solutions_hw4

solutions_hw4 - Solutions to homework 4 1.5#2 Player I...

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Unformatted text preview: Solutions to homework 4 1.5#2 Player I holds a black Ace and a red 8. Player II holds a red 2 and a black 7. If the chosen cards are of the same color Player I wins, if they differ Player II does. The amount won is the sum of the cards. The payoff function is given by the following matrix: A =- 3 8 10- 15 If we denote the probability of Player I putting the Ace by p , then the equalizing strategy tells us that- 3 p + 10(1- p ) = 8 p- 15(1- p ) . 10- 13 p = 23 p- 15 . From which p = 25 36 . Hence, Player I’s optimal strategy is choosing the Ace with probability 25 36 and choosing 8 with probability 11 36 . Similarly, if we denote the probability of Player II chosing the red 2 by q , we get- 3 q + 8(1- q ) = 10 q- 15(1- q ) . 8- 11 q = 25 q- 15 . Hence, q = 23 36 . Therefore, Player II’s optimal strategy is choosing the Ace with probability 23 36 and putting 8 with probability 13 36 . The value of the game is v = 25 36 · (- 3) + 11 36 · 10 =- 75+110 36 = 35 36 . 1.5#3 Sherlock Holmes The payoff function is given by the matrix: A = 100- 50 100 . Let the probability for Moriarty to stop at Canterbury (i.e., selecting row 1) be p . Then we can find the optimal strategy by solving 100 p =- 50 p + 100(1- p ) , which gives us p = 2 5 . And similarly, Sherlock Holmes’ optimal strategy (given that the probability of him getting off at Canterbury is...
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This note was uploaded on 08/24/2009 for the course MATH 262447221 taught by Professor Weisbart during the Spring '09 term at UCLA.

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solutions_hw4 - Solutions to homework 4 1.5#2 Player I...

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