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ps1(2010)Sol

# ps1(2010)Sol - MS&E 246 Game Theory with Engineering...

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MS&E 246: Game Theory with Engineering Applications Problem Set 1 1. (a) (3pts) When s - i = 1, any strategy s i yields a zero payoff. So, there are no strictly dominated strategies. (b) (3pts) s i = 0 is the only weakly dominated strategy. It is weakly dominated by any strategy s i (0 , 1]. (c) (2+2pts) Any ( s 1 , s 2 ) with s 1 + s 2 = 1 is a pure strategy NE. The payoff of player i is s i . Given the strategy s - i , player i has no incentive to deviate from the strategy s i = 1 - s - i . If he chooses s 0 i < s i his payoff will be only s 0 i . If he chooses s 0 i > s i , his payoff will be zero. (1,1) is also a Nash equilibrium. 2. (a) (6pts) Suppose player i has two weakly dominant strategies, s 0 i and s 00 i . Then u i ( s 0 i , s - i ) u i ( s i , s - i ) for all s - i and for all s i . In particular, u i ( s 0 i , s - i ) u i ( s 00 i , s - i ) for all s - i . Since s 00 i is weakly dominant, there is some s - i such that u i ( s 00 i , s - i ) > u i ( s 0 i , s - i ), which is a contradiction. Therefore, a player can not have two weakly dominant strategies. (b) (4pts) According to (a), there can be no such game. 3. (a) (3pts) Playing LL, L or R is quite risky, since we do not know what player 1 will be playing. The risk of obtaining a playoff of -49 is very large compared to the payoff of 1 if player 2 player L, and the risk of obtaining a payoff of -100 is very large compared to the payoff of 2 if player 2 played LL or R. Therefore, it seems ”reasonable” to play M. (b) (4+6pts) The two pure NE of this game are (U,LL) and (D,R). To check for mixed strategy NE, player 1 must mix between U (with probability p) and D (with probability 1-p). Player 2 has 11 possible mixing combinations: { LL,L } , { LL,M } , { LL,R } , { L,M } , { L,R } , { M,R } , { M,L,R } , { LL,M,R } , { LL,L,R } , { LL,L,M } and { LL,L,M,R } . We will show that only the { LL,L } is part of a mixed NE. For player 2 to mix between LL and L (with probabilities q and 1-q respectively), we must have p · (2) + (1 - p ) · ( - 100) = p · (1) + (1 - p ) · ( - 49) which yields p = 51 52 . The utility of player 2 from each strategy is then: u 2 ( LL ) = u 2 ( L ) = 1 26 , u 2 ( M ) = 0 and u 2 ( R ) < 0. Then for player 1 to mix between U and D, we 1

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must have: q · (100)+(1 - q ) · ( - 100) = q · ( - 100)+(1 - q ) · (100) which yields q = 1 2 and u 1 ( U ) = u 1 ( D ) = 0. Therefore, p = 50 51 and q = 1 2 is a mixed NE. We now show that no other mixing combination of player 2 can be part of a mixed NE.
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ps1(2010)Sol - MS&E 246 Game Theory with Engineering...

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