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Non-cooperative Games Elon Kohlberg February 2, 2015

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Matching Pennies HT H1, -1-1, 1 T 1, -1 No Nash Equilibrium…

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LR T 1, 00, 3 B 0, 12, 0

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Mixed strategies are randomizations over the pure strategies – the rows or the columns 1-r L rRrR t T1, 00, 3 1-t B0, 12, 0 Player 1 chooses T with probability t Player 2 chooses R with probability r

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LR T1, 00, 3 B0, 12, 0 Expected Payoffs to Player I: T:.4 * 1 +.6 * 0 =.4 B:.4 * 0 +.6 * 2 = 1.2 Best response to r=.6 is t=0, not t=.4 Try:.3 r=.6

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2/3 L 1/3 R ¼ T1, 00, 3 ¾ B0, 12, 0 Expected payoff to Player I: T: 2/3 B: 2/3 T=1/4 is a best response Try: t=1/4 r=1/3 Expected payoff to Player II: L: 3/4 R: 3/4 R=1/3 is a best response

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Indifference Principle: In a Nash equilibrium, all strategies that are assigned positive probability must have the same expected payoff. Proof If not, switch some weight from the lower expected payoff to the higher.

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Indifference Principle: In a Nash equilibrium, all strategies that are assigned positive probability must have the same expected payoff. Q.Is this a sufficient condition for Nash equilibrium? No. One must also verify that no strategy assigned zero probability yields a higher payoff.

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Indifference Principle: In a Nash equilibrium, all strategies that are assigned positive probability must have the same expected payoff. Q.Is this a sufficient condition for Nash equilibrium? No. One must also verify that no strategy assigned zero probability yields a higher payoff. Q. Are the two conditions sufficient for Nash equilibrium? Yes.

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StrategyPayoff 15.1 27.3 3 42.3 57.1 67.3 75.8

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Equations, based on the indifference principle: t > 0 and 1-t > 0 1-r = 2 r r = 1/3 r > 0 and 1-r > 0 1-t = 3t t = 1/4

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Graphical Analysis Consider the best response graphs: As r goes from 0 to 1, draw the best response, t. As t goes from 0 to 1, draw the best response, r. Intersection?

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Rationales for Mixed-strategy Equilibrium: 1.Self-enforcing beliefs 2.Frequency in a population 3.Intentional randomization

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Battle of the Sexes FootballConcert Football2, 10, 0 Concert0, 01, 2 W M

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Battle of the Sexes FootballConcert Football2, 10, 0 Concert0, 01, 2 W M Mixed-strategy equilibrium: M: 2/3 F; 1/3 C W: 1/3 F; 2/3 C

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Battle of the Sexes FootballConcert Football4, 10, 0 Concert0, 01, 4 W M

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Battle of the Sexes FootballConcert Football4, 10, 0 Concert0, 01, 4 W M Mixed-strategy equilibrium: M: 4/5 F; 1/5 C W: 1/5 F; 4/5 C

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SwerveForward Swerve6, 64, 7 Forward7, 40, 0 Chicken

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SwerveForward Swerve6, 64, 7 Forward7, 40, 0 Mixed-Strategy Equilibrium S=.8 F=.2 (Makes sense: Pure strategy equilibria require breaking the symmetry…) Probability of collision: 4% Chicken

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SwerveForward Swerve6, 62, 12 Forward12, 20, 0 Chicken

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SwerveForward Swerve6, 62, 12 Forward12, 20, 0 Chicken Mixed-Strategy Equilibrium S=.25 F=.75 Probability of collision: 56%

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Stag Hunt StagHare Stag5, 50, 3 Hare3, 03, 3 The norm could be to go after the stag. The norm could be to go after the hare. Q.What is the meaning of the mixed-strategy equilibrium? Q. What probability does it assign to Stag? Q. How would this probability change if the value of a stag were 10 rather than 5?

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Pollution Game CN C2, 2, 22, 3, 2 N3, 2, 20, 0, -1 CN C2, 2, 3-1, 0, 0 N0, -1, 00, 0, 0 II I I III C N

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Pollution Game CN C2, 2, 22, 3, 2 N3, 2, 20, 0, -1 Pure- strategy Nash Equilibria (i)No one cleans (ii)Two out of three clean CN C2, 2, 3-1, 0, 0 N0, -1, 00, 0, 0 II I I III C N

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Pollution Game CN C2, 2, 22, 3, 2 N3, 2, 20, 0, -1 Mixed- strategy Nash Equilibria (iii) One cleans for sure; other two clean with probability 2/3. (iv) Two symmetric equilibria. CN C2, 2, 3-1, 0, 0 N0, -1, 00, 0, 0 II I I III C N

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Mixed Strategies CMPT 882 Computational Game Theory Simon Fraser University Spring 2010 Instructor: Oliver Schulte.

Mixed Strategies CMPT 882 Computational Game Theory Simon Fraser University Spring 2010 Instructor: Oliver Schulte.

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