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Simultaneous- Move Games with Mixed Strategies Zero-sum Games

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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.

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Expected payoff Suppose E plays DL with p=0.75, when N plays DL, E ’ s expected payoff is 0.75x x90=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. 0.75x x34=63.66 for mixed with p=0.75 Navratilova DLCC Evert DL50, 5080, 20 CC90, 1020, 80

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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

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Minimax Method p E ’ s payoff N ’ s DL N ’ s CC

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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

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

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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

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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.

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DLCCq-mix DL50, 5080, 20 50q+80(1-q), 50q+20(1-q) CC90, 1020, 80 90q+20(1-q), 10q+80(1-q) p-mix 50p+90(1-p), 50p+10(1-p) 80p+20(1-p), 20p+80(1-p)

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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) 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

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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

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

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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.

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When one has 3 strategies DLCCq-mix DL q+80(1 -q) CC q+20(1 -q) Lob q+60(1 -q) p-mix 50p1+90p 2+70(1- p1-p2) 80p1+20p 2+60(1- p1-p2)

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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.

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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.

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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.

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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.

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Homework 1, Question 3 of Exercise 2. Find the N.E in the following strategic form game LMR U1, 33, 41, 0 D0, 52, 23, 6

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