# Theoretical analysis for repeated trust game

I am studying trust game (Berg, 1995). There are 2 players in this game: A and B.

A moves first. A sends an amount between 0 and 10 to B. The amount is tripled in B's side. B sends back an amount between 0 and what she got to A.

Suppose A sent 5 to B. B receives 3*5 = 15. B sent back 9 to A. A receives 9.

In this case, the profit of A is 9 - 5 = 4, and the profit of B is 15 - 9 = 6.

(it is a turn-based game).

In fact, I am focusing on repeated trust game, i.e. A and B play again and again. In some rounds A moves first, and in some rounds, B moves first.

I wonder if there exist some studies that analyze the behavior of two players in this repeated game? I tried to look on Google Scholar, but I only found the analysis for repeated prisoner-dilemma or repeated sequential prisoner dilemma (i.e. A and B have only two choices: cooperate or deficit, and they play by turn).

I found a lot of studies that analyze the behavior of players in game empirically, but I did not find a study that analyzes the behavior theoretically, e.g. analyze the Nash equilibrium for two players.

I note that the game sometimes is called investment game.

I appreciate any help.

References:

[1] Berg, Joyce, John Dickhaut, and Kevin McCabe. "Trust, reciprocity, and social history." Games and economic behavior 10.1 (1995): 122-142.

• What is the mathematical question here? -- This site is not about sociology. – Stefan Kohl Aug 3 '17 at 15:20

I'll assume that turns alternate, there are many turns, and it is unknown when the game will end. The rational cooperative optimum is that each player sends $10$ and gets back $x.$ It doesn't matter what $x$ is as long as it is always the same, so it may as well be $0.$

I could adopt the policy that I will send you $10$ if I go first and otherwise , if you sent me $x$ on the previous turn, I will keep the whole $3x$ and send you $x$ next time. If you know for sure I will do that then your optimal selfish strategy is to to always send me $10$ and keep what you get (which will be $10$ tripled).

I suppose there may be no harm in my taking into account what you returned to me last time I was the sender, but also questionable competitive advantage. It is simpler, and efficient, to establish a mutual "keep it simple" no send backs policy. Still, if two turns ago I sent you $x$ and you returned $y \leq 3x$ and now you just sent me $z,$ then perhaps I should plan to send you $w=\min(10,y+z)$ on my next turn and return to you now $\max(0,y+z-10).$ If $y+z-10 \gt 3z$ then the best I can do is send you back $3z.$ As I said, no send back is simpler.

This is pretty much like tit for tat in serial prisoners dilemma. The same considerations arise, If periodically a error happens like you send $x$ but I receive only $x-1$ (or vice-versa) then eventually no-one will send anything. Some forgiveness would avoid such a death spiral, but also could be exploited by a selfish opponent.

• Two comments: 1) it is not sufficient that the length of the game be unknown, but in addition that at every stage the probability that the game continues is sufficiently high. 2) I did not see why the amount x that the receivers sends back to the sender should be constant over time. – Eilon Aug 4 '17 at 3:43

This game is a standard repeated game, hence its analysis follows the standard analysis of repeated games.

Let me assume that in odd stages one player is the sender and in even stages it is the other. For the infinitely repeated game: the minmax value of each player in the stage game is 0 (each player can send 0, lowering the other player's payoff to 0). The set of equilibrium payoffs of the infinitely repeated game is the set of feasible and individually rational payoffs, which is the intersection of the nonnegative orthant and the triangle whose extreme points are (0,0), (-10,30), and (10,20).

For the finitely repeated game, you should check whether there are end effects. In this case, as in the repeated prisoner's dilemma, there are. The receiver in the last period has no incentive to send anything to the sender. Hence the sender in the last period has no incentive to send anything to the receiver, and this propagates back to the first stage, so that no player will ever send anything to the other player, and the unique equilibrium payoff is (0,0).

If the order by which players play is dictated by some random process, then the set of equilibrium payoffs may be different than above.

• Great. Could you point me out some papers/books that analyze this (otherwise about sequential prisoner dilemma should be okay)? – mommomonthewind Aug 5 '17 at 7:44
• So after the intersection one has the triangle with points $(0,0)\ (0,25)\ (10,20)?$ How do you get those? Maybe you meant $(0,30)$ instead of $(-10,30)?$ And why is it not rational that one always sends $10$ and gets back $15?$ Rational or not, would you describe one turn as $(15,15)$ or $(5,15)?$ – Aaron Meyerowitz Aug 6 '17 at 3:59
• Chapter 13 in the textbook "Game Theory" by Maschler-Solan-Zamir describe the folk theorem in repeated games. There you can read how to implement any payoff in the set of feasible and individually rational payoffs. Any other textbook in Game Theory that has a chapter on repeated games will do as well. – Eilon Aug 6 '17 at 8:05