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In $n=2$ person (say $A$ & $B$) prisoner's dilemma, the possible outcomes are $AB, CC,CD,DC,DD$ and the payoffs are $(1,1), (0,3), (3,0), (2,2)$ where $C$ is "cooperation" and $D$ is "defection".

Now for $n=3$ $(A,B,C)$, $A$ has two choices (for $B,C$) and the same for $B$ and $C$, therefore the outcomes are

$$\begin{eqnarray*} CC-CC-CC, \\ CC-CC-CD, \\ CC-CC-DC, \\ CC-CC-DD, \\ \dots \end{eqnarray*}$$

and $64$ such others.

What will be the payoff then? I don't intend to get the full payoff but the methodology to calculate the payoff for say $CD-DD-DC$.

PS: I thought of breaking it into $2$ person prisoner's dilemma like the $CD-DD-DC$ as $CD, DD, DC$, evaluate the individual, divide by 3 and sum it up. Is it wrong to do that?

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  • $\begingroup$ I would assume the game is still symmetric and so the payoffs should (within symmetry) depend only on the number of players who defect: For example: * if CCC, payoffs = (1,1,1); * if DCC payoffs = (3,0,0) (the lone defector has all the bargaining power); * if DDC payoffs = (2,2,0) (while the cooperator loses, neither defector has all the leverage); * if DDD payoffs = (2,2,2). You could argue that the payoffs for DCC and DDC should be different to what I suggested above - maybe defector gets $> 3$ for DCC, as he implicates two others? $\endgroup$ – Marconius Jul 5 '15 at 16:04
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I think you keep the same payoffs, i.e. 1 for everyone if they all co-operate, 2 for everyone if they all defect, 3 if the player defects and isn't defected on and zero otherwise.

Then B defects on both A and C, but both A and C co-operate with B. So we have (0, 3, 0).

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