Commuting probability

Remarks

Counting the pairs by their first entry gives $\mathrm{cp}(G) = \frac{1}{\abs{G}^2} \sum_{x \in G} \abs{C_G(x)}$, where $C_G(x)$ is the centraliser of $x$. Since the conjugacy class of $x$ has $\abs{G}/\abs{C_G(x)}$ elements, this sum gives $\mathrm{cp}(G) = k(G)/\abs{G}$, where $k(G)$ is the number of conjugacy classes of $G$. Example: $S_3$ has $3$ conjugacy classes, so $\mathrm{cp}(S_3) = 3/6 = 1/2$. Example: the quaternion group $Q_8$ has $5$ conjugacy classes, $\{1\}$, $\{-1\}$, $\{\pm i\}$, $\{\pm j\}$, $\{\pm k\}$, so $\mathrm{cp}(Q_8) = 5/8$. The value $\mathrm{cp}(G) = 1$ holds exactly when $G$ is abelian; if $G$ is not abelian, then $\mathrm{cp}(G) \le 5/8$.