# Difference between revisions of "The Prisoner's Dilemma"

The Prisoner's Dilemma
Field: Algebra
Image Created By: Greenmantis

The Prisoner's Dilemma

This 2X2 matrix shows the possible actions and resultant outcomes for an instance of the Prisoner's Dilemma. In each outcome box, Robber #1's payoffs are listed to the left, while Robber #2's are on the right.

# Basic Description

The Problem:

The dilemma is set up as follows: two criminals caught while trying to rob a bank together. The interrogating officer visits each separately and offers each a choice: confess and testify against the other, or hold out and refuse to cooperate with the police. If the first criminal (Robber #1) confesses while the other holds out, he will be sentenced to community service, while the other (Robber #2), will get five years in prison. However, Robber #2 has also been made the same offer. Without a confession, the law only has enough evidence on the two to give them each a year in jail. Should they both confess, on the other hand, they will each get three years in prison. Assuming that both robbers act rationally as they consider their options, what will they do?

The Solution: The robbers will both confess and they will both be sentenced to three years. Most people would agree that the outcome in which they both hold out is better. Unfortunately, this outcome does not meet the criterion for a stable Nash equilibrium - an outcome in which neither player will gain by changing his strategy, and thus both are content to continue their current choices[1]. Each robber would rather confess than not, in order to escape jail, so in the absence of other factors, this position will not be maintained.

# A More Mathematical Explanation

## Logic Behind the Solution

At the beginning of the problem we assumed the robbers were behaving [...]

## Logic Behind the Solution

At the beginning of the problem we assumed the robbers were behaving rationally – formally defined, rational players are those that always act to maximize their own benefit, given the information they possess[2]. Immediately after the police officer makes them the offer, the two can be said to be jointly holding out, so they can expect a year in jail. Robber #1, considering his options, notices that if he confesses while his partner stays loyal, he can go free. Thus, he can improve his payoff by moving from lower right to upper right. Likewise, Robber #2 will find it advantageous to go from lower right to lower left. Now if the first thief has already decided to confess (upper right box), the other should certainly not remain with his initial course of holding out. He should confess as well, to improve his payoff from -5 to -3. Thus, the two will move from the outcome at upper right to the one at upper left. The same logic applies of Robber #2 decides to confess first, leading the two inexorably to the outcome in which the both confess. Once the two have both decided to confess, neither can improve their situation by switching to a “hold out” strategy. For this reason, no arrows lead away from “both confess” outcome. On the other hand, for any of the other outcomes, one of the other of the criminals can improve his situation by changing his choice, so arrows lead away from all the other outcome boxes. It should be noted that these results are not affected by the information available to the robbers.

## Minimax Theory

Von Neumann's research led to the concept of the maximin strategy. The maximin strategy for Player a is the best that he can do if other players are specifically making choices to decrease his payoff. If the strategy player a picks can be represented as Sa, then the maximin strategy Sa* is the one that optimizes his payoffs. Likewise, the minimax strategy is the one chosen by Players b, c, d, etc. to minimize Player a's payoff regardless of his actions[3].

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