# Expected payoff for Iterated Prisoners Dilemma (IPD) strategies

The two strategies playing the game are Tit-for-Tat (TFT) and Psycho (Psy). I am asked to show that if TFT and Psy play IPD with random n, then on average the expected difference in payoffs is positive for Psy and negative for TFT. I have included strategy descriptions below.

In the previous part of the question, I showed by induction that TFT can do no better than tie with Psy. I am assuming that expected difference in payoffs is just the difference in expected probability for n rounds, but I am not quite sure how to calculate it. I know $$E(X) = \sum_{1}^{n} xP(X)$$ but how do you expand it to take in varying values of two strategies in a game. Also, would the probabilities be 1/4 or would they be strategy-dependent (i.e. because, for example, we know that Psy will do the opposite of TFT, would the probability be 1/2 instead of 1/4)?

I've looked online and in the textbook, but I cannot seem to find a formula.

TFT strategy:

• Nice: always cooperates on first round
• Provocative: always defects if opponent defects in previous round
• Forgiving: always cooperates if opponent cooperates again
• Simple: other strategies can adapt to it

Psy strategy:

• Always defects on first round
• Does opposite of what opponent did last round
• Please give www reference for "Tit-for-Tat (TFT) and Psycho (Psy)" – Jean Marie Feb 2 '17 at 6:17

• Yes, it would mean that on average it is equally likely for PSY to be ahead as it is for them to be tied. That means the difference in payoff when you stop (since you stop randomly) is equally likely to be $+x$ or $0$ for PSY (where $x$ is the amount PSY gets ahead by when they initially defect while TFT cooperates). So the average difference in payoff is $+x/2$ in favor of PSY. – spaceisdarkgreen Feb 2 '17 at 6:41