Two-stage Game and Credible Punishment Strategies

I was working through the example presented in this video, https://youtu.be/EZlzB0ujoP8?t=141, discussing a two-stage game consisting of a prisoner's dilemma in the first stage and a free money game in the second stage.

Stage 1 game:

P2
c     d
-------------
c | 3,3 | 1,4 |
P1      -------------
d | 4,1 | 2,2 |
-------------

Stage 2 game:

P2
r     p
-------------
r | 3,3 | 0,0 |
P1      -------------
p | 0,0 | 0,0 |
-------------

Note that the unique NE in the stage 1 game is (defect, defect) resulting in a payoff of (2,2). The stage 2 game has two NE, (reward, reward) resulting in a payoff of (3,3) and (punish,punish) resulting in a reward of (0,0). The author of the video illustrates that one subgame perfect equilibrium of the two-stage game is playing the NE (d,d) in the first stage then playing (r,r) in the second stage, resulting in a total payoff of (3,3)+(2,2)=(5,5).

The author then states that it is possible to do strictly better by using the punishment NE in the second stage to enforce cooperation in the first stage. In the proposed subgame perfect equilibrium, we only reward (that is, we only play (r,r)) if we cooperate in the first stage. This results in a payoff of (3,3)+(3,3)=(6,6).

My issue is, how is this a credible threat? Wouldn't the players know that even if a player defected in the first stage, then they are better off choosing (r,r) regardless, instead of (p,p)?

Your objection reads almost like an objection to (Pareto) dominated equilibria.

In the one-shot prisoner's dilemma (imagine players play only the stage $1$ game once), players are better off choosing $(C,C)$ than the NE $(D,D)$. Similarly, in the stage $2$ game, players are better off choosing $(R,R)$ than $(P,P)$, but $(P,P)$ is still a NE.

Recall that we've fixed the strategies of both players to be "play $C$ in stage $1$, then play $R$ in stage $2$ if the moves in stage $1$ were $(C,C)$, and play $P$ otherwise".

Playing $P$ in stage $2$ when the stage $1$ outcome was not $(C,C)$ is credible because it is a best response to the strategy described above. In particular, a player is willing to play $P$ because they expect their opponent to also play $P$.

They would be better off if they both chose $R$ instead, but this is a non-cooperative game: players choose their strategies individually taking as given the other player's strategy. (Hence the emergence of dominated equilibria that I mention in the second paragraph above.)

• Thanks for the answer. It seems like cheating to restrict the strategies in order to get a better outcome (I don't really understand what's preventing one from restricting the strategies to the point where the only choice is the best outcome, but I guess that's outside the scope of this question).I suppose credibility of the punishment strategy arises because (P,P) is a NE? – jonem Apr 2 '18 at 17:15
• @jonem I'm not sure what you mean when you say "restrict[ing] the strategies" is cheating. This is standard in game theory -- we fix the strategies of the opponent(s), and calculate what the best response is. But regarding credibility: that is exactly right. If $(P,P)$ were not a NE (say payoffs from it were $(-1,-1)$ instead of $(0,0)$), then the punishment strategy would not be credible. – Theoretical Economist Apr 2 '18 at 17:21
• I guess what I mean is, if we're in the setting where we are allowed to fix strategies of both players to some arbitrary rule (the rule we used here is punish unless cooperation in the first stage), why don't we just avoid the roundabout analysis and just say, by looking at the payoffs, players must play (C,C) in stage 1 and (R,R) in stage 2, as in those are their only choices in each respective stage game. Perhaps I'm missing something more subtle. – jonem Apr 2 '18 at 17:31
• @jonem I think I was a little unclear -- we fix the strategies of the players individually. Suppose that the column player's strategy was to play $C$ in stage $1$ and then $R$ in stage $2$. The row player's best response to this strategy would then be to play $D$ in stage $1$ and then $R$ in stage $2$, so your proposed strategy profile would not be an equilibrium. – Theoretical Economist Apr 3 '18 at 0:44