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Some N-player games problems. Congestion Game. Cost of taking toll road is $10 +congestion cost Congestion cost on toll road is t where t is the number of people who take the toll road No toll on back road, cost of taking back road is congestion cost only

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## Some N-player games problems

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**Congestion Game**• Cost of taking toll road is $10 +congestion cost • Congestion cost on toll road is t where t is the number of people who take the toll road • No toll on back road, cost of taking back road is congestion cost only • Congestion cost of back road is 2b where b people take back road. • There are 100 people in all, so b+t= 100**Nash Equilibrium**• In equilibrium if both roads are used, drivers who take either road must be at least as well off as if they switched roads. • Let’s look for when the roads are indifferent. • 10+t=2b and t+b=100. • Solution is 10+t=2(100-t), 3t=190, t=63.33 • Now what? We can’t have fractional drivers. • Is t=64 a Nash equilibrium? • Is t=63 a Nash equilibrium?**Protest game**• N citizens, different ones value protesting differently. • Order them by value of protest v1>v2>…vN • We can draw a “demand curve” for protesting: How many people would protest if cost is p. • We also have something like a “supply curve”. What does it cost to protest if x people are protesting.**Lets draw them**• Two downward-sloping curves. Where is equilibrium? There can be more than one equilibrium.**Smallest unique integer game**• Three players. Each chooses a number 1, 2, or 3. Winner is the person who chose the smallest number that was chosen by nobody else. • Find a symmetric mixed strategy Nash equilibrium in which each number is played with some probability. • Are there any symmetric mixed strategy Nash equilibria in which nobody chooses 3? What about nobody chooses 2?**Volunteers’ Dilemma**N people observe a mugging. Someone needs to call the police. Only one call is needed. Cost of calling is c. Cost of knowing that the person is not helped is T. Should you call or not call? T>c>0. Many asymmetric pure strategy equilibria. Also one symmetric mixed strategy equilibrium.**Mixed strategy payoffs**• Suppose everybody uses a mixed strategy with probability p of calling. • In equilibrium, everyone is indifferent about calling or not calling if expected cost from not calling equals cost from calling. Expected Cost of of not calling is T(1-p)N-1 • Expected cost of calling is c.**A sad result**• Equilibrium has T(1-p)N-1 =c • Then so 1-p=(c/T)1/N-1 • The probability that nobody calls is then (1-p)N=(c/T)N/N-1 • This is an increasing function of N that approaches c/T as N gets large. • The more people who observe the crime, the less likely it is that someone reports it.**Baseball mixed strategy**• Batter chooses row, Pitcher chooses column**Finding mixed strategy equilibrium**• A player will used a mixed strategy only if he is indifferent between the pure strategies that are being mixed. • Let p be the probability that pitcher throws fastball. When is batter indifferent between his two strategies? • .35 p+.3(1-p) is expected payoff from setting up for fastball • .2p+.5(1-p) is expected payoff from setting up for curveball.**Finishing it up**• Solve for p that makes batter indifferent. • Also, pitcher will use a mixed strategy if and only if his expected payoffs from the two strategies are equal. • Let q be probability batter sets for fastball. • Find q.**Stop and Go Game**• Three players, simultaneous move, Two strategies—Stop, Go • Payoff to Stop is 55 • Payoff to Go is 120/m where m is the number of players who choose Go. • 3 pure strategy equilibria where two Go and one Stops. • What about mixed strategy equilibria?**Symmetric equilibrium**• Suppose each plays Go with probability p. • For what p are the expected payoffs from Stop and Go the same? • Payoff from Go is 120 if neither of the others Go, payoff is 60 of exactly one goes and zero if none go. • Expected payoff from Go is 120(1-p)2+ 60x2p(1-p)+40p2 When is this the same as the payoff from Stop?**Solving for equilibrium**• 120(1-p)2+ 60x2p(1-p)+40p2=55 • 120-240p+120 p2+ 120p-120 p2+40p2=55 • 65-120p+40 p2=0 • Solve this quadratic. • I find (6- √10 )/4 which is about .71**Three more mixed strategy equilibria**• Suppose one player plays Go for sure and other two play Go with probability p. • For the two with mixed strategies, payoff from Go will be 40 with probability p and 60 with probability 1-p. • When is 40p+60(1-p)=55? • 20p-5, and p=1/4.

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