LECTURE 2 MIXED STRATEGY GAME May 19, 2003 LECTURE 2 MIXED STRATEGY GAME Notes modified from Yongqin Wang @ Fudan University

Matching pennies Hence, NO Nash equilibrium -1 , 1 1 , -1 Player 1 Player 2 Tail Head -1 , 1 1 , -1 Head is Player 1’s best response to Player 2’s strategy Tail Tail is Player 2’s best response to Player 1’s strategy Tail Tail is Player 1’s best response to Player 2’s strategy Head Head is Player 2’s best response to Player 1’s strategy Head Hence, NO Nash equilibrium

Solving matching pennies Player 2 Head Tail Player 1 -1 , 1 1 , -1 q 1-q r 1-r Randomize your strategies Player 1 chooses Head and Tail with probabilities r and 1-r, respectively. Player 2 chooses Head and Tail with probabilities q and 1-q, respectively. Mixed Strategy: Specifies that an actual move be chosen randomly from the set of pure strategies with some specific probabilities.

Mixed strategy A mixed strategy of a player is a probability distribution over player’s (pure) strategies. A mixed strategy for Chris is a probability distribution (p, 1-p), where p is the probability of playing Opera, and 1-p is that probability of playing Prize Fight. If p=1 then Chris actually plays Opera. If p=0 then Chris actually plays Prize Fight. Battle of sexes Pat Opera Prize Fight Chris Opera (p) 2 , 1 0 , 0 Prize Fight (1-p) 0 , 0 1 , 2

Solving matching pennies Player 2 Head Tail Player 1 -1 , 1 1 , -1 Expected payoffs r 1-2q 1-r 2q-1 q 1-q Player 1’s expected payoffs If Player 1 chooses Head, -q+(1-q)=1-2q If Player 1 chooses Tail, q-(1-q)=2q-1

Solving matching pennies Player 2 Head Tail Player 1 -1 , 1 1 , -1 Expected payoffs r 1-2q 1-r 2q-1 q 1-q 1 q r 1/2 Player 1’s best response B1(q): For q<0.5, Head (r=1) For q>0.5, Tail (r=0) For q=0.5, indifferent (0r1)

Solving matching pennies Player 2 Head Tail Player 1 -1 , 1 1 , -1 Expected payoffs r 1-2q 1-r 2q-1 q 1-q Expected payoffs 2r-1 1-2r Player 2’s expected payoffs If Player 2 chooses Head, r-(1-r)=2r-1 If Player 2 chooses Tail, -r+(1-r)=1-2r

Solving matching pennies Player 2 Head Tail Player 1 -1 , 1 1 , -1 1-2q 2q-1 Expected payoffs r 1-r q 1-q Expected payoffs 2r-1 1-2r 1 q r 1/2 Player 2’s best response B2(r): For r<0.5, Tail (q=0) For r>0.5, Head (q=1) For r=0.5, indifferent (0q1)

Solving matching pennies Player 2 Head Tail Player 1 -1 , 1 1 , -1 Player 1’s best response B1(q): For q<0.5, Head (r=1) For q>0.5, Tail (r=0) For q=0.5, indifferent (0r1) Player 2’s best response B2(r): For r<0.5, Tail (q=0) For r>0.5, Head (q=1) For r=0.5, indifferent (0q1) Check r = 0.5  B1(0.5) q = 0.5  B2(0.5) r 1-r q 1-q Mixed strategy Nash equilibrium 1 q r 1/2

Mixed strategy: example Matching pennies Player 1 has two pure strategies: H and T ( 1(H)=0.5, 1(T)=0.5 ) is a Mixed strategy. That is, player 1 plays H and T with probabilities 0.5 and 0.5, respectively. ( 1(H)=0.3, 1(T)=0.7 ) is another Mixed strategy. That is, player 1 plays H and T with probabilities 0.3 and 0.7, respectively.

Mixed strategy: example Player 2 L (0) C (1/3) R (2/3) Player 1 T (3/4) 0 , 2 3 , 3 1 , 1 M (0) 4 , 0 0 , 4 2 , 3 B (1/4) 3 , 4 5 , 1 0 , 7 Player 1: (3/4, 0, ¼) is a mixed strategy. That is, 1(T)=3/4, 1(M)=0 and 1(B)=1/4. Player 2: (0, 1/3, 2/3) is a mixed strategy. That is, 2(L)=0, 2(C)=1/3 and 2(R)=2/3.

Expected payoffs: 2 players each with two pure strategies s21 ( q ) s22 ( 1- q ) Player 1 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) Player 1 plays a mixed strategy (r, 1- r ). Player 2 plays a mixed strategy ( q, 1- q ). Player 1’s expected payoff of playing s11: EU1(s11, (q, 1-q))=q×u1(s11, s21)+(1-q)×u1(s11, s22) Player 1’s expected payoff of playing s12: EU1(s12, (q, 1-q))= q×u1(s12, s21)+(1-q)×u1(s12, s22) Player 1’s expected payoff from her mixed strategy: v1((r, 1-r), (q, 1-q))=rEU1(s11, (q, 1-q))+(1-r)EU1(s12, (q, 1-q))

Expected payoffs: 2 players each with two pure strategies s21 ( q ) s22 ( 1- q ) Player 1 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) Player 1 plays a mixed strategy (r, 1- r ). Player 2 plays a mixed strategy ( q, 1- q ). Player 2’s expected payoff of playing s21: EU2(s21, (r, 1-r))=r×u2(s11, s21)+(1-r)×u2(s12, s21) Player 2’s expected payoff of playing s22: EU2(s22, (r, 1-r))= r×u2(s11, s22)+(1-r)×u2(s12, s22) Player 2’s expected payoff from her mixed strategy: v2((r, 1-r),(q, 1-q))=qEU2(s21, (r, 1-r))+(1-q)EU2(s22, (r, 1-r))

Expected payoffs: example Player 2 H (0.3) T (0.7) Player 1 H (0.4) -1 , 1 1 , -1 T (0.6) Player 1: EU1(H, (0.3, 0.7)) = 0.3×(-1) + 0.7×1=0.4 EU1(T, (0.3, 0.7)) = 0.3×1 + 0.7×(-1)=-0.4 v1((0.4, 0.6), (0.3, 0.7))=0.40.4+0.6(-0.4)=-0.08 Player 2: EU2(H, (0.4, 0.6)) = 0.4×1+0.6×(-1) = -0.2 EU2(T, (0.4, 0.6)) = 0.4×(-1)+0.6×1 = 0.2 v2((0.4, 0.6), (0.3, 0.7))=0.3×(-0.2)+0.7×0.2=0.08

Expected payoffs: example Player 2 L (0) C (1/3) R (2/3) Player 1 T (3/4) 0 , 2 3 , 3 1 , 1 M (0) 4 , 0 0 , 4 2 , 3 B (1/4) 3 , 4 5 , 1 0 , 7 Mixed strategies: p1=( 3/4, 0, ¼ ); p2=( 0, 1/3, 2/3 ). Player 1: EU1(T, p2)=3(1/3)+1(2/3)=5/3, EU1(M, p2)=0(1/3)+2(2/3)=4/3 EU1(B, p2)=5(1/3)+0(2/3)=5/3. v1(p1, p2) = 5/3 Player 2: EU2(L, p1)=2(3/4)+4(1/4)=5/2, EU2(C, p1)=3(3/4)+3(1/4)=5/2, EU2(R, p1)=1(3/4)+7(1/4)=5/2. v1(p1, p2) = 5/2

Mixed strategy equilibrium A probability distribution for each player The distributions are mutual best responses to one another in the sense of expected payoffs It is a stochastic steady state

Mixed strategy vs. Correlated strategy

Mixed strategy equilibrium: 2-player each with two pure strategies s21 ( q ) s22 ( 1- q ) Player 1 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

Find mixed strategy equilibrium in 2-player each with two pure strategies Find the best response correspondence for player 1, given player 2’s mixed strategy Find the best response correspondence for player 2, given player 1’s mixed strategy Use the best response correspondences to determine mixed strategy Nash equilibria.

Employee Monitoring Employees can work hard or shirk Salary: $100K unless caught shirking Cost of effort: $50K Managers can monitor or not Value of employee output: $200K Profit if employee doesn’t work: $0 Cost of monitoring: $10K

Employee Monitoring Employee’s best response B1(q): 50 , 90 50 , 100 Shirk (r=0) if q<0.5 Work (r=1) if q>0.5 Any mixed strategy (0r1) if q=0.5 Expected payoffs 50 100(1-q) Expected payoffs 100r-10 200r-100 Manager Monitor ( q ) Not Monitor (1-q) Employee Work ( r ) 50 , 90 50 , 100 Shirk (1-r ) 0 , -10 100 , -100

Employee Monitoring Manager’s best response B2(r): 50 , 90 50 , 100 Monitor (q=1) if r<0.9 Not Monitor (q=0) if r>0.9 Any mixed strategy (0q1) if r=0.9 Expected payoffs 50 100(1-q) Expected payoffs 100r-10 200r-100 Manager Monitor ( q ) Not Monitor (1-q) Employee Work ( r ) 50 , 90 50 , 100 Shirk (1-r ) 0 , -10 100 , -100

Employee Monitoring r q Employee’s best response B1(q): Shirk (r=0) if q<0.5 Work (r=1) if q>0.5 Any mixed strategy (0r1) if q=0.5 Manager’s best response B2(r): Monitor (q=1) if r<0.9 Not Monitor (q=0) if r>0.9 Any mixed strategy (0q1) if r=0.9 Mixed strategy Nash equilibrium ((0.9,0.1),(0.5,0.5)) 1 q r 0.5 0.9

Battle of sexes Chris’ expected payoff of playing Opera: 2q Chris’ expected payoff of playing Prize Fight: 1-q Chris’ best response B1(q): Prize Fight (r=0) if q<1/3 Opera (r=1) if q>1/3 Any mixed strategy (0r1) if q=1/3 Pat Opera (q) Prize Fight (1-q) Chris Opera ( r ) 2 , 1 0 , 0 Prize Fight (1-r) 0 , 0 1 , 2

Battle of sexes Pat’s expected payoff of playing Opera: r Pat’s expected payoff of playing Prize Fight: 2(1-r) Pat’s best response B2(r): Prize Fight (q=0) if r<2/3 Opera (q=1) if r>2/3 Any mixed strategy (0q1) if r=2/3, Pat Opera (q) Prize Fight (1-q) Chris Opera ( r ) 2 , 1 0 , 0 Prize Fight (1-r) 0 , 0 1 , 2

Battle of sexes Chris’ best response B1(q): Pat’s best response B2(r): Prize Fight (r=0) if q<1/3 Opera (r=1) if q>1/3 Any mixed strategy (0r1) if q=1/3 Pat’s best response B2(r): Prize Fight (q=0) if r<2/3 Opera (q=1) if r>2/3 Any mixed strategy (0q1) if r=2/3 Three Nash equilibria: ((1, 0), (1, 0)) ((0, 1), (0, 1)) ((2/3, 1/3), (1/3, 2/3)) 1 q r 2/3 1/3

Battle of sexes The limitation of best-response approach Can only deal with simple games Comments of mixed strategies Randomization vNM expected utility Choice under uncertainty

2-player each with two strategies Lecture 6 May 27, 2003 2-player each with two strategies Player 2 s21 ( q ) s22 ( 1- q ) Player 1 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*))  EU1(s11, (q*, 1-q*)) v1((r*, 1-r*), (q*, 1-q*))  EU1(s12, (q*, 1-q*)) v2((r*, 1-r*), (q*, 1-q*))  EU2(s21, (r*, 1-r*)) v2((r*, 1-r*), (q*, 1-q*))  EU2(s22, (r*, 1-r*))

Expected payoffs: 2 players each with two pure strategies s21 ( q ) s22 ( 1- q ) Player 1 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) Player 1 plays a mixed strategy (r, 1- r ). Player 2 plays a mixed strategy ( q, 1- q ). Player 1’s expected payoff of playing s11: EU1(s11, (q, 1-q))=q×u1(s11, s21)+(1-q)×u1(s11, s22) Player 1’s expected payoff of playing s12: EU1(s12, (q, 1-q))= q×u1(s12, s21)+(1-q)×u1(s12, s22) Player 1’s expected payoff from her mixed strategy: v1((r, 1-r), (q, 1-q))=rEU1(s11, (q, 1-q))+(1-r)EU1(s12, (q, 1-q))

Expected payoffs: 2 players each with two pure strategies s21 ( q ) s22 ( 1- q ) Player 1 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) Player 1 plays a mixed strategy (r, 1- r ). Player 2 plays a mixed strategy ( q, 1- q ). Player 2’s expected payoff of playing s21: EU2(s21, (r, 1-r))=r×u2(s11, s21)+(1-r)×u2(s12, s21) Player 2’s expected payoff of playing s22: EU2(s22, (r, 1-r))= r×u2(s11, s22)+(1-r)×u2(s12, s22) Player 2’s expected payoff from her mixed strategy: v2((r, 1-r),(q, 1-q))=qEU2(s21, (r, 1-r))+(1-q)EU2(s22, (r, 1-r))

Theorem 1: illustration Lecture 6 May 27, 2003 Theorem 1: illustration Matching pennies Player 2 H (0.5) T (0.5) Player 1 -1 , 1 1 , -1 Player 1: EU1(H, (0.5, 0.5)) = 0.5×(-1) + 0.5×1=0 EU1(T, (0.5, 0.5)) = 0.5×1 + 0.5×(-1)=0 v1((0.5, 0.5), (0.5, 0.5))=0.50+0.50=0 Player 2: EU2(H, (0.5, 0.5)) = 0.5×1+0.5×(-1) =0 EU2(T, (0.5, 0.5)) = 0.5×(-1)+0.5×1 = 0 v2((0.5, 0.5), (0.5, 0.5))=0.5×0+0.5×0=0

Theorem 1: illustration Matching pennies Player 2 H (0.5) T (0.5) Player 1 -1 , 1 1 , -1 Player 1: v1((0.5, 0.5), (0.5, 0.5))  EU1(H, (0.5, 0.5)) v1((0.5, 0.5), (0.5, 0.5))  EU1(T, (0.5, 0.5)) Player 2: v2((0.5, 0.5), (0.5, 0.5))  EU2(H, (0.5, 0.5)) v2((0.5, 0.5), (0.5, 0.5))  EU2(T, (0.5, 0.5)) Hence, ((0.5, 0.5), (0.5, 0.5)) is a mixed strategy Nash equilibrium by Theorem 1.

Theorem 1: illustration Employee’s expected payoff of playing “work” EU1(Work, (0.5, 0.5)) = 0.5×50 + 0.5×50=50 Employee’s expected payoff of playing “shirk” EU1(Shirk, (0.5, 0.5)) = 0.5×0 + 0.5×100=50 Employee’s expected payoff of her mixed strategy v1((0.9, 0.1), (0.5, 0.5))=0.950+0.150=50 Employee Monitoring Manager Monitor (0.5) Not Monitor (0.5) Employee Work (0.9) 50 , 90 50 , 100 Shirk (0.1) 0 , -10 100 , -100

Theorem 1: illustration Manager’s expected payoff of playing “Monitor” EU2(Monitor, (0.9, 0.1)) = 0.9×90+0.1×(-10) =80 Manager’s expected payoff of playing “Not” EU2(Not, (0.9, 0.1)) = 0.9×100+0.1×(-100) = 80 Manager’s expected payoff of her mixed strategy v2((0.9, 0.1), (0.5, 0.5))=0.5×80+0.5×80=80 Employee Monitoring Manager Monitor (0.5) Not Monitor (0.5) Employee Work (0.9) 50 , 90 50 , 100 Shirk (0.1) 0 , -10 100 , -100

Theorem 1: illustration Employee v1((0.9, 0.1), (0.5, 0.5))  EU1(Work, (0.5, 0.5)) v1((0.9, 0.1), (0.5, 0.5))  EU1(Shirk, (0.5, 0.5)) Manager v2((0.9, 0.1), (0.5, 0.5))  EU2(Monitor, (0.9, 0.1)) v2((0.9, 0.1), (0.5, 0.5))  EU2(Not, (0.9, 0.1)) Hence, ((0.9, 0.1), (0.5, 0.5)) is a mixed strategy Nash equilibrium by Theorem 1. Employee Monitoring Manager Monitor (0.5) No Monitor (0.5) Employee Work (0.9) 50 , 90 50 , 100 Shirk (0.1) 0 , -10 100 , -100

Theorem 1: illustration Use Theorem 1 to check whether ((2/3, 1/3), (1/3, 2/3)) is a mixed strategy Nash equilibrium. Battle of sexes Pat Opera (1/3) Prize Fight (2/3) Chris Opera (2/3 ) 2 , 1 0 , 0 Prize Fight (1/3) 0 , 0 1 , 2

Mixed strategy equilibrium: 2-player each with two strategies s21 ( q ) s22 ( 1- q ) Player 1 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 strategies. Significance: it gives conditions for a mixed strategy NE in terms of each player’s expected payoffs only to her pure strategies

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*))

Use Theorem 2 to find mixed strategy Nash equilibrium: illustration Matching pennies Player 2 H ( q ) T ( 1–q ) Player 1 H ( r ) -1 , 1 1 , -1 T ( 1–r ) Player 1 is indifferent between playing Head and Tail. EU1(H, (q, 1–q)) = q×(-1) + (1–q)×1=1–2q EU1(T, (q, 1–q)) = q×1 + ×(1–q) (-1)=2q–1 EU1(H, (q, 1–q)) = EU1(T, (q, 1–q)) 1–2q = 2q–1 4q = 2 This give us q = 1/2

Use Theorem 2 to find mixed strategy Nash equilibrium: illustration Matching pennies Player 2 H ( q ) T ( 1–q ) Player 1 H ( r ) -1 , 1 1 , -1 T ( 1–r ) Player 2 is indifferent between playing Head and Tail. EU2(H, (r, 1–r)) = r ×1+(1–r)×(-1) =2r – 1 EU2(T, (r, 1–r)) = r×(-1)+(1–r)×1 = 1 – 2r EU2(H, (r, 1–r)) = EU2(T, (r, 1–r)) 2r – 1= 1 – 2r 4r = 2 This give us r = 1/2 Hence, ((0.5, 0.5), (0.5, 0.5)) is a mixed strategy Nash equilibrium by Theorem 2.

Use Theorem 2 to find mixed strategy Nash equilibrium: illustration Employee’s expected payoff of playing “work” EU1(Work, (q, 1–q)) = q×50 + (1–q)×50=50 Employee’s expected payoff of playing “shirk” EU1(Shirk, (q, 1–q)) = q×0 + (1–q)×100=100(1–q) Employee is indifferent between playing Work and Shirk. 50=100(1–q) q=1/2 Employee Monitoring Manager Monitor ( q ) Not Monitor (1–q ) Employee Work (r) 50 , 90 50 , 100 Shirk (1–r) 0 , -10 100 , -100

Use Theorem 2 to find mixed strategy Nash equilibrium: illustration Manager’s expected payoff of playing “Monitor” EU2(Monitor, (r, 1–r)) = r×90+(1–r)×(-10) =100r–10 Manager’s expected payoff of playing “Not” EU2(Not, (r, 1–r)) = r×100+(1–r)×(-100) =200r–100 Manager is indifferent between playing Monitor and Not 100r–10 =200r–100 implies that r=0.9. Hence, ((0.9, 0.1), (0.5, 0.5)) is a mixed strategy Nash equilibrium by Theorem 2. Employee Monitoring Manager Monitor ( q ) Not Monitor (1–q ) Employee Work (r) 50 , 90 50 , 100 Shirk (1–r) 0 , -10 100 , -100

Use Theorem 2 to find mixed strategy Nash equilibrium: illustration Use Theorem 2 to find a mixed Nash equilibrium Battle of sexes Pat Opera (q) Prize Fight (1-q) Chris Opera ( r ) 2 , 1 0 , 0 Prize Fight (1-r) 0 , 0 1 , 2

Use Theorem 2 to find mixed strategy Nash equilibrium: illustration Use Theorem 2 to find a mixed Nash equilibrium Example Player 2 L (q) R (1-q) Player 1 T ( r ) 6 , 4 2 , 6 B (1-r) 3 , 3 6 , 1

Mixed strategy Nash equilibrium 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

Use Theorem 2 to find mixed strategy Nash equilibrium: illustration Lecture 6 May 27, 2003 Use Theorem 2 to find mixed strategy Nash equilibrium: illustration 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 Battle of sexes Pat Opera (q) Prize Fight (1-q) Chris Opera ( r ) 2 , 1 0 , 0 Prize Fight (1-r) 0 , 0 1 , 2

Use Theorem 2 to find mixed strategy Nash equilibrium: illustration Lecture 6 May 27, 2003 Use Theorem 2 to find mixed strategy Nash equilibrium: illustration 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 Battle of sexes Pat Opera (q) Prize Fight (1-q) Chris Opera ( r ) 2 , 1 0 , 0 Prize Fight (1-r) 0 , 0 1 , 2

Use Theorem 2 to find mixed strategy Nash equilibrium: illustration Lecture 6 May 27, 2003 Use Theorem 2 to find mixed strategy Nash equilibrium: illustration 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. Battle of sexes Pat Opera (q) Prize Fight (1-q) Chris Opera ( r ) 2 , 1 0 , 0 Prize Fight (1-r) 0 , 0 1 , 2

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.

Example 1 Bruce’s expected payoff of playing Head Sheila H ( q ) T ( 1–q ) Bruce H ( r ) 1 , 4 4 , 1 T ( 1–r ) 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

Example 1 1 , 4 4 , 1 Sheila’s expected payoff of playing Head H ( q ) T ( 1–q ) Bruce H ( r ) 1 , 4 4 , 1 T ( 1–r ) 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.

Example 2 Player 1’s expected payoff of playing T L ( q ) R ( 1–q ) Player 1 T ( r ) 6 , 0 0 , 6 B ( 1–r ) 3 , 2 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

Example 2 6 , 0 0 , 6 3 , 2 Player 2’s expected payoff of playing L L ( q ) R ( 1–q ) Player 1 T ( r ) 6 , 0 0 , 6 B ( 1–r ) 3 , 2 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.

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

Example 3:Market entry game Firm 2 Enter ( q ) Not Enter ( 1–q ) Firm 1 Enter ( r ) -100 , -100 500 , 0 Not Enter ( 1–r ) 0 , 500 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

Example 4 How many Nash equilibria can you find? Player 2 L ( q ) R ( 1–q ) Player 1 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

Example: Rock, paper and scissors Each of two players simultaneously announces either Rock, or Paper, or Scissors. Paper beats (wraps) rock Rock beats (blunts) scissors Scissors beats (cuts) paper The player who names the winning object receives $1 from her opponent If both players name the same choice then no payment is made

Example: Rock, paper and scissors Player 2 Rock Paper Scissors Player 1 0 , 0 -1 , 1 1 , -1 Can you guess a mixed strategy Nash equilibrium?

Mixed strategy Nash equilibrium: 2-player each with two pure strategies s21 ( q ) s22 ( 1- q ) Player 1 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

2-player each with a finite number of pure strategies Set of players: {Player 1, Player 2} Sets of strategies: player 1: S1= { s11, s12, ..., s1J } player 2: S2= { s21, s22, ..., s2K } Payoff functions: player 1: u1(s1j, s2k) player 2: u2(s1j, s2k) for j = 1, 2, ..., J and k = 1, 2, ..., K

2-player each with a finite number of pure strategies s21 (p21) s22 (p22) ....... s2K (p2K) s11 (p11) u2(s11, s21) u1(s11, s21) u2(s11, s22) u1(s11, s22) u2(s11, s2K) u1(s11, s2K) s12 (p12) u2(s12, s21) u1(s12, s21) u2(s12, s22) u1(s12, s22) u2(s12, s2K) u1(s12, s2K) .... ...... s1J (p1J) u2(s1J, s21) u1(s1J, s21) u2(s1J, s22) u1(s1J, s22) u2(s1J, s2K) u1(s1J, s2K) Player 1 Player 1’s mixed strategy: p1=(p11, p12, ..., p1J ) Player 2’s mixed strategy: p2=(p21, p22, ..., p2K )

Expected payoffs: 2-player each with a finite number of pure strategies Player 1’s expected payoff of pure strategy s11: EU1(s11, p2)=p21×u1(s11, s21)+p22×u1(s11, s22)+...+p2k×u1(s11, s2k)+...+p2K×u1(s11, s2K) Player 1’s expected payoff of pure strategy s12: EU1(s12, p2)=p21×u1(s12, s21)+p22×u1(s12, s22)+...+p2k×u1(s12, s2k)+...+p2K×u1(s12, s2K) ......... Player 1’s expected payoff of pure strategy s1J: EU1(s1J, p2)=p21×u1(s1J, s21)+p22×u1(s1J, s22)+...+p2k×u1(s1J, s2k)+...+p2K×u1(s1J, s2K) Player 1’s expected payoff from her mixed strategy p1: v1(p1, p2)=p11EU1(s11, p2)+p12EU1(s12, p2)+...+p1jEU1(s1j, p2)+... +p1JEU1(s1J, p2)

Expected payoffs: 2-player each with a finite number of pure strategies Player 2’s expected payoff of pure strategy s21: EU2(s21, p1)=p11×u2(s11, s21)+p12×u2(s12, s21)+...+p1j×u2(s1j, s21)+...+p1J×u2(s1J, s21) Player 2’s expected payoff of pure strategy s22: EU2(s22, p1)=p11×u2(s11, s22)+p12×u2(s12, s22)+...+p1j×u2(s1j, s22)+...+p1J×u2(s1J, s22) ........... Player 2’s expected payoff of pure strategy s2K: EU2(s2K, p1)=p11×u2(s11, s2K)+p12×u2(s12, s2K)+...+p1j×u2(s1j, s2K)+...+p1J×u2(s1J, s2K) Player 2’s expected payoff from her mixed strategy p2: v2(p1, p2)=p21EU2(s21, p1)+p22EU2(s22, p1) +...+p2kEU2(s2k, p1)+.... +p2KEU2(s2K, p1)

Mixed strategy Nash equilibrium: 2-player each with a finite number of pure strategies A pair of mixed strategies (p1*, p2*), where p1*=(p11*, p12*, ..., p1J* ) p2*=(p21*, p22*, ..., p2K* ) is a mixed strategy Nash equilibrium if player 1’s mixed strategy p1* is a best response to player 2’s mixed strategy p2*, and p2* is a best response to p1*. Or, v1(p1*, p2*) v1(p1, p2*), for all player 1’s mixed strategy p1, and v2(p1*, p2*)  v2(p1*, p2), for all player 2’s mixed strategy p2. That is, given player 2’s mixed strategy p2*, player 1 cannot be better off if she deviates from p1*. Given player 1’s mixed strategy p1*, player 2 cannot be better off if she deviates from p2*.

2-player each with a finite number of pure strategies Theorem 3 (property of mixed Nash equilibrium) A pair of mixed strategies (p1*, p2*), where p1*=(p11*, p12*, ..., p1J* ) p2*=(p21*, p22*, ..., p2K* ) is a mixed strategy Nash equilibrium if and only if v1(p1*, p2*)  EU1(s1j, p2*), for j = 1, 2, ..., J and v2(p1*, p2*)  EU2(s2k, p1*), for k= 1, 2, ..., K

2-player each with a finite number of pure strategies Theorem 4 A pair of mixed strategies (p1*, p2*), where p1*=(p11*, p12*, ..., p1J* ) p2*=(p21*, p22*, ..., p2K* ) is a mixed strategy Nash equilibrium if and only if they satisfies the following conditions: player 1: for any m and n, if p1m*>0 and p1n*>0 then EU1(s1m, p2*) = EU1(s1n, p2*); if p1m*>0 and p1n*=0 then EU1(s1m, p2*)  EU1(s1n, p2*) player 2: for any i and k, if p2i*>0 and p2k*>0 then EU2(s2i, p1*) = EU2(s2k, p1*); if p2i*>0 and p2k*=0 then EU2(s2i, p1*)  EU2(s2k, p1*)

2-player each with a finite number of pure strategies What does Theorem 4 tell us? A pair of mixed strategies (p1*, p2*), where p1*=(p11*, p12*, ..., p1J* ), p2*=(p21*, p22*, ..., p2K* ) is a mixed strategy Nash equilibrium if and only if they satisfies the following conditions: Given player 2’s p2*, player 1’s expected payoff of every pure strategy to which she assigns positive probability is the same, and player 1’s expected payoff of any pure strategy to which she assigns positive probability is not less than the expected payoff of any pure strategy to which she assigns zero probability. Given player 1’s p1*, player 2’s expected payoff of every pure strategy to which she assigns positive probability is the same, and player 2’s expected payoff of any pure strategy to which she assigns positive probability is not less than the expected payoff of any pure strategy to which she assigns zero probability.

2-player each with a finite number of pure strategies Theorem 4 implies that we have a mixed strategy Nash equilibrium in the following situation Given player 2’s mixed strategy, Player 1 is indifferent among the pure strategies to which she assigns positive probabilities. The expected payoff of any pure strategy she assigns positive probability is not less than the expected payoff of any pure strategy she assigns zero probability. Given player 1’s mixed strategy, Player 2 is indifferent among the pure strategies to which she assigns positive probabilities. The expected payoff of any pure strategy she assigns positive probability is not less than the expected payoff of any pure strategy she assigns zero probability.

Theorem 4: illustration Player 2 L (0) C (1/3) R (2/3) Player 1 T (3/4) 0 , 2 3 , 3 1 , 1 M (0) 4 , 0 0 , 4 2 , 3 B (1/4) 3 , 4 5 , 1 0 , 7 Check whether ((3/4, 0, 1/4), (0, 1/3, 2/3)) is a mixed strategy Nash equilibrium Player 1: EU1(T, p2) = 00+3(1/3)+1(2/3)=5/3, EU1(M, p2) = 40+0(1/3)+2(2/3)=4/3 EU1(B, p2) = 30+5(1/3)+0(2/3)=5/3. Hence, EU1(T, p2) = EU1(B, p2) > EU1(M, p2)

Theorem 4: illustration Player 2 L (0) C (1/3) R (2/3) Player 1 T (3/4) 0 , 2 3 , 3 1 , 1 M (0) 4 , 0 0 , 4 2 , 3 B (1/4) 3 , 4 5 , 1 0 , 7 Player 2: EU2(L, p1)=2(3/4) + 00 + 4(1/4)=5/2, EU2(C, p1)=3(3/4) + 40 + 1(1/4)=5/2, EU2(R, p1)=1(3/4) + 30 + 7(1/4)=5/2. Hence, EU2(C, p1)=EU2(R, p1)EU2(L, p1) Therefore, ((3/4, 0, 1/4), (0, 1/3, 2/3)) is a mixed strategy Nash equilibrium by Theorem 4.

Example: Rock, paper and scissors Player 2 Rock (p21) Paper (p22) Scissors (p23) Player 1 Rock (p11) 0 , 0 -1 , 1 1 , -1 Paper (p12) Scissors (p13) Check whether there is a mixed strategy Nash equilibrium in which p11>0, p12>0, p13>0, p21>0, p22>0, p23>0.

Example: Rock, paper and scissors Player 2 Rock (p21) Paper (p22) Scissors (p23) Player 1 Rock (p11) 0 , 0 -1 , 1 1 , -1 Paper (p12) Scissors (p13) If each player assigns positive probability to every of her pure strategy, then by Theorem 4, each player is indifferent among her three pure strategies.

Example: Rock, paper and scissors Player 2 Rock (p21) Paper (p22) Scissors (p23) Player 1 Rock (p11) 0 , 0 -1 , 1 1 , -1 Paper (p12) Scissors (p13) Player 1 is indifferent among her three pure strategies: EU1(Rock, p2) = 0p21+(-1) p22+1 p23 EU1(Paper, p2) = 1 p21+0 p22+(-1) p23 EU1(Scissors, p2) = (-1) p21+1 p22+0 p23 EU1(Rock, p2)= EU1(Paper, p2)= EU1(Scissors, p2) Together with p21+ p22+ p23=1, we have three equations and three unknowns.

Example: Rock, paper and scissors Player 2 Rock (p21) Paper (p22) Scissors (p23) Player 1 Rock (p11) 0 , 0 -1 , 1 1 , -1 Paper (p12) Scissors (p13) 0p21+(-1) p22+1 p23= 1 p21+0 p22+(-1) p23 0p21+(-1) p22+1 p23 = (-1) p21+1 p22+0 p23 p21+ p22+ p23=1 The solution is p21= p22= p23=1/3

Example: Rock, paper and scissors Player 2 Rock (p21) Paper (p22) Scissors (p23) Player 1 Rock (p11) 0 , 0 -1 , 1 1 , -1 Paper (p12) Scissors (p13) Player 2 is indifferent among her three pure strategies: EU2(Rock, p1)=0p11+(-1) p12+1 p13 EU2(Paper, p1)=1 p11+0 p12+(-1) p13 EU2(Scissors, p1)=(-1) p11+1 p12+0 p13 EU2(Rock, p1)= EU2(Paper, p1)=EU2(Scissors, p1) Together with p11+ p12+ p13=1, we have three equations and three unknowns.

Example: Rock, paper and scissors Player 2 Rock (p21) Paper (p22) Scissors (p23) Player 1 Rock (p11) 0 , 0 -1 , 1 1 , -1 Paper (p12) Scissors (p13) 0p11+(-1) p12+1 p13=1 p11+0 p12+(-1) p13 0p11+(-1) p12+1 p13=(-1) p11+1 p12+0 p13 p11+ p12+ p13=1 The solution is p11= p12= p13=1/3

Example: Rock, paper and scissors Player 2 Rock (1/3) Paper (1/3) Scissors (1/3) Player 1 0 , 0 -1 , 1 1 , -1 Player 1: EU1(Rock, p2) = 0(1/3)+(-1)(1/3)+1(1/3)=0 EU1(Paper, p2) = 1(1/3)+0(1/3)+(-1)(1/3)=0 EU1(Scissors, p2) = (-1)(1/3)+1(1/3)+0(1/3)=0 Player 2: EU2(Rock, p1)=0(1/3)+(-1)(1/3)+1(1/3)=0 EU2(Paper, p1)=1(1/3)+0(1/3)+(-1)(1/3)=0 EU2(Scissors, p1)=(-1)(1/3)+1(1/3)+0(1/3)=0 Therefore, (p1=(1/3, 1/3, 1/3), p2=(1/3, 1/3, 1/3)) is a mixed strategy Nash equilibrium by Theorem 4.

Example: Rock, paper and scissors Player 2 Rock (p21) Paper (p22) Scissors (p23) Player 1 Rock (p11) 0 , 0 -1 , 1 1 , -1 Paper (p12) Scissors (p13) Check whether there is a mixed strategy Nash equilibrium in which one of p11, p12, p13 is positive, and at least two of p21, p22, p23 are positive. The answer is No.

Example: Rock, paper and scissors Player 2 Rock (p21) Paper (p22) Scissors (p23) Player 1 Rock (p11) 0 , 0 -1 , 1 1 , -1 Paper (p12) Scissors (p13) Check whether there is a mixed strategy Nash equilibrium in which two of p11, p12, p13 is positive, and at least two of p21, p22, p23 are positive. The answer is No.

Example: Rock, paper and scissors Player 2 Rock (p21) Paper (p22) Scissors (p23) Player 1 Rock (p11) 0 , 0 -1 , 1 1 , -1 Paper (p12) Scissors (p13) Therefore, (p1=(1/3, 1/3, 1/3), p2=(1/3, 1/3, 1/3)) is the unique mixed strategy Nash equilibrium by Theorem 4.

Lecture 6 May 27, 2003 Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 T (p11) 2 , 2 0 , 3 1 , 3 B (p12) 3 , 2 1 , 1 0 , 2 We first consider pure-strategy Nash equilibria. How many can you find? In order to find all Nash equilibria, we need to consider 15 more cases by Theorem 4. We first consider complicated cases. Some cases are very easy.

Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 Lecture 6 May 27, 2003 Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 T (p11) 2 , 2 0 , 3 1 , 3 B (p12) 3 , 2 1 , 1 0 , 2 Case 1: check whether there is a mixed strategy in which p11>0, p12>0, p21>0, p22>0, p23>0 By theorem 4, we should have 2p11+2 p12= 3p11+1 p12 = 3p11+2 p12 and p11+p12=1. We should have 2p21+0 p22+1 p23 =3p21+1 p22+0 p23 and p21+ p22+ p23 = 1 Solve these equations. If we can get a solution that satisfies p11>0, p12>0, p21>0, p22>0, p23>0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11>0, p12>0, p21>0, p22>0, p23>0, then we do not have such a mixed strategy Nash equilibrium.

Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 Lecture 6 May 27, 2003 Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 T (p11) 2 , 2 0 , 3 1 , 3 B (p12) 3 , 2 1 , 1 0 , 2 Case 2: check whether there is a mixed strategy in which p11>0, p12>0, p21>0, p22>0, p23=0 By theorem 4, we should have 2p11+2 p12= 3p11+1 p12  3p11+2 p12 and p11+p12=1. We should have 2p21+0 p22+1 p23 =3p21+1 p22+0 p23 and p21+ p22+ p23 = 1 Solve these. If we can get a solution that satisfies p11>0, p12>0, p21>0, p22>0, p23=0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11>0, p12>0, p21>0, p22>0, p23=0, then we do not have such a mixed strategy Nash equilibrium.

Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 Lecture 6 May 27, 2003 Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 T (p11) 2 , 2 0 , 3 1 , 3 B (p12) 3 , 2 1 , 1 0 , 2 Case 3: check whether there is a mixed strategy in which p11>0, p12>0, p21>0, p22=0, p23>0 By theorem 4, we should have 2p11+2 p12 = 3p11+2 p12  3p11+1 p12 and p11+p12=1. We should have 2p21+0 p22+1 p23 =3p21+1 p22+0 p23 and p21+ p22+ p23 = 1 Solve these. If we can get a solution that satisfies p11>0, p12>0, p21>0, p22=0, p23>0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11>0, p12>0, p21>0, p22=0, p23>0, then we do not have such a mixed strategy Nash equilibrium.

Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 Lecture 6 May 27, 2003 Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 T (p11) 2 , 2 0 , 3 1 , 3 B (p12) 3 , 2 1 , 1 0 , 2 Case 4: check whether there is a mixed strategy in which p11>0, p12>0, p21=0, p22>0, p23>0 By theorem 4, we should have 2p11+2 p12 3p11+1 p12 = 3p11+2 p12 and p11+p12=1. We should have 2p21+0 p22+1 p23 =3p21+1 p22+0 p23 and p21+ p22+ p23 = 1 Solve these. If we can get a solution that satisfies p11>0, p12>0, p21=0, p22>0, p23>0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11>0, p12>0, p21=0, p22>0, p23>0, then we do not have such a mixed strategy Nash equilibrium.

Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 Lecture 6 May 27, 2003 Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 T (p11) 2 , 2 0 , 3 1 , 3 B (p12) 3 , 2 1 , 1 0 , 2 Case 5: check whether there is a mixed strategy in which p11>0, p12>0, p21>0, p22=0, p23=0 (Note this implies p21=1) By theorem 4, we should have 2p11+2 p12 3p11+1 p12 and 2p11+2 p12 3p11+2 p12 and p11+p12=1. We should have 2p21+0 p22+1 p23 =3p21+1 p22+0 p23 and p21+ p22+ p23 = 1 Solve these. If we can get a solution that satisfies p11>0, p12>0, p21>0, p22=0, p23=0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11>0, p12>0, p21>0, p22=0, p23=0, then we do not have such a mixed strategy Nash equilibrium.

Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 Lecture 6 May 27, 2003 Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 T (p11) 2 , 2 0 , 3 1 , 3 B (p12) 3 , 2 1 , 1 0 , 2 Case 6: check whether there is a mixed strategy in which p11>0, p12>0, p21=0, p22>0, p23=0 (Note this implies p22=1) By theorem 4, we should have 3p11+1 p12  2p11+2 p12 and 3p11+1 p12  3p11+2 p12 and p11+p12=1. We should have 2p21+0 p22+1 p23 =3p21+1 p22+0 p23 and p21+ p22+ p23 = 1 Solve these. If we can get a solution that satisfies p11>0, p12>0, p21=0, p22>0, p23=0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11>0, p12>0, p21=0, p22>0, p23=0, then we do not have such a mixed strategy Nash equilibrium.

Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 Lecture 6 May 27, 2003 Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 T (p11) 2 , 2 0 , 3 1 , 3 B (p12) 3 , 2 1 , 1 0 , 2 Case 7: check whether there is a mixed strategy in which p11>0, p12>0, p21=0, p22=0, p23>0 (Note this implies p23=1) By theorem 4, we should have 3p11+2 p12  2p11+2 p12 and 3p11+2 p12  3p11+1 p12 and p11+p12=1. We should have 2p21+0 p22+1 p23 =3p21+1 p22+0 p23 and p21+ p22+ p23 = 1 Solve these. If we can get a solution that satisfies p11>0, p12>0, p21=0, p22=0, p23>0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11>0, p12>0, p21=0, p22=0, p23>0, then we do not have such a mixed strategy Nash equilibrium.

Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 Lecture 6 May 27, 2003 Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 T (p11) 2 , 2 0 , 3 1 , 3 B (p12) 3 , 2 1 , 1 0 , 2 Case 8: check whether there is a mixed strategy in which p11>0, p12=0, p21>0, p22>0, p23>0 (Note this implies p11=1) By theorem 4, we should have 2p11+2 p12= 3p11+1 p12 = 3p11+2 p12 and p11+p12=1. We should have 2p21+0 p22+1 p23 3p21+1 p22+0 p23 and p21+ p22+ p23 = 1 Solve these. If we can get a solution that satisfies p11>0, p12=0, p21>0, p22>0, p23>0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11>0, p12=0, p21>0, p22>0, p23>0, then we do not have such a mixed strategy Nash equilibrium.

Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 Lecture 6 May 27, 2003 Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 T (p11) 2 , 2 0 , 3 1 , 3 B (p12) 3 , 2 1 , 1 0 , 2 Case 9: check whether there is a mixed strategy in which p11>0, p12=0, p21>0, p22>0, p23=0 (Note this implies p11=1) By theorem 4, we should have 2p11+2 p12= 3p11+1 p12  3p11+2 p12 and p11+p12=1. We should have 2p21+0 p22+1 p23 3p21+1 p22+0 p23 and p21+ p22+ p23 = 1 Solve these. If we can get a solution that satisfies p11>0, p12=0, p21>0, p22>0, p23=0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11>0, p12=0, p21>0, p22>0, p23=0, then we do not have such a mixed strategy Nash equilibrium.

Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 Lecture 6 May 27, 2003 Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 T (p11) 2 , 2 0 , 3 1 , 3 B (p12) 3 , 2 1 , 1 0 , 2 Case 10: check whether there is a mixed strategy in which p11>0, p12=0, p21>0, p22=0, p23>0 (Note this implies p11=1) By theorem 4, we should have 2p11+2 p12 3p11+1 p12  3p11+2 p12 and p11+p12=1. We should have 2p21+0 p22+1 p23 3p21+1 p22+0 p23 and p21+ p22+ p23 = 1 Solve these. If we can get a solution that satisfies p11>0, p12=0, p21>0, p22=0, p23>0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11>0, p12=0, p21>0, p22=0, p23>0, then we do not have such a mixed strategy Nash equilibrium.

Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 Lecture 6 May 27, 2003 Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 T (p11) 2 , 2 0 , 3 1 , 3 B (p12) 3 , 2 1 , 1 0 , 2 Case 11: check whether there is a mixed strategy in which p11>0, p12=0, p21=0, p22>0, p23>0 (Note this implies p11=1) By theorem 4, we should have 2p11+2 p12 3p11+1 p12 = 3p11+2 p12 and p11+p12=1. We should have 2p21+0 p22+1 p23 3p21+1 p22+0 p23 and p21+ p22+ p23 = 1 Solve these. If we can get a solution that satisfies p11>0, p12=0, p21=0, p22>0, p23>0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11>0, p12=0, p21=0, p22>0, p23>0, then we do not have such a mixed strategy Nash equilibrium.

Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 Lecture 6 May 27, 2003 Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 T (p11) 2 , 2 0 , 3 1 , 3 B (p12) 3 , 2 1 , 1 0 , 2 Case 12: check whether there is a mixed strategy in which p11=0, p12>0, p21>0, p22>0, p23>0 (Note this implies p12=1) By theorem 4, we should have 2p11+2 p12= 3p11+1 p12 = 3p11+2 p12 and p11+p12=1. We should have 2p21+0 p22+1 p23 3p21+1 p22+0 p23 and p21+ p22+ p23 = 1 Solve these. If we can get a solution that satisfies p11=0, p12>0, p21>0, p22>0, p23>0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11=0, p12>0, p21>0, p22>0, p23>0, then we do not have such a mixed strategy Nash equilibrium.

Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 Lecture 6 May 27, 2003 Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 T (p11) 2 , 2 0 , 3 1 , 3 B (p12) 3 , 2 1 , 1 0 , 2 Case 13: check whether there is a mixed strategy in which p11=0, p12>0, p21>0, p22>0, p23=0 (Note this implies p12=1) By theorem 4, we should have 2p11+2 p12= 3p11+1 p12  3p11+2 p12 and p11+p12=1. We should have 2p21+0 p22+1 p23 3p21+1 p22+0 p23 and p21+ p22+ p23 = 1 Solve these. If we can get a solution that satisfies p11=0, p12>0, p21>0, p22>0, p23=0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11=0, p12>0, p21>0, p22>0, p23=0, then we do not have such a mixed strategy Nash equilibrium.

Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 Lecture 6 May 27, 2003 Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 T (p11) 2 , 2 0 , 3 1 , 3 B (p12) 3 , 2 1 , 1 0 , 2 Case 14: check whether there is a mixed strategy in which p11=0, p12>0, p21>0, p22=0, p23>0 (Note this implies p12=1) By theorem 4, we should have 2p11+2 p12 3p11+1 p12  3p11+2 p12 and p11+p12=1. We should have 2p21+0 p22+1 p23  3p21+1 p22+0 p23 and p21+ p22+ p23 = 1 Solve these. If we can get a solution that satisfies p11=0, p12>0, p21>0, p22=0, p23>0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11=0, p12>0, p21>0, p22=0, p23>0, then we do not have such a mixed strategy Nash equilibrium.

Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 Lecture 6 May 27, 2003 Exercise 138.1 of Osborne Player 2 L (p21) M (p22) R (p23) Player 1 T (p11) 2 , 2 0 , 3 1 , 3 B (p12) 3 , 2 1 , 1 0 , 2 Case 15: check whether there is a mixed strategy in which p11=0, p12>0, p21=0, p22>0, p23>0 (Note this implies p12=1) By theorem 4, we should have 2p11+2 p12 3p11+1 p12 = 3p11+2 p12 and p11+p12=1. We should have 2p21+0 p22+1 p23  3p21+1 p22+0 p23 and p21+ p22+ p23 = 1 Solve these. If we can get a solution that satisfies p11=0, p12>0, p21=0, p22>0, p23>0 then we have a mixed strategy Nash equilibrium. Otherwise, if we can not find a solution or we find a solution that does not satisfies p11=0, p12>0, p21=0, p22>0, p23>0, then we do not have such a mixed strategy Nash equilibrium.