Abstract:In a previous note, a two-player game between a random player and a deterministic player was introduced, and it was shown analytically that the winning probability of the deterministic player is the derangement probability dn/n!. The natural question left open was to explain this coincidence combinatorially. This paper gives such an explanation and extends it to a larger family of games. In the generalized game, the deterministic player removes q elements per turn, or all remaining elements if fewer than q remain. We couple the game exactly to the cycle decomposition of a uniformly random permutation. Under this coupling, the random player wins precisely when the first cycle of length at most q, read in canonical cycle order, is a fixed point. The case q = 1 recovers derangements, while the general case is governed by the first short cycle of the permutation.
From: Yehonatan Fridman [view email]
[v1]
Wed, 17 Jun 2026 11:38:08 UTC (6 KB)
此内容由惯性聚合(RSS阅读器)自动聚合整理，仅供阅读参考。 原文来自 — 版权归原作者所有。