Lecture20

# Lecture20 - Lecture#20 Recall the matching pennies game...

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Lecture #20 ecall the matching pennies game from last time Recall the matching pennies game from last time… MATCHING PENNIES wo friends take a penny in their hand and place it either “heads up” or “tails up” Two friends take a penny in their hand and place it either heads up or tails up . They have agreed to payoffs as follows: If, when they open their hands, the pennies are the same side up then player 1 ets a dollar from player 2 gets a dollar from player 2. If, when they open their hands, the pennies show opposite sides up then player 2 gets a dollar from player 1. Let’s represent this game in both the extensive form and the normal form… Player 1 1 HT HH TT Player 2 1 1 2 H T H (1, -1) (-1, 1) T (-1, 1) (1, -1) (1) (-1) (-1) (1) (-1) (1) (1) (-1)

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Remember that we said that this game has no pure strategy Nash equilibrium. However, if these players mix strategies (and the game is finite) there will exist a mixed strategy Nash equilibrium. MIXED STRATEGY NASH EQUILIBRIA If a player is going to mix on a set of pure strategies then all of the strategies in the set must have the same payoff. Hence, the player’s probability distribution over these strategies is “arbitrary”. Of course, the other player’s payoffs do depend upon the probability distribution and hence, a player’s mixed strategy is completely determined by the other players. Let’s clarify how this works by solving the mixed strategy Nash equilibrium for the matching pennies game and then the (slightly) more complex Battle of the Sexes game. irst let’s consider Player 1’s payoffs if Player 2 plays H with a probability of β First, let s consider Player 1s payoffs if Player 2 plays H with a probability of and plays T with a probability of (1 – β ). 1 2 H (prob. = β ) T (prob. = (1 - β )) H (1, -1) (-1, 1) T (-1, 1) (1, -1)
Denote Player 1’s payoffs from playing H given that Player 2 mixes as π 1 (H, β ) and also denote Player 1’s payoffs from playing T given that Player 2 mixes as π 1 (T, β ). Now, let’s consider Player 1’s payoff from playing H… π 1 (H, β ) = (1) · β + (-1) · (1 – β ) π 1 (H, β ) = β ±–1 ±+ ± β π 1 (H, β ) = 2 β and what about Player 1’s payoff from playing T… π 1 β ) = (-1) · β + (1) · (1 – β ) π β ) = - β + 1 – β 1 π 1 β ) = 1 – 2 β Now, for Player 2 to make Player 1 indifferent between playing H or T they must choose the β that makes Player 1’s payoffs the same for H and T (i.e. π yp y ( 1 (H, β ) = π 1 β )). So…

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π 1 (H, β ) = π 1 (T, β ) 2 β – 1 = 1 – 2 β 4 β = 2 β = ½ Now, let’s consider Player 2’s payoff from playing H if Player 1 mixes by playing H with a probability of α and plays T with a probability of (1 – α ). 1 2 H T H (prob. = α ) (1, -1) (-1, 1) Denote Player 2’s payoffs from playing H given that Player 1 mixes as π ( α , H) T (prob. = (1 - α )) (-1, 1) (1, -1) yp y p y g g y 2 ,) and also denote Player 2’s payoffs from playing T given that Player 1 mixes as π 2 ( α , T).
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Lecture20 - Lecture#20 Recall the matching pennies game...

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