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An extended "iterated" version of the game also exists. In this version, the classic game is played repeatedly between the same prisoners, who continuously have the opportunity to penalize the other for previous decisions. If the number of times the game will be played is known to the players, then (by backward induction) two classically rational players will betray each other repeatedly, for the same reasons as the single-shot variant. In an infinite or unknown length game there is no fixed optimum strategy, and prisoner's dilemma tournaments have been held to compete and test algorithms for such cases.

An extended "iterated" version of the game also exists. In this version, the classic game is played repeatedly between the same prisoners, who continuously have the opportunity to penalize the other for previous decisions. If the number of times the game will be played is known to the players, then (by backward induction) two classically rational players will betray each other repeatedly, for the same reasons as the single-shot variant. In an infinite or unknown length game there is no fixed optimum strategy, and prisoner's dilemma tournaments have been held to compete and test algorithms for such cases.

一个扩展的“迭代”版本的博弈由此衍生出来。在这个版本中，经典博弈会在在同一组囚犯之间重复进行，他们不断有机会因为以前的决定而惩罚另一个。如果参与者知道博弈的次数，那么(通过逆向归纳法)两个经典理性的玩家就会因为和在单次博弈相同的原因反复背叛对方。在无限次或未知次数的博弈中，没有固定的最优策略，囚徒困境竞赛因而能被用来竞争并且检验这种情况下的算法。

一个扩展的“迭代”版本的博弈由此衍生出来。在这个版本中，经典博弈会在在同一组囚犯之间重复进行，他们不断有机会因为以前的决定而惩罚另一个。如果参与者知道博弈的次数，那么(通过逆向归纳法)两个经典理性的玩家就会因为和在单次博弈相同的原因反复背叛对方。在无限次或未知次数的博弈中，没有固定的最优策略，囚徒困境竞赛因而能被用来竞赛并且检验这种情况下的算法。

The normal game is shown below:

The normal game is shown below:

正常的游戏如下:

正常的博弈如下:

It is assumed that both prisoners understand the nature of the game, have no loyalty to each other, and will have no opportunity for retribution or reward outside the game. Regardless of what the other decides, each prisoner gets a higher reward by betraying the other ("defecting"). The reasoning involves an argument by dilemma: B will either cooperate or defect. If B cooperates, A should defect, because going free is better than serving 1 year. If B defects, A should also defect, because serving 2 years is better than serving 3. So either way, A should defect. Parallel reasoning will show that B should defect.

It is assumed that both prisoners understand the nature of the game, have no loyalty to each other, and will have no opportunity for retribution or reward outside the game. Regardless of what the other decides, each prisoner gets a higher reward by betraying the other ("defecting"). The reasoning involves an argument by dilemma: B will either cooperate or defect. If B cooperates, A should defect, because going free is better than serving 1 year. If B defects, A should also defect, because serving 2 years is better than serving 3. So either way, A should defect. Parallel reasoning will show that B should defect.

假设两个囚犯都了解游戏的本质，对彼此没有忠诚，在游戏之外没有机会得到报复或奖励。不管对方怎么决定，每个犯人背叛对方都会得到更高的奖励(“叛逃”)。推理涉及一个进退两难的论点: b 要么合作，要么缺陷。如果 b 合作，a 应该叛变，因为免费服刑总比服刑1年好。如果 b 有缺陷，a 也应该有缺陷，因为服刑2年总比服刑3年好。所以不管怎样，a 都应该叛变。并行推理将表明 b 应该缺陷。

假设两个囚犯都了解博弈的本质，对彼此没有忠诚，且在博弈之外没有机会得到报复或奖励。那么不管对方怎么决定，每个犯人背叛对方都会得到更高的奖励(“叛变”)。推理涉及一个进退两难的论点:B 要么合作，要么叛变。如果B合作，A 应该叛变，因为得到释放总比服刑1年好。如果 B叛变，A也应该叛变，因为服刑2年总比服刑3年好。所以不管怎样，a 都应该叛变。并行推理表明B应该选择叛变。

Because defection always results in a better payoff than cooperation regardless of the other player's choice, it is a dominant strategy. Mutual defection is the only strong Nash equilibrium in the game (i.e. the only outcome from which each player could only do worse by unilaterally changing strategy). The dilemma, then, is that mutual cooperation yields a better outcome than mutual defection but is not the rational outcome because the choice to cooperate, from a self-interested perspective, is irrational.

Because defection always results in a better payoff than cooperation regardless of the other player's choice, it is a dominant strategy. Mutual defection is the only strong Nash equilibrium in the game (i.e. the only outcome from which each player could only do worse by unilaterally changing strategy). The dilemma, then, is that mutual cooperation yields a better outcome than mutual defection but is not the rational outcome because the choice to cooperate, from a self-interested perspective, is irrational.

因为不管对方的选择如何，背叛总是比合作带来更好的回报，所以这是一个占优势的策略。相互背叛是游戏中唯一强大的纳什均衡点。唯一的结果是，每个参与者只能通过单方面改变战略来做得更糟)。因此，困境在于，相互合作比相互背叛产生更好的结果，但不是理性的结果，因为从自我利益的角度来看，合作的选择是非理性的。

因为不管对方的选择如何，背叛总是比合作带来更好的回报，所以这是一个占优势的策略。相互背叛是游戏中唯一的强纳什均衡点（每个参与者单方面的改变策略只能使自己的情况变糟)。因此，困境在于，虽然相互合作比相互背叛产生更好的结果，但它却不是理性的结果，因为从自我利益的角度来看，合作的选择是非理性的。

If both players cooperate, they both receive the reward ''R'' for cooperating. If both players defect, they both receive the punishment payoff ''P''. If Blue defects while Red cooperates, then Blue receives the temptation payoff ''T'', while Red receives the "sucker's" payoff, ''S''. Similarly, if Blue cooperates while Red defects, then Blue receives the sucker's payoff ''S'', while Red receives the temptation payoff ''T''.

If both players cooperate, they both receive the reward ''R'' for cooperating. If both players defect, they both receive the punishment payoff ''P''. If Blue defects while Red cooperates, then Blue receives the temptation payoff ''T'', while Red receives the "sucker's" payoff, ''S''. Similarly, if Blue cooperates while Red defects, then Blue receives the sucker's payoff ''S'', while Red receives the temptation payoff ''T''.

If both players cooperate, they both receive the reward R for cooperating. If both players defect, they both receive the punishment payoff P. If Blue defects while Red cooperates, then Blue receives the temptation payoff T, while Red receives the "sucker's" payoff, S. Similarly, if Blue cooperates while Red defects, then Blue receives the sucker's payoff S, while Red receives the temptation payoff T.

If both players cooperate, they both receive the reward R for cooperating. If both players defect, they both receive the punishment payoff P. If Blue defects while Red cooperates, then Blue receives the temptation payoff T, while Red receives the "sucker's" payoff, S. Similarly, if Blue cooperates while Red defects, then Blue receives the sucker's payoff S, while Red receives the tem punishment payoffptation payoff T.

The payoff relationship  implies that mutual cooperation is superior to mutual defection, while the payoff relationships  and  imply that defection is the dominant strategy for both agents.

The payoff relationship  implies that mutual cooperation is superior to mutual defection, while the payoff relationships  and  imply that defection is the dominant strategy for both agents.

The "donation game" is a form of prisoner's dilemma in which cooperation corresponds to offering the other player a benefit b at a personal cost c with b > c. Defection means offering nothing. The payoff matrix is thus

The "donation game" is a form of prisoner's dilemma in which cooperation corresponds to offering the other player a benefit b at a personal cost c with b > c. Defection means offering nothing. The payoff matrix is thus

 

Note that  (i.e. ) which qualifies the donation game to be an iterated game (see next section).

Note that  (i.e. ) which qualifies the donation game to be an iterated game (see next section).

请注意(即)这使得捐赠游戏成为一个迭代游戏(见下一节)。

请注意(即)这使得捐赠博弈成为一个迭代游戏(见下一节)。

 

The donation game may be applied to markets. Suppose X grows oranges, Y grows apples. The marginal utility of an apple to the orange-grower X is b, which is higher than the marginal utility (c) of an orange, since X has a surplus of oranges and no apples. Similarly, for apple-grower Y, the marginal utility of an orange is b while the marginal utility of an apple is c. If X and Y contract to exchange an apple and an orange, and each fulfills their end of the deal, then each receive a payoff of b-c. If one "defects" and does not deliver as promised, the defector will receive a payoff of b, while the cooperator will lose c. If both defect, then neither one gains or loses anything.

The donation game may be applied to markets. Suppose X grows oranges, Y grows apples. The marginal utility of an apple to the orange-grower X is b, which is higher than the marginal utility (c) of an orange, since X has a surplus of oranges and no apples. Similarly, for apple-grower Y, the marginal utility of an orange is b while the marginal utility of an apple is c. If X and Y contract to exchange an apple and an orange, and each fulfills their end of the deal, then each receive a payoff of b-c. If one "defects" and does not deliver as promised, the defector will receive a payoff of b, while the cooperator will lose c. If both defect, then neither one gains or loses anything.

捐赠游戏可能适用于市场。假设 x 种橘子，y 种苹果。苹果对橙子种植者 x 的边际效用是 b，这比橙子的边际效用高，因为 x 只有橙子而没有苹果。同样，对于苹果种植者 y 来说，橙子的边际效用是 b，苹果的边际效用是 c。 如果 x 和 y 签约交换一个苹果和一个橙子，并且每个人都完成了交易，那么每个人都会得到 b-c 的回报。如果一个“缺陷”没有按照承诺交货，那么这个“缺陷”将得到 b 的回报，而合作者将失去 c。 如果两者都有缺陷，那么谁也不会得到或失去任何东西。

捐赠博弈可能适用于市场。假设 X种橘子，Y 种苹果。苹果对橙子种植者 X 的边际效用是 b，这比橙子的边际效用c高，因为 x 有橙子剩余而没有苹果。同样，对于苹果种植者 y 来说，橙子的边际效用是 b，而苹果的边际效用是 c。 如果 x 和 y 签约交换一个苹果和一个橙子，并且每个人都完成了交易，那么每个人都会得到 从b到c 的效用收益。如果一方违约没有按照承诺交货，那么这个违约者将得到 b 效用的收益，而合作者将失去 c的效用收益。 如果两者都违约，那么谁也不会得到或失去任何东西。

If two players play prisoner's dilemma more than once in succession and they remember previous actions of their opponent and change their strategy accordingly, the game is called iterated prisoner's dilemma.

If two players play prisoner's dilemma more than once in succession and they remember previous actions of their opponent and change their strategy accordingly, the game is called iterated prisoner's dilemma.

In addition to the general form above, the iterative version also requires that , to prevent alternating cooperation and defection giving a greater reward than mutual cooperation.

In addition to the general form above, the iterative version also requires that , to prevent alternating cooperation and defection giving a greater reward than mutual cooperation.

The iterated prisoner's dilemma game is fundamental to some theories of human cooperation and trust. On the assumption that the game can model transactions between two people requiring trust, cooperative behaviour in populations may be modeled by a multi-player, iterated, version of the game. It has, consequently, fascinated many scholars over the years. In 1975, Grofman and Pool estimated the count of scholarly articles devoted to it at over 2,000. The iterated prisoner's dilemma has also been referred to as the "peace-war game".

The iterated prisoner's dilemma game is fundamental to some theories of human cooperation and trust. On the assumption that the game can model transactions between two people requiring trust, cooperative behaviour in populations may be modeled by a multi-player, iterated, version of the game. It has, consequently, fascinated many scholars over the years. In 1975, Grofman and Pool estimated the count of scholarly articles devoted to it at over 2,000. The iterated prisoner's dilemma has also been referred to as the "peace-war game".

重复囚徒困境博弈是人类合作与信任理论的基础。假设博弈可以为两个需要信任的人之间的交易建模，群体中的合作行为可以由多个参与者迭代的博弈模型来建模。因此，这些年来，它吸引了许多学者。1975年，葛夫曼和普尔估计关于它的学术文章超过2000篇。重复的囚徒困境也被称为“和平-战争游戏”。

迭代囚徒困境博弈是一些人类合作与信任理论的基础。假设博弈可以为两个需要信任的人之间的交易建模，那么群体中的合作行为也可以由多个参与者迭代的博弈模型来建模。因此，这些年来，它吸引了许多学者。1975年，葛夫曼和普尔估计关于它的学术文章超过2000篇。迭代的的囚徒困境也被称为“和平-战争博弈”。

If the game is played exactly N times and both players know this, then it is optimal to defect in all rounds. The only possible Nash equilibrium is to always defect. The proof is inductive: one might as well defect on the last turn, since the opponent will not have a chance to later retaliate. Therefore, both will defect on the last turn. Thus, the player might as well defect on the second-to-last turn, since the opponent will defect on the last no matter what is done, and so on.  The same applies if the game length is unknown but has a known upper limit.

If the game is played exactly N times and both players know this, then it is optimal to defect in all rounds. The only possible Nash equilibrium is to always defect. The proof is inductive: one might as well defect on the last turn, since the opponent will not have a chance to later retaliate. Therefore, both will defect on the last turn. Thus, the player might as well defect on the second-to-last turn, since the opponent will defect on the last no matter what is done, and so on.  The same applies if the game length is unknown but has a known upper limit.

如果这个游戏正好玩了 n 次，并且两个玩家都知道这一点，那么在所有回合中最佳的策略就是变节。唯一可能的纳什均衡点就是永远背叛。证据是归纳的: 一个人不妨在最后一回合投降，因为对手以后没有机会反击。因此，双方都会在最后一个回合脱节。因此，玩家不妨在倒数第二回合时变节，因为无论做什么，对手都会在倒数第三回合变节，依此类推。如果游戏的长度是未知的，但有一个已知的上限，这同样适用。

如果这个游戏正好玩了 n 次，并且两个玩家都知道这一点，那么在所有回合中最佳的策略就是叛变。唯一可能的纳什均衡点就是永远叛变。证据是归纳的: 一个人不妨在最后一回合叛变，因为对手以后没有机会反击。因此，双方都会在最后一个回合叛变。所以玩家同样也会在倒数第二回合时变节，因为无论做什么，对手都会在倒数第三回合变节，依此类推。同样方法适用于博弈次

 

 

 

Unlike the standard prisoner's dilemma, in the iterated prisoner's dilemma the defection strategy is counter-intuitive and fails badly to predict the behavior of human players. Within standard economic theory, though, this is the only correct answer.  The [[superrational]] strategy in the iterated prisoner's dilemma with fixed ''N'' is to cooperate against a superrational opponent, and in the limit of large ''N'', experimental results on strategies agree with the superrational version, not the game-theoretic rational one.

Unlike the standard prisoner's dilemma, in the iterated prisoner's dilemma the defection strategy is counter-intuitive and fails badly to predict the behavior of human players. Within standard economic theory, though, this is the only correct answer.  The [[superrational]] strategy in the iterated prisoner's dilemma with fixed ''N'' is to cooperate against a superrational opponent, and in the limit of large ''N'', experimental results on strategies agree with the superrational version, not the game-theoretic rational one.

Unlike the standard prisoner's dilemma, in the iterated prisoner's dilemma the defection strategy is counter-intuitive and fails badly to predict the behavior of human players. Within standard economic theory, though, this is the only correct answer.  The superrational strategy in the iterated prisoner's dilemma with fixed N is to cooperate against a superrational opponent, and in the limit of large N, experimental results on strategies agree with the superrational version, not the game-theoretic rational one.

Unlike the standard prisoner's dilemma, in the iterated prisoner's dilemma the defection strategy is counter-intuitive and fails badly to predict the behavior of human players. Within standard economic theory, though, this is the only correct answer.  The superrational strategy in the iterated prisoner's dilemma with fixed N is to cooperate against a superrational opponent, and in the limit of large N, experimental results on strategies agree with the superrational version, not the game-theoretic rational one.

与标准的囚徒困境不同，在重复的囚徒困境中，叛逃策略是违反直觉的，严重不能预测人类玩家的行为。然而，在标准的经济理论中，这是唯一正确的答案。具有固定 n 的重复囚徒困境中的超理性策略是对超理性对手的合作，在大 n 的限制下，实验结果与超理性的结果一致，而不是博弈论的理性结果。

与标准的囚徒困境不同，在迭代的囚徒困境中，叛变策略是严重违反直觉的，以至于不能预测人类玩家的行为。然而，在标准的经济理论中，这是唯一正确的答案。具有固定次数 n 的迭代囚徒困境中的超理性策略是与超理性对手进行合作，在 n 很大的限制下，实验结果的策略与超理性结果的策略一致，而不是博弈论的理性结果。

For cooperation to emerge between game theoretic rational players, the total number of rounds N must be unknown to the players. In this case "always defect" may no longer be a strictly dominant strategy, only a Nash equilibrium. Amongst results shown by Robert Aumann in a 1959 paper, rational players repeatedly interacting for indefinitely long games can sustain the cooperative outcome.

For cooperation to emerge between game theoretic rational players, the total number of rounds N must be unknown to the players. In this case "always defect" may no longer be a strictly dominant strategy, only a Nash equilibrium. Amongst results shown by Robert Aumann in a 1959 paper, rational players repeatedly interacting for indefinitely long games can sustain the cooperative outcome.

为了使博弈论理性参与者之间出现合作，参与者必须不知道 n 回合的总数。在这种情况下，“总是缺陷”可能不再是一个严格占主导地位的策略，而只是一个纳什均衡点。罗伯特 · 奥曼在1959年的一篇论文中表明，理性参与者在无限长的博弈中反复互动可以维持合作的结果。

为了使博弈论理性参与者之间出现合作，参与者必须不知道 n 回合的总数。在这种情况下，“总是叛变”可能不再是一个严格占主导地位的策略，而只是一个纳什均衡点。罗伯特 · 奥曼在1959年的一篇论文中表明，理性参与者在无限多次的博弈中通过反复互动可以维持合作的结果。

According to a 2019 experimental study in the American Economic Review which tested what strategies real-life subjects used in iterated prisoners' dilemma situations with perfect monitoring, the majority of chosen strategies were always defect, tit-for-tat, and Grim trigger. Which strategy the subjects chose depended on the parameters of the game.

According to a 2019 experimental study in the American Economic Review which tested what strategies real-life subjects used in iterated prisoners' dilemma situations with perfect monitoring, the majority of chosen strategies were always defect, tit-for-tat, and Grim trigger. Which strategy the subjects chose depended on the parameters of the game.

美国经济评论》(American Economic Review)2019年的一项实验研究测试了现实生活中的实验对象在完全监控的情况下在反复出现的囚徒困境中使用的策略，结果显示，大多数选择的策略都是缺陷、针锋相对和残酷的触发。受试者选择的策略取决于游戏的参数。

《美国经济评论》(American Economic Review)2019年的一项实验研究测试了现实生活中的实验对象在完全监控的情况下在迭代的囚徒困境中使用的策略，结果显示，大多数选择的策略都是叛变、针锋相对抑或是触发策略。受试者选择的策略取决于博弈的参数。

Interest in the iterated prisoner's dilemma (IPD) was kindled by Robert Axelrod in his book The Evolution of Cooperation (1984). In it he reports on a tournament he organized of the N step prisoner's dilemma (with N fixed) in which participants have to choose their mutual strategy again and again, and have memory of their previous encounters. Axelrod invited academic colleagues all over the world to devise computer strategies to compete in an IPD tournament. The programs that were entered varied widely in algorithmic complexity, initial hostility, capacity for forgiveness, and so forth.

Interest in the iterated prisoner's dilemma (IPD) was kindled by Robert Axelrod in his book The Evolution of Cooperation (1984). In it he reports on a tournament he organized of the N step prisoner's dilemma (with N fixed) in which participants have to choose their mutual strategy again and again, and have memory of their previous encounters. Axelrod invited academic colleagues all over the world to devise computer strategies to compete in an IPD tournament. The programs that were entered varied widely in algorithmic complexity, initial hostility, capacity for forgiveness, and so forth.

罗伯特 · 阿克塞尔罗德在他的著作《合作的进化》(1984)中点燃了人们对反复出现的囚徒困境(IPD)的兴趣。在这篇文章中，他报道了一个关于 n 步囚徒困境的锦标赛，参赛者必须一次又一次地选择他们共同的策略，并且要记住他们之前的遭遇。阿克塞尔罗德邀请世界各地的学术界同仁设计计算机策略，以参加知识产权促进发展锦标赛。输入的程序在算法复杂性、最初的敌意、宽恕能力等方面差异很大。

罗伯特 · 阿克塞尔罗德在他的著作《合作的进化》(1984)中激起了了人们对迭代的囚徒困境(IPD)的兴趣。在这篇文章中，他报道了一个关于 n 次囚徒困境的比赛，参赛者必须一次又一次地选择他们共同的策略，并且要记住他们之前的遭遇。阿克塞尔罗德邀请世界各地的学术界同仁设计计算机策略来参加此次比赛。输入的程序在算法复杂性、最初敌意、宽恕能力等方面差异很大。

Axelrod discovered that when these encounters were repeated over a long period of time with many players, each with different strategies, greedy strategies tended to do very poorly in the long run while more altruistic strategies did better, as judged purely by self-interest. He used this to show a possible mechanism for the evolution of altruistic behaviour from mechanisms that are initially purely selfish, by natural selection.

Axelrod discovered that when these encounters were repeated over a long period of time with many players, each with different strategies, greedy strategies tended to do very poorly in the long run while more altruistic strategies did better, as judged purely by self-interest. He used this to show a possible mechanism for the evolution of altruistic behaviour from mechanisms that are initially purely selfish, by natural selection.

The winning deterministic strategy was tit for tat, which Anatol Rapoport developed and entered into the tournament. It was the simplest of any program entered, containing only four lines of BASIC, and won the contest. The strategy is simply to cooperate on the first iteration of the game; after that, the player does what his or her opponent did on the previous move. Depending on the situation, a slightly better strategy can be "tit for tat with forgiveness". When the opponent defects, on the next move, the player sometimes cooperates anyway, with a small probability (around 1–5%). This allows for occasional recovery from getting trapped in a cycle of defections. The exact probability depends on the line-up of opponents.

The winning deterministic strategy was tit for tat, which Anatol Rapoport developed and entered into the tournament. It was the simplest of any program entered, containing only four lines of BASIC, and won the contest. The strategy is simply to cooperate on the first iteration of the game; after that, the player does what his or her opponent did on the previous move. Depending on the situation, a slightly better strategy can be "tit for tat with forgiveness". When the opponent defects, on the next move, the player sometimes cooperates anyway, with a small probability (around 1–5%). This allows for occasional recovery from getting trapped in a cycle of defections. The exact probability depends on the line-up of opponents.

决定性的胜利策略是以牙还牙，这是阿纳托尔 · 拉波波特开发并参加锦标赛的策略。这是所有参赛程序中最简单的一个，只有四行 BASIC 语言，并且赢得了比赛。策略很简单，就是在游戏的第一次迭代中进行合作; 在此之后，玩家做他或她的对手在前一步中所做的事情。根据具体情况，一个稍微好一点的策略可以是“以牙还牙，以牙还牙”。当对手出现缺陷时，在下一步棋中，玩家有时还是会合作，概率很小(大约1-5%)。这允许偶尔从陷入叛逃循环中恢复过来。确切的概率取决于对手的阵容。

最终获胜的决定性策略是以牙还牙，这是阿纳托尔 · 拉波波特开发并参加比赛的策略。这是所有参赛程序中最简单的一个，只有四行 BASIC 语言，并且赢得了比赛。策略很简单，就是在游戏的第一次迭代中进行合作; 在此之后，玩家做他或她的对手在前一步中所做的事情。根据具体情况，一个稍微好一点的策略可以是“带着宽恕的心以牙还牙”。当对手叛变时，在下一次博弈中，玩家有时还是会合作，但概率很小(大约1-5%)。这允许博弈能偶尔从陷入叛变循环中恢复过来。确切的概率取决于对手的安排。

  Nice: The most important condition is that the strategy must be "nice", that is, it will not defect before its opponent does (this is sometimes referred to as an "optimistic" algorithm). Almost all of the top-scoring strategies were nice; therefore, a purely selfish strategy will not "cheat" on its opponent, for purely self-interested reasons first.

  Nice: The most important condition is that the strategy must be "nice", that is, it will not defect before its opponent does (this is sometimes referred to as an "optimistic" algorithm). Almost all of the top-scoring strategies were nice; therefore, a purely selfish strategy will not "cheat" on its opponent, for purely self-interested reasons first.

Nice: 最重要的条件是策略必须是“ Nice” ，也就是说，它不会在对手之前叛变(这有时被称为“ optimistic”算法)。几乎所有得分最高的策略都很好; 因此，一个纯粹自私的策略不会出于纯粹自身利益的原因而“欺骗”对手。

友好： 最重要的条件是策略必须是好的 ，也就是说，它不会在对手之前叛变(这有时被称为“乐观”算法)。几乎所有得分最高的策略都是友好的; 因此，一个纯粹自私的策略不会出于纯粹自身利益的原因而“欺骗”对手。

; Retaliating: However, Axelrod contended, the successful strategy must not be a blind optimist. It must sometimes retaliate. An example of a non-retaliating strategy is Always Cooperate. This is a very bad choice, as "nasty" strategies will ruthlessly exploit such players.

; Retaliating: However, Axelrod contended, the successful strategy must not be a blind optimist. It must sometimes retaliate. An example of a non-retaliating strategy is Always Cooperate. This is a very bad choice, as "nasty" strategies will ruthlessly exploit such players.