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m110-a

# m110-a - 14.12 Game Theory Midterm I Prof Muhamet Yildiz...

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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 fi nd an equilibrium. Answer: At the last node, Player 1 chooses ; at the right node, Player 2 then 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.) R S P R 0 0 2 2 2 3 S 2 2 0 0 2 1 P 3 1 1 2 1 1 Answer: (Because of a typo, the question became asymmetric, making the answer longer. We gave nearly full credit to the students who thought the game was symmet- ric.) 1

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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 ff erent between his strategies. Indi ff erence between R and S yields 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 ff erence between S and P yields 2 ( 2 2 ) = 1+2 ( 2 2 ) . Substituting 2 = 1 3 , we obtain 4 2 2 2 = 1 3 . Together with 2 + 2 = 2 3 , this yields 2 = 1 6 2 = 1 2 Similarly, indi ff erence between R and S for player 2 yields 2 ( 1 1 ) = 2 1 1 , i.e., 3 1 = 2 ( 1 + 1 ) . Since 1 + 1 = 1 1 , this yields 1 = 2 5 The indi ff erence between S and P yields 2 ( 1 1 ) = 1+2 (
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m110-a - 14.12 Game Theory Midterm I Prof Muhamet Yildiz...

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