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Unformatted text preview: ECE 496 SOLUTIONS TO HOMEWORK ASSIGNMENT I Spring 2007 The numbers correspond with the numbering of the Thought Problems at the end of Chapter 1 of the book. 2. In parts (a) through (c) we list the outcomes of the games in the standard way assuming that Player 1 is playing TITFORTAT. The resulting payoff pairs are in parentheses. Part (d) is a bit trickier. (a) CD (0 , 5) DD (1 , 1) DD (1 , 1) DD (1 , 1) DD (1 , 1) DD (1 , 1) DD (1 , 1) DD (1 , 1) DD (1 , 1) DD (1 , 1) Total payoff to Player 1 = 9. Total payoff to Player 2 = 14. (b) CC (3 , 3) CC (3 , 3) CC (3 , 3) CC (3 , 3) CC (3 , 3) CC (3 , 3) CC (3 , 3) CC (3 , 3) CC (3 , 3) CC (3 , 3) Total payoff to Player 1 = 30. Total payoff to Player 2 = 30. (c) CD (0 , 5) DD (1 , 1) DC (5 , 0) CC (5 , 5) CD (0 , 5) DD (1 , 1) DC (5 , 0) CC (5 , 5) CD (0 , 5) DD (1 , 1) Total payoff to Player 1 = 23. Total payoff to Player 2 = 28 (d) You are Player 1 and are playing TFT. The other player is playing random moves. That description is a bit vague, but Ill assume what Melanie Mitchell intended was for the random player to flip a fair coin before each move i.e., to choose C or D with probability 1/2 in each repetition of the game, with all the choices independent of each other. You play C in the first game. Player 2 moves at random; hence the outcome is either CC or CD with probability 1/2 each. So your expected payoff from that first game is 1 / 2(3 + 0) = 3 / 2, and Player 2s expected payoff is 1 / 2(3 + 5) = 4. Now, what about subsequent games? Note that Player 2s move in the first stage determines not only the payoffs for that stage, but also what move you make in the second stage. Hence you will play C with probability 1/2 and play D with probability 1/2 in that stage. If you play C, you have an expected payoff of 3/2 again; if you play D, you have an expected payoff of 1 / 2(5 + 1) = 3. (Note that Player 2s move in the second stage is independent of yours, since the move you make is determined solely by Player 2s coin flip in the first stage.) Since these eventualities occur with equal probability 1/2, your overall expected payoff in the second stage is 1 / 2(3 / 2 + 3) = 9 / 4. Similarly, Player 2 plays C or D with probability 1/2 in the second stage; he faces a C or a D from you with equal probability independent of his choice here (again because what determines your move is what Player 2 did in the first stage). By the same reasoning as for you, Player 2s expected payoff in the second stage is also 9/4. You can reason similarly about the rest of the stages and conclude both players have expected payoff 9/4 in every stage after the first. Conclusion: Player 1s expected payoff is 3 / 2 + 81 / 4 = 87 / 4 and Player 2s expected payoff is 4 + 81 / 4 = 97 / 4....
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This note was uploaded on 02/01/2010 for the course ECE 496 taught by Professor Delchamps during the Spring '07 term at Cornell University (Engineering School).
 Spring '07
 DELCHAMPS
 Algorithms

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