The Iterated Prisoner’s Dilemma: Press-Dyson Interactive
The Game
In each round, each of us can choose to cooperate or to defect. We play repeated rounds. In each round:
- If we both cooperate, we both get 3 points.
- If you cooperate and I defect, I get 5 points and you get none.
- If you defect and I cooperate, you get 5 points and I get none.
- If we both defect, we both get 1 point.
- Fixed score
- Extortion
- Generous
This is possibly the strategy that slightly bettered generous Tit for Tat in Stewart and Plotkin’s simulation. It is fair in the sense that if you always cooperate then I will too. It is a self-sacrificing strategy: if you do not always cooperate I will do worse than you. In fact on average I lose precisely twice as much as you: if your average score is less than 3, mine will be less by as much again.
How does it work?
My play is based purely on how both of us played in the previous move:
- If we both cooperated last time, then I cooperate.
- If I cheated you last time (you cooperated and I defected), then I cooperate with probability 8/10.
- If you cheated me last time (I cooperated and you defected), then I cooperate with probability 3/10.
- If we both defected last time, I cooperate with probability 2/10.
These probabilities were chosen to establish the relation SX − 2SY + 3 = 0 using Press and Dyson’s results, where SX is my score and SY is yours.
This may or may not be the same as the strategy called ZDGTFT-2 by Stewart and Plotkin. Unfortunately they don’t describe their strategy explicitly, only saying that it satisfies the relation above; but there are infinitely many such strategies.