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Wstęp do Teorii Gier. Consider the following game. 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
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Security level • MrRaw’sgameis a zero-sum game with MrRaw’spayoffsunchaged • MrRaw’ssecuritylevelis the value of MrRaw’sgame: • No saddle point, somixedstrategy • x=5/6, expectedpayoff=10/3=value of MrRaw’sgame.
Security level • MrsColumn’sgameis a zero-sum game with MrsColumn’spayoffsunchaged • MrsColumn’ssecuritylevelis the value of MrsColumn’sgame: • Thereis a saddle point 6 • So6=value of MrsColumn’sgame.
NasharbitrationScheme • 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 strategicpositioninthegame. (Mrs. Columnpositioninthegameaboveisstrong)
Nasharbitrationscheme • Challenge: to find a method of arbitratinggameswhich: • Does not involveillegitimatemanipulation of utilities • Doestakeintoaccountstrategicineqalities • Has a claim to fairness • First good idea: von Neumann, Morgenstern (1944): anyreasonablesolution to a non-zero-sum gameshould be: • Pareto-optimal • At orabove the securitylevel for bothplayers • The set of suchoutcomes (pureormixed) iscalledthenegotiation set of thegame.
Payoff polygon and Pareto-efficient points Mrs Column’s payoff (4,8) Pareto-efficient frontier (2,6) (10,5) MrRaw’s payoff (0,0)
Status Quo and negotiation set Mrs Column’s payoff (4,8) Negotiation set (2,6) 6 SQ (10,5) MrRaw’s payoff (0,0) 10/3
Nash 1950 arbitration scheme • What point in the negotiation set do we choose? • Nash arbitration scheme: • Axioms: • Rationality: The solution lies in the negotiation set • Linear invariance: If we linearly transform all the utilities, the solutions will also be linearly transformed • 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 • 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 N Q SQ P 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.
Nash 1950 Theorem • Theorem: There is one and only one arbitration scheme which satisfies Axioms 1-4. It is this: if SQ=(x0,y0), then the arbitrated solution point N is the point (x,y) in the polygon with x ≥ x0, and y ≥ y0, which maximizes the product (x-x0)(y-y0).
Nash solution – our example • In our example Nash solution is (5 2/3, 7 1/6) Mrs Column’s payoff (4,8) (5 2/3, 7 1/6) (2,6) 6 SQ (10,5) MrRaw’s payoff (0,0) 10/3
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.
