Static (or SimultaneousMove) Games of Complete Information Mixed Strategy Nash Equilibrium F. Valognes - Game Theory - Chp 9 1 Outline of Static Games of Complete Information Introduction to games Normal-form (or strategic-form) representation Iterated elimination of strictly dominated strategies Nash equilibrium Review of concave functions, optimization Applications of Nash equilibrium Mixed strategy Nash equilibrium F. Valognes - Game Theory - Chp 9 2 Today’s Agenda Review of previous class Examples F. Valognes - Game Theory - Chp 9 3 Mixed strategy Nash equilibrium Mixed Strategy: A mixed strategy of a player is a probability distribution over the player’s strategies. Mixed strategy Nash equilibrium A probability distribution for each player The distributions are mutual best responses to one another in the sense of expected payoffs F. Valognes - Game Theory - Chp 9 4 Mixed strategy Nash equilibrium: 2player each with two pure strategies Player 2 Player 1 s21 ( q ) s22 ( 1- q ) s11 ( r ) u1(s11, s21), u2(s11, s21) u1(s11, s22), u2(s11, s22) s12 (1- r ) u1(s12, s21), u2(s12, s21) u1(s12, s22), u2(s12, s22) Mixed strategy Nash equilibrium: A pair of mixed strategies ((r*, 1-r*), (q*, 1-q*)) is a Nash equilibrium if (r*,1-r*) is a best response to (q*, 1-q*), and (q*, 1-q*) is a best response to (r*,1-r*). That is, v1((r*, 1-r*), (q*, 1-q*)) ≥ v1((r, 1-r), (q*, 1-q*)), for all 0≤ r ≤1 v2((r*, 1-r*), (q*, 1-q*)) ≥ v2((r*, 1-r*), (q, 1-q)), for all 0≤ q ≤1 F. Valognes - Game Theory - Chp 9 5 2-player each with two strategies Player 2 Player 1 s21 ( q ) s22 ( 1- q ) s11 ( r ) u1(s11, s21), u2(s11, s21) u1(s11, s22), u2(s11, s22) s12 (1- r ) u1(s12, s21), u2(s12, s21) u1(s12, s22), u2(s12, s22) Theorem 1 (property of mixed Nash equilibrium) A pair of mixed strategies ((r*, 1-r*), (q*, 1-q*)) is a Nash equilibrium if and only if v1((r*, 1-r*), (q*, 1-q*)) v1((r*, 1-r*), (q*, 1-q*)) v2((r*, 1-r*), (q*, 1-q*)) v2((r*, 1-r*), (q*, 1-q*)) ≥ ≥ ≥ ≥ EU1(s11, (q*, 1-q*)) EU1(s12, (q*, 1-q*)) EU2(s21, (r*, 1-r*)) EU2(s22, (r*, 1-r*)) F. Valognes - Game Theory - Chp 9 6 Mixed strategy equilibrium: 2-player each with two strategies Player 2 Player 1 s21 ( q ) s22 ( 1- q ) s11 ( r ) u1(s11, s21), u2(s11, s21) u1(s11, s22), u2(s11, s22) s12 (1- r ) u1(s12, s21), u2(s12, s21) u1(s12, s22), u2(s12, s22) Theorem 2 Let ((r*, 1-r*), (q*, 1-q*)) be a pair of mixed strategies, where 0 <r*<1, 0<q*<1. Then ((r*, 1-r*), (q*, 1-q*)) is a mixed strategy Nash equilibrium if and only if EU1(s11, (q*, 1-q*)) = EU1(s12, (q*, 1-q*)) EU2(s21, (r*, 1-r*)) = EU2(s22, (r*, 1-r*)) That is, each player is indifferent between her two pure strategies. F. Valognes - Game Theory - Chp 9 7 Use indifference to find mixed Nash equilibrium (2-player each with 2 strategies) Use Theorem 2 to find mixed strategy Nash equilibria Solve EU1(s11, (q*, 1-q*)) = EU1(s12, (q*, 1-q*)) Solve EU2(s21, (r*, 1-r*)) = EU2(s22, (r*, 1-r*)) F. Valognes - Game Theory - Chp 9 8 Use Theorem 2 to find mixed strategy Nash equilibrium: illustration Battle of sexes Pat Opera (q) Chris Opera ( r ) Prize Fight (1-r) 2 , 0 Prize Fight (1-q) 1 0 , 0 , 0 1 , 2 Chris’ expected payoff of playing Opera EU1(O, (q, 1–q)) = q×2 + (1–q)×0 = 2q Chris’ expected payoff of playing Prize Fight EU1(F, (q, 1–q)) = q×0 + (1–q)×1 = 1–q Chris is indifferent between playing Opera and Prize EU1(O, (q, 1–q)) = EU1(F, (q, 1–q)) 2q = 1–q 3q = 1 This give us q = 1/3 F. Valognes - Game Theory - Chp 9 9 Use Theorem 2 to find mixed strategy Nash equilibrium: illustration Battle of sexes Pat Opera (q) Chris Opera ( r ) Prize Fight (1-r) 2 , 0 Prize Fight (1-q) 1 0 , 0 , 0 1 , 2 Pat’s expected payoff of playing Opera EU2(O, (r, 1–r)) = r ×1+(1–r)×0 = r Pat’s expected payoff of playing Prize Fight EU2(F, (r, 1–r)) = r×0+(1–r)×2 = 2 – 2r Pat is indifferent between playing Opera and Prize EU2(O, (r, 1–r)) = EU2(F, (r, 1–r)) r = 2 – 2r 3r = 2 This give us r = 2/3 F. Valognes - Game Theory - Chp 9 10 Use Theorem 2 to find mixed strategy Nash equilibrium: illustration Battle of sexes Pat Opera (q) Chris Opera ( r ) Prize Fight (1-r) 2 , 0 Prize Fight (1-q) 1 0 , 0 , 0 1 , 2 Hence, ( (2/3, 1/3), (1/3, 2/3) ) is a mixed strategy Nash equilibrium. That is, Chris chooses Opera with probability 2/3 and Prize Fight with probability 1/3. Pat chooses Opera with probability 1/3 and Prize Fight with probability 2/3. F. Valognes - Game Theory - Chp 9 11 Example 1 Bruce and Sheila determine whether to go to the opera or to a pro wrestling show. Sheila gets utility of 4 from going to the opera and 1 from pro wrestling. Bruce gets utility of 1 from going to the opera and 4 from pro wrestling. They agree to decide what to do in the following way: Bruce and Sheila each puts a penny below an issue of the TV guide on the coffee table (assume they don’t cheat by looking at the other). They count to 3 and simultaneously reveal which side of their penny is up. If the pennies match (both heads, or both tails), Sheila decides what to watch, while if the pennies don’t match (heads, tails or tails, heads) then Bruce decides. F. Valognes - Game Theory - Chp 9 12 Example 1 Sheila H (q) Bruce T ( 1–q ) H (r) 1 , 4 4 , 1 T ( 1–r ) 4 , 1 1 , 4 Bruce’s expected payoff of playing Head EU1(H, (q, 1–q)) = q×1 + (1–q)×4 = 4–3q Bruce’s expected payoff of playing Tail EU1(T, (q, 1–q)) = q×4 + (1–q)×1 = 1+3q Bruce is indifferent between playing Head and Tail EU1(H, (q, 1–q)) = EU1(T, (q, 1–q)) 4–3q = 1+3q 6q = 3 This give us q = 1/2 F. Valognes - Game Theory - Chp 9 13 Example 1 Sheila H (q) Bruce T ( 1–q ) H (r) 1 , 4 4 , 1 T ( 1–r ) 4 , 1 1 , 4 Sheila’s expected payoff of playing Head EU2(H, (r, 1–r)) = r ×4+(1–r)×1 = 3r + 1 Sheila’s expected payoff of playing Tail EU2(T, (r, 1–r)) = r×1+(1–r)×4 = 4 – 3r Sheila is indifferent between playing Head and Tail EU2(H, (r, 1–r)) = EU2(T, (r, 1–r)) 3r + 1 = 4 – 3r 6r = 3 This give us r = ½ ( (1/2, 1/2), (1/2, 1/2) ) is a mixed strategy Nash equilibrium. F. Valognes - Game Theory - Chp 9 14 Example 2 Player 2 L (q) Player 1 R ( 1–q ) T (r) 6 , 0 0 , 6 B ( 1–r ) 3 , 2 6 , 0 Player 1’s expected payoff of playing T EU1(T, (q, 1–q)) = q×6 + (1–q)×0 = 6q Player 1’s expected payoff of playing B EU1(B, (q, 1–q)) = q×3 + (1–q)×6 = 6-3q Player 1 is indifferent between playing T and B EU1(T, (q, 1–q)) = EU1(B, (q, 1–q)) 6q = 6-3q 9q = 6 This give us q = 2/3 F. Valognes - Game Theory - Chp 9 15 Example 2 Player 2 L (q) Player 1 R ( 1–q ) T (r) 6 , 0 0 , 6 B ( 1–r ) 3 , 2 6 , 0 Player 2’s expected payoff of playing L EU2(L, (r, 1–r)) = r ×0+(1–r)×2 =2- 2r Player 2’s expected payoff of playing R EU2(R, (r, 1–r)) = r×6+(1–r)×0 = 6r Player 2 is indifferent between playing L and R EU2(L, (r, 1–r)) = EU2(R, (r, 1–r)) 2- 2r = 6r 8r = 2 This gives us r = ¼ ( (1/4, 3/4), (2/3, 1/3) ) is a mixed strategy Nash equilibrium. F. Valognes - Game Theory - Chp 9 16 Example 3:Market entry game Two firms, Firm 1 and Firm 2, must decide whether to put one of their restaurants in a shopping mall simultaneously. Each has two strategies: Enter, Not Enter If either firm plays “Not Enter”, it earns 0 profit If one plays “Enter” and the other plays “Not Enter” then the firm plays “Enter” earns $500K If both plays “Enter” then both lose $100K because the demand is limited F. Valognes - Game Theory - Chp 9 17 Example 3:Market entry game Firm 2 Enter ( q ) Firm 1 Enter ( r ) Not Enter ( 1–r ) -100 , 0 , Not Enter ( 1–q ) -100 500 500 , 0 0 , 0 How many Nash equilibria can you find? Two pure strategy Nash equilibrium (Not Enter, Enter) and (Enter, Not Enter) One mixed strategy Nash equilibrium ((5/6, 1/6), (5/6, 1/6)) That is r=5/6 and q=5/6 F. Valognes - Game Theory - Chp 9 18 Example 4 Player 2 L (q) Player 1 R ( 1–q ) T (r) 1 , 1 1 , 2 B ( 1–r ) 2 , 3 0 , 1 How many Nash equilibria can you find? Two pure strategy Nash equilibrium (B, L) and (T, R) One mixed strategy Nash equilibrium ((2/3, 1/3), (1/2, 1/2)) That is r=2/3 and q=1/2 F. Valognes - Game Theory - Chp 9 19 Summary Mixed strategies Mixed Nash equilibrium Find mixed Nash equilibrium Next time 2-player game each with a finite number of strategies Reading lists Chapter 1.3 of Gibbons and Cha 4.3 of Osborne F. Valognes - Game Theory - Chp 9 20
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