ECO290E Game Theory GRIPSeco290e博弈夹子.pptVIP

ECO290E: Game Theory Lecture 5 Mixed Strategy Equilibrium Zero-Sum Game Matching Pennies No Nash Equilibrium? The distinguishing feature of zero-sum game is that each player would like to outguess the other since there is no “win-win” situation. Example: Porker (bluff or not), Battle (by land or by sea), Tennis (left or right to serve) When each player would always like outguess the other(s), there is no Nash equilibrium. Contradicting Nash’s theorem?? Mixed Strategy A mixed strategy for a player is a probability distribution over some (or all) of her strategies. The strategy we have studied so far, i.e., taking some action for sure, is called a pure-strategy. When the outcome of the game is uncertain, we assume that each player maximizes expected value of her payoff. Expected utility theory (von Neumann and Morgenstern, 1944) Matching Pennies Again Introducing mixed strategies How to Find Equilibrium? If a player takes both “Heads” and “Tails” with positive probability, she must be indifferent between these two pure strategies, i.e., the expected payoff derived by choosing Heads must be equal to that by choosing Tails. -p+(1-p)=p-(1-p), hence p=0.5. q-(1-q)=-q+(1-q), hence q=0.5. How to Verify Equilibrium? Note that if p=0.5, Player 1 does not have a strict incentive to change her strategy from q=0.5. Similarly, Player 2 does not have a strict incentive to change his strategy from p=0.5, if q=0.5. Therefore, p=q=0.5 constitutes a mixed-strategy equilibrium. Q: Is this equilibrium reasonable/stable? A: Yes (each player ends up randomizing two strategies equally if the rival is smart enough). Modified Matching Pennies Suppose the payoffs in the up-left cell changes as the following: Indifference Property Under mixed-strategy NE, Player 1 must be indifferent between choosing H and T: -2p+(1-p)=p-(1-p), hence p=0.4. Similarly, Player 2 must be indifferent between choosing H and T: 2q-(1-q)=-q+(1-q), hence q=0.4. You can easily verify that (p,q)=(0.4,0.4) indee