m110-a

m110-a - 14.12 Game Theory - Midterm I 10/19/2010 Prof....

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14.12 Game Theory — Midterm I 10/19/2010 Prof. Muhamet Yildiz Instructions. This is an open book exam; you can use any written material. You have one hour and 20 minutes. Each question is 25 points. Good luck! 1. Consider the following game. (a) Using backward induction f nd an equilibrium. Answer: At the last node, Player 1 chooses ;a tther igh tnode ,P layer2then chooses . At the left node, she chooses . Hence, at the beginning, Player 1 chooses . The equilibrium is (   ) . (b) Write the game in normal form. Answer: The game in normal form is 1 \ 2      2,1 2,1 1,2 1,2  2,1 2,1 1,2 1,2  2,1 1,0 2,1 1,0  2,1 0,1 2,1 0,1 2. Compute a Nash equilibrium of the following game. (This is a version of Rock-Scissors- Paper with preference for Paper.) RS P R 0 0 2 2 2 3 S 2 2 0 0 2 1 P 3 1 1 2 1 1 Answer:
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This game has a unique Nash equilibrium, which is in mixed strategies. Write , , and for the probabilities with which player plays strategies R, S, and P, respectively. Of course, + + =1 (1) Now, ( 2  2  2 ) must make Player 1 indi f erent between his strategies. Indi f erence betweenRandSy ie lds 2( 2 2 )=2( 2 2 ) ,i .e . , 2 2 = 2 + 2 Since 2 + 2 =1 2 by (1), this yields 2 =1 3 On the other hand, the indi f erence between S and P yields 2( 2 2 )=1+2( 2 2 ) . Substituting 2 =1 3 ,weobta in 4 2 2 2 = 1 3 .T oge the rw i th 2 + 2 =2 3 ,th is yields 2 =1 6 2 =1 2 Similarly, indi f erence between R and S for player 2 yields
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m110-a - 14.12 Game Theory - Midterm I 10/19/2010 Prof....

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