Chapter 11: Repeated Games, Cooperation, and Enforcement. The Encyclopedia of Economic Principals, interactive companion

Every example here is a constructed teaching example: it uses the hypothetical numbers of the chapter's worked examples, plus a few values added for comparison and labelled as such in each panel. Nothing here measures a real market, firm or household.

Demonstration 1 of 4

A pollution charge that flips the dilemma

How large a charge on dirty operation makes filtering each mill's best choice?

Without a charge, dirty operation pays more whatever the rival does, so both pollute and earn 3 instead of 6. A charge larger than the temptation 8 - 6 = 2 reverses both comparisons.

Equation, written in LaTeX: T=8>R=6>P=3>S=1,

$Equation, written in LaTeX: 6>5 \text{and} 1>0.$

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Payoffs in millions: both filter 6 each, both dirty 3 each, a dirty mill facing a filtering one 8 and the filtering mill 1. The charge f is subtracted from a dirty mill's payoff.

Predict first. What is the smallest whole-number charge that makes filtering strictly dominant?

Your prediction 3 2 1

Charge on a dirty mill

Figure: A pollution charge that flips the dilemma. Payoff matrix of two mills with a charge of 0 on dirty operation; equilibria (D, D). — Charge on a dirty mill: 0
Constructed example: the chapter's hypothetical mills (charges 0 and 3); charges 1 and 2 are added for comparison.

Dominance: Dirty strictly dominates
Nash equilibria: (D, D)
Total at (C, C): 12
Total at the equilibrium: 6
Gain from defecting at (C, C): 2

Against a filtering rival: Filter 6, Dirty 8 - 0 = 8; Against a dirty rival: Filter 1, Dirty 3 - 0 = 3. So Dirty strictly dominates. The unique equilibrium is (D, D). Defecting from (C, C) gains 8 - 0 - 6 = 2.

Worked steps

Against a filtering rival: Filter 6, Dirty 8 - 0 = 8
Against a dirty rival: Filter 1, Dirty 3 - 0 = 3
Equilibria: (D, D)

Use the idea

Size a penalty to exceed the gain from defecting in every case, not only on average.

Where the conclusion applies

A one-shot game, a charge that is enforced with certainty and payoffs in owner value only.

Check your understanding: With f = 2, what are the equilibria?

Dirty pays 8 - 2 = 6 against 6 and 3 - 2 = 1 against 1: each mill is indifferent, so all four profiles are Nash equilibria.

Chapter 11 source: section "Prisoner's dilemma".

Demonstration 2 of 4

Who clears the drain?

When any one merchant can clear the drain, how often does nobody do it?

Each merchant must be indifferent between acting and waiting, so as the group grows each one acts less, and the chance that nobody acts rises.

$Equation, written in LaTeX: p^*=1-(\frac{30}{120})^{1/3}=1-0.25^{1/3}\approx0.370.$

$Equation, written in LaTeX: (1-p^*)^4\approx0.630^4\approx0.157,$

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n merchants each avoid damage B = 120 if at least one acts; the one who acts pays the private cost. In the symmetric equilibrium each acts with probability p*.

Predict first. With more merchants, does the drain get cleared more or less reliably?

Your prediction Less reliably More reliably

Merchants

Private cost of acting

Each acts with p*: 0.370
Nobody acts: 0.157
Drain cleared: 84.3%
Expected volunteers: 1.48
Expected payoff each: 90.0

p* = 1 - (30.0 / 120)^(1/3) = 1 - 0.2500^(1/3) = 0.370. Nobody acts with probability (1 - 0.370)^4 = 0.157, so the drain is cleared 84.3% of the time. Expected volunteers 4 x 0.370 = 1.48; each merchant expects 120 - 30.0 = 90.0.

Worked steps

p* = 1 - 0.2500^(1/3) = 0.370
Failure = (1 - p*)^4 = 0.157
Volunteers = 4 x 0.370 = 1.48
Payoff = 120 - 30.0 = 90.0

Use the idea

Assign a duty to one named person, or cut the private cost, rather than relying on a larger group.

Where the conclusion applies

Identical merchants, simultaneous choices and the symmetric mixed equilibrium.

Check your understanding: With 6 merchants and cost 30, what is the failure probability?

p = 1 - 0.25^(1/5) = 0.242; failure 0.758^6 = 0.189, higher than with four.

Chapter 11 source: section "Volunteer's dilemma".

Demonstration 3 of 4

A known last round vs an uncertain end

Why does cooperation unravel with a known last delivery but survive an uncertain end?

With a known last round, nothing can reward cooperation then, and the logic runs backward. A random end keeps a future in every round.

$Equation, written in LaTeX: V_C=\frac{4}{1-0.8}=20.$

$Equation, written in LaTeX: V_D=6+\frac{0.8(1)}{1-0.8}=10.$

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Mutual checking pays 4 each, a lone shirker 6 and the checker 0, mutual shirking 1. With a random end the venture continues after each delivery with probability q.

Predict first. What continuation probability makes checking just incentive compatible?

Your prediction 0.4 0.3 0.8

Horizon

Continuation probability

Figure: A known last round vs an uncertain end. Three deliveries all played as shirk, paying 1 each, against 4 from mutual checking. — Horizon: Three known deliveries, Continuation probability: 0.8
Constructed example: the chapter's hypothetical suppliers (three deliveries, q = 0.8); q of 0.3 and 0.4 is added for comparison. The continuation probability is used only with a random end.

Total if both shirk: 3
Total if both check: 12
Checking sustained: no
Continuation probability: not used (fixed end)

On delivery 3 shirking dominates because 6 > 4 and 1 > 0, and nothing follows. Delivery-2 play cannot change delivery 3, so shirking dominates there too, and so on back to delivery 1. Each earns 1 + 1 + 1 = 3 instead of 4 + 4 + 4 = 12.

Worked steps

Delivery 3: shirk (6 > 4, 1 > 0)
Delivery 2: shirk, delivery 3 is fixed
Delivery 1: shirk
Total 1 + 1 + 1 = 3 against 12

Use the idea

Long relationships without a fixed end date support cooperation better than fixed-term ones.

Where the conclusion applies

Perfect detection, grim reversion to shirking and stable payoffs.

Check your understanding: With q = 0.3, is checking sustained?

V_C = 4 / 0.7 = 5.71; V_D = 6 + 0.3 / 0.7 = 6.43; it fails (the threshold is q = 0.4).

Chapter 11 source: section "Finite-horizon unraveling".

Demonstration 4 of 4

How long must punishment last?

How many months of mutual cutting deter a carrier from cutting capacity?

A deviation gains 4 now and loses 4 a month during the punishment. The punishment must last long enough, given patience, for the discounted losses to exceed the gain.

$Equation, written in LaTeX: \delta\geq\frac{10-6}{10-2}=0.5.$

$Equation, written in LaTeX: V_C=\frac{6}{1-0.75}=24,$

$Equation, written in LaTeX: 10+0.75(2)+0.75^2(2)+0.75^3(\frac{6}{1-0.75})=22.75.$

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Each month both maintaining pays 6 each, a lone cutter 10, mutual cutting 2 (millions). After a cut both cut for the punishment length, then return to maintaining.

Predict first. With delta = 0.9, is a two-month punishment enough?

Your prediction Yes No

Discount factor delta

Punishment length

Figure: How long must punishment last? Monthly payoff streams for complying and cutting once with a permanent (grim) punishment at delta 0.75. — Discount factor delta: 0.75, Punishment length: Permanent (grim)
Constructed example: the chapter's hypothetical carriers (delta 0.75 and 0.40, punishments of 1, 2 months and grim); delta 0.5 is the book's threshold and 0.9 is added for comparison.

V_C: 24.00
V_D: 16.00
Punishment: permanent (grim)
Deterred: yes
Grim-trigger threshold: 0.50

Compliance is worth 6 / (1 - 0.75) = 24.00. With a permanent (grim) punishment, V_D = 10 + 0.75 x 2 / (1 - 0.75) = 10 + 6.00 = 16.00. So deviation is deterred. Under grim trigger cooperation needs delta of at least (10 - 6) / (10 - 2) = 0.50.

Worked steps

V_C = 6 / 0.25 = 24.00
V_D = 10 + 0.75 x 2 / (1 - 0.75) = 10 + 6.00 = 16.00
Margin = 24.00 - 16.00 = 8.00

Use the idea

Match the length of a sanction to how much the parties value the future.

Where the conclusion applies

Perfect monitoring, punishments that are carried out, and a return to cooperation afterwards.

Check your understanding: At delta 0.9 with two months of punishment, how do V_C and V_D compare?

V_C = 60; V_D = 10 + 0.9 x 2 + 0.81 x 2 + 0.729 x 60 = 57.16 < 60, so deviation is deterred.

Chapter 11 source: section "Trigger-strategy enforcement in repeated games".

A pollution charge that flips the dilemma

Choose an example

Calculated values

Worked steps

Use the idea

Where the conclusion applies

Who clears the drain?

Choose an example

Calculated values

Worked steps

Use the idea

Where the conclusion applies

A known last round vs an uncertain end

Choose an example

Calculated values

Worked steps

Use the idea

Where the conclusion applies

How long must punishment last?

Choose an example

Calculated values

Worked steps

Use the idea

Where the conclusion applies