In the description of an iterated prisoner's dilemma on Wikipedia, it states that in order for a game to be considered iterable, it must conform to the rule 2R > T + S, where R is the reward for cooperation, T is the temptation payoff, and S is the sucker's payoff. Could someone provide an example of a dilemma in which this would not be the case, i.e. it would be non-iterable?


The payoff depends entirely on how you set the game up. Here's one example of the case where $2R \leq T + S$:

                                  Player B
                           Cooperate  |    Defect
Player     Cooperate        (1, 1)    |  (-2, 100)
A          Defect          (100, -2)  |  (-1, -1)

Here, $R = 1$, $T = 100$, and $S = -1$, and $2R < T + S$.

Edit: To add a bit more value to the answer... any game could be played repeatedly (and thus is "iterable" in that sense.) However, my understanding is that the Wikipedia page writer means that for (Cooperate, Cooperate) to be sustained, you at least need this condition to be true.

  • $\begingroup$ I am perplexed. Was the question really "give me three numbers $T,R,S$ for which $S < R < T$ and $2R < T + S$"? $\endgroup$
    – Giskard
    Oct 21 '19 at 7:16
  • $\begingroup$ Maybe I was wrong, but the question: "Could someone provide an example of a dilemma in which this would not be the case, i.e. it would be non-iterable?" suggests so (with standard PD conditions, of course). Couldn't think of anything else :( $\endgroup$
    – Art
    Oct 21 '19 at 7:17
  • $\begingroup$ I agree with your interpretation, I just cannot imagine someone being unable to solve the question. If you start with $S = 0$, pick any $R$, and then pick $T = 2R - S + 1$ you have such a triple. Seems like 30-40 seconds of experimentation would yield a solution to anyone interested. $\endgroup$
    – Giskard
    Oct 21 '19 at 7:19
  • 1
    $\begingroup$ I guess the OP misinterpreted the concept of the entire game a bit. We'll just have to see what he/she says. $\endgroup$
    – Art
    Oct 21 '19 at 7:20
  • $\begingroup$ From the OP-cited Wikipedia: "In addition to the general form above, the iterative version also requires that 2R > T + S, to prevent alternating cooperation and defection giving a greater reward than mutual cooperation." OP just misinterpreted what this was saying - in fairness, it's not really explained why one needs sustained cooperation. $\endgroup$
    – heh
    Oct 21 '19 at 14:38

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