Let $G$ be the symmetric group $S_n$ or the projective general linear group $PGL_2(n)$.

Let $X$ be a cyclically reduced word in the abstract variables $x_1, x_2, \ldots,x_k$, i.e. $X$ is a product containing $x_1, x_2, \ldots,x_k$ and their inverses, without any element appearing next to its own inverse in any cyclic permutation. (Only words with length $4$, $6$, $8$ are needed in my research.)

Consider the probability $P$ that the word sums to $1$, with each $x_i$ chosen uniformly and independently from $G$.

**Question:**

What are the upper bounds of $\log_{|G|}P$?

If $\log_{|G|}P$ converges when $n→\infty$, what's the value?

Answers are acceptable for either $G=S_n$ or $G=PGL_2(n)$.

**Known:**

If there's a variable occuring only once in $X$, then $P$ is exactly $1/|G|$.

If $X=x_1^k$, then the limit is $-1/k$ for symmetric groups by David E Speyer's argument.

As Richard Stanley pointed out, if $X=x_1x_2x_1^{-1}x_2^{-1}$, then $P=|Conj(G)|/|G|$. ($|Conj(G)|$ is the number of conjugacy classes of $G$)

The formula $P=|Conj(G)|/|G|$ holds for the words $x_1x_1x_2x_2$ and $x_1x_2x_1x_2^{-1}$ if all the characters of $G$ are real, and that's exactly the case for $S_n$ and $PGL_2(n)$.

Random Structures & Algorithms5(1994), 703-730. An unrelated comment: for $X=xyx^{-1}y^{-1}$, the limit is 1, since the number of commuting pairs in any finite group $G$ is $|G|\cdot k(G)$, where $k(G)$ is the number of conjugacy classes, and the number of conjugacy classes in $S_n$ is around $e^{c\sqrt{n}}$. $\endgroup$ – Richard Stanley Jun 9 '19 at 13:15Enumerative Combinatorics, vol. 2. Possibly part (i) can solve the problem for $xy^kxy^{-k}$ and $xy^kx^{-1}y^k$, but I have not tried to do this. See also (f) for a possible approach to $X=x_1^{a_1}\cdots x_m^{a_m}$. $\endgroup$ – Richard Stanley Jun 9 '19 at 22:22