<p>Imagine you have played the first 19 rounds. Now a chance event decides on whether there will be another, final, round. What's your optimal action in this last round, in case it actually occurs? Defection of course. And the same is true for your opponent. Knowing this, what is the optimal choice in round 19? ... As you can see, the game unravels just as it did with a deterministic 20th round. Backward induction works just as well as with a deterministic number of rounds. This is true as long as there is a round which is definitely the last one, so even &quot;stochastic continuation with a hard limit&quot; (as suggested in your comment above) wouldn't change the outcome.</p>

CC BY-SA 4.0
2022-12-01T17:54:32.267Z
2022-12-01T17:54:32.267Z
VARulle
<p>In a repeated prisoner's dilemma with some probability δ of continuing after each round, a Subgame Perfect Nash Equilibrium may be found which induces cooperation instead of defection in each round.  However, in a repeated prisoner's dilemma with a fixed, finite number of rounds, defection occurs in the last round and unravels all the way back to the first round via backward induction.</p>
<p>How minimal can the uncertainty about the future be while still permitting such a cooperative strategy?  For example, if prisoner's dilemma will be played for at least 19 rounds, and then a random event with probability δ will occur to determine whether a 20th round will be played, is that enough to break the backward induction conclusion that defection is inevitable?  How fickle is the backward induction conclusion to the inclusion of randomness at single points of the game?</p>

CC BY-SA 4.0
2022-12-01T07:28:22.913Z
If the 20th round is played, is this definitely the last round? Or may it go on again?
VARulle
CC BY-SA 4.0
2022-12-01T17:00:43.460Z
@VARulle The idea is that if the 20th round is played, it is definitely the last round.  More generally I'm trying to figure out when the game has enough uncertainty to break the inevitability of defection, so another question on the opposite end of the spectrum might be, "What if there is some probability δ of continuing after each round, but also a hard limit of 10 rounds?"
user10478
<game-theory><nash-equilibrium><probability><repeated-games><cooperative-game-theory>
Slight Uncertainty of Continuation in Repeated Prisoner's Dilemma
user10478
# Slight Uncertainty of Continuation in Repeated Prisoner's Dilemma

(Asked by: username_0 on 2022-11-30)

In a repeated prisoner's dilemma with some probability δ of continuing after each round, a Subgame Perfect Nash Equilibrium may be found which induces cooperation instead of defection in each round. However, in a repeated prisoner's dilemma with a fixed, finite number of rounds, defection occurs in the last round and unravels all the way back to the first round via backward induction.

How minimal can the uncertainty about the future be while still permitting such a cooperative strategy? For example, if prisoner's dilemma will be played for at least 19 rounds, and then a random event with probability δ will occur to determine whether a 20th round will be played, is that enough to break the backward induction conclusion that defection is inevitable? How fickle is the backward induction conclusion to the inclusion of randomness at single points of the game?

**Tags:** game-theory, nash-equilibrium, probability, repeated-games, cooperative-game-theory
> **Comments:**
> **username_1**: If the 20th round is played, is this definitely the last round? Or may it go on again?  
> **username_0**: @VARulle The idea is that if the 20th round is played, it is definitely the last round.  More generally I'm trying to figure out when the game has enough uncertainty to break the inevitability of defection, so another question on the opposite end of the spectrum might be, "What if there is some probability δ of continuing after each round, but also a hard limit of 10 rounds?"  

---

## 1 answers

### Answer by username_1 on 2022-12-01. Score: 3 | Accepted answer

Imagine you have played the first 19 rounds. Now a chance event decides on whether there will be another, final, round. What's your optimal action in this last round, in case it actually occurs? Defection of course. And the same is true for your opponent. Knowing this, what is the optimal choice in round 19? ... As you can see, the game unravels just as it did with a deterministic 20th round. Backward induction works just as well as with a deterministic number of rounds. This is true as long as there is a round which is definitely the last one, so even "stochastic continuation with a hard limit" (as suggested in your comment above) wouldn't change the outcome.


---

