Got the idea from skew symmetry of matrice rings, where when you swap 2 elements you need to add a minus sign. But I was wondering to what extend is this possible for only a finite group. I guess I can ask this question about that!
More precisely if $a, b$ and $c$ are group elements in a group with $100$ elements then following equation should hold for at least $80\%$ of the time, where you choose $a, b$ uniformly (so probability of $1$ specific ordered pair is $1/10000$): $ab = cba$, where $c$ is fixed but not the identity. So how many groups with such a $c$ exists?
By rewriting I got that $aba^{-1}b^{-1}=c$. So the commutator $[a, b] = c$ holds in $80\%$ of the cases. So the groups must be not abelian.
It can't be $100\%$ since when $a=c$ then $cb=cbc$ which leads to $c =$ identity. Furthermore $cx=dxc$ but $d$ is not equal to $c$. But how many exceptions exist? Or what is not allowed to be $c$?
Also $c=b^{-1}$ is not allowed since $ab = b^{-1}ba$, so $ab = a$, so $b = 1$ so $c = 1$.
As well as $c=b$ is not allowed: $ab = bba \implies b^{-1}ab = ba \implies b^{-1}ab = bab$ by changing the order from $ba$ to $ab$ but because of rule you need to add the b to the left. So that I got that $b^{-1} = b$. So $bb=b^2=1$. Using that in the commutator relation: $aba^{-1}b^{-1}=b$, we get to $aba^{-1} = bb$, to $ab=a$ so that $b=1$ and so $c=1.$
Or at least only one of the expressions is allowed: Either $ab = bba$ or $ba = bab.$ IDK, I am getting confused.
Also I do not understand how more complicated rules/relations affects the combinatorics? Or how different rules combined has effect in the combinatorics?
How would YOU tackle that question? I would also be happy to get answers where the size of the group is less than $100$ or where the probability is more than $80\%.$ I am just curious!
Let $G$ be a finite group throughout. I will follow your convention of $[a,b]=aba^{-1}b^{-1}$ (though it is the opposite of the one I usually follow).
I claim the following:
Lemma. Let $a\in G$. Then for any $x,y\in G$, the following are equivalent:
Proof. $(1)\implies(2)$: If $[a,x]=[a,y]$, then $axa^{-1}x^{-1} = aya^{-1}y^{-1}$, hence $xa^{-1}x^{-1}=ya^{-1}y^{-1}$. Therefore, $xax^{-1}=yay^{-1}$, so $y^{-1}x$ centralizes $a$. Thus, $y^{-1}x\in C_G(a)$, so $xC_G(a)=yC_G(a)$, as claimed.
$(2)\implies(1)$: if $xC_G(a) = yC_G(a)$, then $y^{-1}x$ centralizes $a$, so $y^{-1}xax^{-1}y = a$. This gives $xax^{-1}=yay^{-1}$, hence $[a,x]=[x,a]^{-1}=[y,a]^{-1}=[a,y]$ as desired. $\Box$
So, let $G$ be a nonabelian group, and fix an element $a\in G$. For any fixed $c$, the number of pairs $(a,b)$ such that $[a,b]=c$ is at most the order of $C_G(a)$ (in fact, it is either zero or $|C_G(a)|$). For $c\neq e$, this is at most $\frac{1}{2}|G|$ (since it would require $a$ to be noncentral, and hence $C_G(a)\neq G$).
So we have that for any $c\neq e$,
$$\begin{align*} \#\{(a,b)\in G\times G\mid [a,b]=c\} &= \sum_{a\notin Z(G)}\#\{b\in G\mid [a,b]=c\}\\ &\leq \sum_{a\notin Z(G)}|C_G(a)| \\ &\leq \sum_{a\notin Z(G)}\frac {1}{2}|G|\\ &= \frac{1}{2}|G|(|G|-|Z(G)|)\\ &\lt \frac{1}{2}|G|^2. \end{align*}$$ The last inequality is strict because $|Z(G)|\geq 1$, so $|G|-|Z(G)|\lt |G|$.
Therefore, $$\frac{\#\{(a,b)\in G\times G\mid [a,b]=c\}}{|G\times G|} \lt \frac{\frac{1}{2}|G|^2}{|G|^2} = \frac{1}{2}.$$ This proves the theorem.
Note. It is well known that for a finite group, if $G$ is nonabelian then the probability that two random elements commute is at most $\frac{5}{8}$, with the bound achieved in the quaternion group. See here for sundry references to this and improvements thereof. So even if we remve the condition $c\neq e$, we would only get "at least $80\%$ of pairs $(a,b)$ have $[a,b]=c$" when $c=e$ and $G$ is abelian, in which case we get $100\%$ of the pairs. For nonabelian $G$, we'll get fewer than $50\%$ when $c\neq e$, and at most $62.5\%$ when $c=e$ and $G$ is nonabelian.