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Why is the tat-for-tat strategy a Nash equilibrium in infinitely repeated games, but not a Nash equilibrium in a finite scenario? Specifically for this matrix:

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Assume higher payoffs reflect higher utility. It's a prisoner's dilemma situation.

Since tit-for-tat assumes we start at (Honor, Honor), and play the strategy that the other player last played in future rounds, I don't really see why it's a Nash equilibrium in an infinite scenario and not a finite scenario.

In a finite scenario (e.g. one round), wouldn't the players end up at the NE (Cheat, Cheat) because they follow their self-interest? And in an infinite scenario, wouldn't they end up at (Honor, Honor) (not a NE) assuming the discount factor is high enough?

Any clarity appreciated!

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(i) In the 1 round case, tit-for-tat is not a NE. To see this notice that the tit-for-tat strategy, as you describe, dictates that the players play $(H,H)$ in the first (and only) round---as you point out, this is clearly not a NE, since either player can increase her payoff by changing her strategy from $H$ to $C$. Perhaps what you missed is that the strategy includes proscription in first round, which does not cohere from the 1-shot Nash behavior.

(ii) In the $N$-round case, again tit-for-tat is not a NE. This is a little more subtle that point (i) but not much. If both players play according to the strategy, then they will make it to the $N-1$ round playing $H$. At this point, the strategy dictates they play $H$ in the $N^{th}$, and final, round. For the same reason as in (i) this is not a best response, hence not an equilibrium (note that forward looking agents would anticipate this, so the strategy would break down immediately, but analyzing the last round is sufficient to see it is not an equilibrium).

(iii) In the infinitely-repeated case, tit-for-tat can persist, but it depends on how agents discount future utility in comparison with current utility. The general logic is that since there is no final period where things begin to unravel, players are always willing to forgo current utility to maintain good standing (the tit part of the tit-for-tat?) and therefore a higher future (continuation) payoff. Of course, if the players care about today's payoff far more than tomorrow's, they will defect and play $C$. See here for more about NE in infinitely-repeated games.

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  • $\begingroup$ thanks for your helpful reply! I'm still slightly confused on something. In the finite case, players continue with (H, H), which is not a NE as you've pointed out. But in the infinite case, assuming the discount factor is high enough, then firms have no incentive to deviate from cooperation (H, H). Is this correct? If so, is (H, H) a NE in the infinite case because now players can't be better off due to the discount factor? Just struggling to reconcile why (H, H) is not a NE in the finite case whilst it is in the infinite case. $\endgroup$ – user28065 May 21 at 13:43
  • $\begingroup$ (H, H) is not a NE in the infinite case, it is just the strategy profile played in each round under the Tit-for-Tat strategy profile. The NE (for high enough discount factor) is (TfT, TfT). Likewise, (TfT, TfT) is not a NE in the finite case because it prescribes to Honor in the final round, which is not optimal, given that the round is the last one. It's the finiteness of repetition (with both players knowing the number of rounds) that destroys the NE-property of (TfT, TfT). $\endgroup$ – VARulle May 22 at 9:48
  • $\begingroup$ Sorry, forgot to mention that my last comment is a response to @user28065. $\endgroup$ – VARulle May 22 at 13:12
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Edit: This is an incorrect answer. Please disregard and see comments below.


I don't think tit-for-tat is NE for infinitely repeated games. Tit-for-tat simply is, on average, the best strategy against many other strategies.

Tit-for-tat will lose out, for example, if the only other strategy is to deviate all the time.

Read Dawkin's The Selfish Gene for more on this.

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    $\begingroup$ I think OP implicitly assumes, as is common in the literature, that both players adopt the tit-for-tat strategy. In that case, the pair of tit-for-tat strategies can form a NE in an infinitely repeated prisoner's dilemma when the players' discount factor is sufficiently large. $\endgroup$ – Herr K. May 21 at 4:31
  • $\begingroup$ @HerrK. Great point. I was focusing on the nature of the experiment (as in, why is tit-for-tat better than grim reaper) and forgot about the NE notion. $\endgroup$ – Art May 21 at 4:48
  • $\begingroup$ @HerrK thanks for your comment. I left one above in response to 201p - do you have any thoughts on it? $\endgroup$ – user28065 May 21 at 13:47
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Let's first consider the case with the discount factor $\rho = 1$.

In any finitely repeated case, the NE is to play (cheat, cheat) simply because of backward induction. A sophisticated rational player would think that, in the last round, the other player has no chance to punish him by playing cheat in the next round because there's no next round, therefore, he will play cheat. Another sophisticated player of course can also anticipate this and play cheat also in the last round. Given this, in the second-last round, no player can punish the other one by playing cheat in the next round because they are both playing cheat in the next round, and thus, they will both play cheat in the second last round. The same process goes on and on until the first round, and both players will also play cheat. The key thing here is there is a last round to begin with in a finitely repeated case.

In the infinitely repeated case, there is no last round to begin with, so each player can punish the other player by playing cheat in the next round if the other player plays cheat in this round. The presence of punishment makes the "tit-for-tat" strategy become an equilibrium in this case.

However, when the discount factor $\rho$ is sufficiently small, the "tit-for-tat" strategy can not be an equilibrium any more. The discount factor $\rho$ is usually interpreted as how much one discounts future utility. In an infinitely repeated game, $\rho$ can also be interpreted as how likely one player is going to play the same game with the other player in the next round. And straightforwardly, the less likely they are going to interact again, the less credible the threat is of playing cheat in the next round.

PS: I previously thought the TfT strategy was the Grim strategy. Thanks to @VARulle for pointing this out.

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  • $\begingroup$ The explanation is generally correct, but it's not true that with TfT a player can "credibly threaten" to punish the other player. The TfT-NE is not subgame perfect, so the threat implicit in this strategy is actually non-credible. $\endgroup$ – VARulle May 22 at 10:01
  • $\begingroup$ Thanks for the comment. However, based on my understanding, a subgame in an infinitely repeated game is defined by the history of play. With $\rho > 5/13$, TfT shoud be SP. There can only be two possible histories. 1, either player plays NC once, and they both play NC in the remaining game. This is SP for sure. 2, both players always play C, so the TfT strategy is still in effect and thus define an NE. Correct me if I am wrong, thanks. $\endgroup$ – Lin Jing May 22 at 10:53
  • $\begingroup$ What is NC? Did you change notation? $\endgroup$ – VARulle May 22 at 11:18
  • $\begingroup$ Sorry, my bad. NC stands for Non-cooperative, which is Cheat in the OP's example, and C is Honor. $\endgroup$ – Lin Jing May 22 at 11:19
  • $\begingroup$ Consider the subgame that follows after player 1 honored and player 2 cheated. If both players follow TfT after that, they will start an infinite sequence of alternating unilateral cheating, resulting in a total payoff lower than the payoff from mutual honoring. This gives player 1 an incentive to "forgive" the cheating and honor instead of cheat. So TfT doesn't induce a NE in that subgame and is therefore not subgame perfect. $\endgroup$ – VARulle May 22 at 11:34

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