Given $k$ conjugacy classes $C_1,C_2,\dots,C_k$ in a finite group $G$ with unity $e$, we may ask what is the probability $p(C_1,\dots,C_k)$ that $g_1\dots g_k=e$ if $g_i\in C_i$ is chosen uniformly at random. In the (say, complex) group algebra $\mathbb{C}[G]$ it is just the coefficient of $e$ in the product $\prod_{i=1}^k \varphi_i$, where we denote $$\varphi_i=|C_i|^{-1}\sum\limits_{g\in С_i} g.$$ It is natural to expand central functions $\varphi_i$ as linear combinations of irreducible characters (we identify an irreducible character $\chi$ on the group $G$ and the element $\sum \chi(g) g$ of $\mathbb{C}[G]$, which we again denote $\chi$.) These elements $\chi$ satisfy simple algebraic relations in the group algebra: $\chi_1\chi_2=0$ for $\chi_1\ne \chi_2$, $\chi^2=\frac{|G|}{\dim \chi} \chi$. This is bit more delicate property of characters than what is usually called `orthogonality relations'. We have $$\varphi_i=|G|^{-1}\sum_{\chi} \bar{\chi}(c_i)\cdot \chi,$$where summation is taken over all irreducible characters and $c_i\in C_i$ are arbitrary fixed elements. This is just the orthogonality relation itself. Well, now substitute this expression of $\varphi_i$ and expand brackets in $\prod_{i=1}^k \varphi_i$. Summands which contain at least two different characters disappear. Thus we have
$$p(C_1,\dots,C_k)=[e]\prod_{i=1}^k \varphi_i=|G|^{-k}\sum_{\chi} [e]\frac{|G|^{k-1}\prod_{i=1}^k \bar{\chi}(c_i)}{(\dim \chi)^{k-1}}\chi=|G|^{-1}\sum_{\chi}\frac{\prod_{i=1}^k \bar{\chi}(c_i)}{(\dim \chi)^{k-2}}.$$This formula is particularly useful for the symmetric group $G=S_n$, where we may use combinatorial formulae of Murnaghan - Nakayama for the values of characters. For example, if $C_1$ is the class of a long cycle, there exist only $n$ characters for which $\chi(c_1)\ne 0$. The following is Boccara's nice formula for the case $C_1=C_2=$(long cycles), $C_3$=class with cyclic type $(\lambda_1,\dots,\lambda_m)$:$$p(C_1,C_2,C_3)=\frac1{(n-1)!}\int_0^1 \prod_{i=1}^m (x^{\lambda_i}-(x-1)^{\lambda_i})dx.$$So, for $C_3=C_2=C_1$ (and odd $n$) we get that a probability that product of two long cycles is again a long cycle equals $2/(n+1)$.
See more on product of cycles in the slides by Richard Stanley.
This allows to get a number of combinatorial results on product of classes is the symmetric group $S_n$.
