什么是帕累托最优？
What is Pareto optimality?
一种状态若无法在不使任何人变差的前提下让某人变好，就是帕累托最优。它衡量分配效率，但不关心公平。
A state is Pareto optimal if no one can be made better off without harming another; it measures efficiency, not fairness.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-11-02
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美国数学会(AMS)《博弈论与经济模型》
American Mathematical Society (AMS), Game Theory and Economic Modeling
独立核验依据 · Independent verification: 核验依据 1 ISBN 9787100047876（豆瓣书目条目: 博弈论与经济模型）
#博弈论 #最优
极小化极大（minimax）定理是什么？
What is the minimax theorem?
冯·诺依曼证明：在零和博弈中，存在一种策略使一方的最大可能损失最小化，且与另一方的最小可能收益最大化相等（博弈值）。
Von Neumann showed in zero-sum games a strategy minimizes the opponent's best payoff, equal to the game's value—the minimax principle.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-11-02
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美国数学会(AMS)《博弈论与经济模型》
American Mathematical Society (AMS), Game Theory and Economic Modeling
独立核验依据 · Independent verification: 核验依据 1 ISBN 9787100047876（豆瓣书目条目: 博弈论与经济模型）
#博弈论 #零和
冯·诺依曼极小极大定理对二人零和博弈给出了什么保证？
What guarantee does von Neumann's minimax theorem give for two-player zero-sum games?
定理表明 $\max_{x}\min_{y} u(x,y)=\min_{y}\max_{x} u(x,y)$，即先手最大化最小收益等于后手最小化最大损失。双方在最优混合策略下的期望收益相等，博弈存在唯一「值」，为二人零和博弈奠定了数学基础。
The theorem states $\max_{x}\min_{y} u(x,y)=\min_{y}\max_{x} u(x,y)$: the maximizer's best guaranteed payoff equals the minimizer's best defense. Under optimal mixed strategies both players' expected payoffs coincide, giving the game a unique value and a mathematical foundation for zero-sum games.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-11-02
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美国数学会(AMS)《博弈论与经济模型》
American Mathematical Society (AMS), Game Theory and Economic Modeling
独立核验依据 · Independent verification: 核验依据 1 ISBN 9787100047876（豆瓣书目条目: 博弈论与经济模型）
#极小极大定理 #零和 #混合策略
演化博弈论研究什么？
What does evolutionary game theory study?
它把博弈策略看作可遗传的“类型”，用复制子方程描述哪种策略在群体中随时间长盛不衰，解释合作、利他等现象如何稳定存在。
It treats strategies as heritable types, using replicator dynamics to show which prevail—explaining how cooperation and altruism stabilize.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 来源待核验 · Source pending review
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出处备注：本题原挂出处与答案内容不符，已撤回；具体书目待补，暂不展示来源。
本题原挂出处与答案内容不符，已撤回；具体书目待补。
Previous citation withdrawn as unsupported for this answer; a specific source is pending.
#博弈论 #演化
夏普利值（Shapley value）是什么？
What is the Shapley value?
夏普利值公平地衡量每个参与者对合作总收益的平均边际贡献，是合作博弈中分配功劳的标准方法。它现在也用于机器学习特征重要性分析。
The Shapley value fairly measures each player's average marginal contribution to a coalition's payoff, the standard solution in cooperative games. It now also ranks ML feature importance.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 来源待核验 · Source pending review
来源 · Source
出处备注：作者 / 书名 / 版次 / 出版社 / 年份齐备，仅缺独立比对证据；已在可信度报告单列披露，不再重复核验。
L. S. Shapley《A Value for n-Person Games》，载《Contributions to the Theory of Games II》, Annals of Mathematics Studies 28, Princeton University Press, 1953
L. S. Shapley, 'A Value for n-Person Games', in Contributions to the Theory of Games II, Annals of Mathematics Studies 28, Princeton University Press, 1953
#博弈论 #合作 #分配
贝叶斯博弈与信号博弈是什么？
What are Bayesian games and signaling games?
贝叶斯博弈中参与者对他人类型只有概率信念，用期望效用做决策；信号博弈研究一方如何发出可信信号（如教育作为能力信号）。它们是信息经济学的基础。
In Bayesian games players hold probabilistic beliefs about others' types and maximize expected utility; signaling games study credible signals (e.g. education as ability signal). They ground information economics.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-11-02
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美国数学会(AMS)《博弈论与经济模型》
American Mathematical Society (AMS), Game Theory and Economic Modeling
独立核验依据 · Independent verification: 核验依据 1 ISBN 9787100047876（豆瓣书目条目: 博弈论与经济模型）
#博弈论 #不完全信息 #信号
博弈论中的“民间定理”是什么？
What is the folk theorem in game theory?
在重复博弈中，只要参与者足够看重未来、能长期互动，许多看似不可信的合作结果都可成为均衡。民间定理解释了声誉如何维持合作。
In repeated games, if players value the future enough, many cooperative outcomes become equilibria. The folk theorem explains how reputation sustains cooperation.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-11-02
来源 · Source
美国数学会(AMS)《博弈论与经济模型》
American Mathematical Society (AMS), Game Theory and Economic Modeling
独立核验依据 · Independent verification: 核验依据 1 ISBN 9787100047876（豆瓣书目条目: 博弈论与经济模型）
#博弈论 #重复博弈 #合作
在囚徒困境的重复博弈中，“以牙还牙”策略为何能在无限重复情形下支撑合作？Axelrod 的锦标赛揭示了它的什么弱点？
In repeated Prisoner's Dilemma, how does the tit-for-tat strategy sustain cooperation under infinitely repeated play, and what weakness of it did Axelrod's tournaments reveal?
当未来收益的贴现因子足够高时，“以牙还牙”通过立即报复背叛、立即原谅回归合作的机制，使背叛的一次性收益无法抵消后续合作破裂的损失，从而构成均衡。Axelrod 的锦标赛同时显示该策略易受噪声干扰：一次误判即可引发双方交替报复的死循环，因此在含噪声环境中表现明显下降。
When the discount factor on future payoffs is high enough, tit-for-tat punishes defection immediately and forgives immediately upon return to cooperation, so a one-shot gain from betraying cannot outweigh the collapse of future cooperation—making it an equilibrium. Axelrod's tournaments also exposed its fragility to noise: a single misread move can lock both players into alternating retaliation, sharply degrading performance in noisy settings.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 来源待核验 · Source pending review
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出处备注：此条未指向带版次的可引用著作（通识性内容、机构文件或未展开的英文书名），本库不提供独立书目出处；如需精确引用，需由编辑补全具体篇目与版次。
Axelrod, R. (1984). The Evolution of Cooperation. Basic Books, Chapters 1–3.
Axelrod, R. (1984). The Evolution of Cooperation. Basic Books, Chapters 1–3.
#重复博弈 #以牙还牙策略
在多堆 Nim 游戏中,为什么把各堆石子数按二进制异或后结果为零就意味着即将行动的一方注定失败?
In multi-pile Nim, why does a zero bitwise XOR across all pile sizes imply that the player about to move is doomed to lose?
异或为零意味着每个二进制位上出现奇数次该位 1 的堆数都是偶数;对手的任何一次取子都至少翻转一位的奇偶性,从而把异或值推离零。反过来,只要异或非零,总能找到某一堆并从中取出合适数目的石子,让所有二进制位重新配平,使异或回到零。零态不可能经一步操作仍保持为零,于是面对零态的一方只能不断把非零态交还给对手,直到最后一颗石子被取走而落败——这正是 Bouton 在 1901 年完成的完整证明。
A zero XOR means each binary digit appears in an even number of piles; any move the opponent makes flips at least one digit's parity and pushes the XOR away from zero. Conversely, whenever the XOR is nonzero, some pile exists from which removing the right number of stones rebalances every digit and returns the XOR to zero. Since no single move preserves a zero XOR, whoever faces such a position keeps handing non-zero positions back until the last stone falls and they lose — exactly Bouton's complete proof from 1901.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-10-04
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C. L. Bouton《A Strategy in the Game of Nim》,Annals of Mathematics 第4卷,1901年;另见 Berlekamp、Conway 与 Guy《取胜之道》(Academic Press, 1982)
C. L. Bouton, "A Strategy in the Game of Nim," Annals of Mathematics 4, 1901; also Berlekamp, Conway & Guy, Winning Ways for Your Mathematical Plays, Academic Press, 1982
独立核验依据 · Independent verification: 核验依据 1 ISBN 9780120911011 (1985)
#尼姆游戏 #必胜策略
拍卖的收益等价定理是什么？
What is the revenue equivalence theorem?
在买家估值独立且风险中性的标准假设下，许多常见拍卖（英式、荷式、第一/第二价格密封）给卖家带来的期望收益相同。它奠定了拍卖机制设计的理论基础。
Under standard independent risk-neutral valuations, many common auctions (English, Dutch, 1st/2nd-price sealed) yield the same expected revenue to the seller.
难度 进阶 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-11-02
来源 · Source
美国数学会(AMS)《博弈论与经济模型》
American Mathematical Society (AMS), Game Theory and Economic Modeling
独立核验依据 · Independent verification: 核验依据 1 ISBN 9787100047876（豆瓣书目条目: 博弈论与经济模型）
#博弈论 #拍卖
零和博弈是什么？
What is a zero-sum game?
零和博弈中一方所得恰为另一方所失，总收益恒为零，如棋类与扑克。它与非零和博弈（可能双赢）相对，是博弈论最早研究的类型。
In a zero-sum game one player's gain is another's loss, total payoff zero, like chess or poker. Opposed to nonzero-sum games, it was the first type studied.
难度 进阶 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-11-02
来源 · Source
美国数学会(AMS)《博弈论与经济模型》
American Mathematical Society (AMS), Game Theory and Economic Modeling
独立核验依据 · Independent verification: 核验依据 1 ISBN 9787100047876（豆瓣书目条目: 博弈论与经济模型）
#博弈论 #零和 #对抗
演化博弈论如何用复制者方程描述策略在种群中的兴衰？
How does evolutionary game theory use the replicator equation to describe the rise and fall of strategies in a population?
复制者方程 $\dot x_i=x_i(f_i-\bar f)$ 表示策略 $i$ 的频率 $x_i$ 随时间变化正比于其相对适应度（$f_i$ 为该策略收益，$\bar f$ 为种群平均收益）。收益高于平均的策略会被「复制」增多，低于平均的则衰退，从而把博弈均衡与生物演化动力学联系起来。
The replicator equation $\dot x_i=x_i(f_i-\bar f)$ says the frequency $x_i$ of strategy $i$ changes proportionally to its relative fitness ($f_i$ the strategy's payoff, $\bar f$ the population average). Strategies above average grow, those below shrink, linking game equilibria with evolutionary dynamics.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 已核验 · Verified 2026-10-04
来源 · Source
Nowak《Evolutionary Dynamics: Exploring the Equations of Life》（Harvard University Press）
Nowak, Evolutionary Dynamics: Exploring the Equations of Life, Harvard University Press
独立核验依据 · Independent verification: 核验依据 1 ISBN 9780674023383 (2006)
#演化博弈 #复制者方程 #适应度
什么是纳什均衡？
What is a Nash equilibrium?
在纳什均衡中，每个参与者在别人策略不变时都不再愿意单独改变自己的选择。它是非合作博弈预测结果的核心概念。
In a Nash equilibrium, no player can benefit by unilaterally changing strategy; the central prediction of non-cooperative games.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 已核验 · Verified 2026-11-02
来源 · Source
美国数学会(AMS)《博弈论与经济模型》
American Mathematical Society (AMS), Game Theory and Economic Modeling
独立核验依据 · Independent verification: 核验依据 1 ISBN 9787100047876（豆瓣书目条目: 博弈论与经济模型）
#博弈论 #均衡
囚徒困境是什么？
What is the prisoner's dilemma?
两名嫌犯各自选择“招供”或“沉默”，个人最优是招供，但都招供的后果比都沉默更糟——说明个人理性可能导致集体不利。
Each suspect betters themselves by confessing, yet both confessing is worse than both staying silent—individual rationality can hurt the group.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 已核验 · Verified 2026-11-02
来源 · Source
美国数学会(AMS)《博弈论与经济模型》
American Mathematical Society (AMS), Game Theory and Economic Modeling
独立核验依据 · Independent verification: 核验依据 1 ISBN 9787100047876（豆瓣书目条目: 博弈论与经济模型）
#博弈论 #困境
什么是机制设计？
What is mechanism design?
机制设计是“逆向博弈论”：先定下想要的结果（如真实报价），再设计规则使参与者出于自身利益去实现它。拍卖设计是其典型应用。
Mechanism design is reverse game theory: choose a desired outcome, then build rules so self-interested agents achieve it; used in auction design.
难度 进阶 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-11-02
来源 · Source
美国数学会(AMS)《博弈论与经济模型》
American Mathematical Society (AMS), Game Theory and Economic Modeling
独立核验依据 · Independent verification: 核验依据 1 ISBN 9787100047876（豆瓣书目条目: 博弈论与经济模型）
#博弈论 #机制
公地悲剧是什么？
What is the tragedy of the commons?
公地悲剧指每个使用者为私利过度消耗共享资源，最终使所有人受损，如过度捕捞。它不是单人博弈，而是多人互动导致的集体失败。
The tragedy of the commons occurs when individuals overuse a shared resource for private gain, ruining it for all, like overfishing. It is collective failure from self-interest.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 已核验 · Verified 2026-11-03
来源 · Source
埃莉诺·奥斯特罗姆《治理公地：集体行动制度的演进》(1990)
Elinor Ostrom, Governing the Commons: The Evolution of Institutions for Collective Action (1990)
独立核验依据 · Independent verification: 核验依据 1 ARCHIVE 标识 governing-the-commons：Governing The Commons:
#博弈论 #公共资源 #合作
零和博弈的核心特征是什么？
What is the defining feature of a zero-sum game?
在零和博弈中，所有参与者的收益之和恒为零（一方所得正是他方所失）。国际象棋、扑克等属此类。冯·诺依曼的极小极大定理证明，此类博弈在混合策略下存在价值确定的均衡。
In a zero-sum game the sum of all players' payoffs is always zero (one's gain is another's loss), as in chess or poker. Von Neumann's minimax theorem shows such games have a well-defined equilibrium value under mixed strategies.
难度 进阶 · 适合 大众
CC BY 4.0 · 已核验 · Verified 2026-11-02
来源 · Source
美国数学会(AMS)《博弈论与经济模型》
American Mathematical Society (AMS), Game Theory and Economic Modeling
独立核验依据 · Independent verification: 核验依据 1 ISBN 9787100047876（豆瓣书目条目: 博弈论与经济模型）
#零和博弈 #极小极大 #混合策略
颤抖手完美均衡是什么？
What is a trembling-hand perfect equilibrium?
它要求策略即使在每个参与者都“以极小概率手抖犯错的”扰动下，仍是最优反应。它排除了那些依赖对手“绝不出错”的脆弱纳什均衡，使预测更稳健。
It requires a strategy to remain optimal even if every player 'trembles' with tiny probability; it rules out fragile Nash equilibria that rely on opponents never erring.
难度 进阶 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-11-02
来源 · Source
美国数学会(AMS)《博弈论与经济模型》
American Mathematical Society (AMS), Game Theory and Economic Modeling
独立核验依据 · Independent verification: 核验依据 1 ISBN 9787100047876（豆瓣书目条目: 博弈论与经济模型）
#博弈论 #颤抖手
混合策略纳什均衡中的「随机化」起到什么作用？
What role does randomization play in a mixed-strategy Nash equilibrium?
当纯策略无法形成稳定均衡时，参与者以一定概率分布选择动作，使对手在多种应对间无差异。例如在匹配硬币中，各以 $1/2$ 概率出正反面可使对方期望收益为零，从而构成均衡，随机化在此消除了可被对手利用的确定性规律。
When no pure-strategy equilibrium is stable, players randomize over actions with a probability distribution that makes opponents indifferent among responses. In matching pennies, playing heads/tails each with probability $1/2$ yields the opponent zero expected gain, forming an equilibrium; randomization removes exploitable deterministic patterns.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 已核验 · Verified 2026-11-02
来源 · Source
美国数学会(AMS)《博弈论与经济模型》
American Mathematical Society (AMS), Game Theory and Economic Modeling
独立核验依据 · Independent verification: 核验依据 1 ISBN 9787100047876（豆瓣书目条目: 博弈论与经济模型）
#混合策略 #随机化 #无差异
博弈论中「纳什均衡」指怎样的策略组合？
What kind of strategy profile does a Nash equilibrium describe in game theory?
纳什均衡是一组策略，使得在给定他人策略不变时，没有任何单个参与者能通过单方面改变自己的策略而获益。它是非合作博弈中最核心的稳定性概念，例如在囚徒困境中「都坦白」即为一纳什均衡。
A Nash equilibrium is a strategy profile where no player can benefit by unilaterally changing their own strategy given others' strategies are fixed. It is the central stability concept in noncooperative games; in the prisoner's dilemma both confess is a Nash equilibrium.
难度 进阶 · 适合 大众
CC BY 4.0 · 已核验 · Verified 2026-11-02
来源 · Source
美国数学会(AMS)《博弈论与经济模型》
American Mathematical Society (AMS), Game Theory and Economic Modeling
独立核验依据 · Independent verification: 核验依据 1 ISBN 9787100047876（豆瓣书目条目: 博弈论与经济模型）
#纳什均衡 #非合作博弈 #稳定性
博弈论中「帕累托最优」是什么意思？
What does Pareto optimal mean in game theory?
一个结果若在不使任何人境况变差的前提下无法让任何参与者变得更好，便是帕累托最优（有效）。纳什均衡未必帕累托最优——囚徒困境的均衡就不是，而合作结果才是帕累托更优的。
An outcome is Pareto optimal (efficient) if no participant can be made better off without making someone else worse off. A Nash equilibrium need not be Pareto optimal—the prisoner's dilemma equilibrium is not, while the cooperative outcome is Pareto superior.
难度 进阶 · 适合 大众
CC BY 4.0 · 已核验 · Verified 2026-11-02
来源 · Source
美国数学会(AMS)《博弈论与经济模型》
American Mathematical Society (AMS), Game Theory and Economic Modeling
独立核验依据 · Independent verification: 核验依据 1 ISBN 9787100047876（豆瓣书目条目: 博弈论与经济模型）
#帕累托最优 #效率 #纳什均衡
零行列式（ZD）策略是什么？
What are zero-determinant (ZD) strategies?
在重复囚徒困境中，ZD 策略利用矩阵行列式为零的性质，能强行把自己的收益与对手收益按固定比例绑定，甚至“迫使”对手接受不利分配。它改写了人们对合作的认识。
In the repeated prisoner's dilemma, ZD strategies use a zero determinant to unilaterally enforce a fixed linear relation between one's payoff and the opponent's—even extorting them.
难度 进阶 · 适合 高中(15-18)
CC BY 4.0 · 来源待核验 · Source pending review
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出处备注：本题原挂出处与答案内容不符，已撤回；具体书目待补，暂不展示来源。
本题原挂出处与答案内容不符，已撤回；具体书目待补。
Previous citation withdrawn as unsupported for this answer; a specific source is pending.
#博弈论 #零行列式
囚徒困境为何说明个体理性可能导致集体非最优？
Why does the prisoner's dilemma show that individual rationality can lead to a collectively suboptimal outcome?
双方各自选择「坦白」以规避更重刑期（占优策略），结果都得中等刑期；但若都保持沉默，则都能获更轻判决。个体明智的自保选择反而使整体福利低于合作情形，揭示了个人理性与集体最优的冲突。
Each player confesses to avoid the harsher sentence (a dominant strategy), yielding moderate sentences for both; yet mutual silence would give lighter ones. Individually rational self-protection leaves the group worse off than cooperation, exposing the clash between personal and collective optimality.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 已核验 · Verified 2026-11-02
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美国数学会(AMS)《博弈论与经济模型》
American Mathematical Society (AMS), Game Theory and Economic Modeling
独立核验依据 · Independent verification: 核验依据 1 ISBN 9787100047876（豆瓣书目条目: 博弈论与经济模型）
#囚徒困境 #占优策略 #集体最优
相关均衡相比纳什均衡有何更强的解集概念？
What stronger solution concept does a correlated equilibrium offer over Nash equilibrium?
相关均衡允许存在一个公共信号装置，使参与者在收到建议后都愿意遵从，此时联合分布下的期望收益不低于任何单方面偏离。它由 Aumann 提出，解集严格包含纳什均衡，且常能达成比纳什更高效的合作结果。
A correlated equilibrium permits a public signal device whose recommendations each player then willingly follows, with no incentive to deviate unilaterally. Introduced by Aumann, its set strictly contains the Nash equilibria and can achieve more efficient outcomes.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 已核验 · Verified 2026-11-02
来源 · Source
美国数学会(AMS)《博弈论与经济模型》
American Mathematical Society (AMS), Game Theory and Economic Modeling
独立核验依据 · Independent verification: 核验依据 1 ISBN 9787100047876（豆瓣书目条目: 博弈论与经济模型）
#相关均衡 #Aumann #信号
囚徒困境中“双方背叛”为何是唯一的纳什均衡却非帕累托最优?重复博弈如何改变这一结论?
In the prisoner's dilemma, why is mutual defection the unique Nash equilibrium yet not Pareto optimal, and how does repeated play change this conclusion?
无论对方选什么,单个参与者背叛的收益都严格更高,故(背叛,背叛)是唯一均衡;但双方合作收益更大,均衡帕累托劣于合作结果。当博弈无限次重复且参与人有足够耐心时,无名氏定理表明通过以牙还牙等触发策略,合作可成为子博弈精炼均衡的结果。
Each player's payoff is strictly higher by defecting regardless of the opponent's action, so (defect, defect) is the unique equilibrium, yet both players would be better off cooperating — the equilibrium is Pareto dominated. When the game repeats indefinitely and players are sufficiently patient, the folk theorem shows cooperation can be sustained as a subgame-perfect outcome via trigger strategies such as tit-for-tat.
难度 进阶 · 适合 大众
CC BY 4.0 · 已核验 · Verified 2026-11-04
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Fudenberg 与 Maskin 1986 年 Econometrica 论文“The Folk Theorem in Repeated Games with Discounting”,以及 Myerson《Game Theory》相关章节
Fudenberg and Maskin, "The Folk Theorem in Repeated Games with Discounting", Econometrica 54 (1986); see also Myerson, Game Theory, on equilibrium refinements
独立核验依据 · Independent verification: 核验依据 1 Crossref DOI 10.2307/1911307（本条出处含多部著作，按段登记，核到的是这一段点名的这部
#纳什均衡 #重复博弈
两堆棋子分别有3枚和5枚，两人轮流取子：每次任选一堆并从中取走至少一枚，取到最后一枚者获胜。先手是否有必胜策略？若有，第一步应如何取？
Two piles hold 3 and 5 tokens. Two players alternate moves; each move removes at least one token from a single pile chosen by the mover, and taking the last token wins. Does the first player have a winning strategy, and if so, what should the opening move be?
把各堆数量写成二进制并求异或和（Nim和）：3⊕5=6≠0，由Bouton定理，异或和非零的局面是先手必胜的N局面。必胜首着是从5枚那堆取走2枚，使两堆各剩3枚，Nim和变为0。此后对手的任何一步都会使Nim和非零，先手总能再次将其归零，最终拿走最后一枚。
Write the pile sizes in binary and take their bitwise XOR (the nim-sum): 3 ⊕ 5 = 6 ≠ 0, so by Bouton's theorem the position is an N-position won by the player to move. The winning opening removes 2 tokens from the pile of 5, leaving two piles of 3 whose nim-sum is 0. Every move from a zero nim-sum position makes it nonzero, so the first player can always restore the balance and secure the last token.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 已核验 · Verified 2026-11-02
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Bouton, C. L.《Nim, a Game with a Complete Mathematical Theory》, Annals of Mathematics 第二辑第3卷第6期, 1902年, 35-39页
C. L. Bouton, 'Nim, a Game with a Complete Mathematical Theory', Annals of Mathematics, ser. 2, vol. 3, no. 6, pp. 35-39, 1902.
独立核验依据 · Independent verification: 核验依据 1 CROSSREF 标识 10.2307/1967631：Nim, A Game with a Complete
#组合博弈论 #Nim游戏
石头剪刀布有没有必胜策略?数学家怎么研究这个游戏?
Is there a winning strategy for rock paper scissors? How do mathematicians study it?
对付会观察你的对手,最佳策略是随机出——三种选择各占三分之一,对手无法预测,长期就不输不赢。但如果对手有偏好,统计习惯再针对性出拳就能占优:浙江大学科学家2014年实验发现,赢家爱重复上一招,输家爱换招。博弈论的启示:对抗随机,唯一解也是随机。
Against an observant foe the best play is random — each choice one third of the time, unpredictable and unbeatable long-run. But exploit habits: a 2014 Zhejiang University experiment found winners tend to repeat their last move and losers tend to switch. Game theory's lesson: against randomness, randomness is the only answer.
难度 进阶 · 适合 小学高段(9-12)
CC BY 4.0 · 来源待核验 · Source pending review
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出处备注：著录含书名，但缺版次 / 出版社 / 年份中的一部分，需编辑补全后才能做独立比对。
J. Li, F. Qi 与 Y. Ma《Influences of Previous Outcomes and the Players' Skill on the Behavior in the Iterated Rock-Paper-Scissors Game》, PLOS ONE 9(8):e105293, 2014
J. Li, F. Qi & Y. Ma, 'Influences of Previous Outcomes and the Players' Skill on the Behavior in the Iterated Rock-Paper-Scissors Game', PLOS ONE 9(8):e105293, 2014
#博弈论 #混合策略 #随机性
玩石头剪刀布时怎样才算公平？
What makes rock-paper-scissors fair?
如果两种出法出现机会一样、互相克制，游戏就公平。每种手势都能赢一种、输一种、平一种，所以没有永远不败的招数。
If each move is equally likely and beats one/loses to one/ties one, the game is fair—no move always wins.
难度 入门 · 适合 小学低段(6-8)
CC BY 4.0 · 已核验 · Verified 2026-11-02
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美国数学会(AMS)《博弈论与经济模型》
American Mathematical Society (AMS), Game Theory and Economic Modeling
独立核验依据 · Independent verification: 核验依据 1 ISBN 9787100047876（豆瓣书目条目: 博弈论与经济模型）
#博弈论 #游戏