两人彩票收集竞赛中的 ballot 事件：一种 renewal-Catalan 渐近分析
The Ballot Event for Two-Player Coupon Collection: A Renewal--Catalan Asymptotic
AI总结：
本文研究两人彩票收集竞赛，证明最终胜者从未落后概率渐近为 2/d，利用 renewal 分解和 Catalan 哈尔米特分析。
摘要：
我们研究两人彩票收集竞赛，其中两个独立收集者每轮各从 d 种等可能的彩票类型中抽取一个彩票。Myers 和 Wilf 给出了若干两人事件的有限公式，并明确留出 ballot 类问题，即求最终胜者从未落后的概率。我们证明该概率满足 b_d ~ 2/d，当 d 趋近于无穷大时。证明使用了在平局边界处的 renewal 分解。第一个单侧平局打破有显式入口分布；其水平，按 d^{1/2} 缩放，收敛于瑞利分布；平局后，领先者的生存概率由 Catalan 或赌徒 ruin 哈尔米特支配。主要估计显示，此比较哈尔米特在精确同时轮链中的累积缺陷是可忽略的。
英文摘要：
We study the two-player coupon-collector competition in which two independent collectors draw one coupon each per round from a set of $d$ equally likely coupon types. Myers and Wilf gave finite formulae for several two-player events and explicitly left open the ballot-type problem of finding the probability that the ultimate winner was never behind. We prove that this probability satisfies $$ b_d \sim \frac{2}{d}, \qquad d\to\infty .$$ The proof uses a renewal decomposition at the tie boundary. The first one-sided tie-break has an explicit entrance distribution; its level, scaled by $d^{1/2}$, converges to a Rayleigh law; and, after the break, the leader's survival probability is governed by a Catalan, or gambler's-ruin, harmonic. The main estimate shows that the accumulated defect of this comparison harmonic in the exact simultaneous-round chain is negligible.
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