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Poor Bart.. Always Rock.. Nothing beats plays Rock! I’ll play good old Rock!! Paper! Class 3 Simultaneous-move games • Analytic tool: Game Table • Nash equilibrium: mutual best response Prisoner’sdilemma games Dominant strategies Ways to solve the dilemma Games with no pure-strategy equilibrium Mixed-strategy equilibrium Graphic and algebraic solution Sensibility of being unpredictable Mixed Strategy : A strategy in which a player plays his available (pure) strategies with certain probabilities. Nash’s existence theorem : If each player in a game has a finite number of pure strategies, then there exists at least one equilibrium in (possibly) mixed strategies. ∴If there are no pure strategy equilibria, there must be a unique mixed strategy equilibrium. Player 2 Heads Tails Player 1 Heads 1,-1 -1,1 Tails -1,1 2,-1 • Players either place heads or tails • Player 1 wins if they match; player 2 wins if they don’t Player 2 Heads Tails Player 1 Heads 1,-1 -1,1 Tails -1,1 2,-1 • Player 1’s best strategy is to match Player 2’s strategy; Player 2’s best strategy is to place the opposite strategy of Player 1. • There is no equilibrium where both players adopting one (pure) strategy is optimal. I.e., Given the strategy of the other, the player always wants to do something else. Randomize: play each strategy with some probability Play a Mixed Strategy NE : A mixed strategy NE is an outcome in which players adopt each strategy with some non-negative probability Note: A pure strategy NE is a special case of a mixed strategy NE But how much should you mix? Up to the point where the opponent is indifferent between their own alternatives. Why? If they are not indifferent, then you are too predictable and therefore they will use it to their own advantage (so to your own disadvantage) Suppose that Player 1 plays H w/ probability p and T w/ probability 1-p Payoff of Player 2 from playing H: -1 p + 1 (1-p) = 1 - 2p Payoff of Player 2 from playing T: 1 p +[ -1 (1-p)] = 2p - 1 Player 2 will be indifferent between H and T if and only if player 1 plays H with the following probability: 1 – 2p = 2p – 1 4p = 2 p = ½ So player 1 should play H and T w/ probability ½ each. Player 2’s payoff will then be 1 – 2p = 1 – 2 (½) = 0. Suppose that Player 2 plays H w/ probability q and T w/ probability 1-q Payoff of Player 1 from playing H: 1 q + [-1 (1-q)] = 2q - 1 Payoff of Player 1 from playing T: [ -1 q] + 2 (1-q) = 2 – 3q Player 1 will be indifferent between H and T if and only if Player 2 plays H with the following probability: 2q - 1 = 2 – 3q 5q = 3 q = 3/5 = 0.6. So Player 2 should play H w/ prob. 0.6 and T w/ prob. 0.4. Player 1’s payoff will be 2q – 1 = 2(0.6) – 1 = 0.2 Payoffs: • Player 1 = 0.2 • Player 2 = 0 Whatif Player 1 plays H with any other probability? • Example: p = 0.45 (play tails more often because matches on tails pay 2 instead of 1). • Player 2 will always play H, and player 1’s expected payoff will be -0.1, less than 0.2 before. Sometimes no pure strategy exists • Must randomize actions How to randomize? • Make other player indifferent • Otherwise, they will take preferred action, and we should respond to that Examples of when to randomize • Sports • Auditing • Most of the time when keeping secrets Venus and Serena Williams play tennis. Each can hit the ball in different ways. How should they play in order to win? Players: Venus, Serena Williams Strategies: • Serena: can pass down-the-line (DL) or cross-court (CC) • Venus: can position herself to receive either DL or CC Information: Serena’s turn to return the ball Payoffs • Fraction(%) of times that each player wins the point Preferences: Better if wins the point Venus DL CC DL 50, 50 80, 20 Serena CC 90, 10 20, 80 No pure strategy Nash Equilibrium. Suppose Serena can mix her strategies p: probability that Serena chooses DL 1-p: probability that Serena chooses CC Suppose Venus can mix strategies as well q: probability that Venus positions herself for DL 1-q: probability that Venus positions herself for CC Venus DL CC q-mix 50q+80(1-q), DL 50, 50 80, 20 50q+20(1-q) 90q+20(1-q), Serena CC 90, 10 20, 80 10q+80(1-q) p- 50p+90(1-p), 80p+20(1-p), mix 50p+10(1-p) 20p+80(1-p) Note: When p=1, Serena always plays DL When p=0, Serena always plays CC Use best response analysis to find optimal mixing probabilities Need to know how one’s expected utility from playing a certain strategy changes by different probability mixes of the other Venus’s payoff, success probability, as a function of p. 100 20p+80(1-p) [CC] Venus’s Success % 80 50p+10(1-p) [DL] p=.70 0 1 Serena’s p-mix Serena’s payoff, in turn, depends on the choices of q 50q+80(1-q) [DL] 90q+20(1-q) [CC] 100 Serena’s Success % q = 0.6 80 0 1 Venus’s q-mix • Serena’s Best Response • Venus’s Best Response Function Function 1 1 0.7 p p 0 0.6 1 0 1 q q 1 Mixed Strategy 0.7 NE Occurs at p=.70, q=.60 p 0 q 0.6 1 Equilibria, pure or mixed, obtain whenever the best response functions intersect Mutual Best Responses! Players are indifferent between choosing either strategy Opponent’s indifference property : An equilibrium mixed strategy of one player in a two-person game has to be such that the other player is indifferent among all the pure strategies that are actually used in his mixture Choose p so as to make Venus indifferent between playing either pure strategy Why? ∵Serena is always worse off if Venus knows how Serena will return the ball. By making Venus indifferent between the two strategies, Serena can keep Venus guessing Randomizing just right takes away any ability to be taken advantage of Serena solves for p that equates Venus’s payoffs from DL and CC: 50p+10(1-p) = 20p+80(1-p) ∴p = .70 Suppose Serena plays DL with probability .7 and CC with probability .3 Then, Venus’s success rate from • DL: 50(.70)+10(.30) = 38% • CC: 20(.70)+80(.30) = 38% Serena’s success rate: 62% Venus solves for q that equates Serena’s payoffs from DL and CC: 50q+80(1-q) = 90q+20(1-q) ∴q = .60 Suppose Venus plays DL with probability .6 and CC with probability .4 Then, Serena’s success rate from • DL: 50(.60)+80(.40) = 62% • CC: 90(.60)+20(.40) = 62% Venus’s success rate: 38% NE in mixed strategies : A probability distribution for each player, where the distributions are mutual best responses to one another in the sense of expectations NE in pure strategies : A special case of mixed strategies, where the probabilities are chosen from the set {0, 1} Why is this a useful objective? • Making your opponent indifferent in expected terms is equivalent to minimizing your opponents’ ability to recognize and exploit systematic patterns of your own behavior • In constant-sum games, keeping your opponent indifferent is equivalent to keeping yourself indifferent Mixed Strategies: • If opponent knows what I will do, I will always lose! • Randomizing just right takes away any ability to be taken advantage of Implications (strangely): • A player chooses his strategy so as to make his opponent indifferent • If done right, the other player earns the same payoff from either of his strategies Employees can either work or shirk. Managers want to monitor them so that they work, but monitoring is costly. What is the optimal monitoring strategy? Players: Employee, Manager Strategies: • Employee: Work or Shirk • Manager: Monitor or Not Monitor Information : Both players know that working costs effort and monitoring costs money Payoffs: • Employee - Salary is $100K unless caught shirking - Working hard costs effort worth $50K • Manager - Value of employee output is $200K - $0 if employee doesn’t work - Cost of monitoring is $10K Preferences: The higher the monetary value, the better Manager Monitor Not Monitor Work 50K, 90K 50K, 100K Employee Shirk 0, -10K 100K, -100K ∴ No pure strategy equilibria Let p be the probability the employee shirks Let q be the probability the manager monitors find the expected payoffs using p and q First, Then, calculate the best response Wherever the best responses cross is the mixed-strategy equilibrium If employee works: • E (payoff/work) = 50q + 50(1-q) = 50 If employee shirks: • E (payoff/shirk) = 0q + 100(1-q) = 100 – 100q E (payoff/work) = E (payoff/shirk) 50 = 100 – 100q Indifferent when q=1/2 Best strategy for all possible strategies of the manager • If q < ½: Shirk • If q > ½: Work • If q = ½ : Indifferent If manager monitors: • E (payoff/monitor) = 90(1-p) -10p = 90 -100p If manager not monitors: • E (payoff/not monitor) = 100(1-p)-100p = 100 – 200p E (payoff/monitor) = E (payoff/not monitor) 90 -100p = 100 – 200p Indifferent when p = 1/10 Best strategy for all possible strategies of the manager • If p < 1/10: Not Monitor • If p > 1/10: Monitor • If p = 1/10 : Indifferent • Employee’s Best • Manager’s Best Response Function Response Function 1 1 p p 0.1 0 0.5 1 0 1 q q 1 Shirk Mixed Strategy NE Occurs at p p=.1, q=.5 Work 0.1 0 1 q 0.5 Not Monitor Monitor Employee shirks with probability 1/10 Manager monitors with probability ½ Employee’s expected payoff: (1/10)[½*0+½*100]+(9/10)[½*50+½*50]=50 Manager’s expected profit: ½[(9/10)*90-(1/10)*10] +½[(9/10)*100-(1/10)*100]=80 Bothplayers are indifferent between any mixture over their strategies E.g. Employee receives a payoff of $50K whether he works (½0+½100=50) or shirks (½50+½50=50) Regardless of what employee does, expected payoff is the same Since a player does not care what mixture she uses, she picks the mixture that will make her opponent indifferent! Motivate compliance at lower monitoring cost Parking Tickets Audits Drug Testing Increasein cost to all parties should not change the optimal “mix” for that party “Not surprisingly, it was not intuitive to them that a 10% increase in enforcement costs should not be met with an equal decrease in enforcement. By reducing the ticketing rate, incidence of illegal parking increased by over 40%.” IRS Commissioner Charles Rossotti: • Audits more expensive now than in ’97 • Number of audits decreased • Offshore evasion alone increased to $70billion dollars! Recommends: As audits get more expensive, need to increase budget to keep the number of audits constant! Manager’s strategy of monitor ½ of the time must mean that there is a 50% chance of monitoring in every round Cannot just monitor every other day Humans are bad at this! Exploit patterns! McDonald and Burger King decide whether to put their restaurants in a shopping mall or not. Their profits depend on the actions of each other. Players: McDonald, Burger King Strategies: Enter, Not Enter Information: The market size is such that only one firm entering yields positive profits. Payoffs: If only one firm enters, it earns $300K If both firms enter, each earns -$100K Preferences: Better the higher the payoff BK Enter Not Enter Enter -1, -1 3, 0 MD Not Enter 0, 3 0, 0 There are two pure strategy equilibria Are there any mixed strategy equilibria as well? MD chooses probability p of entering, so that BK is indifferent between entering and not entering BK’s payoff from entering: p(-1)+(1-p)(3) = 3-4p BK’s payoff from not entering: p(0)+(1-p)(0) = 0 ∴BK indifferent when p=3/4 BK chooses probability q of entering, so that MD is indifferent between entering and not entering MD’s payoff from entering: q(-1)+(1-q)(3) = 3-4q MD’s payoff from not entering: q(0)+(1-q)(0) = 0 ∴MD indifferent when q=3/4 Both MD and BK choose Enter with probability ¾ and Not Enter with probability ¼ Note the symmetry in equilibrium since the payoff structure are the same Expected Payoff of BK Best Response Function of BK 3 1 p=¾ q 0 1 -1 p=¾ 1 Payoff from Enter Payoff from Not Enter Expected Payoff of MD Best Response Function of MD 3 1 q=¾ p 0 1 -1 q=¾ 1 Payoff from Enter Payoff from Not Enter PSE#1 1 MSE q=¾ PSE#2 p=¾ 1 BK’s best response MD’s best response What if one of the firms has a competitive advantage? Suppose if MD is the sole entrant, it earns $400K. Otherwise, the game is the same as before. BK Enter Not Enter Enter -1. -1 4, 0 MD Not Enter 0, 3 0, 0 The pure strategy equilibria remains the same. What about the mixed strategy equilibrium? MD’s choice of p remains the same since BK’s payoffs have not changed What about BK’s choice of q? BK chooses q so that MD is indifferent between entering and not entering: q(-1)+(1-q)(4) = 0 q = 4/5 > 3/4 ∴BK’s probability of entering increases Suppose BK does not adjust its probability of entering What would MD do? 1 MD would q = 4/5 choose p = 1!! q=¾ MD’s new b.r. p=¾ 1 If MD chooses p = 1, then BK incurs negative profits by keeping q at ¾ i.e. BK’s profit (q=¾) = ¾(-1) + 0 = -¾ By increasing q to 4/5, BK induces MD’s expected profit when enter to become (4/5)(-1) + (1/5)(4) = 0, which is equal to the expected profit when not enter Again, make the competitor indifferent! BK is in a disadvantageous position. But the MSE is for BK to increase the probability of entering. Why? Unreasonable predictors of one-time human interaction Reasonablepredictors of long-term proportions or multimarket contacts

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posted: | 1/27/2011 |

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