Behavioral Mistakes Support Cooperation
in an N-Person Repeated Public Goods Game

Let us assume that only conditional cooperators make mistakes (to make a contribution, a cooperator needs to take an action, whereas a defector simply does nothing). For simplicity, a player who made a mistake does not know that it is he/she that made the mistake. Cooperators only know the total number of cooperations in their group, and they determine whether they would cooperate or defect in the next round depending on this number.

Because some cooperators defect by mistake, the payoff to individuals now depends on the number of errors made by the cooperators as well as the number of cooperators in a group. Let $j$ be the number of cooperators among $n-1$ other members in a group. Let $\psi(j,q,\varepsilon)$ be the probability that $q$ individuals among the $j$ other cooperators in a group make mistakes when the probability of making a mistake is $\varepsilon$.

The payoffs from the one-shot game, given that $j$ among $n-1$ other individuals choose cooperation, now become

Suppose that the game is repeated with probability $\delta$. Let $V^{\varepsilon}(T_{k}|j)$ be the payoff for the $T_{k}$ strategy when $j$ among the $n-1$ other members in his or her group are the conditional cooperators and the conditional cooperators play defection by mistake with probability $\varepsilon$. Then, the behavioral mistakes affect $V^{\varepsilon}(T_{k}|j)$ via whether the conditional cooperators maintain their cooperation in the following stages as well as via a one shot payoff, $F^{\varepsilon}(C|j)$ and $F^{\varepsilon}(D|j)$.

We showed in Proposition 1 that the $T_{n-1}$ strategy is neutrally stable (not evolutionarily stable). Now we will show how behavioral mistakes affect the stability of the $T_{n-1}$ strategy. Consider a population that is homogeneously composed of individuals using the $T_{n-1}$ strategy. A player with the $T_{n-1}$ strategy receives $F^{\varepsilon}(C|n-1)$ from the first round and maintains cooperation in the next round only if there is no defection in the current round. Therefore, the mutual cooperation continues only if there are neither defectors’ defecting nor cooperators’ mistakenly defecting. In this environment, the payoff for the $T_{n-1}$ strategy from the repeated game is

Rearranging the above condition gives

where $(1-\varepsilon)^{n}$ is the probability that no conditional cooperators make mistakes, that is, the probability that the periods of mutual cooperation continue without triggering retaliation by any mistakes. Suppose that there appears a mutant with the $T_{n}$ strategy in this population. This strategy will immediately trigger the $T_{n-1}$ players’ retaliation. Therefore, the payoff for the $T_{n}$ strategy is

Suppose that there appears a mutant with the $T_{k}$ strategy ($k<n-1$) in the $T_{n-1}$ population. This $k$-cooperator plays $C$ in the first round and keeps cooperating as long as the number of defections that are mistakenly played by the conditional cooperators is less than or equal to $n-k-1$. The payoff for the $T_{k}$ strategy from the repeated game in a population where all of the other players are $T_{n-1}$ becomes

where $k<n-1$. The last term on the right-hand side expresses the payoff for the $T_{k}$ strategy in the next round when $T_{n-1}$ players have already withdrawn their cooperation and $T_{k<n-1}$ players continue cooperating. This situation occurs when the number of mistakes in one stage is greater than 1 and less than or equal to $n-k-1$, and $k$ cooperators will also withdraw cooperation thereafter. The rearrangement gives

where $k<n-1$. Here, the behavioral mistakes have two effects on the evolution of the conditional cooperative strategies. First, the mistakes reduce the period of mutual cooperation and have a detrimental effect on the evolution of cooperation. Suppose that a group is entirely composed of $T_{n-1}$ players. Without mistakes, all of the members continue cooperating until the game ends, and the expected length of the period of mutual cooperation is $\frac{1}{1-\delta}$. When the players make mistakes with probability $\varepsilon$, even one mistake will trigger retaliation from the other $T_{n-1}$ players, and the mutual cooperation ends. Therefore, the period of mutual cooperation is reduced to $\frac{1}{1-(1-\varepsilon)^{n}\delta}$ and the value of $V(T_{n-1}|n-1,\varepsilon)$ in (6) becomes smaller.

Second, in an error-prone environment, each conditional cooperator reacts differently to the others’ mistakes according to his/her tolerance level. For example, $T_{n-1}$ players immediately begin retaliation by withdrawing cooperation, $T_{n-2}$ players tolerate one mistake, and $T_{n-h}$ players tolerate $h-1$ mistakes. Note that the behavioral indistinguishability between the $T_{k}$ strategies (where $k<n$) disappears in the absence of $T_{n}$, and the payoffs for the $T_{k}$ strategies are no longer the same, even in a situation where the $T_{n}$ strategy does not exist. As mentioned in the previous section, the dynamic instability occurs because the $T_{n-1}$ strategy is only neutrally stable, not evolutionarily stable. In other words, the dynamic problem occurs because both $T_{n-1}$ and $T_{k<n-1}$ receive the same payoff in the absence of a defection strategy in an error-free environment. As soon as the possibility of making errors is introduced, the problem of indistinguishability between all of the conditional cooperative strategies in the absence of a defect strategy can be solved because each conditional cooperative strategy reacts differently to the error that the other players make.

The following proposition shows that mistakes can eliminate the indistinguishability between the $T_{k<n}$ strategies so that the state where all of the individuals are $T_{n-1}$ players is evolutionarily stable.

As long as there is a probability of making mistakes, the indistinguishability between the hardest cooperators and the other softer cooperators disappears, which makes the $T_{n-1}$ equilibrium state evolutionarily stable under certain parameter values. The $T_{k}$ strategies ($k<n-1$) are now behaviorally distinguishable from the $T_{n-1}$ strategy because the conditional cooperative strategies are, depending on the level of $k$, different in their responding to the mistakes, even though there are no defectors in the population. In other words, in an environment where the players make mistakes, the $T_{k}$ strategies cannot invade a population that is homogeneously composed of individuals using the $T_{n-1}$ strategy.

Mistakes also affect the evolutionary stability of the softer strategies. Note that the $T_{k}$ strategies when $k<n-1$ are not stable at all in an error-free environment because these strategies allow the universal defection strategy (i.e., the $T_{n}$ strategy) to enjoy benefits from free riding on the cooperation of these strategies (see Corollary 1).

To see the effect of mistakes on the stability of the $T_{k}$ strategy, consider a population that entirely composed of players who have a $T_{k}$ strategy with the same hardness level $k$. A player with the $T_{k}$ strategy receives $F^{\varepsilon}(C|n-1)$ from the first round and maintains cooperation in the next round as long as the number of defections is less than $n-k-1$. Therefore, the mutual cooperation continues when

• •

  the number of mistakes made by the $n-1$ other conditional cooperators is less than or equal to $n-k-2$ (in this case, the maintenance of mutual cooperation does not depend on whether the focal individual makes a mistake or not), or

• •

  the number of mistakes made by the $n-1$ other conditional cooperators is exactly equal to $n-k-1$, and the focal player does not make a mistake.

The first case occurs with probability $\sum_{q=0}^{n-k-2}\psi(n-1,q,\varepsilon)$ and the second case with probability $\psi(n-1,n-k-1,\varepsilon)\times(1-\varepsilon)$. By the same token, the mutual cooperation breaks when

• •

  the number of mistakes made by the $n-1$ other conditional cooperators is exactly equal to $n-k-1$, and the focal player makes a mistake, or

• •

  the number of mistakes made by the $n-1$ other conditional cooperators is greater than $n-k-1$.

Each case occurs with probability $\psi(n-1,n-k-1,\varepsilon)\times\varepsilon$ and $\sum_{q=n-k}^{n-1}\psi(n-1,q,\varepsilon)$, respectively. Summing up all of the possible cases, the expected payoff for the $T_{k}$ strategy when the population is entirely composed of individuals following the $T_{k}$ strategy ($k<n-1$) is written in the following way.

if $k<n-1$.

Suppose that there appears one defector (i.e., a player with strategy $T_{n}$) in the homogenous population of a $T_{k}$ strategy where every player has the same tolerance level $k<n-1$. When the game is repeated with probability $\delta$ , the payoff to the $T_{n}$ player from the repeated game is calculated by

As long as the number of defections, either due to mistakes by the $T_{k}$ players or the defection of the $T_{n}$ player, is less than $n-k-1$, $T_{k}$ players will tolerate defections and maintain cooperation. In this case, the $T_{n}$ player keeps enjoying a free rider’s advantage from the repeated interaction (see (i) in Eq (3.3)). When the number of defections exceeds $n-k-1$, all of the $T_{k}$ players trigger retaliation, which gives every member a 0 payoff in the following rounds (see (ii) in Eq (3.3)).

It will be shown that it is crucial to have $V^{\varepsilon}(T_{k}|n-1)-V^{\varepsilon}(T_{n}|n-1)>0$ to prove the evolutionarily stability of the $T_{k}$ strategy. Rearranging Eq (20) and Eq (3.3) gives

and

respectively. The condition that the payoff for the $T_{k}$ strategy is greater than the payoff for the $T_{n}$ strategy is

where $\Delta(\varepsilon;k)=\frac{\delta(1-\varepsilon)\psi(n-1,n-k-1,\varepsilon)}{1-\delta\sum_{q=0}^{n-k-2}\psi(n-1,q,\varepsilon)}$. Note that the sign of $V^{\varepsilon}(T_{k}|n-1)-V^{\varepsilon}(T_{n}|n-1)$ depends on the sign of $\Delta(\varepsilon;k)-(1-\frac{F^{\varepsilon}(C|n-1)}{F^{\varepsilon}(D|n-1)})$. Because it holds that $0<1-\frac{F^{\varepsilon}(C|n-1)}{F^{\varepsilon}(D|n-1)}<1$, and $\frac{F^{\varepsilon}(C|n-1)}{F^{\varepsilon}(D|n-1)}$ is fixed by the payoff structure of the public goods game, it is crucial to have a sufficiently large $\Delta(\varepsilon;k)$.

A $T_{k}$ strategy (where $k\in\{1,\cdots,n-2\}$) is evolutionarily stable if

In the following proposition, we will show that all of the conditional cooperative strategies are evolutionarily stable when $\varepsilon$ is positive and within a proper open interval and when $\delta$ is sufficiently large. We will show that $W(T_{k}|p_{k}=1)>W(T_{k^{\prime}}|p_{k}=1)$ for $k\in\{1,\ldots,n-2\}$ in all of the following three cases where (1) a mutant is a conditional cooperator who has a softer strategy (i.e., $k^{\prime}<k$), (2) a mutant is a conditional cooperator who has a harder strategy (i.e., $k<k^{\prime}<n$), or (3) a mutant is a defector (i.e., $k^{\prime}=n$).

$T_{k}$ strategies with ($0<k<n-1$) are not stable at all in an error-free environment but become evolutionarily stable when mistakes are introduced, i.e., they do not allow the invasion of other strategies into their population because the mistakes made by the conditional cooperators produce unintended defections among themselves and leave little room for the defector’s free riding. The proof of the above proposition consists of three steps. First, when conditional cooperators make mistakes the expected payoff for the $T_{k}$ strategy in the repeated game is greater than the expected payoff to the mutant $T_{n}$ strategy (see Appendix A, Lemma 1). Behavioral mistakes, if they occur with a sufficiently low probability, no longer allow the defectors to invade the population that homogeneously consists of individuals using the $T_{k}$ strategy (where $0<k<n-1$). Behavioral mistakes produce unintended defections among the cooperators (i.e., even without a defector’s invasion), so that softer conditional cooperators are less likely to tolerate a defector’s invasion. Second, if the expected payoff for the $T_{k}$ strategy in the repeated game is greater than that for the mutant $T_{n}$ strategy, then it is also true that the expected payoff for the $T_{k}$ strategy in the repeated game is greater than the mutant $T_{k^{\prime}}$ strategy’s payoff if $k^{\prime}$ is greater than $k$, i.e., if the mutants are harder conditional cooperators (see Appendix A, Lemma 2). Third, behavioral mistakes also make the expected payoff to the $T_{k}$ strategy in the repeated game when one of these strategies invades $T_{k}$ population greater than the other conditional cooperative strategies that have$k^{\prime}$ lower than $k$ (see Appendix A, Lemma 3). Error makes those strategies distinguishable and makes the payoff for the $T_{k}$ strategy higher (same logic as proof for Proposition 2).

In an error-free environment, no conditional cooperative strategies are evolutionarily stable. The $T_{n-1}$ strategy is neutrally stable and is subject to the drift ultimately destroying the cooperative equilibrium, and none of the $T_{k<n-1}$ strategies are stable at all. Proposition 2 and Proposition 3 say that behavioral errors, if they occur with sufficiently low probability, can make all conditional cooperative strategies evolutionarily stable.

In the above section, we showed that conditional cooperative strategies, $T_{k}$ for $k\in\{1,\cdots,n-1\}$, are evolutionarily stable when $\varepsilon$ is positive but sufficiently low. Furthermore, each $T_{k}$ strategy has its own critical value of $\varepsilon_{k}$ such that the strategy is evolutionary stable in the range of $\varepsilon\in(0,\varepsilon_{k})$. If the critical value of each $T_{k}$ strategy differs according to the hardness level $k$, the magnitude of the probability that a behavioral error occurs could serve as a criterion for equilibrium selection.

In this section, we will show that the range of $\varepsilon$ supporting the evolutionary stability of the $T_{k}$ strategy varies according to the tolerance of the conditional cooperative strategies. Let $\varepsilon_{k}$ be the supremum of $\varepsilon$ that supports the evolutionary stability of a $T_{k}$ strategy (where $k\in\{1,\cdots,n-1\}$). We will show that $\varepsilon_{k}$ increases as $k$ decreases, that is, $0<\varepsilon_{n-1}<\varepsilon_{n-2}<\cdots<\varepsilon_{1}$. For example, if the probability of making an error is sufficiently low so that $0<\varepsilon<\varepsilon_{n-1}$, then all of the conditional cooperative strategies are evolutionarily stable. If the error rate is in the range of $\varepsilon_{n-1}\leq\varepsilon<\varepsilon_{n-2}$, then all of the conditional cooperative strategies except for $T_{n-1}$ are evolutionarily stable, and so on. Lastly, if the error rate is higher than or equal to $\varepsilon_{1}$, no conditional cooperative strategies are evolutionarily stable, in which case only the universal defection strategy, $T_{n}$, is evolutionarily stable. In other words, all conditional cooperative strategies are evolutionarily stable when $0<\varepsilon<\varepsilon_{n-1}$ and fewer of the conditional cooperative strategies remain evolutionarily stable as $\varepsilon$ increases.

In other words, as $\varepsilon$ increases, the $T_{k}$ strategies with a lower $k$ (i.e., with a higher tolerance level) remain evolutionarily stable. To understand the role of behavioral mistakes in the equilibrium selection in the above proposition, we need to examine the two effects that the mistakes produce. First, the mistakes lower the probability that the mutually cooperative phase continues because they trigger retaliation towards cooperators’ unintended defections. Secondly, they also reduce the possibility that the defectors enjoy benefits from free riding on tolerant cooperators. If the incumbent conditional cooperators have a lower $k$, the first effect is less detrimental because the cooperative equilibrium based on the lower $k$ is less vulnerable to a breakdown when a mistake occurs with higher probability.