Wstęp do Teorii Gier. Consider the following game Mrs Column AB Mr Raw A(2,6)(10,5) B(4,8)(0,0)

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Presentation transcript:

Wstęp do Teorii Gier

Consider the following game Mrs Column AB Mr Raw A(2,6)(10,5) B(4,8)(0,0)

Security level Mr Raw’s game is a zero-sum game with Mr Raw’s payoffs unchaged Mr Raw’s security level is the value of Mr Raw’s game: – No saddle point, so mixed strategy – x=5/6, expected payoff=10/3=value of Mr Raw’s game. X AB Mr Raw A(2,-2)(10,-10) B(4,-4)(0,0) Mrs Column AB Mr Raw A(2,6)(10,5) B(4,8)(0,0)

Security level Mrs Column’s game is a zero-sum game with Mrs Column’s payoffs unchaged Mrs Column’s security level is the value of Mrs Column’s game: – There is a saddle point 6 – So 6=value of Mrs Column’s game. Mrs Column AB X A(-6,6)(-5,5) B(-8,8)(0,0) Mrs Column AB Mr Raw A(2,6)(10,5) B(4,8)(0,0)

Nash arbitration Scheme Up to this point, no cooperation btw. players Now different approach: what is a reasonable outcome to the game? Take again our previous game Egalitarian proposal: choose the outcome with the largest total payoff. Then split it equally. (15/2 for each) Two major flaws of this approach: – Payoffs are utilities. They cannot be meaningfully added or transferred across individuals. – Neglects the asymmetries of strategic position in the game. (Mrs. Column position in the game above is strong) Mrs Column AB Mr Raw A(2,6)(10,5) B(4,8)(0,0)

Nash arbitration scheme Challenge: to find a method of arbitrating games which: – Does not involve illegitimate manipulation of utilities – Does take into account strategic ineqalities – Has a claim to fairness First good idea: von Neumann, Morgenstern (1944): any reasonable solution to a non-zero-sum game should be: – Pareto-optimal – At or above the security level for both players The set of such outcomes (pure or mixed) is called the negotiation set of the game.

Payoff polygon and Pareto-efficient points (4,8) (10,5) (2,6) (0,0) Mr Raw’s payoff Mrs Column’s payoff Pareto-efficient frontier

Status Quo and negotiation set (4,8) (10,5) (2,6) (0,0) Mr Raw’s payoff Mrs Column’s payoff Negotiation set 6 10/3 SQ

Nash 1950 arbitration scheme What point in the negotiation set do we choose? – Nash arbitration scheme: Axioms: 1.Rationality: The solution lies in the negotiation set 2.Linear invariance: If we linearly transform all the utilities, the solutions will also be linearly transformed 3.Symmetry: If the polygon happens to be symmetric about the line of slope +1 through SQ, then the solution should be on this line as well 4.Independence of Irrelevant Alternatives: Suppose N is the solution point for a polygon P with status quo point SQ. Suppose Q is another polygon which contains both SQ and N, and is totally contained in P. Then N should also be the solution point for Q with status quo point SQ.

Independence of Irrelevant Alternatives graphically SQ N Suppose N is the solution point for a polygon P with status quo point SQ. Suppose Q is another polygon which contains both SQ and N, and is totally contained in P. Then N should also be the solution point for Q with status quo point SQ. P Q

Nash 1950 Theorem Theorem: There is one and only one arbitration scheme which satisfies Axioms It is this: if SQ=(x 0,y 0 ), then the arbitrated solution point N is the point (x,y) in the polygon with x ≥ x 0, and y ≥ y 0, which maximizes the product (x-x 0 )(y-y 0 ).

Nash solution – our example In our example Nash solution is (5 2 / 3, 7 1 / 6 ) (4,8) (10,5) (2,6) (0,0) Mr Raw’s payoff Mrs Column’s payoff 6 10/3 SQ (5 2 / 3, 7 1 / 6 )

Application The management of a factory is negotiating a new contract with the union representing its workers The union demands new benefits: – One dollar per hour across-the-board raise (R) – Increased pension benefits (P) Managements demands concessions: – Eliminate the 10:00 a.m. coffee break (C) – Automate one of the assembly checkpoints (reduction necessary) (A) You have been called as an arbitrator.