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Explore evolutionary stability in finite automata playing the Prisoner's Dilemma, Nash equilibrium structure, and invasions in repeated games. Detailed insights from notable studies.
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Evolution & Economics No. 4
Evolutionary Stability in Repeated Games Played by Finite Automata Automata K. Binmore & L. SamuelsonJ.E.T. 1991
C C C,D D D D C D C D C Grim Tit For Tat (TFT) C C C C D D D D C C D D Tweedledum Tat For Tit (TAFT) Finite Automata playing the Prisoners’ Dilemma transitions states (& actions)
C C C,D C,D D D C D C D Tweedledee CA C,D C,D C D C D Automata playing the Prisoners’ Dilemma
Two Automata playing together, eventually follow a cycle • (handshake) • The payoff is the limit of the means. • The cost of an automaton is the number of his states. • The cost enters the payoffs lexicographically.
The Structure of Nash Equilibrium in Repeated Games with Finite Automata Dilip Abreu & Ariel Rubinstein Econometrica,1988
(-1,3) (2,2) (0,0) (3,-1) The Structure of Nash Equilibrium in Repeated Games with Finite Automata Dilip Abreu & Ariel Rubinstein Econometrica,1988 N.E. of repeated Game N.E in Repeated Games with Finite Automata (Abreu Rubinstein)
x x ? y a x x x x ? y y ? b a If then:
x x ? y a x x x x ? y ? y b a If then:
x x ? y a Q.E.D.
C D D C D C Tit For Tat (TFT) C,D C,D C D C D Cis not an ESS, it can be invaded byD. Dis not an ESS, it can be invaded byTit For Tat.
C D C C,D D D C D C D C Grim Tit For Tat (TFT) Q.E.D. In the P.D. Tit For Tat and Grim are not MESS (they do not use one state against themselves)
For a general, possibly non symmetric game G. Define the symmetrized version of G: G # #. A player is player 1with probability 0.5 and player 2 with probability 0.5 • The previous lemmas apply to (a1,a2) • An ESS has a single state │a1│=│a2│=1 • If (a1,a2) is a MESS it uses all its states when playing against itself, i.e. a1,a2 use all their states when playing against the other.
C C,D D C D C C,D D C CA C,D D AC C,D D C C C AA D C C C D C D D D C D C D Tat For Tit (TAFT) D C,D CC CD It can be invaded by:
C C,D D C D AC C C D D C D Tat For Tit (TAFT) It can be invaded by:
C C,D D D C CA C C D D C D Tat For Tit (TAFT) It can be invaded by:
C C D C C D C D D C D Tat For Tit (TAFT) D CC It can be invaded by:
C C D D C D Tat For Tit (TAFT) No other (longer and more sophisticated) automaton can invade. Any exploitation of TAFT (playing D against his C) makes TAFT play D, so the average of these two periods is (3+0)/2 = 1.5 < 2, the average of cooperating.
C C,D C C,D D C D D D C AC CA C C D C C C D C D D D C D C D Tat For Tit (TAFT) D C,D CC CD A population consisting of: can be invaded only by: If AC invaded, it does not do well against CD D C D C ……. C D C D …….
C C,D C,D D D C CA C,D D C C C AA D C C C D C D D D C D C D Tat For Tit (TAFT) D C,D CC CD A population consisting of: can be invaded only by CA If AA invaded, it does not do well against CC D C C C C……. C D D D D…….
C C,D D D C CA C C D C C C D C D D D C D C D Tat For Tit (TAFT) D C,D CC CD A population consisting of: can be invaded only by CA but if CA invaded then a sophisticated automaton S can invade and exploit CA . S starts with C. if it saw C it continues with C forever (the opponent must be CD or CC ). If it saw D, it plays D again, if the other then plays D it must be TAFT. S plays another D and then C forever. If, however, after 2x D, the other played C, then it must be CA, and S should play D forever.
C C,D D D C CA C C D C C C D C D D D C D C D Tat For Tit (TAFT) D C,D CC CD A population consisting of: can be invaded only by CA When S invades, CA will vanish, and then S which is a complex automaton will die out.