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.
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 1- 4. 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 ).
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.