jdixon/Prgrpth.pdf>, Section 1.2. I don't know of any reason it should become obviously false for $k>0$, so I guess it's open.


