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Suppose for an arbitrary group word $w$ ower the alphabet of $n$ symbols $\mathfrak{U_w}$ is a variety of all groups $G$, that satisfy an identity $\forall a_1, … , a_n \in G$ $w(a_1, … , a_n) = e$. Is it true, that for any group word $w$ there exists a positive real number $\epsilon (w) > 0$, such that any finite group $G$ is in $\mathfrak{U_w}$ iff $$\frac{\lvert\{(a_1, … , a_n) \in G^n : w(a_1, … , a_n) = e\}\rvert}{{|G|}^n} > 1 - \epsilon(w)?$$

How did this question arise? There is a widely known theorem proved by P. Erdős and P. Turán that states:

A finite group $G$ is abelian iff $$\frac{|\{(a, b) \in G^2 : [a, b] = e\}|}{{|G|}^2} > \frac{5}{8}.$$

This theorem can be rephrased using aforementioned terminology as $\epsilon([a, b]) = \frac{3}{8}$.

There also is a generalisation of this theorem, stating that a finite group $G$ is nilpotent of class $n$ iff $$\frac{|\{(a_0, a_1, … , a_n) \in G^{n + 1} : [ … [[a_0, a_1], a_2]… a_n] = e\}|}{{|G|}^{n + 1}} > 1 - \frac{3}{2^{n + 2}},$$ thus making $\epsilon([ … [[a_0, a_1], a_2]… a_n]) = \frac{3}{2^{n + 2}}$.

However, I have never seen similar statements about other one-word varieties being proved or disproved, despite such question seeming quite natural . . .

Actually, I doubt that the conjecture in the main part of question is true. However, I failed to find any counterexamples myself.

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    $\begingroup$ I one wrote out an answer which is about generalising the Erdos-Turan result to infinite groups: math.stackexchange.com/a/2809964/10513 You might find it interesting/relevant. $\endgroup$ – user1729 Jan 18 at 11:54
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    $\begingroup$ Not mentioned in a 2015 survey Farrokhi, D. G. (2015). ON THE PROBABILITY THAT A GROUP SATISFIES A LAW: A SURVEY (Research on finite groups and their representations, vertex operator algebras, and algebraic combinatorics), muroran-it.ac.jp/mathsci/danwakai/past/articles/201404-201503/…. Mentioned as open in a note by John D. Dixon, "Probabilistic Group Theory", people.math.carleton.ca/~jdixon/Prgrpth.pdf $\endgroup$ – Dap Jan 18 at 18:00
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    $\begingroup$ Is the $n=1$ case obviously true? Or is even that case difficult? $\endgroup$ – Mees de Vries Jan 23 at 15:48
  • $\begingroup$ @MeesdeVries, for the case $n = 1$ only three results are currently known: $\epsilon(x) = \frac{1}{2}$, $\epsilon(x^2) = \frac{1}{4}$ and $\epsilon(x^3) = \frac{2}{9}$. $\endgroup$ – Yanior Weg Aug 2 at 11:19
  • $\begingroup$ @MeesdeVries, there is also a conjecture, that $\epsilon(x^p) = \frac{p-1}{p^2}$ for prime $p$, however it remains unproven. $\endgroup$ – Yanior Weg Aug 2 at 12:02

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