Coalescence Probabilities of Cycle Products
Holden Mui
TL;DR
The paper provides a purely combinatorial proof for the probability that $1,\dots,k$ lie in the same cycle of a product of two random $n$-cycles, giving the explicit formula $P = \frac{1}{k} + \frac{4 (-1)^n}{ \binom{2k}{k}} \sum_{\substack{1 \le i \le k-1 \\ i \not\equiv n \bmod 2}} \binom{2k-1}{k+i} \left(\frac{1}{n+i+1} - \frac{1}{n-i}\right)$. The authors develop a diagrammatic, bijective framework by mapping colored $n$-cycles to pairs of Eulerian tours on degree-$(s_1,\dots,s_r)$ digraphs and extend it to count $t$-colored $k$-subsets of $r$-colored $n$-cycles via colored strips, yielding exact counting formulas. They then transform these counts to a sum suitable for partial fraction decomposition, derive auxiliary expressions $A(n,k)$ and $B(n,k)$, and, using four key identities, obtain the closed form that exhibits the probability as a rational function in $n(n+1)$ with a parity dependence. This combinatorial approach corroborates known results from character theory, clarifies parity-driven structure, and suggests avenues for generalizing to more refined cycle-type partitions, with potential implications for understanding cycle-coalescence phenomena in random permutations.
Abstract
Generalizing a formula of Stanley, we prove combinatorially that the probability that $1, 2, \dots, k$ are contained in the same cycle of a product of two random $n$-cycles is \[\frac{1}{k} + \frac{4 (-1)^n}{ \binom{2k}{k}} \sum_{\substack{1 \leq i \leq k-1 \\ i \not\equiv n \bmod 2}} \binom{2k-1}{k+i} \left(\frac{1}{n+i+1} - \frac{1}{n-i}\right).\]
