# Simple hawk-dove type super-game

We have the following super-game: $$\begin{array}{cc} & a_1 & a_2 \\ a_1 & 0,0 & -1,1 \\ a_2 & 1,-1 & -2,-2 \end{array}$$ I want to show that both the trigger (or grim) strategy and the mutual punishment, give the cooperative solution $$(a_1,a_1)=(0,0)$$ for the discount factor $$\delta>\dfrac{1}{2}$$. In the mutual punishment case, if players have observed $$(a_1,a_1)$$ or $$(a_2,a_2)$$ in the preivous round, they choose $$a_1$$. In case they have observed $$(a_1,a_2)$$ or $$(a_2,a_1)$$, they choose $$a_2$$ for one round and then we asumme that they cooperate again.

My thoughts for the second is (player 1 (p1) is the line player and player 2 the column player (p2)): Strategy: For as long as we cooperate we both gain a zero payoff, so the value function is: $$V_1^{C}= 0 + 0 \cdot \delta + 0 \cdot \delta^{2} + ... = 0$$ Suppose that p2, deviates unilaterally from the cooperation strategy in round 1 by playing $$a_2$$, so both players in the second round play the mutual punishment strategy, that is, $$-2$$ looses for everyone. From then onward, we assume that they both play according to the cooperative strategy, namely, the value function is: $$V_2^{NC}=1 - 2 \cdot \delta + 0 \cdot \delta + 0 \cdot \delta^{2} + ... = 1 - 2 \cdot \delta$$ They will cooperate if: $$V_1^{C}>V_2^{NC}\Rightarrow \delta>\dfrac{1}{2}$$ In the case of grim/trigger strategy, the players stop to cooperate in perpetuity, how does the solution differs?

Note: The game is the hawk-dove type super-game.

Finally, I believe it is easy to answer it. Well, in case where some player will not cooperate and the triger strategy is enabled, we have the following soloutions: Soppuse that p2 does not cooperate in the first round so, he will gain a payoof of 1, but in the second round p1 plays $$a_2$$ and he won't cooperate with p2 in perpetual. Thus, we have that p2 will play $$a_1$$, because by playing $$a_2$$ he increases his looses and the value function is: $$V_2^{NC}=1-1\cdot\delta-1\cdot\delta^2-1\cdot\delta^3+...=1-\delta\sum_{j=1}^{+\infty}\delta^{j}=1-\dfrac{\delta}{1-\delta}$$ Hence, they will cooperate if: $$0>1-\dfrac{\delta}{1-\delta}\Rightarrow \delta>\dfrac{1}{2}$$ Differently, if both players choose trigger strategies, so p2 also chooses $$a_2$$, then both will make looses of -2 and the value function is: $$V_2^{NC}=1-2\cdot\delta-2\cdot\delta^2-2\cdot\delta^3+...=1-2\delta\sum_{j=1}^{+\infty}\delta^{j}=1-\dfrac{2\delta}{1-\delta}$$ and hence they will cooperate if: $$0>1-\dfrac{2\delta}{1-\delta}\Rightarrow \delta>\dfrac{1}{3}$$