Conditions for a Prisoner’s Dilemma.
In order for a game to be a variation of Prisoner’s Dilemma it must satisfy two conditions:
-
\(\displaystyle T>R>P>S\)
-
\(\displaystyle (T+S)/2 \lt R\)
Section 4.4 What Makes a Prisoner’s Dilemma?
In this section we give a mathematical description of Prisoner’s Dilemma and compare it to some similar games.
The Class-wide Prisoner’s Dilemma game we played in Section 4.3 has the payoff matrix given in Table 4.4.1 for each pair of players.
| Player 2 | |||
| Cooperate | Defect | ||
| Player 1 | Cooperate | \((3, 3)\) | \((0, 5)\) |
| Defect | \((5, 0)\) | \((1, 1)\) | |
We can classify each of the values for the payoffs as follows:
-
Reward for Mutual Cooperation: \(R=3.\)
-
Punishment for Defecting: \(P=1.\)
-
Temptation to Defect: \(T=5.\)
-
Sucker’s Payoff: \(S=0.\)
Let’s apply this description of Prisoner’s Dilemma to a few games we’ve seen. We can use Conditions for a Prisoner’s Dilemma to check if a game is really a Prisoner’s Dilemma.
Activity 4.4.1. Description of conditions.
Describe conditions (1) and (2) in Conditions for a Prisoner’s Dilemma in words.
Activity 4.4.2. The conditions for Classwide Prisoner’s Dilemma.
Show the Conditions for a Prisoner’s Dilemma hold for the Class-wide Prisoner’s Dilemma in Table 4.4.1.
Activity 4.4.3. The conditions for Prisoner’s Dilemma.
Recall the matrix for Prisoner’s Dilemma from Example 4.2.3.
| Prisoner 2 | |||
| Confess | Don’t Confess | ||
| Prisoner 1 | Confess | \((8, 8)\) | \((0.25, 10)\) |
| Don’t Confess | \((10, 0.25)\) | \((1, 1)\) | |
Determine \(R, P, T,\) and \(S\) for this game. Be careful, think about what cooperating versus defecting should mean. Show the Conditions for a Prisoner’s Dilemma are satisfied.
Activity 4.4.4. The conditions for Chicken.
Recall the matrix for Chicken from Example 4.2.5.
| Driver 2 | |||
| Swerve | Straight | ||
| Driver 1 | Swerve | \((0, 0)\) | \((-1, 10)\) |
| Straight | \((10, -1)\) | \((-100, -100)\) | |
Determine \(R, P, T,\) and \(S\) for this game. Again, think about what cooperating and defecting mean in this game. Determine if the Conditions for a Prisoner’s Dilemma are satisfied. If not, which condition(s) fail?
Activity 4.4.5. The conditions on another game.
Consider the cooperate-defect game where the first row/column is C and the second row/column is D:
\begin{equation*}
\left[\begin{matrix}
(3, 3) \amp (0, 50)\\
(50, 0) \amp (.01, .01)
\end{matrix}\right].
\end{equation*}
Determine \(R, P, T,\) and \(S\) for this game. Determine if the Conditions for a Prisoner’s Dilemma are satisfied. If not, which condition(s) fail?
Activity 4.4.6. A little more practice.
Consider the cooperate-defect game where the first row/column is C and the second row/column is D:
\begin{equation*}
\left[\begin{matrix}
(1000, 1000) \amp (0, 100)\\
(100, 0) \amp (100, 100)
\end{matrix}\right].
\end{equation*}
Determine \(R, P, T,\) and \(S\) for this game. Determine if Conditions for a Prisoner’s Dilemma are satisfied. If not, which condition(s) fail?
Activity 4.4.7. Compare the games.
The games in Activity 4.4.4, Activity 4.4.5, and Activity 4.4.6 are not true Prisoner’s Dilemmas. For each game, how do the changes in payoffs affect how you play? In particular, in Prisoner’s Dilemma, a player will generally choose to defect. This results in a non-optimal payoff for each player. Is this still true in Activity 4.4.4, Activity 4.4.5, and Activity 4.4.6? If possible, use canges in the Conditions for a Prisoner’s Dilemma to help explain any differences in how one should play.
We can now define defection as the idea that if everyone did it, things would be worse for everyone. Yet, if only one (or a small) number did it, life would be sweeter for that individual. We can define cooperation as the act of resisting temptation for the betterment of all players.
Activity 4.4.8. Example from real life.
Give an example of defection and cooperation from real life. Explain how your example of defection make things worse for everyone if everyone did it, but would benefit the defctor. Explain how cooperation is improves things for all, even if the payoff is smaller for the individual.
Reading Questions Check Your Understanding
1.
2.
True.
- \(T>R>P>S; (T+S)/2 < R.\)
False.
- \(T>R>P>S; (T+S)/2 < R.\)
True or False: the game
\begin{equation*}
\left[
\begin{matrix}(5, 5) \amp (15, -1)\\
(-1, 15) \amp (10, 10)\\
\end{matrix}
\right]
\end{equation*}
satisfies the Conditions for a Prisoner’s Dilemma.
3.
4.
True.
- \(T\ngtr R; (T+S)/2 < R.\)
False.
- \(T\ngtr R; (T+S)/2 < R.\)
True or False: the game
\begin{equation*}
\left[
\begin{matrix}(10, 10) \amp (-1, 5)\\
(5, -1) \amp (2, 2)\\
\end{matrix}
\right]
\end{equation*}
satisfies the Conditions for a Prisoner’s Dilemma.
5.
6.
True.
- \(T>R>P>S; (T+S)/2 < R.\)
False.
- \(T>R>P>S; (T+S)/2 < R.\)
True or False: the game
\begin{equation*}
\left[
\begin{matrix}(2, 2) \amp (-5, 5)\\
(5, -5) \amp (-2, -2)\\
\end{matrix}
\right]
\end{equation*}
satisfies the Conditions for a Prisoner’s Dilemma.
7.
The game given in
\begin{equation*}
\left[
\begin{matrix}(2, 2) \amp (-5, 5)\\
(5, -5) \amp (-2, -2)\\
\end{matrix}
\right]
\end{equation*}
has equilibrium point(s).
- \(0\)
- \(1\)
- \((-2, -2)\) is the only equilibrium point.
- \(2\)
- \(3\)
8.
9.
True.
- \(P\ngtr S; (T+S)/2 \nless R.\)
False.
- \(P\ngtr S; (T+S)/2 \nless R.\)
True or False: the game
\begin{equation*}
\left[
\begin{matrix}(-2, -2) \amp (5, 0)\\
(0, 5) \amp (2, 2)\\
\end{matrix}
\right]
\end{equation*}
satisfies the Conditions for a Prisoner’s Dilemma.
10.
The game given in
\begin{equation*}
\left[
\begin{matrix}(-2, -2) \amp (5, 0)\\
(0, 5) \amp (2, 2)\\
\end{matrix}
\right]
\end{equation*}
has equilibrium point(s).
- \(0\)
- \(1\)
- \(2\)
- \((0, 5)\) and \((5, 0)\) is the two equilibrium points.
- \(3\)
11.
12.
True.
- \(T>R>P>S; (T+S)/2 \nless R.\)
False.
- \(T>R>P>S; (T+S)/2 \nless R.\)
True or False: the game
\begin{equation*}
\left[
\begin{matrix}(0, 0) \amp (-15, 20)\\
(20, -15) \amp (-10, -10)\\
\end{matrix}
\right]
\end{equation*}
satisfies the Conditions for a Prisoner’s Dilemma.
13.
The game given in
\begin{equation*}
\left[
\begin{matrix}(0, 0) \amp (-15, 20)\\
(20, -15) \amp (-10, -10)\\
\end{matrix}
\right]
\end{equation*}
has equilibrium point(s).
- \(0\)
- \(1\)
- \((-10, -10)\) is the only equilibrium point.
- \(2\)
- \(3\)
