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Eponine Lupo

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Questions from last time 3 player games Games larger than 2x2—rock, paper, scissors Review/explain Nash Equilibrium Nash Equilibrium in R Instability of NE—move towards pure strategy Prisoner’s Dilemma, Battle of the Sexes, 3 rd Game Application to Life

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14, 24, 328, 30, 27 30, 16, 2413, 12, L R R L 16, 24, 3030, 16, 24 30, 23,1414, 24, L R R L LR 3 Strategy Profile: {R,L,L} is the Solution to this Game

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0, 0-1, 11, -1 0, 0-1, 1 1, -10, 0 Player 2 R P S Player 1 R P S No pure strategy NE Only mixed NE is {(1/3,1/3,1/3),(1/3,1/3,1/3)}

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“A strategy profile is a Nash Equilibrium if and only if each player’s prescribed strategy is a best response to the strategies of others” Equilibrium that is reached even if it is not the best joint outcome 4, 60, 44, 4 5, 30, 01, 7 1, 13, 52, 3 Player 2 L C R Player 1 U M D Strategy Profile: {D,C} is the Nash Equilibrium **There is no incentive for either player to deviate from this strategy profile

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Sometimes there is NO pure Nash Equilibrium, or there is more than one pure Nash Equilibrium In these cases, use Mixed Strategy Nash Equilibriums to solve the games Take for example a modified game of Rock, Paper, Scissors where player 1 cannot ever play “Scissors” What now is the Nash Equilibrium? Put another way, how are Player 1 and Player 2 going to play?

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Once Player 1’s strategy of S is taken away, Player 2’s strategy R is iteratively dominated by strategy P. 0, 0-1, 11, -1 0, 0-1, 1 Player 2 R P S Player 1 R P

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Now the game has been cut down from a 3x3 to 2x2 game There are still no pure strategy NE From here we can determine the mixed strategy NE -1, 11, -1 0, 0-1, 1 Player 2 q 1-q P S Player 1 p R 1-p P Player 1 wants to have a mixed strategy (p, 1-p) such that Player 2 has no advantage playing either pure strategy P or S. u 2 ((p, 1-p),P)=u 2 ((p, 1-p),S) 1p+0(1-p) = (-1)p+1(1-p) 1p = -2p+1 3p = 1 p=1/3 S 1 = (1/3, 2/3)

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-1, 11, -1 0, 0-1, 1 Player 1 p R 1-p P Likewise, Player 2 wants to have a mixed strategy (q, 1-q) such that Player 1 has no advantage playing either pure strategy R or P. u 1 (R,(q, 1-q))=u 1 (P,(q, 1-q)) -1q+1(1-q) = 0q+(-1)(1-q) -2q+1 = q-1 3q = 2 q=2/3 S 2 = (2/3, 1/3) Player 2 q 1-q P S

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Therefore the mixed strategy: Player 1: (1/3Rock, 2/3Paper) Player 2: (2/3Paper, 1/3Scissors) is the only one that cannot be “exploited” by either player. The values of p and q are such that if Player 1 changes p, his payoff will not change but Player 2’s payoff may be affected Thus, it is a Mixed Strategy Nash Equilibrium.

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The Nash Equilibrium is a very unstable point If you do not begin exactly at the NE, you cannot stochastically find the NE Theoretically you will “shoot off” to a pure strategy: (0,0) (0,1) (1,0) or (1,1) (similar for n players) Consider the following: 2 players randomly choose values for p and q Knowing player 2’s mixed strategy (q, 1-q), player 1 adjusts his mixed strategy of (p,1-p) in order to maximize his payoffs With player 1’s new mixed strategy in mind, player 2 will adjust his mixed strategy in order to maximize his payoffs This see-saw continues until both players can no longer change their strategies to increase their payoffs

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Unfortunately, I was unable to find a way to discover a mixed strategy NE in R for any number of players Is my code wrong? Is there simply no way to find the NE in R? I don’t know

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In life, we react to other people’s choices in order to increase our utility or happiness Ignoring a younger sibling who is irritating Accepting an invitation to go to a baseball game Once we react, the other person reacts to our reaction and life goes on One stage games are rare in life Very rarely are we in a “NE” for any aspect of our lives There is almost always a choice that can better our current utility

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