$\bb X$上的Alice与Bob：反转、耦合、更新
Alice and Bob on $\Bbb X$: reversal, coupling, renewal
AI总结：
针对$\bb X$上引发热议的抛硬币得分胜负问题，本文运用反转、耦合、更新的概率技术定量解答了Alice与Bob谁更易获胜的问题，解释了问题背景的有趣性。
摘要：
一个涉及抛硬币的简洁问题出现在$\bb X$上，引发了一场激烈的“社交数学”热潮。在一系列公平硬币抛掷中，每出现连续两次正面Alice得一分，每出现正面后紧接着反面Bob得一分。谁更有可能赢得这场游戏？后续讨论展现了相互冲突的直觉，并得出了正确答案（胜负差距极小）。本文解释了该问题的背景为何有趣，以及如何运用反转、耦合和更新这些概率技术对其进行定量解答。
英文摘要：
A neat question involving coin flips surfaced on $\Bbb X$, and generated an intensive `storm' of `social mathematics'. In a sequence of flips of a fair coin, Alice wins a point at each appearance of two consecutive heads, and Bob wins a point whenever a head is followed immediately by a tail. Who is more likely to win the game? The subsequent discussion illustrated conflicting intuitions, and concluded with the correct answer (it is a close thing). It is explained here why the context of the question is interesting and how it may be answered in a quantitative manner using the probabilistic techniques of reversal, coupling, and renewal.
机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。
