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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.

Arnaud D.
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Chain Markov
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    I one wrote out an answer which is about generalising the Erdos-Turan result to infinite groups: https://math.stackexchange.com/a/2809964/10513 You might find it interesting/relevant. – user1729 Jan 18 '19 at 11:54
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    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), http://www.muroran-it.ac.jp/mathsci/danwakai/past/articles/201404-201503/07-20141203-farrokhi.pdf. Mentioned as open in a note by John D. Dixon, "Probabilistic Group Theory", http://people.math.carleton.ca/~jdixon/Prgrpth.pdf – Dap Jan 18 '19 at 18:00
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    Is the $n=1$ case obviously true? Or is even that case difficult? – Mees de Vries Jan 23 '19 at 15:48
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    @MeesdeVries, for the case $n = 1$ only three results are currently known: $\epsilon(x) = \frac{1}{2}$, $\epsilon(x^2) = \frac{1}{4}$ and $\epsilon(x^3) = \frac{2}{9}$. – Chain Markov Aug 02 '19 at 11:19
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    @MeesdeVries, there is also a conjecture, that $\epsilon(x^p) = \frac{p-1}{p^2}$ for prime $p$, however it remains unproven. – Chain Markov Aug 02 '19 at 12:02
  • For some recent progress, see the paper [Delizia et al., Gaps in probabilities of satisfying some commutator-like identities (2019)](https://arxiv.org/abs/1809.02997), where the conjecture is proved for the metabelian and $2$-Engel word. – darko Oct 07 '19 at 10:16
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    @YaniorWeg This conjecture is false for $p=5$; see my answer at https://mathoverflow.net/a/337483/297 . – David E Speyer Feb 14 '20 at 15:36
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    I proved analogous results about the equations $xy^2=y^2x$, $xy^3=y^3x$ and $xy=yx^{-1}$ (and infinitely many others derived from these by adding further variables) earlier this year: [preprint](https://arxiv.org/abs/2002.01773). – Z. A. K. Oct 16 '20 at 13:23
  • @Z. A. K. (Well, technically I proved them during my PhD, but I wrote them up earlier this year) – Z. A. K. Oct 16 '20 at 16:57

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