Second Thoughts of Social Dilemma in Mechanism Design

Abstract

This paper shows that second thoughts are not an innocent device in our daily life, but it is human wisdom that plays an important role in resolving problems such as social dilemmas. We design a simple mechanism to achieve Pareto efficiency in social dilemmas and then compare the performance of this mechanism with and without second thoughts. First, second thoughts change the payoff structure of the game in favor of cooperation. Second, this mechanism is robust even when players deviate from a payoff maximizing behavior.

Appendices

The author would like to thank Yoshitaka Okano for supporting the proof.

Appendix 1: The Proof of Proposition 2

(i) Let m and l be the numbers of C and D players in the first stage, respectively. If m = n, the outcome is (αnw ,αnw ,…,αnw ). If l = n, the outcome is (w,w,…,w).

Suppose 1 ≤ l < n. Consider the choice of players who chose C in the first stage after observing the choices of D players in the second thought stage. Let 0 ≤ l′ < l be the number of D players who change their choices from D to C in the second thought stage. Since α(m′ + l′ + 1)w < α(m′ + l′)w + w for all 1 ≤ m′ ≤ m − 1, where m′ is the number of C players in the first stage who remains to choose C in the subgame after the second thought stage, D is better than C for any C player in the third stage after observing the choices in the second thought stage. That is, all players who chose C in the first stage choose D after the second thought stage.

Consider any strategy path on which at least one D player chose D again in the second thought stage. If this were the case, every C player after the second thought stage would choose D. Keeping this fact in mind, let us choose the youngest D player (e.g., by names or numbers assigned to players) who chose D in the second thought stage. Then, the subgame after the choice of the youngest D player is a sequential social dilemma game, and hence, every D player after the choice chooses D.

We now identify the payoff outcome of every subgame constructed by the end nodes in the first stage. Choose any subgame except for the cases where all players chose C or all players chose D in the first stage. Suppose that every D player except for the last one changed his or her choice from D to C in the second thought stage. Consider next the choice of the last D player. If the player chooses C, the payoff is αnw , whereas if the player chooses D, the payoff is α(l − 1)w + w. Since αnw  − α(l − 1)w – w = α{n − (l − 1)}w − w and n − (l − 1) = m + 1:

• If α > 1/(m + 1), then the last D player chooses C.

• If α = 1/(m + 1), then the last D player is indifferent between C and D.

• If α < 1/(m + 1), then the last D player chooses D.

Suppose α > 1/(m + 1). If the penultimate player chooses D, then the payoff is α(l − 2)w + w since the last D player chooses D in the second thought stage and every C player in the first stage chooses D in the third stage. If the player chooses C, then it is αnw since the last player chooses C. Since αnw  − {α(l − 2)w + w} = {α(n – l + 2) − 1} w = {α(m + 2) − 1)}w > 0, the player chooses C. Since αnw  − {α(l − 2)w + w} > 0, αnw  − {α(l − k)w + w} > 0 for all 2 ≤ k ≤ l. That is, the k-th player to last chooses C, and hence, all D players choose C in the second thought stage, and the payoff outcome is (αnw ,…,αnw ).

Suppose α < 1/(m + 1). Then, the last D player chooses D, and hence, the payoff of the penultimate player is α(l − 1)w if the player chooses C. If the player chooses D, then the payoff is α(l − 2)w + w. Since α(l − 2)w + w – α(l − 1)w = (1 − α)w > 0, the player chooses D. Since α(l − k)w + w – α(l – k + 1)w = (1 − α)w > 0 for all 2 ≤ k ≤ l, the k-th player to last chooses D, and hence, no D players in the first stage change their decisions in the second thought stage, and the payoff outcome is (w,…,w).

Take any α satisfying 1/ n < α < 1 and α ∉ {1/(n − 1),1/(n − 2),…,1/2}. Consider the case of α > 1/2. Then, α > 1/2 ≥ 1/(m + 1) for all m ≥ 1, and hence, the payoff outcome of every subgame other than (D,D,…,D) is (αnw ,…,αnw ): without loss of generality, consider player 1. The payoff in subgame (C,D,D,…,D) is αnw and that in subgame (D,D,…,D) is w. Since αnw  > w, C is better than D. Since α > 1/(m + 1) for all m ≥ 1, the outcome of the two subgames (C,·) and (D,·) is (C,C,…,C) where “·” shows that at least one player’s choice is C. That is, player 1 is indifferent between the outcomes of subgames (C,·) and (D,·). Therefore, C weakly dominates D for all players, and hence, (C,C,…,C) is the SPEWDS outcome.

Consider next the case of 1/2 ≥ 1/(k + 1) > α > 1/(k + 2) ≥ 1/ n. Consider player 1. Let “·” indicate that the number of C is k. Then, the payoff in subgame (C,·) is αnw since α > 1/{(k + 1) + 1}, and that in subgame (D,·) is w since 1/(k + 1) > α. That is, C is better than D. Since the outcome of the two subgames (C,·) and (D,·) is the same where “·” indicates that the number of C is not k, C weakly dominates D for all players, and hence, (C,C,…,C) is the SPEWDS outcome.

Thus, if α ∉ {1/(n − 1),1/(n − 2),…,1/2}, the SAMST implements cooperation in SPEWDS.

(ii) Suppose α = 1/(m + 1). Then, the last D player is indifferent between C and D since αnw  = α(l − 1)w + w. Suppose that the penultimate player chooses C. Then, the payoff of the penultimate player is αnw if the last D player chooses C and is α(l − 1)w if the last D player chooses D. If the penultimate player chooses D, then the payoff is α(l − 2)w + w. Since αnw  – {α(l − 2)w + w} = αw  > 0, αnw  > α(l − 2)w + w > α(l − 1)w. That is, both C and D survive by using the elimination of weakly dominated strategies. Since αnw  > α(l − k − 1)w + w > α(l − k)w for all $k=1,\dots,l-1$, both C and D survive by using the elimination of weakly dominated strategies for all D players.

Let ${\underset{\raise2pt\hbox{\kern-1pt--}}{m}}(\alpha)=\left\lceil \left(1/\alpha \right)-1\right\rceil$ where ⌈a⌉ is the smallest integer not less than a. Since 1/ n < α < 1, $1\le {\underset{\raise2pt\hbox{\kern-1pt--}}{m}}(\alpha)\le n-2.$ Suppose that (1/ α) − 1 is an integer. Then, ${\underset{\raise2pt\hbox{\kern-1pt--}}{m}}=\left(1/\alpha \right)-1.$ The following case shows that there exists a player who is indifferent between C and D when the number of C players is ${\underset{\raise2pt\hbox{\kern-1pt--}}{m}}$ or ${\underset{\raise2pt\hbox{\kern-1pt--}}{m}}-1.$ Consider two cases:

• Case 1: Suppose that the number of C players is ${\underset{\raise2pt\hbox{\kern-1pt--}}{m}}.$ Choose any player who is not a member of the C players. If the player chooses C, the payoff outcome is αnw . If the player chooses D, the maximum possible payoff is that all D players other than the player change from D to C and the player is the last D player since all C players change from C to D after the second thought stage. Then, the payoff is $\alpha(\overline{l}-1)w+w$, and hence, $\alpha\kern-.2pt nw-\left\{\alpha(\overline{l}-1)w+w\right\}=\{\alpha n-\alpha(\overline{l}-1)-1\}w=\{\alpha({\underset{\raise2pt\hbox{\kern-1pt--}}{m}}+1)-1\}w=0$, where $\overline{l}=n-{\underset{\raise2pt\hbox{\kern-1pt--}}{m}}$ and $\overline{l}$ ≥ 2 since n ≥ ${\underset{\raise2pt\hbox{\kern-1pt--}}{m}}+2.$ That is, the payoff of C is the same as the payoff of D for the player.

• Case 2: Suppose that the number of C players is ${\underset{\raise2pt\hbox{\kern-1pt--}}{m}}-1.$ Choose any player who is not a member of the C players. If the player chooses C in the first stage, we show that the payoff outcome should be at least w. Since $\overline{l}\ge 2,$ there must be at least one D player. If all D players change from D to C in the second thought stage, the C player obtains αnw . If at least one D player chooses D in the second thought stage, the C player obtains at least w by changing from C to D after the second thought stage. If the player chooses D, the payoff is w. That is, the payoff of C can be the same as the payoff of D for the player.

Thus, there is a possibility that C and D are indistinguishable for some players. Let player 1 be such a player and suppose that the first ${\underset{\raise2pt\hbox{\kern-1pt--}}{m}}$ players choose C. Then, since C and D are indistinguishable,

$$ {\begin{array}{l}\mathrm{the}\ \mathrm{payoff}\ \mathrm{outcome}\ \mathrm{of}\ \mathrm{subgame}\ (\underset{{\underset{\raise2pt\hbox{\kern-1pt--}}{m}}}{\underbrace{C,C,\dots, C}};\underset{\overline{l}}{\underbrace{D,\dots, D}})\\ {}=\mathrm{the}\ \mathrm{payoff}\ \mathrm{outcome}\ \mathrm{of}\ \mathrm{subgame}\ (D,\underset{{\underset{\raise2pt\hbox{\kern-1pt--}}{m}}-1}{\underbrace{C,\dots, C}};\underset{\overline{l}}{\underbrace{D,\dots, D}}).\end{array}} $$

Since the payoff outcome of the latter should be (w,w,…,w), each of the last $\overline{l}$ players in the former can obtain αnw by changing from D to C. That is, C weakly dominates D for the last $\overline{l}$ players.

In contrast, compare $\mathrm{the}\ \mathrm{payoff}\ \mathrm{outcome}\ \mathrm{of}\ \mathrm{subgame}\ (D,\underset{{\underset{\raise2pt\hbox{\kern-1pt--}}{m}}}{\underbrace{C,\dots, C}};\underset{\overline{l}-1}{\underbrace{D,\dots, D}})$ with $\mathrm{the}\ \mathrm{payoff}\ \mathrm{outcome}\ \mathrm{of}\ \mathrm{subgame}\ (\underset{{\underset{\raise2pt\hbox{\kern-1pt--}}{m}}+1}{\underbrace{C,C,\dots, C}};\underset{\overline{l}-1}{\underbrace{D,\dots, D}}).$ The latter payoff outcome should be (αnw ,…,αnw ), and hence, the payoff of player 1 should be αnw . Since player 1 in the former should obtain αnw , which is more than w, at least one player changes from D to C in the second thought stage, and hence, all C players who change from C to D after the second thought stage should obtain strictly more than w. Then, each of the same ${\underset{\raise2pt\hbox{\kern-1pt--}}{m}}$ players obtains w by changing from C to D. That is, C weakly dominates D for the ${\underset{\raise2pt\hbox{\kern-1pt--}}{m}}$ players. Thus, C and D are indistinguishable for player 1, and C weakly dominates D for the rest.

Suppose that C and D are indistinguishable for player 1. Then, there exists another reduced normal form game in the first stage where C weakly dominates D for player 1. Since C and D are indistinguishable for player 1, the payoff of player 1 in subgame (C,C,…,C;D,…,D) is either αnw or w. Since the payoff outcome in this subgame can be either (αnw ,…,αnw ) or (w,…,w), there is another reduced normal form game where C weakly dominates D for player 1.

Since the choice of a player who faces indistinguishability is arbitrary, C weakly rules D for all players. That is, the SAMST implements cooperation in SPEWRS.█

Appendix 2: The Proof of Proposition 3

The author would like to thank Yoshitaka Okano for supporting the proof.

(i) Let$f(q)={\left(1-q\right)}^n,g(q)={\sum}_{k=0}^{\overline{l}}\ {{}_nC}_k{q}^k\ \mathrm{and}\ h(q)={\sum}_{k=\overline{l}+1}^{n-1}\ {{}_nC}_k{\left(1-q\right)}^{n-k}\break {q}^{2k}.$ Then, f(0) = 1. Since $f^\prime(q){\,=\,}{\,-\,}n(1-q)^{n-1},f^\prime(0){\,=\,}{\,-\,}n$. Since $\overline{l}\in \left\{1,\dots, n-1\right\}$ and $g(q){\,=\,}{{}_nC}_0{q}^0+{{}_nC}_1{q}^1+{\sum}_{k=2}^{\overline{l}}\ {{}_nC}_k{q}^k{\,=\,}1+ nq+{\sum}_{k=2}^{\overline{l}}\ {{}_nC}_k{q}^k,$ g(0) = 1 and $g^{\prime}(0){\,=\,}n$. Since h(q) = q ² r(q), where $r(q){\,=\,}{\sum}_{k=\overline{l}+1}^{n-1}\ {{}_nC}_k{\left(1-q\right)}^{n-k}{q}^{2\left(k-1\right)},{h}^{\prime}(0){\,=\,}0.$ Since ${C}_{\mathit{SAMST}}(n,q,\overline{l}){\,=\,}f(q)\break g(q)+h(q),$

$$ \frac{\partial {C}_{\mathit{SAMST}}(n,0,\overline{l})}{\partial q}={f}^{\prime }(0)g(0)+f(0){g}^{\prime}(0)+{h}^{\prime}(0)=-n+n+0=0. $$

(ii) By definition, since ${C}_{\mathit{SAMST}}(n,q,\overline{l})$ has a positive part in addition to C _SAM(n,q) on (0,1), we have the result.█