Collusion, deviation from equilibrium Two firms can produce either low (L), medium (M) or high (H) quantity. The payoff matrix is given by:




What is the outcome of this game if firms play only once?
Suppose the game is played for infinite periods and $\delta = 1/2$. Is it possible for firms to collude and choose $(L, L)$? Is it possible for firms to collude and choose $(M, M)$?


From the picture we can see that if the firms play once then the outcome is $(H,H)$. However I don't really know how to solve the second part and would really appreciate some help. What I do know in repeated infinite games that may be useful for this problem is the following.

Lets take the case of deviating from $(H,H)$ to $(M,M)$ for firm 1. 

Equilibrium payoff = $3+3\delta+3\delta^2...=\frac{3}{1-\delta}$

Deviation payoff (I think that if firm 1 deviated and played $M$ then firm 2 would play L since they have higher payoff $(10)$ and so we would have)= $5+2\delta+2\delta^2... = 5+\frac{2\delta}{(1-\delta) }  $

Where $\delta$ is the interest rate after the first round. I feel that this portion is irrelevant, however, since we are talking about collusion and should consider both firms, instead of just firm 1's strategy.
Converting my comments into an answer... 

In the infinitely repeated game, if the players want to collude on $(L,L)$, then there needs to be an incentive for each to stick to the plan, i.e. a punishment if one of them deviates from playing $L$.

As you have figured out, this amounts comparing the discounted payoffs from playing $(L,L)$ with a one-shot deviation of playing $H$ against $L$, and then playing $(H,H)$ forever.