# Finding subgame-perfect Nash equilibrium in the Trust game

I am facing a game theory problem which is as follows:

An experiment was designed to study individuals' propensity to be trusting and to be trustworthy in a task called the investment game. In this game two subjects are each given 3 dollars as a show up fee.

One of the subjects is told that he or she will be decision maker one (DM1), and can send an amount (0 dollar, 1 dollar, 2 dollar or 3 dollar) to the other subject (decision maker two, DM2). The rules are simple. Every dollar sent will be doubled by the experimenter before it reaches DM2, who then gets to decide how much of the doubled money to keep and how much to send back to DM1. After DM2’s decision the game is over and subjects leave. The experiment is double-blind meaning that neither the subjects nor the experimenter knew who was matched together, or what subject made what decision.

In order to analyze this problem, I have drawn the extensive and normal forms which are showed [here][1] and [here][2].

I find no matter what kind of choice each subject choose, they all choose to play either (0,0) or (3,3) in the Nash equilibrium.

Now the question is know how to find subgame-perfect Nash equilibrium in terms of normal form and extensive form?

Is the normal form of the Trust game a symmetric game?

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In order for there to be a problem, there needs to be something to solve? What are the two players expected to use as a criterion to choose whether they should give more or less money? Do they want to maximize their common profit or their individual profit? It is not clear from here what should be considered... or maybe I am too tired. – Patrick Da Silva Aug 19 '12 at 12:45
oh, in this problem, each subject want to maximize their profit – user824624 Aug 19 '12 at 12:49

To find any Nash equilibria, fix one player's choice and choose the option that maximizes the other player's outcome. For instance, suppose DM1 chooses 0, corresponding to the first row of your table. Then DM2 is looking at pay-offs of 3 (by choosing 0), 2 (by choosing 1), 1 (by choosing 2), and 0 (by choosing 3). In this situation, DM2 chooses 0, giving the maximum payoff of 3, assuming DM1 chose 0.

In the same way, for each choice that DM1 makes, DM2's payoff is optimized by choosing 0 (that is, in each row the second number of each ordered pair is greatest in the first column). We say DM2 has a dominated strategy of choosing 0.

From the perspective of the other player, suppose DM2 chooses 0. The DM1 chooses 0 since the resulting 3 is the greatest payoff in that column. For every fixed choice of DM2, DM1's payoff is optimized by choosing 0 (in each column the first number of each ordered pair is greatest in the first row). So DM1 also has a dominated strategy of choosing 0.

That means the Nash equilibrium occurs when both players choose 0, giving them each a payoff of 3.

The concept of Nash equilibrium follows from assuming that you have no say over what the other player chooses. This is similar to the prisoner's dilemma, which also has a sub-optimal Nash equilibrium. Here, if the players could trust each other, they could each do much better by both choosing 3, giving each of payoff of 6. But if you believe your opponent will choose 3, you can get a payoff of 9 by choosing 0 instead, which gives your opponent 0 -- that's the trust aspect.

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Thank you, Brian for your quick response. what about the Subgame-­‐perfect Nash equilibrium – user824624 Aug 19 '12 at 19:11
Hmm, certainly the table you came up with describes a symmetric game, and your table makes sense to me from the problem description. Not sure what to tell you. – Brian Hopkins Aug 19 '12 at 19:13
strange, because the question followed by the trust game, it says that The normal formof the trust game is clearly not symmetric and consequently the notion of Evolutionary Stable Strategy does not (immediately) apply. Yet we can make the game symmetric. That makes me very confused – user824624 Aug 19 '12 at 19:19
Maybe it's that we were assuming money sent from DM2 to DM1 is also doubled. If not, consider the choices (1,2). From payoffs (3,3), after DM1 sends the dollar we have (2,5), and if DM2 send 2 back undoubled the final payoff is (4,3), not the (6,3) you have. – Brian Hopkins Aug 19 '12 at 19:42
I think the money sent from DM2 to DM1 is also doubled – user824624 Aug 20 '12 at 0:54