Word evaluating to a group element and its inverse with different frequency I'm supervising an undergraduate research project. Among other things, I've got the student to look at this paper of Gene Kopp and John Wiltshire-Gordon. This question arose from a missing complex conjugate in something the student wrote.
Let $g$ be an element of a finite group $G$, and $w$ a word in $n$ variables. If you evaluate the word on all $n$-tuples of elements of $G$, does it give $g$ and $g^{-1}$ the same number of times?
I thought the answer must be "no", but found it frustratingly difficult to come up with an example. After tapping local knowledge it seems that I was right. There's a recent paper of Alexander Lubotzky that proves that if $G$ is a finite simple group then the only restriction on a subset $A\subseteq G$ for it to be the image of the word map for some word in 2 variables is that it contains the identity and is fixed by $\operatorname{Aut}(G)$. Since there are finite simple groups (e.g., the Mathieu group $M_{11}$) with elements that are not sent to their inverses by any automorphism, this answers the question. 
However, my real question is whether there's a relatively simple example? 
Lubotzky's paper doesn't give an explicit word, although it does show that, for $M_{11}$, there's a word that works with length at most about $1.7\times 10^{244552995}$. 
Presumably one can do a bit better than that?
There are obvious restrictions on $g\in G$ and $w$ that rule out really small examples. There can't be any automorphism of $G$ sending $g$ to $g^{-1}$ or any automorphism of the free group $F_n$ sending $w$ to $w^{-1}$. 
 A: Yes, one can do much better than $1.7 \times 10^{244552995}$
(not surprisingly, because we're asking less than Lubotzky:
one of the two counts must be less than the other, 
but not necessarily zero).
In fact a word of length $10$ suffices.
I tried $G = M_{11}$ and $g$ an element of order $11$, and took $n=2$,
which makes exhaustive computation easily feasible (the first variable 
can be assumed to lie in one of the $10$ conjugacy classes so 
there's only $10 \, |M_{11}| = 79200$
group elements to compute given $w$).
None of the words $w(x,y) = x^a y^b x^c y^d$ seems to work, 
but several of the form $w(x,y) = x^a y^b x^c y^d x^e$ solve the problem. 
The first one (in lexicographic order) with all exponents at most $3$ is
$(a,b,c,d,e) = (1,2,1,3,3)$, i.e. $w(x,y) = x y^2 x y^3 x^3$,
for which $w(x,y) = g$ has $7491$ solutions 
but $w(x,y) = g^{-1}$ has only $7458$.
A: Check out Corollary 1 in
Winter, David L.
The automorphism group of an extraspecial p-group. 
Rocky Mountain J. Math. 2 (1972), no. 2, 159–168. 
20B25 
It gives the example you need.
EDIT A further search reveals this (not so old) MO discussion: element algebraically distinguishable from its inverse
A: They are solving for $\gamma_G(w) = \# \{ t \in G^n : w(t)=1 \} $.  For words that define surfaces they get a count in terms of the characters of $G$:
$$ \gamma_G(w)= \big|G\big|^{n-1} \underbrace{\sum_{\rho \in \mathrm{Irr}(G)} (\dim \rho)^k \langle \rho|g\rangle}_{\zeta_G(-k)} $$
This formula appears in many places, e.g. arXiv:0905.0731:Topological Quantum Field Theories from Compact Lie Groups. They are using the fact the characters of a group form a TQFT.
I don't really understand why people don't study group statistics using this type of result.  You can generalize the bound $\gamma_G(aba^{-1}b^{-1})\leq 5/8$ easily.

Have you tried a word like $a^2 b$ ?  I think this proves that $a^2 b$ does not work:
$$\gamma_G(a^2 bc)= \big| \big\{ a,b: a^2bc = 1  \big\}\big|= \big| \big\{ a,b: abca^{-1} = 1  \big\}\big| = |G| \cdot\big| \big\{ b: bc = 1  \big\}\big| = |G| $$
but then $\gamma_G(a^2 bc) = \gamma_G(a^2 bc^{-1})$.
More succinctly, $(a^2b)^{-1} = b^{-1}a^{-2}$ however this is induced by automorphism of the free group: $$a^2 b \mapsto b^{-1}a^2 \mapsto b^{-1}a^{-2} $$
Also, as you mentioned $g, g^{-1}$ should not be conjugate (e.g. alternating groups have an outer automorphism, $Aut(A_n) = S_n$).
