Simultaneous-Move Games with Mixed Strategies

1. Simultaneous-Move Games with Mixed Strategies Zero-sum Games

2. Mixed Strategy • Random choice of the pure strategies • Pure strategy of probability distribution • Suppose Player one has 2 strategies, A and B. Let p in [0,1] be the probability for Player one to play A, then 1-p is the probability to play B. • When p=1, one plays A purely and when p=0, one plays B purely.

3. Expected payoff • Suppose E plays DL with p=0.75, when N plays DL, E’s expected payoff is 0.75x50+0.25x90=60. • Suppose N also plays mixed strategies with prob. q=0.2 for playing DL. Then E’s expected payoff is • 0.2x50+0.8x80=74 for playing DL • 0.2x90+0.8x20=34 for CC. • 0.75x74+0.25x34=63.66 for mixed with p=0.75

4. For any p, E’s expected payoff is • 50p+90(1-p)=90-40p when N plays DL • 80p+20(1-p)=20+60p when N plays CC • Note 90-40p>20+60p when p<0.7

5. Minimax Method E’s payoff 90 N’s DL N’s CC 20 p 0.7

6. If N mixes her strategies, E’s payoff is between two lines. • For p<=0.7, N playing CC purely will minimize E’s payoff (Note E can guarantee a payoff equal to 20+60p) • For p>=0.7, N playing DL purely will minimize E’s payoff (Note E can guarantee a payoff equal to 90-40p) • E can choose p=0.7 to maximin to get 62 no matter N picks DL or CC

7. N’s payoff 80 E’s CC, N’s payoff=80-70q E’s DL, N’s payoff=20+30q 20 q 0.6

8. Maximin strategy (p, q)=(0.7,0.6) • With N playing q=0.6, E’s expected payoff is 50x.6+80x.4=62 for playing DL purely 90x.6+20x.4=62 for playing CC purely • With E playing p=0.7, N’s expected payoff is 50x.7+10x.3=38 for playing DL purely 20x.7+80x.3=38 for playing CC purely

9. N.E. in mixed strategy • Theorem If a player would mix two or more strategies as the N.E strategy, the expected payoffs from playing those strategies purely (given opponents’ equilibrium strategies) should be the same, as the equilibrium payoff under the mixed strategy.

11. For Evert 50q+80(1-q)>90q+20(1-q) if q<0.6 Pure DL (p=1) when q<0.6 50q+80(1-q)<90q+20(1-q) if q>0.6 Pure CC (p=0) when q>0.6 50q+80(1-q)=90q+20(1-q) if q=0.6 Any mix (p=0~1) when q=0.6

12. For Navratilova 50p+10(1-p)>20p+80(1-p) if p>0.7 Pure DL (q=1) when p>0.7 50p+10(1-p)<20p+80(1-p) if p<0.7 Pure CC (q=0) when p<0.7 50p+10(1-p)=20p+80(1-p) if p=0.7 Any mix (q=0~1) when p=0.7

13. q 1 N.E. (0.7, 0.6) 0.6 E’s best response p 0 1 N’s best response 0.7

14. Follow the theorem, E would mix if 50q+80(1-q)=90q+20(1-q) or q=0.6, so that she’s indifferent between DL and CC (both 62). And N would mix if p=0.7. • Opponent’s indifference property. • N.E. as a system of beliefs. • Consider mixed strategy, then N.E. will be the same as minimax method.

15. When one has 3 strategies

16. E will mix the three if there’s a q such that 50q+80(1-q)= 90q+20(1-q)= 70q+60(1-q). However the answer is NO. It means E will only mix the two out of the three.

17. Case1: E mixes DL & CC only. From previous argument p=0.7 and q=0.6, and payoffs are 62 for E, 38 for N. When playing Lob purely with q=0.6, expected payoff=66, thus E will deviate. (0.7, 0.3, 0) cannot be E’s equilibrium strategy.

18. Case2: E mixes DL & Lob only. • E will mix the two when 50q+80(1-q)=70q+60(1-q) or q=0.5. For N to mix, when 50p+30(1-p)=20p+40(1-p), or p=0.25. • Payoffs are 65 and 35 respectively When playing CC purely with q=0.5, expected payoff=55, thus E will NOT deviate. • [(0.25, 0, 0.75), (0.5, 0.5)] is a N.E.

19. Case3: E mixes CC & Lob only. • E will mix the two when 90q+20(1-q)=70q+60(1-q) or q=2/3. For N to mix, when 10p+30(1-p)=80p+40(1-p), never. Actually N will play CC purely since CC dominates DL when E mixes only CC & Lob. • In equilibrium, E will not just mix CC & Lob.

20. Homework • 1, Question 3 of Exercise • 2. Find the N.E in the following strategic form game