Why a Known Ending Changes the Incentives
Repeated interaction can make cooperation seem attractive: a player who values future rewards may avoid a profitable one-time deviation to preserve good relations. But a known final round creates a problem. Once that round arrives, there is no later interaction to protect. If defection is the best response in the one-round game, cooperation cannot be sustained then. Backward induction carries that conclusion into the preceding rounds.
The result is called the unraveling of cooperation. It is not a claim that people never cooperate in finite relationships. It is a conclusion from a particular model: a known finite horizon, a repeated Prisoner’s Dilemma with unchanged payoffs, and players who choose strategies to maximize those payoffs. The assumptions matter as much as the conclusion.
Write the two actions as \(C\) (cooperate) and \(D\) (defect). A standard symmetric Prisoner’s Dilemma has the following payoffs for a player: \(T\) when that player defects against cooperation, \(R\) when both cooperate, \(P\) when both defect, and \(S\) when that player cooperates against defection. The usual strict inequalities are \(T>R>P>S\), together with \(2R>T+S\). The first set makes defection strictly better for each player in the one-round game; the second says that mutual cooperation produces a higher combined payoff than alternating who defects.
Because \(T>R\), a player facing \(C\) prefers \(D\) to \(C\). Because \(P>S\), a player facing \(D\) also prefers \(D\). Thus \(D\) strictly dominates \(C\) in the one-round game, and \((D,D)\) is its unique Nash equilibrium. Here, a Nash equilibrium is an action profile in which each player’s action is a best response to the other player’s action.
The Backward-Induction Argument
A finite repeated game can be represented as a sequence of decision points. At each point, players know the round and the observed history. Backward induction starts at the last decision point and works back, using the best actions in later rounds to evaluate earlier choices. In this setting, every history matters: the final-round conclusion must hold whether earlier play was cooperative, uncooperative, or mixed.
In round \(T\), there are no future payoffs to influence. Since \(D\) strictly dominates \(C\) in the stage game, both players choose \(D\) after every possible history.
In round \(T-1\), a player cannot induce cooperation in round \(T\) by choosing \(C\), nor prevent defection by choosing \(D\). The final-round equilibrium is \((D,D)\) after every history.
With the next round’s equilibrium actions fixed, the current round again has the same stage-game incentives. Continue until the first round is reached.
The logic can be stated as an induction. Suppose that after every possible history, the equilibrium continuation from round \(t+1\) through round \(T\) is the same regardless of the actions chosen in round \(t\). Then, when comparing \(C\) and \(D\) in round \(t\), a player receives the same equilibrium continuation payoff after either action. The comparison therefore reduces to the current round’s payoffs. Since \(D\) strictly dominates \(C\), both players choose \(D\). The final round supplies the base case: there is no continuation after it, and \((D,D)\) is the unique equilibrium. Applying the induction backward proves that both players choose \(D\) in every round.
This reasoning identifies the subgame-perfect equilibrium: in every subgame, including one that begins after an off-path history, players’ continuation strategies must be optimal. In the standard finite repeated Prisoner’s Dilemma, the strategy “choose \(D\) in every round, after every history” is subgame-perfect. A proposed plan to cooperate early and punish a deviation later is not credible if carrying out that punishment would be against a player’s own interest when the later round arrives.
Worked Example: Three Rounds with a Known Final Round
Two competing firms choose whether to maintain a costly shared service standard (\(C\)) or cut that standard (\(D\)). In each round, their payoffs, in points, are \((C,C)=(3,3)\), \((D,C)=(5,0)\), \((C,D)=(0,5)\), and \((D,D)=(1,1)\). They play three rounds. The final round is common knowledge, payoffs are additive and undiscounted, and each firm maximizes its total points. Find the subgame-perfect outcome and each firm’s total payoff.
Method. Use backward induction, checking the incentives at every possible history. The stage-game values are \(T=5\), \(R=3\), \(P=1\), and \(S=0\). They satisfy \(5>3>1>0\), and \(2R=6>T+S=5\). Also, against \(C\), defecting gives \(5\) rather than \(3\); against \(D\), defecting gives \(1\) rather than \(0\). Defection is strictly dominant.
Round 3. Whatever happened in the first two rounds, each firm chooses \(D\). If its rival chooses \(C\), it earns \(5\) rather than \(3\); if its rival chooses \(D\), it earns \(1\) rather than \(0\). The final-round equilibrium is \((D,D)\).
Round 2. A firm’s round-3 payoff is \(1\) whether it chooses \(C\) or \(D\) in round 2, because round 3 ends in \((D,D)\) after either choice and every other history. If both firms choose \(C\) in round 2, each receives \(3+1=4\) over rounds 2 and 3. A unilateral switch to \(D\) against \(C\) gives \(5+1=6\). If both choose \(D\), each receives \(1+1=2\). Thus the round-2 equilibrium is \((D,D)\).
Round 1. The equilibrium continuation in rounds 2 and 3 is \((D,D)\) after every round-1 history, so each firm’s continuation total is \(1+1=2\). Mutual cooperation in round 1 would give \(3+2=5\) points to each firm. A unilateral deviation against \(C\) gives \(5+2=7\), so cooperation is not optimal. The round-1 equilibrium is also \((D,D)\).
The equilibrium path is therefore \((D,D)\) in all three rounds. Each firm receives \(1+1+1=3\) points. The calculation uses round payoffs as points summed across the three rounds; it does not discount a first-round payoff as though it were received at a different date.
Check and common mistake. The result is consistent with strict dominance in each round and with backward induction. A tempting but incorrect argument is that firms will cooperate in rounds 1 and 2 because they can threaten to punish a defector in round 3. That threat cannot change the final-round outcome: in round 3, defection is optimal regardless of the history. Treating a non-credible punishment as an extra cost of deviation gives the wrong result.
Why Cooperation Unravels One Round at a Time
The argument is not that future outcomes never matter in repeated games. Rather, with these assumptions, future equilibrium outcomes cannot be changed by a player’s current action in a way that sustains cooperation. In the last round, there is no future to influence. Once that round’s behavior is determined, the same logic applies to the round before it: the continuation is fixed at defection regardless of the current action. The reasoning repeats all the way to the beginning.
A useful way to test a proposed cooperative strategy is to ask two separate questions. First, is cooperation optimal in the final round after every history? If not, a player may defect when that round arrives. Second, if the proposed strategy prescribes punishment, would the player actually prefer to carry it out in the subgame where punishment is due? A threat that is not optimal when the time comes cannot support an equilibrium strategy.
Worked Example: Four Rounds and a Non-Credible Punishment
Two hypothetical retailers repeatedly decide whether to honor a costly joint return-service standard (\(C\)) or avoid that cost by weakening it (\(D\)). The stage payoffs are \((C,C)=(4,4)\), \((D,C)=(7,1)\), \((C,D)=(1,7)\), and \((D,D)=(2,2)\). There are four known rounds, and total payoffs are the undiscounted sum of round payoffs. Evaluate the proposal “cooperate for three rounds, then punish any earlier deviation by defecting in round 4,” and find the equilibrium outcome.
Step 1: Check the one-round incentives. Here \(T=7\), \(R=4\), \(P=2\), and \(S=1\), so \(7>4>2>1\). Also, \(2R=8>T+S=8\) is false as a strict inequality: the two totals are equal. That condition is often included in the standard definition to say mutual cooperation strictly raises the combined payoff relative to alternating exploitation. Its failure does not affect the unraveling proof here, because the required incentive condition, strict dominance of \(D\), does hold: \(7>4\) against \(C\), and \(2>1\) against \(D\).
Step 2: Solve round 4. After any history, a retailer facing \(C\) gets \(7\) from \(D\) instead of \(4\) from \(C\). Facing \(D\), it gets \(2\) from \(D\) instead of \(1\) from \(C\). Thus round 4 ends in \((D,D)\), including when no earlier deviation occurred. A promised round-4 punishment is not a separate choice that can force a different outcome; each retailer still chooses its best response in that round.
Step 3: Work backward. Since round 4’s equilibrium is \((D,D)\) after every history, it cannot be altered by the actions in round 3. In round 3, \(D\) strictly dominates \(C\), so both choose \(D\). The same argument applies in round 2 and round 1. Hence the equilibrium path is \((D,D)\) in all four rounds, giving each retailer \(2+2+2+2=8\) points.
Step 4: Check the proposed cooperative path. If both retailers cooperated in round 1, the continuation prescribed by the proposed plan would give \(4\) points in round 1 and \(4\) in round 2, but round 3 already has a one-round incentive to defect, and round 4’s defection is unavoidable in equilibrium. More directly, even at the final round of the proposed plan, cooperation against a cooperating rival yields \(4\), while defecting yields \(7\). The proposal therefore fails the final-round best-response test, so it cannot be subgame-perfect.
The method avoids assuming that a stated punishment will be carried out merely because it appears in a plan. A strategy must specify optimal behavior at every history, not only on the path where everyone cooperates.
Assumptions Behind the Result
The unraveling conclusion is strong, but it is not assumption-free. In an exam analysis, state the model before applying backward induction. The standard result relies on the following conditions:
- A finite, commonly known horizon. Both players know the exact final round. Without a known endpoint, the final-round argument does not apply in this form.
- A Prisoner’s Dilemma stage game. Defection is strictly better against either action, so the one-round game has a unique equilibrium at \((D,D)\). If the stage game instead has multiple equilibria or different incentives, the backward steps may produce a different outcome.
- Unchanged incentives across rounds. The same actions yield the same payoffs in every round, and current actions do not change the payoff structure of a later round.
- Payoff-maximizing behavior with common knowledge of rationality. Players understand the game and reason that the other player will also choose optimal strategies at later decision points.
- No binding agreement or commitment device that changes incentives. A promise to cooperate or punish is not enough. If a contract, enforceable transfer, or other institution changes the payoffs or restricts actions, the stage game being analyzed is no longer the same one.
The result also depends on what payoffs represent. If the stated payoffs leave out a genuine concern for fairness, a relationship, or a future consequence, then they do not fully describe players’ preferences. Such considerations may change behavior, but they must be incorporated into the model rather than silently added to the backward-induction calculation.
Worked Example: A Changed Final-Round Incentive
Consider a two-round game with the standard stage payoffs \((C,C)=(3,3)\), \((D,C)=(5,0)\), \((C,D)=(0,5)\), and \((D,D)=(1,1)\). In round 2 only, a binding program pays each player a bonus of \(6\) points for choosing \(C\), regardless of the rival’s action. The bonus is part of the payoff and is guaranteed. Find the equilibrium actions in each round.
Round 2 with the bonus. If the rival chooses \(C\), choosing \(C\) gives \(3+6=9\), while choosing \(D\) gives \(5\). If the rival chooses \(D\), choosing \(C\) gives \(0+6=6\), while choosing \(D\) gives \(1\). Thus \(C\) strictly dominates \(D\) in round 2, and its unique equilibrium is \((C,C)\).
Round 1. The guaranteed round-2 bonus and its equilibrium payoff do not depend on the actions in round 1. Each player therefore compares only round-1 stage payoffs when choosing a round-1 action. In that round, \(D\) strictly dominates \(C\), so round 1’s equilibrium is \((D,D)\).
Each player’s total equilibrium payoff is \(1+9=10\): one point from round 1 and nine points from round 2. Cooperation does not unravel in the final round here because the final-round incentives were changed. This example does not contradict the standard result; it violates the assumption that the unchanged Prisoner’s Dilemma stage game is played in every round.
Common Exam Trap
The most common error is to reason forward from a desirable cooperative path instead of backward from the final decision. For example, a candidate may say, “If a player defects in round 1, the other player will defect in round 3, so the deviation is not worthwhile.” That conclusion is valid only if round-3 defection is optimal in the relevant subgame. In the standard Prisoner’s Dilemma it is optimal, but it happens after every history, including a history with no deviation. It therefore cannot distinguish a cooperative round-1 choice from a defection in round 1.
A related mistake is to stop after finding the equilibrium action profile on the path. Subgame perfection requires optimal actions after off-path histories too. When the final-round incentives are unchanged, checking all histories confirms the same result: \((D,D)\) is the final-round equilibrium regardless of what happened before.
Check Your Understanding
For each problem, apply the one-round incentives first, then work backward from the last round.
- A two-round game has payoffs \((C,C)=(4,4)\), \((D,C)=(6,1)\), \((C,D)=(1,6)\), and \((D,D)=(2,2)\). The horizon is known, payoffs are unchanged, and totals are undiscounted. What is the subgame-perfect action in each round and each player’s total payoff? (Answer: \((D,D)\) in both rounds; 4 points per player.)
- A three-round game has \(T=8\), \(R=5\), \(P=2\), and \(S=0\), with the standard symmetric payoff arrangement. Is this a Prisoner’s Dilemma, and what is each player’s equilibrium total? (Answer: Yes; \(8>5>2>0\) and \(2R=10>T+S=8\); both choose \(D\) each round and each receives 6 points.)
- In a four-round game, the stage payoffs are \((C,C)=(3,3)\), \((D,C)=(6,0)\), \((C,D)=(0,6)\), and \((D,D)=(1,1)\). The players propose to cooperate for the first three rounds and defect only in the last. Find the subgame-perfect path and each player’s total. (Answer: \((D,D)\) in all four rounds; 4 points per player.)
- In a two-round Prisoner’s Dilemma, the first-round payoffs are \((C,C)=(5,5)\) and \((D,C)=(8,1)\). Whatever happens in round 1, round 2 has a unique equilibrium payoff of \(2\) to each player. If both are expected to cooperate in round 1, compare one player’s payoff from cooperating with the payoff from deviating to \(D\), assuming equilibrium continuation. (Answer: Cooperating yields \(5+2=7\); deviating yields \(8+2=10\), a gain of 3.)
- A two-round game has the usual Prisoner’s Dilemma payoffs in round 1. In round 2, a guaranteed bonus of 4 is paid to each player who chooses \(C\), regardless of the rival’s action. What is the equilibrium action profile in each round? (Answer: \((D,D)\) in round 1; in round 2, \(C\) strictly dominates \(D\), so \((C,C)\).)