<p>The payoff depends entirely on how you set the game up. Here's one example of the case where <span class="math-container">$2R \leq T + S$</span>:</p>

<pre><code>                                  Player B
                           Cooperate  |    Defect
Player     Cooperate        (1, 1)    |  (-2, 100)
A          Defect          (100, -2)  |  (-1, -1)
</code></pre>

<p>Here, <span class="math-container">$R = 1$</span>, <span class="math-container">$T = 100$</span>, and <span class="math-container">$S = -1$</span>, and <span class="math-container">$2R &lt; T + S$</span>.</p>

<p>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.</p>

CC BY-SA 4.0
2019-10-21T07:16:24.930Z
I am perplexed. Was the question really "give me three numbers $T,R,S$ for which $S < R < T$ and $2R < T + S$"?
Giskard
CC BY-SA 4.0
2019-10-21T07:17:35.433Z
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 :(
Art
CC BY-SA 4.0
2019-10-21T07:19:32.780Z
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.
Giskard
CC BY-SA 4.0
2019-10-21T07:20:52.617Z
I guess the OP misinterpreted the concept of the entire game a bit. We'll just have to see what he/she says.
Art
CC BY-SA 4.0
2019-10-21T14:38:51.043Z
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.
heh
CC BY-SA 4.0
2019-10-21T17:07:49.783Z
Yes, thanks for the help.   Basically my misunderstanding relates to the comment above.  If I am understanding it correctly now, two players who wanted to "cooperate" if 2R < T + S would have to choose a strategy where they alternated between getting S and T over two turns to optimize their cooperation.
PRGJ
CC BY-SA 4.0
2019-10-21T07:14:35.260Z
2019-10-21T07:24:05.063Z
2019-10-21T07:24:05.063Z
Art
<p>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?   </p>

<game-theory><repeated-games>
Clarification of iterated prisoners dilemma
PRGJ
# Clarification of iterated prisoners dilemma

(Asked by: username_0 on 2019-10-21)

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?

**Tags:** game-theory, repeated-games

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## 1 answers

### Answer by username_1 on 2019-10-21. Score: 3 | Accepted answer

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.

> **Comments:**
> **username_2**: I am perplexed. Was the question really "give me three numbers $T,R,S$ for which $S < R < T$ and $2R < T + S$"?  
> **username_1**: 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 :(  
> **username_2**: 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.  
> **username_1**: I guess the OP misinterpreted the concept of the entire game a bit. We'll just have to see what he/she says.  
> **username_3**: 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.  
> **username_0**: Yes, thanks for the help.   Basically my misunderstanding relates to the comment above.  If I am understanding it correctly now, two players who wanted to "cooperate" if 2R < T + S would have to choose a strategy where they alternated between getting S and T over two turns to optimize their cooperation.  

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