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Unformatted text preview: GAME THEORY Lecture 13­14 Instructor: Nuno Limão Nash Equilibrium with Mixed Strategies slides_414_13_14_2011_no_solutions 1 OUTLINE Error! Not a valid bookmark self­reference. Preliminaries: Probability and expected payoffs Nash Equilibrium in mixed strategies Other results Applications: entry game Applications: predicting goals Maximin strategies and Nash equilibrium slides_414_13_14_2011_no_solutions 2 MOTIVATION Games analyzed thus far have stable outcomes, i.e. Nash equilibria that involve pure strategies: either enter or not, price high or low, etc Do all games have pure strategy NE? No. Many cases where unpredictability must be part of an optimal strategy Penalty kicks: http://www.youtube.com/watch?v=yZ4vmY5QKwo&NR=1&feature=fvwp Outguessing games Other? Symmetric congestion games (entry) Do such games have any equilibria? Try experiment below slides_414_13_14_2011_no_solutions 3 Setup Police officer: (row) can patrol streets (1 ) or park (2) Dealer: (column) has similar choices Payoffs Experiment Login: veconlab.econ.virginia.edu/login.htm Session: Ngl8 slides_414_13_14_2011_no_solutions 4 Questions Is there a pure strategy NE? No. Police gains if they match and dealer if they mismatch Given that predictability by any is a bad strategy, want to randomize, how? Symmetric payoffs suggest that incentives are similar for both and perhaps street given probability = ½. Is there an optimal strategy? How can we find it even if payoffs are not symmetric? slides_414_13_14_2011_no_solutions 5 Findings from experiment w/ symmetric payoffs: towards end converge to probabilities close to 1/2 for each player slides_414_13_14_2011_no_solutions 6 Experiment w/ asymmetric payoffs Dealer now has higher payoff if goes to park (and police goes to street) Suppose that we know that ½ is a NE for each player in the initial (symmetric game). Can it still be an equilibrium? What did we observe? Police more often at the Park (71%) and Dealer less often (29%)! How do results compare to NE prediction? Object of this lecture. Preview: only the police got it right but Winner: Mark Bruno, $520 slides_414_13_14_2011_no_solutions 7 PRELIMINARIES: PROBABILITY AND EXPECTED PAYOFFS Definitions and basic properties related to probabilities Random event: one that is unpredictable, e.g. # hit on a particular roulette spin Probability of a random event: frequency with which that event occurs if we had an infinite number of trials, e.g. the probability of “17” over n trials = # times “17” hit/n as n‐>infinity If roulette is “fair” (= probability of each #) then the fact that it has 38 possible outcomes (0,00, 1,…36) means that this probability is 1/38 for any number Probability distribution: collection of p for each possible outcome i of a random event, e.g. {p00,p0,p1,…p36} slides_414_13_14_2011_no_solutions 8 Definitions and basic properties related to probabilities (continued) Properties of probability distributions Each pi is between 0 and 1 ∑pi = 1 over all possible outcomes i Independent random events: when the outcome of one event does not affect the probability of the other. E.g. number hit on a US roulette and a European roulette Joint probability of two independent events: probability that both occur, this is simply the product of their probabilities, p*q slides_414_13_14_2011_no_solutions 9 Expected value A random variable is a random event that takes numerical values, e.g. value of a company’s stock or your happiness the following day The expected value of a random variable is the weighted sum of the realizations of that random variable, where the weights attached to the realization of each possible outcome are their probabilities. Example I: Payoff to roulette Bet $10 on #14 and $5 on 21 Casino pays out only on positive numbers: 1‐36 and gives “fair odds” if these were the only numbers and you hit, i.e. $36 for each $1 bet Expected value of your winnings: (1/38)*(10)*36+(1/38)*(5)*36+(36/38)*0=540/38~14.21 Note that your actual winnings in any given bet are either 360, 180 or 0 but if you repeat it many times then the probability associated with each of these outcomes is 1/38,1/38,36/38 so your average winnings are 14.21 (<$15, hence the casino profits) slides_414_13_14_2011_no_solutions 10 Expected value (continued) Example II: Rain or shine vi: your utility tomorrow under alternative weather i=rain or shine. Random because weather is a random event, p= probability of rain Expected value of your utility tomorrow: V = p vrain +(1‐p) vshine slides_414_13_14_2011_no_solutions 11 Preferences over uncertain options We assume that individuals are expected utility maximizers When face with a choice between a certain payoff of 100 and an uncertain one of 60 and 180 Choose certain if 100>p180+(1‐p)60 , i.e. if p is sufficiently low Note that now the level of the payoffs matters to rank preferences over outcomes, i.e. suppose p= 1/3 so indifferent in the example above. If payoff for certain outcome was slightly higher (but it was still between 60 and 80 so the ordinal ranking is maintained) then prefer the certain. Thus intensity of preferences will affect the equilibrium choices when we randomize whereas it did not when we used pure strategies. Implies we need to be more careful about specifying payoffs Note that if ALL payoffs are multiplied by the same number (e.g. convert from $ to cents) this does not affect the choice. Neither does adding the same amount to each payoffs slides_414_13_14_2011_no_solutions 12 NASH EQUILIBRIUM IN MIXED STRATEGIES Pure strategy Definition: one that does not require randomization e.g. in police vs. dealer game the pure strategies were either street or park. Mixed strategy Definition: a probability distribution over the set of a player’s pure strategies, e.g.: a mixed strategy for the police is simply p, the probability of patrolling the streets Randomized version of a game Replaces pure strategies with their mixed strategy counterpart Replaces payoff for each strategy profile with the expected payoff Nash Equilibrium in mixed strategies Informal definition: It is a mixed strategy profile that maximizes the expected payoff for each player when taking the other’s actions as given. Formal definition (see appendix) slides_414_13_14_2011_no_solutions 13 Finding mixed strategy NE: Police example A drug dealer may set up his shop at the street corner or at the park. The police can patrol either the street corner or the park. What are the pure strategy Nash equilibria? None. How should we find mixed strategy equilibrium? Set up expected payoffs Determine best response for police, p=BRPO(d), (i.e. p as a function of d, the probability of dealer in street) Determine best response for dealer, d=BRDD(p)) (i.e. d as a function of p) The NE will be the (p*=BRPO(d*),d*=BRDD(p*)) values that satisfy those BR slides_414_13_14_2011_no_solutions 14 General expected payoff for police: VPO (p,d) = p x d x 80 + p x (1‐d) x 0+ (1‐p) x d x 10 + (1‐p) x (1‐ d)x 60 = p x [d x 80 + (1‐d) x 0]+ (1‐p) x [d x 10 + (1‐ d)x 60] Note that VPO (p,d) = p x VPO (1,d)+ (1‐p) VPO (0,d) i.e. we can rewrite general expected payoff as a weighted average of two other expected payoffs from each pure strategy: patrolling streets with probability 1: VPO (p=1,d) or patrolling park with probability 1 (so streets w/ prob. 0): VPO (p=0,d) slides_414_13_14_2011_no_solutions 15 Thus if VPO (1,d)> VPO (0,d) for all d then VPO (p,d) is maximized by p=1, but can’t hold since if d‐>0 then VPO (1,d)= 0 < 60 =VPO (0,d) VPO (1,d)< VPO (0,d) for all d then VPO (p,d) is maximized by p=0 but can’t hold since if d‐>1 then VPO (1,d)= 80 > 10 =VPO (0,d) More generally, expect that if d is high then VPO (p,d) is maximized by high p (catch dealer) but if d is low the opposite occurs, indifferent in between slides_414_13_14_2011_no_solutions 16 Deriving BR for police What d leaves police indifferent between pure strategies (VPO (1,d)= VPO (0,d))? d x 80 + (1‐d) x 0 = d x 10 + (1‐ d)x 60 d=6/13 If d=6/13 then VPO (0,d) = VPO (1,d) = VPO (p,d) for all p. So BRPO(d=6/13)=[0,1] If d<6/13 then VPO (0,d) > VPO (p,d) for all p>0 , so BRPO(d<6/13)=0 since then VPO (0,d) > VPO (1,d) VPO (p,d) = p x VPO (1,d)+ (1‐p) VPO (0,d) [higher p places > weight on pure strategy that yields lower expected payoff] If d>6/13 then VPO (0,d) < VPO (p,d) for all p>0 , so BRPO(d>6/13)=1 since in this case VPO (0,d) < VPO (1,d) VPO (p,d) = p x VPO (1,d)+ (1‐p) VPO (0,d) [higher p places > weight on pure strategy that yields higher expected payoff] slides_414_13_14_2011_no_solutions 17 Graphical representation of BR for police slides_414_13_14_2011_no_solutions 18 Deriving BR for dealer Expected Payoff VDD (p,d) = d x [p x 20 + (1‐p) x 90]+ (1‐d) x [p x 100 + (1‐ p)x 40] VDD (p,d) = d x VDD (p,1)+ (1‐d) x VDD (p,0) Indifference: p x 20 + (1‐p) x 90 = p x 100 + (1‐ p)x 40 p=5/13 BRDD(p=5/13)=[0,1] BRDD(p<5/13)= 1 : Police low likelihood to go to street then Dealer moves there BRDD(p>5/13)= 0 : Police high likelihood to go to street then Dealer moves to Park slides_414_13_14_2011_no_solutions 19 Nash equilibrium: (p*=5/13,d*=6/13) slides_414_13_14_2011_no_solutions 20 Finding equilibrium mix using the “opponent’s indifference principle” When would you ever be willing to simply toss a coin or randomize to decide between two actions or pure strategies? When you are perfectly indifferent between them. But this can only happen if your opponent’s equilibrium strategy makes it just so. Therefore in a NE with randomization this must be true. Thus the equilibrium strategy for the dealer can be found by equating the expected payoffs of the police under each pure strategy and vice versa We verified this when we found the equilibrium, e.g. VPO (1,d=6/13)= VPO (0,d=6/13) VDD (p=5/13,1)= VDD (p=5/13,0) Note counter­intuitive result: your own payoffs are irrelevant to decide the mix! slides_414_13_14_2011_no_solutions 21 Exercise: compute the mixed strategy for the two games below slides_414_13_14_2011_no_solutions 22 Solution 1st game … Solution 2nd game ... slides_414_13_14_2011_no_solutions 23 OTHER RESULTS Existence of NE in mixed strategies: Every finite game has a NE in mixed strategies (Nash, 1951) [famous result. If we have time to discuss a proof we will later] The general approach to find the mixed strategy can also be used to find pure strategy since p can be 1 or 0. Of course you would not want to do so but just highlights that a pure strategy NE is simply a special case of a mixed strategy Odd number of NE: In almost all finite games there is a finite, odd number of Nash equilibria in mixed strategies (Wilson) Useful since if you find an even number of pure strategy equilibria then there is an extra one in mixed strategies slides_414_13_14_2011_no_solutions 24 Illustrating last two points: coordination game (again!) Firm 2 Firm 1 Ipad apps Other apps Ipad apps Other apps 1,1 0,0 0,0 1,1 Pure: (I, I), (O,O) Mixed: px1+(1‐p)x0=px0+(1‐p)x1p=1/2 BR: draw and show intersection at p=0, ½, 1 slides_414_13_14_2011_no_solutions 25 Using IDSDS to initially simplify. Suppose there is a game with 3 (or more) pure strategies for each player where at least one is strictly dominated. Then a NE strategy assigns it zero probability. Implication: in practice can first check which strategies survive IDSDS and then search for mixed strategy equilibria in remaining game (See entry example) slides_414_13_14_2011_no_solutions 26 APPLICATIONS: ENTRY GAME Setup Four firms are deciding whether to enter a market. Payoff for each firm i is 0 if it does not enter Gross profit minus entry cost if it enters Their heterogeneous entry costs and symmetric gross profits upon entry are # firms in market Gross profit per entrant 1000 400 250 150 1 2 3 4 slides_414_13_14_2011_no_solutions 27 Exercise: Find all the Nash equilibria of the entry game using the following hints First write out the payoffs for each under each possible situation and look for ways to simplify problem Firm 1 2 3 4 Payoff w/ 0 Payoff w/ 1 Payoff w/ 2 Payoff w/ 3 competitors competitors competitors competitors 900 300 150 50 700 100 ‐50 ‐150 700 100 ‐50 ‐150 500 ‐100 ‐250 ‐350 Is there equilibrium with firm 1 out? Is there equilibrium with firm 4 in? Can we reduce the game to a simple 2x2 matrix? How many pure strategy equilibria are there (if any)? Is there a mixed strategy equilibrium with randomization? slides_414_13_14_2011_no_solutions 28 Solution… slides_414_13_14_2011_no_solutions 29 Illustrates Use of IDSDS to simplify Possibility of some players randomizing and others using pure strategies Usefulness of fact about odd number of equilibria slides_414_13_14_2011_no_solutions 30 APPLICATIONS: PREDICTING GOALS Objectives Solving for mixed strategy NE with >2 pure strategies available Evaluating the predictive power of the theory in a setting that resembles simple game and we have data: penalty kicks Penalty kick setup Simultaneous move Kicker chooses only direction: left, right, center Goalkeeper: left, right, center (from kicker’s perspective) Payoffs (based on data) Kicker: probability of scoring Goalkeeper: probability of not scoring slides_414_13_14_2011_no_solutions 31 No Nash equilibrium in pure strategies Deriving equilibrium with mixed strategies Goalkeeper probability choices for Left, Right and center are gl, gr, gc= 1 – gl – gr. Kicker’s expected payoffs from Left: 0.65 gl + 0.95(1 – gl – gr) + 0.65 gr = Center: 0.95 gl + 0 (1 – gl – gr) + 0.95 gr = 0.95(gl+ gr) Right: 0.95 gl + 0.95 (1 – gl – gr) + 0.65 gr = slides_414_13_14_2011_no_solutions 0.95 ‐ 0.3gl 0.95‐ 0.3 gr 32 In equilibrium the expected payoff for the kicker of any pure strategy is equal to that of the other two. This implies the following Left=Right: 0.95 ‐ 0.3gl = 0.95‐ 0.3 gr gl=gr=g Left (=right)= center: 0.95 ‐ 0.3g = 0.95(g+ g) g≈0.43 Optimal strategy for kicker found similarly and yields similar probabilities (0.43 left, right and center = 1‐0.43‐0.43 = 0.14 Predictions vs. observed outcomes Prediction I: A goal is scored about 82% of the time since using the NE mixed strategy we obtain .43 (.43 x .65 + .14 x .95 + .43 x .95) [gl (kl x p_ll + kc x p_lc + kr x p_lr )] + .14 (.43 x .95 + .14 x 0+ .43 x .95) [gc (kl x p_cl + …)] + .43 (.43 x .95 + .14 x .95 + .43 x .65) [gr (kl x p_rl +…)] Prediction II: In equilibrium kicker is indifferent between kicking left or right so probability of scoring should be equalized slides_414_13_14_2011_no_solutions 33 Observed probabilities (i.e. score frequencies) are as predicted (statistically identical and ~82%) Probability of scoring when kicking right (or center) is 82.7% Probability of scoring when kicking left (or center) is 81.1% slides_414_13_14_2011_no_solutions 34 MAXIMIN STRATEGIES AND NASH EQUILIBRIUM Recall a game of pure conflict is one in which what is good for one player is bad for the other. Games of pure conflict may “suggest” cautious behavior. One extreme case is to believe that opponent will know exactly what you will do and choose his strategy to minimize YOUR payoff (since this is equivalent to maximize his given pure conflict) Von‐Neuman proposed that in such cases we may want to focus on “Maximin” solutions, i.e. strategy profiles where both players use Maximin strategies Definition of Maximin strategies: A strategy that maximizes a player’s expected payoff given that the other players choose strategies to minimize that players expected payoff in light of that player’s strategy (i.e. correctly anticipating it) Questions: How do we find Maximin solutions? What is the relationship between maximin solutions and Nash equilibria? slides_414_13_14_2011_no_solutions 35 Application: Holmes v. Moriarty in The Final Solution Holmes can escape Moriarty to Dover or Canterbury, Moriarty can wait either Maximin strategy for Holmes (ph of going to Dover) Asks what expected payoff is if opponent either goes to Dover : ph20 + (1‐ ph) 70 =70‐50 ph or Canterbury ph90 + (1‐ ph) 10 =10+80 ph slides_414_13_14_2011_no_solutions 36 If Moriarty correctly anticipates Holmes’ strategy (ph) and chooses his strategy to minimize Holmes’s payoff then if Holmes chooses ph < 6/13 then Moriarty chooses Canterbury and otherwise Dover. Thus the Maximin for Holmes is exactly 6/13 since that is the max expected value that he can obtain slides_414_13_14_2011_no_solutions 37 Moriarty’s maximin strategy: found similarly Maximin solution: must satisfy both maximin strategies: ph=6/13 and pm=8/13 Relationship to mixed strategy NE? Same! We can use the earlier method to find directly We can also just verify that at the probabilities above the players are indifferent between their pure strategies, e.g. pm=8/13 implies Holmes’s expected payoff is equalized across pure strategies Dover: (8/13)20+(5/13)90=610/13 Canterbury: (8/13)70+(5/13)10=610/13 slides_414_13_14_2011_no_solutions 38 More general result linking Maximin and NE For any two‐person game of pure conflict, the maximin solution is a Nash equilibrium If a two‐person game of pure conflict has a Nash equilibrium in which both players randomize, then each player’s Nash equilibrium strategy is also his maximin strategy Intuition: Under the Nash both are maximizing their payoffs taking the other’s actions as given and correctly anticipating equilibrium. But by minimizing the opponents expected payoff in maximin players are also maximizing own expected payoff (b/c it is a game of pure conflict) slides_414_13_14_2011_no_solutions 39 ...
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This note was uploaded on 10/25/2011 for the course ECON 414 taught by Professor Staff during the Spring '08 term at Maryland.

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