# Dealing with an infinitely repeated game

I have been playing around with problems related to game theory, and I ran into this issue related to an infinitely repeated game. Consider this game repeated an infinite number of times: $$\begin{array}{|c|c|c|} \hline &A&B \\ \hline C&(1,1)&(-2,2) \\ \hline D&(2,-2)&(0,0) \\ \hline \end{array}$$ Where Player 1 is the rows, Player 2 is the columns. I am wondering, is there potentially a grim trigger nash equilibrium to this game, and if so, what rate of defection would need to exist in order for the specific Nash Equilibrium to be sustained?

Grim trigger is a strategy lacking forgiveness. If an agent initially defects, and the other agent is grim, the grim agent will subsequently defect forever.

• My understanding is that there is no rational basis for a grim strategy in this game, because once the grim trigger is enacted, there will be no benefit for either agent for the remainder of the infinite iterations.

The biggest problem with iterated Dilemmas is that, if the number of iterations is finite, an agent that wants to "win" from a score perspective will want to defect on the last turn to gain some advantage:

If the game is played exactly N times and both players know this, then it is optimal to defect in all rounds. The only possible Nash equilibrium is to always defect. The proof is inductive: one might as well defect on the last turn, since the opponent will not have a chance to later retaliate. Therefore, both will defect on the last turn. Thus, the player might as well defect on the second-to-last turn, since the opponent will defect on the last no matter what is done, and so on. The same applies if the game length is unknown but has a known upper limit.
Iterated Dilemma (wiki)

If there is no upper bound on iterations, there is no reason to preemptively defect.

I find real world grim strategies to be unlikely, since forever is an awfully long time, (and the 30 years war only lasted 30 years;)

It's unclear that mutual defection in infinite Dilemma is truly a Nash equilibrium because a case may be made that a rational player might gain by changing their strategy and cooperating for an iteration or two, based on the chance of the opponent also changing their strategy.