Unit III: The Evolution of Cooperation Can Selfishness Save the Environment? Repeated Games: the Folk Theorem Evolutionary Games A Tournament How to Promote.

Unit III: The Evolution of Cooperation

•Can Selfishness Save the Environment?•Repeated Games: the Folk Theorem•Evolutionary Games•A Tournament•How to Promote Cooperation/Unit Review

4/14

7/284/9

Repeated Games

Some Questions:

• What happens when a game is repeated? • Can threats and promises about the future

influence behavior in the present?• Cheap talk• Finitely repeated games: Backward induction• Indefinitely repeated games: Trigger strategies

Can threats and promises about future actions influence behavior in the present?

Consider the following game, played 2X:

C 3,3 0,5

D 5,0 1,1

Repeated Games

C D

See Gibbons: 82-104.

Repeated Games

Draw the extensive form game:

(3,3) (0,5) (5,0) (1,1)

(6,6) (3,8) (8,3) (4,4) (3,8)(0,10)(5,5)(1,6)(8,3) (5,5)(10,0) (6,1) (4,4) (1,6) (6,1) (2,2)

Repeated Games

Now, consider three repeated game strategies:

D (ALWAYS DEFECT): Defect on every move.

C (ALWAYS COOPERATE): Cooperate on every move.

T (TRIGGER): Cooperate on the first move, then cooperate after the other cooperates. If the other defects, then defect forever.

Repeated Games

If the game is played twice, the V(alue) to a player using ALWAYS DEFECT (D) against an opponent using ALWAYS DEFECT(D) is:

V (D/D) = 1 + 1 = 2 V (C/C) = 3 + 3 = 6V (T/T) = 3 + 3 = 6V (D/C) = 5 + 5 = 10V (D/T) = 5 + 1 = 6V (C/D) = 0 + 0 = 0V (C/T) = 3 + 3 = 6

V (T/D) = 0 + 1 = 1V (T/C) = 3 + 3 = 6

Repeated Games

And 3x:

V (D/D) = 1 + 1 + 1 = 3 V (C/C) = 3 + 3 + 3 = 9V (T/T) = 3 + 3 + 3 = 9V (D/C) = 5 + 5 + 5 = 15V (D/T) = 5 + 1 + 1 = 7V (C/D) = 0 + 0 + 0 = 0V (C/T) = 3 + 3 + 3 = 9

V (T/D) = 0 + 1 + 1 = 2V (T/C) = 3 + 3 + 3 = 9

Repeated Games

Time average payoffs: n=3

V (D/D) = 1 + 1 + 1 = 3 /3 = 1V (C/C) = 3 + 3 + 3 = 9 /3 = 3V (T/T) = 3 + 3 + 3 = 9 /3 = 3V (D/C) = 5 + 5 + 5 = 15 /3 = 5V (D/T) = 5 + 1 + 1 = 7 /3 = 7/3V (C/D) = 0 + 0 + 0 = 0 /3 = 0V (C/T) = 3 + 3 + 3 = 9 /3 = 3

V (T/D) = 0 + 1 + 1 = 2 /3 = 2/3

V (T/C) = 3 + 3 + 3 = 9 /3 = 3

Repeated Games

Time average payoffs: n

V (D/D) = 1 + 1 + 1 + ... /n = 1V (C/C) = 3 + 3 + 3 + ... /n = 3V (T/T) = 3 + 3 + 3 + ... /n = 3V (D/C) = 5 + 5 + 5 + ... /n = 5V (D/T) = 5 + 1 + 1 + ... /n = 1 + V (C/D) = 0 + 0 + 0 + ... /n = 0V (C/T) = 3 + 3 + 3 + … /n = 3

V (T/D) = 0 + 1 + 1 + ... /n = 1 -

V (T/C) = 3 + 3 + 3 + ... /n = 3

Repeated Games Now draw the matrix form of this game:

1x

T 3,3 0,5 3,3

C 3,3 0,5 3,3

D 5,0 1,1 5,0

C D T

Repeated Games

T 3,3 1-1+ 3,3

C 3,3 0,5 3,3

D 5,0 1,1 1+,1-

C D T

If the game is repeated, ALWAYS DEFECTis no longer dominant.

Time Average

Payoffs

Repeated Games

T 3,3 1-1+ 3,3

C 3,3 0,5 3,3

D 5,0 1,1 1+,1-

C D T

… and TRIGGERachieves “a NE with itself.”

Repeated Games

Time Average

Payoffs

T(emptation) >R(eward)>P(unishment)>S(ucker)

T R,R P-P+ R,R

C R,R S,T R,R

D T,S P,P P+,P-

C D T

Discounting

The discount parameter, , is the weight of the next payoff relative to the current payoff.

In a indefinitely repeated game, can also be interpreted as the likelihood of the game continuing for another round (so that the expected number of moves per game is 1/(1-)).

The V(alue) to someone using ALWAYS DEFECT (D) when playing with someone using TRIGGER (T) is the sum of T for the first move, P for the second, 2P for the third, and so on (Axelrod: 13-4):

V (D/T) = T + P + 2P + …

“The Shadow of the Future”

Discounting

Writing this as V (D/T) = T + P + 2P +..., we have the following:

V (D/D) = P + P + 2P + … = P/(1-)

V (C/C) = R + R + 2R + … = R/(1-)

V (T/T) = R + R + 2R + … = R/(1-)

V (D/C) = T + T + 2T + … = T/(1-)

V (D/T) = T + P + 2P + … = T+ P/(1-)

V (C/D) = S + S + 2S + … = S/(1-)

V (C/T) = R + R + 2R + … = R/(1- )

V (T/D) = S + P + 2P + … = S+ P/(1-)

V (T/C) = R + R + 2R + … = R/(1- )

T

C

D

DiscountedPayoffs

T > R > P > S 0 > > 1

R/(1-) S/(1-) R/(1-)

R/(1-) T/(1-) R/(1-)T/(1-) P/(1-) T + P/(1-)

S/(1-) P/(1-) S + P/(1-)

Discounting

C D T

R/(1-) S + P/(1-) R/(1- )

R/(1-) T + P/(1-) R/(1-)

T

C

D

DiscountedPayoffs

T > R > P > S 0 > > 1

T weakly dominates C

R/(1-) S/(1-) R/(1-)

R/(1-) T/(1-) R/(1-)T/(1-) P/(1-) T + P/(1-)

S/(1-) P/(1-) S + P/(1-)

Discounting

C D T

R/(1-) S + P/(1-) R/(1- )

R/(1-) T + P/(1-) R/(1-)

Discounting

Now consider what happens to these values as varies (from 0-1):

V (D/D) = P + P + 2P + … = P/(1-)

V (C/C) = R + R + 2R + … = R/(1-)

V (T/T) = R + R + 2R + … = R/(1-)

V (D/C) = T + T + 2T + … = T/(1-)

V (D/T) = T + P + 2P + … = T+ P/(1-)

V (C/D) = S + S + 2S + … = S/(1-)

V (C/T) = R + R + 2R + … = R/(1- )

V (T/D) = S + P + 2P + … = S+ P/(1-)

V (T/C) = R + R + 2R + … = R/(1- )

Discounting


V (D/D) = P + P + 2P + … = P/(1-)

V (C/C) = R + R + 2R + … = R/(1-)

V (T/T) = R + R + 2R + … = R/(1-)

V (D/C) = T + T + 2T + … = T/(1-)

V (D/T) = T + P + 2P + … = T+ P/(1-)

V (C/D) = S + S + 2S + … = S/(1-)

V (C/T) = R + R + 2R + … = R/(1- )

V (T/D) = S + P + 2P + … = S+ P/(1-)

V (T/C) = R + R + 2R + … = R/(1- )

Discounting


V (D/D) = P + P + 2P + … = P+ P/(1-) V (C/C) = R + R + 2R + … = R/(1-)

V (T/T) = R + R + 2R + … = R/(1-)

V (D/C) = T + T + 2T + … = T/(1-)

V (D/T) = T + P + 2P + … = T+ P/(1-)

V (C/D) = S + S + 2S + … = S/(1-)

V (C/T) = R + R + 2R + … = R/(1- )

V (T/D) = S + P + 2P + … = S+ P/(1-) V (T/C) = R + R + 2R + … = R/(1- )

V(D/D) > V(T/D) D is a best response to D

Discounting


V (D/D) = P + P + 2P + … = P+ P/(1-)

V (C/C) = R + R + 2R + … = R/(1-)

V (T/T) = R + R + 2R + … = R/(1-)

V (D/C) = T + T + 2T + … = T/(1-)

V (D/T) = T + P + 2P + … = T+ P/(1-)

V (C/D) = S + S + 2S + … = S/(1-)

V (C/T) = R + R + 2R + … = R/(1- )

V (T/D) = S + P + 2P + … = S+ P/(1-)

V (T/C) = R + R + 2R + … = R/(1- )

2

1

3

?

Discounting


For all values of : V(D/T) > V(D/D) > V(T/D) V(T/T) > V(D/D) > V(T/D)

Is there a value of s.t., V(D/T) = V(T/T)? Call this *.

If < *, the following ordering hold:

V(D/T) > V(T/T) > V(D/D) > V(T/D)

D is dominant: GAME SOLVED

V(D/T) = V(T/T)T+P(1-) = R/(1-) T-t+P = R T-R = (T-P)

* = (T-R)/(T-P)

?

Discounting


For all values of : V(D/T) > V(D/D) > V(T/D) V(T/T) > V(D/D) > V(T/D)

Is there a value of s.t., V(D/T) = V(T/T)? Call this *.

* = (T-R)/(T-P)

If > *, the following ordering hold:

V(T/T) > V(D/T) > V(D/D) > V(T/D)

D is a best response to D; T is a best response to T; multiple NE.

Discounting

V(T/T) = R/(1-)

* 1

V

TR

Graphically:

The V(alue) to a player using ALWAYSDEFECT (D) against TRIGGER (T), and the V(T/T) as a functionof the discount

parameter ()

V(D/T) = T + P/(1-)

The Folk Theorem

(R,R)

(T,S)

(S,T)

(P,P)

The payoff set of the repeated PD is the convex closure of the points [(T,S); (R,R); (S,T); (P,P)].

The Folk Theorem

(R,R)

(T,S)

(S,T)

(P,P)

The shaded area is the set of payoffs that Pareto-dominate the one-shot NE (P,P).

The Folk Theorem

(R,R)

(T,S)

(S,T)

(P,P)

Theorem: Any payoff that pareto-dominates the one-shot NE can be supported in a SPNE of the repeated game, if the discount parameter is sufficiently high.

The Folk Theorem

(R,R)

(T,S)

(S,T)

(P,P)

In other words, in the repeatedgame, if the future matters “enough”i.e., ( > *),there are zillions of equilibria!

• The theorem tells us that in general, repeated games give rise to a very large set of Nash equilibria. In the repeated PD, these are pareto-rankable, i.e., some are efficient and some are not.

• In this context, evolution can be seen as a process that selects for repeated game strategies with efficient payoffs.

“Survival of the Fittest”

The Folk Theorem

Evolutionary Games

Fifteen months after I had begun my systematic enquiry, I happened to read for amusement ‘Malthus on Population’ . . . It at once struck me that . . . favorable variations would tend to be preserved, and unfavorable ones to be destroyed. Here then I had at last got a theory by which to work.

Charles Darwin

Evolutionary Games

• Evolutionary Stability (ESS)• Hawk-Dove: an example• The Replicator Dynamic• The Trouble with TIT FOR TAT• Designing Repeated Game Strategies• Finite Automata

Evolutionary Games

Biological Evolution: Under the pressure of natural selection, any population (capable of reproduction and variation) will evolve so as to become better adapted to its environment, i.e., will develop in the direction of increasing “fitness.”

Economic Evolution: Firms that adopt efficient “routines” will survive, expand, and multiply; whereas others will be “weeded out” (Nelson and Winters, 1982).

Evolutionary Stability

Evolutionary Stable Strategy (ESS): A strategy is evolutionarily stable if it cannot be invaded by a mutant strategy.

(Maynard Smith & Price, 1973)

A strategy, A, is ESS, if

i) V(A/A) > V(B/A), for all B

ii) either V(A/A) > V(B/A) or V(A/B) > V(B/B), for all B

Hawk-Dove: an example

Imagine a population of Hawks and Doves competing over a scarce resource (say food in a given area). The share of each type in the population changes according to the payoff matrix, so that payoffs determine the number of offspring left to the next generation.

v = value of the resourcec = cost of fighting

H/D: Hawk gets resource; Dove flees (v, 0)D/D: Share resource (v/2, v/2)H/H: Share resource less cost of fighting ((v-c)/2, (v-c)/2)

(See Hargreave-Heap and Varoufakis: 195-214; Casti: 71-75.)


H D

H (v-c)/2,(v-c)/2 v,0

D 0,v v/2,v/2

v = value of resourcec = cost of fighting


H D

H -1,-1 4,0

D 0,4 2, 2

v = value of resource = 4c = cost of fighting = 6


H D

H -1,-1 4,0

D 0,4 2, 2

NE = {(1,0);(0,1);(2/3,2/3)}

unstable stable

The mixed NE corresponds to a population that is 2/3 Hawks and 1/3 Doves


H D

H -1,-1 4,0

D 0,4 2, 2

NE = {(1,0);(0,1);(2/3,2/3)}

unstable stable Is any strategy ESS?

H D

H

D

-1,-1 4,0

0,4 2,2



ii) either V(A/A) > V(B/A)

or V(A/B) > V(B/B), for all B EP2(O) = 3p

EP2(F) = 5-5p

p* = 5/8


NE = {(1,0);(0,1);(2/3,2/3)}

H D

H

D

-1,-1 4,0

0,4 2,2



In other words, to be ESS, a strategy must be a NE with itself.

Neither H nor D is ESS.

(For these payoffs.) EP2(O) = 3p

EP2(F) = 5-5p

p* = 5/8


NE = {(1,0);(0,1);(2/3,2/3)}

H D

H

D

-1,-1 4,0

0,4 2,2



ii) either V(A/A) > V(B/A) or V(A/B) > V(B/B), for all B

What about the mixed NE strategy? = 3p

EP2(F) = 5-5p

p* = 5/8


NE = {(1,0);(0,1);(2/3,2/3)}

H D

H

D

-1,-1 4,0

0,4 2,2

V(H/H) = -1

V(H/D) = 4

V(D/H) = 0

V(D/D) = 2

V(H/M) = 2/3V(H/H)+1/3V(H/D) = 2/3

V(M/H) = 2/3V(H/H)+1/3V(D/H) = -2/3

V(D/M) = 2/3V(D/H)+1/3V(D/D) = 2/3

V(M/D) = 2/3V(H/D)+1/3V(D/D) = 10/3

V(M/M) = 2/3V(D/H)+1/3V(D/D) = 2/3


NE = {(1,0);(0,1);(2/3,2/3)}

Where M is the mixed strategy 2/3 Hawk, 1/3 Dove

H D

H

D

-1,-1 4,0

0,4 2,2

V(H/H) = -1

V(H/D) = 4

V(D/H) = 0

V(D/D) = 2

V(H/M) = 2/3V(H/H)+1/3V(H/D) = 2/3

V(M/H) = 2/3 ( -1 ) +1/3 ( 4 ) = 2/3

V(D/M) = 2/3V(D/H)+1/3V(D/D) = 2/3

V(M/D) = 2/3V(H/D)+1/3V(D/D) = 10/3

V(M/M) = 2/3V(D/H)+1/3V(D/D) = 2/3


NE = {(1,0);(0,1);(2/3,2/3)}

H D

H

D

-1,-1 4,0

0,4 2,2

V(H/H) = -1V(H/D) = 4V(D/H) = 0V(D/D) = 2V(H/M) = 2/3V(H/H)+1/3V(H/D) = 2/3V(M/H) = 2/3V(H/H)+1/3V(D/H) = -2/3V(D/M) = 2/3V(D/H)+1/3V(D/D) = 2/3V(M/D) = 2/3V(H/D)+1/3V(D/D) = 10/3V(M/M) = 4/9V(H/H)+2/9V(H/D)

= 2/9V(D/H)+1/9V(D/D) = 2/3


NE = {(1,0);(0,1);(2/3,2/3)}

H D

H

D

-1,-1 4,0

0,4 2,2

To be an ESS

i) V(M/M) > V(B/M), for all B

ii) either V(M/M) > V(B/M)

or V(M/B) > V(B/B), for all B

(O) = 3p

EP2(F) = 5-5p

p* = 5/8


NE = {(1,0);(0,1);(2/3,2/3)}

H D

H

D

-1,-1 4,0

0,4 2,2

To be an ESS

i) V(M/M) = V(H/M) = V(D/M) = 2/3

ii) either V(M/M) > V(B/M)

or V(M/B) > V(B/B), for all B

(O) = 3p

EP2(F) = 5-5p

p* = 5/8


NE = {(1,0);(0,1);(2/3,2/3)}

V(M/D) > V(D/D)

10/3 > 2

V(M/H) > V(H/H)

-2/3 > -1

Evolutionary Stability in IRPD?



Is D an ESS?

i) V(D/D) > V(STFT/D) ?

ii) V(D/D) > V(STFT/D) or V(D/STFT) > V(STFT/STFT) ?

Consider a mutant strategy called e.g., SUSPICIOUS TIT FOR TAT (STFT). STFT defects on the first round, then plays like TFT




Is D an ESS?

i) V(D/D) = V(STFT/D)

ii) V(D/D) = V(STFT/D) or V(D/STFT) = V(STFT/STFT)

Consider a mutant strategy called e.g., SUSPICIOUS TIT FOR TAT (STFT). STFT defects on the first round, then plays like TFT

D and STFT are “neutral mutants”


Axelrod & Hamilton (1981) demonstrated that D is not an ESS, opening the way to subsequent tournament studies of the game.

This is a sort-of Folk Theorem for evolutionary games: In the one-shot Prisoner’s Dilemma, DEFECT is strictly dominant. But in the repeated game, ALWAYS DEFECT (D) can be invaded by a mutant strategy, e.g., SUSPICIOUS TIT FOR TAT (STFT).

• Many cooperative strategies do better than D, thus they can gain a foothold and grow as a share of the population.

• Depending on the initial population, the equilibrium reached can exhibit any amount of cooperation.

Is STFT an ESS?


It can be shown that there is no ESS in IRPD (Boyd & Lorberbaum, 1987; Lorberbaum, 1994).

There can be stable polymorphisms among neutral mutants, whose realized behaviors are indistinguishable from one another. (This is the case, for example, of a population of C and TFT).

NoiseIf the system is perturbed by “noise,” these behaviors become distinct and differences in their reproductive success rates are amplified.

As a result, interest has shifted from the proof of the existence of a solution to the design of repeated game strategies that perform well against other sophisticated strategies.

Consider a population of strategies competing over a niche that can only maintain a fixed number of individuals, i.e., the population’s size is upwardly bounded by the system’s carrying capacity.

In each generation, each strategy is matched against every other, itself, & RANDOM in pairwise games.

Between generations, the strategies reproduce, where the chance of successful reproduction (“fitness”) is determined by the payoffs (i.e., payoffs play the role of reproductive rates).

Then, strategies that do better than average will grow as a share of the population and those that do worse than average will eventually die-out. . .

Replicator Dynamics

Replicator Dynamics

There is a very simple way to describe this process.Let:

x(A) = the proportion of the population using strategy A in a given generation; V(A) = strategy A’s tournament score; V = the population’s average score.

Then A’s population share in the next generation is:

x’(A) = x(A)

V(A)V

Replicator Dynamics

For any finite set of strategies, the replicator dynamic will attain a fixed-point, where population shares do not change and all strategies are equally fit, i.e., V(A) = V(B), for all B.

However, the dynamic described is population-specific. For instance, if the population consists entirely of naive cooperators (ALWAYS COOPERATE), then x(A) = x’(A) = 1, and the process is at a fixed-point. To be sure, the population is in equilibrium, but only in a very weak sense. For if a single D strategy were to “invade” the population, the system would be driven away from equilibrium, and C would be driven toward extinction.

Simulating Evolution

An evolutionary model includes three components:

Reproduction + Selection + Variation

Population

of

Strategies

SelectionMechanism

VariationMechanism

Mutation orLearning

Reproduction

Competition

Invasion

The Trouble with TIT FOR TAT

TIT FOR TAT is susceptible to 2 types of perturbations:

Mutations: random Cs can invade TFT (TFT is not ESS), which in turn allows exploiters to gain a foothold.

Noise: a “mistake” between a pair of TFTs induces CD, DC cycles (“mirroring” or “echo” effect).

TIT FOR TAT never beats its opponent; it wins because it elicits reciprocal cooperation. It never exploits “naively” nice strategies.

(See Poundstone: 242-248; Casti 76-84.)

Noise in the form of random errors in implementing or perceiving an action is a common problem in real-world interactions. Such misunderstandings may lead “well-intentioned” cooperators into periods of alternating or mutual defection resulting in lower tournament scores.

TFT: C C C C

TFT: C C C D“mistake”


Noise in the form of random errors in implementing or perceiving an action is a common problem in real-world interactions. Such misunderstandings may lead “well-intentioned” cooperators into periods of alternating or mutual defection resulting in lower tournament scores.

TFT: C C C C D C D ….

TFT: C C C D C D C ….“mistake”

Avg Payoff = R (T+S)/2


Nowak and Sigmund (1993) ran an extensive series of computer-based experiments and found the simple learning rule PAVLOV outperformed TIT FOR TAT in the presence of noise.

PAVLOV (win-stay, loose-switch) Cooperate after both cooperated or both defected;otherwise defect.


PAVLOV cannot be invaded by random C; PAVLOV is an exploiter (will “fleece a sucker” once it discovers no need to fear retaliation).

A mistake between a pair of PAVLOVs causes only a single round of mutual defection followed by a return to mutual cooperation.

PAV: C C C C DPAV: C C C D D

“mistake”


PAVLOV cannot be invaded by random C; PAVLOV is an exploiter (will “fleece a sucker” once it discovers no need to fear retaliation).

A mistake between a pair of PAVLOVs causes only a single round of mutual defection followed by a return to mutual cooperation.

PAV: C C C C D C CPAV: C C C D D C C

“mistake”


Pop. Share

0.140

0.100

0.060

0.020

0 200 400 600 800 Generations


1(TFT)326

7,9

10411

5

81814,12,1513

No. = Position after 1st Generation

Source:Axelrod 1984, p. 51.


PAV

TFT

GRIM (TRIGGER)D

R

C

Population shares for 6 RPD strategies (including RANDOM), with noise at 0.01 level.

Pop. Shares 0.50

0.40

0.30

0.20

0.10

0.00Generations

GTFT?

4/16 A Tournament

How to Promote Cooperation?

Axelrod, Ch. 6-9: 104-191.

Next Time

Unit III: The Evolution of Cooperation Can Selfishness Save the Environment? Repeated Games: the Folk Theorem Evolutionary Games A Tournament How to Promote.

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