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2004_midterm_1_s - 14.12 Game Theory Fall 2004 Answers for...

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14.12 Game Theory Fall 2004 Answers for Midterm 1, Fall 2004 1. Consider the following game: A B C a 3 , 0 0 , 3 0 , x b 0 , 3 3 , 0 0 , x c x, 0 x, 0 x, x (a) The question does not ask to fi nd all Nash equilibria. One way is to fi nd all the pure-strategy Nash equilibria and then use the symmetry of the game to guess a symmetric mixed-strategy Nash equilibria and check that it is indeed a Nash equilibria. The Nash equilibria of that game are ( c, C ) ( 1 3 a + 1 3 b + 1 3 c ; 1 3 A + 1 3 B + 1 3 C ) ( 1 2 a + 1 2 b ; 1 2 A + 1 2 B ) In the solutions, we fi nd all the equilibria starting with the pure-strategy ones. The best responses of player 1 are BR 1 ( A ) = { a } BR 1 ( B ) = { b } BR 1 ( C ) = { c } The best responses of player 2 are BR 2 ( a ) = { B } BR 2 ( b ) = { A } BR 2 ( c ) = { C } Hence the only pure-strategy Nash equilibrium is ( c, C ) In a mixed-strategy equilibrium, player 2 is playing A with probability p , B with probability q and C with probability 1 p q . The expected payo ff s of player 1 are EU 1 ( a ) = 3 p EU 1 ( b ) = 3 q EU 1 ( c ) = 1 1
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Hence his best responses are p > 1 3 and p > q = a q > 1 3 and q > p = b 1 3 > p and 1 3 > q = c p = q > 1 3 = { a, b } p = 1 3 > q = { a, c } q = 1 3 > p = { b, c } p = q = 1 3 = { a, b, c } By symmetry, if player 1 is playing a with probability r , b with probability s and c with probability 1 r s , the best responses of player 2 are r > 1 3 and r > s = B s > 1 3 and r > s = A 1 3 > r and 1 3 > s = C r = s > 1 3 = { A, B } r = 1 3 > s = { B, C } s = 1 3 > r = { A, C } r = s = 1 3 = { A, B, C } - Let’s fi rst look at an equilibrium where player 2 is mixing among { A, B, C } with positive probabili- ties. In order for this strategy to be a best response it must be the case that player 1’s probabilities are r = s = 1 3 . Player 1 is playing his three strategies with a positive probability which is a best response to p = q = 1 3 . We conclude that ( 1 3 a + 1 3 b + 1 3 c ; 1 3 A + 1 3 B + 1 3 C ) is a Nash equilibrium. - Then look at an equilibrium where player 2 is mixing among { A, B } . This means that p + q = 1 and the best responses of player 1 are p > 1 2 = 1 a = 2 B cannot be an equilibrium p = 1 2 = 1 { a, b } p < 1 2 = 1 b = 2 A cannot be an equilibrium Player 1 is thus mixing among { a, b } .And using the symmetry, this can be the case in equilibrium if and only if r = s = 1 2 . We conclude that ( 1 2 a + 1 2 b ; 1 2 A + 1 2 B ) is a Nash equilibrium.
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