Math 135: Answers to Practice Problems

Transcription

1 Math 35: Answers to Practice Problems Answers to problems from the textbook: Many of the problems from the textbook have answers in the back of the book. Here are the answers to the problems that don t have answers in the back of the book (f) No, it is not a Nash equilibrium. To see this, we start by assuming that uses the strategy A + B + C + D, and we determine s best response to this strategy. If chooses A, s expected value is: (3)+ ()+ ( 2)+ ( 3) = If chooses B, s expected value is: ( 5)+ ( 3)+ (2)+ (6) = 0 Thus, s best response is B. Thus, would not use the strategy 2 A+ 2 B, so it is not a Nash equilibrium. (g) No, it is not a Nash equilibrium for the same reason as (f). If uses the strategy A + B + C + D, s best response is B, not 5A + 3B. (h) No, it is not a Nash equilibrium. To see this, we start by assuming that uses the strategy A + B, and we determine s best response to this strategy. If chooses A, s expected value is: ( 3) + (5) = If chooses B, s expected value is ( ) + (3) = If chooses C, s expected value is (2) + ( 2) = 0 If chooses D, s expected value is (3) + ( 6) = 3/2 Thus, s best response is A or B, or any combination of A and B. Thus, would not use the strategy B + C, so it is not a Nash equilibrium. (i) Yes, it is a Nash equilibrium. To see this, we start by assuming that uses the strategy B + C, and we determine s best response to this strategy: If chooses A, s expected value is: () + ( 2) = /2 If chooses B, s expected value is: ( 3) + (2) = /2 Thus, s best response is A or B, or any combination of A and B. In particular, 5A + 3 B is a best response. Now, we assume that uses the strategy 5A + 3 B, and we determine s best response to this strategy: If chooses A, s expected value is: 5( 3) + 3(5) = 0 If chooses B, s expected value is: 5( ) + 3 (3) = / If chooses C, s expected value is: 5(2) + 3 ( 2) = / If chooses D, s expected value is: 5(3) + 3 ( 6) = 3/ Thus, s best response is B or C, or any combination of B and C. In particular, B + C is a best response. Since both strategies are a best response to the other s strategy, there is no incentive for either player to switch strategies, so it is a Nash equilibrium.

2 .3.0. Row B dominates Row A, Row C dominates Row D, and Column A dominates Column D, so we can eliminate Row A, Row D, and Column D. Once we eliminate those rows and columns, Column A dominates Column B, so we can eliminate Column B. This leaves us with the following matrix: A C B (2, ) (6, 0) C (0, 0) (0, 3) Now, we try to find pure strategy Nash equilibria by creating the best response diagram: A C B 2, 6, 0 C 0, 0 0, 3 As we can see from the best response diagram, there are no pure strategy Nash equilibria. Now, we try to find mixed strategy Nash equilibria. Suppose that uses the mixed strategy ( p)b + pc. Then, s expected payoffs are: If chooses A, s expected value is: ( p)( ) + p(0) = + p If chooses C, s expected value is: ( p)(0) + p( 3) = 3p We set these equal: + p = 3p Solving, we get that p = /. Thus, if uses the strategy 3B + C, will be indifferent between A and C (so that any mixed strategy involving A and C will be a best response for ). Now, suppose that uses the mixed strategy ( p)a+pc. Then, s expected payoff are: If chooses B, s expected value is: ( p)(2) + p(6) = 2 6p If chooses C, s expected value is: ( p)(0) + p(0) = 0 We set these equal: 2 6p = 0 Solving, we get that p = /3. Thus, if uses the strategy 2A + C, will be 3 3 indifferent between B and C (so that any mixed strategy involving B and C will be a best response for ). Thus, this game has one Nash equilibrium. It is ( 3 B + C, 2 3 A + 3 C). 2

3 (a) We create a table with all possible orderings of A, B, and C, and showing the marginal contributions of each of A, B, and C in the given order: Order A B C ABC ACB BAC BCA CAB CBA The average for A is 360/6 = 60, the average for B is 90/6 = 5, and the average for C is 90/6 = 5. Thus the Shapley allocation is: A B C Answers to problems not from the textbook:. (a) The first player has the winning strategy, because when we write the numbers in binary, some columns do not have an even number of s: 23 = = = = + = (decimal) = 0 (binary) 9 (decimal) = 00 (binary) 3 (decimal) = 00 (binary) 2 (decimal) = 000 (binary) (decimal) = 00 (binary) The s column, the s column, and the 2 s column all have an odd number of s, so the first player has the winning strategy. (b) The first player should remove 0 beans from the pile with 3 beans (this will create an even number of s in each column). 2. (a) s optimal strategy is: Choose B. If chooses, choose E. s optimal strategy is: If chooses A, choose. If chooses B, choose. (b) has four strategies: Choose A. If chooses 2, choose C. Choose A. If chooses 2, choose D. Choose B. If chooses, choose E. Choose B. If chooses, choose F. 3

4 We will abbreviate these strategies as AC, AD, BE, and BF. has four strategies: If chooses A, choose. If chooses B, choose 3. If chooses A, choose. If chooses B, choose. If chooses A, choose 2. If chooses B, choose 3. If chooses A, choose 2. If chooses B, choose. We will abbreviate these strategies as (, 3), (, ), (2, 3), and (2, ). The payoff matrix for this game is: (, 3) (, ) (2, 3) (2, ) AC 5, 3 5, 3 2, 6 2, 6 AD 5, 3 5, 3 7, 2 7, 2 BE, 2 6,, 2 6, BF, 2, 3, 2, 3 3. (a) Yes, it is a Nash equilibrium. To see this, we start by assuming that uses the strategy Rock + Paper + Scissors, and we determine s best response to this strategy: If chooses Rock, s expected value is: 9 (0) + 9 ( ) + 9 () = 0 If chooses Paper, s expected value is: 9 () + 9 (0) + 9 ( ) = 0 If chooses Scissors, s expected value is: 9 ( ) + 9 () + 9 (0) = 0 Thus, s best response is Rock, Paper, Scissors, or any combination of these. In particular, Rock + Paper + Scissors is a best response. Now, we assume that uses the strategy Rock + Paper + Scissors, and we determine s best response to this strategy: If chooses Rock, s expected value is: 9 (0) + 9 ( ) + 9 () = 0 If chooses Paper, s expected value is: 9 () + 9 (0) + 9 ( ) = 0 If chooses Scissors, s expected value is: 9 ( ) + 9 () + 9 (0) = 0 Thus, s best response is Rock, Paper, Scissors, or any combination of these. In particular, Rock+ Paper+ Scissors is a best response. Since both strategies are a best response to the other s strategy, there is no incentive for either player to switch strategies, so it is a Nash equilibrium. (b) No, it is not a Nash equilibrium. To see this, we start by assuming that uses the strategy 5Rock + Paper, and we determine s best response to this 9 9 strategy. If chooses Rock, s expected value is: 5(0) + ( ) = If chooses Paper, s expected value is: 5() + (0) = 5

5 If chooses Scissors, s expected value is: 5( ) + () = Thus, s best response is Paper. Thus, would not use the strategy Rock + Scissors, so it is not a Nash equilibrium. 9 9 (c) If uses the strategy 5Rock + Paper, then: 9 9 If chooses Rock, s expected value is: 5(0) + ( ) = If chooses Paper, s expected value is: 5() + (0) = 5 If chooses Scissors, s expected value is: 5( ) + () = Thus, ( if uses the strategy Rock + Scissors, then s expected payoff 9 9 is ) ( ) 9 9 = 36 =. 9 If uses the strategy Rock + Scissors, then: 9 9 If chooses Rock, s expected value is: (0) + () = If chooses Paper, s expected value is: () + ( ) = If chooses Scissors, s expected value is: ( ) + 32 (0) = Thus, if uses the strategy 5Rock + Paper, then s expected payoff is ( 9 ) + 9 ( 9 ) = 36 9 =. Thus, s expected payoff is and s expected payoff is.. (a) First, we check for pure strategy Nash equilibria by creating the best response diagram: A B A, 6 2, B, 5 (, ) From the best response diagram, we see that there are two pure strategy Nash equilibria: (A, B) and (B, A). Next, we check for mixed strategy Nash equilibria. Suppose that uses the mixed strategy ( p)a + pb. Then, s expected payoffs are: If chooses A, s expected value is: ( p)(6) + p(5) = 6 p If chooses B, s expected value is: ( p)() + p() = p We set these equal: 6 p = p Solving, we get that p = 2/3. Thus, if uses the strategy A + 2 B, will 3 3 be indifferent between A and B (so that any mixed strategy involving A and B will be a best response for ). Now, suppose that uses the mixed strategy ( p)a + pb. Then, s expected payoffs are: 5

6 If chooses A, s expected value is: ( p)() + p(2) = + p If chooses B, s expected value is: ( p)() + p() = 3p We set these equal: + p = 3p Solving, we get that p = 3/. Thus, if uses the strategy A + 3 B, will be indifferent between A and B (so that any mixed strategy involving A and B will be a best response for ). Thus, there are three Nash equilibria: (A, B), (B, A), and ( A + 2B, A + 3B). 3 3 (b) Suppose that uses the mixed strategy ( p)a + pb. Then: If chooses A, s expected value is: ( p)() + p() = + 3p If chooses B, s expected value is: ( p)(2) + p() = 2 p We graph the equations for these lines: 3 2 p From the graph, we can see that s prudential strategy lies at the intersection of the two lines. Thus, we set the two equations equal: + 3p = 2 p Solving, we get that that p = /. We can plug this value for p into either equation to get that s security level is + 3 ( ) = 7/. Thus, s prudential strategy is 3A + B and her secruity level is 7/. (c) Suppose that uses the mixed strategy ( p)a + pb. Then: If chooses A, s expected value is: ( p)(6) + p() = 6 + 2p If chooses B, s expected value is: ( p)(5) + p() = 5 p 6

7 We graph the equations for these lines: 6 2 p From the graph, we can see that s prudential strategy occurs at p = 0, which means his prudential strategy is A. His security level is 5. Thus, s prudential strategy is A and his secruity level is (a) First, we check for pure strategy Nash equilibria by creating the best response diagram: A B A, 2 3, 6 B 5, ) 2, 7 From the best response diagram, we see that there is one pure strategy Nash equilibrium: (A, B). Before checking for mixed strategy Nash equilibria, we should check for rows or columns that dominate (since Nash equilibria only occur in undominated rows and columns). In this matrix, Column B dominates Column A, so we can eliminate Column A. After we eliminate Column A, Row A dominates Row B, so we can eliminate Row B. This leaves us only with the strategy pair (A, B), so there are not any additional mixed strategy Nash equilibria. There is exactly one Nash equilibrium: (A, B). (b) Suppose that uses the mixed strategy ( p)a + pb. Then: If chooses A, s expected value is: ( p)() + p(5) = + p If chooses B, s expected value is: ( p)(3) + p(2) = 3 p 7

8 We graph the equations for these lines: p From the graph, we can see that s prudential strategy lies at the intersection of the two lines. Thus, we set the two equations equal: + p = 3 p Solving, we get that that p = 2/5. We can plug this value for p into either equation to get that s security level is + ( 2 5) = 3/5. Thus, s prudential strategy is 3A + 2 B and her secruity level is 3/ (c) Suppose that uses the mixed strategy ( p)a + pb. Then: If chooses A, s expected value is: ( p)(2) + p(6) = 2 + p If chooses B, s expected value is: ( p)() + p(7) = + 6p We graph the equations for these lines:

9 6 2 p From the graph, we can see that s prudential strategy occurs at p =, which means his prudential strategy is B. His security level is 6. Thus, s prudential strategy is B and his secruity level is First, we check for pure strategy Nash equilibrium by creating the best response diagram: A B A 9, 3 3, B, 2 6, C 5, 7, 2 D, 5 7, 9 From the best response diagram, we see that there are two pure strategy Nash equilibria: (A, A) and (D, B). Next, we look for mixed strategy Nash equilibria. Suppose that uses the mixed strategy ( p)a + pb. Then, s expected payoffs are: If chooses A, s expected value is: ( p)(9) + p(3) = 9 6p If chooses B, s expected value is: ( p)() + p(6) = 2p If chooses C, s expected value is: ( p)(5) + p() = 5 p If chooses D, s expected value is: ( p)() + p(7) = + 3p We graph the equations for these lines: 9

10 Potential Nash Equilibria D B C A p From the graph, we can see that there are two potential mixed strategy Nash equilibria (not counting pure strategy Nash equilibria). One potential Nash equilibrium comes from using strategies A and B; the other comes from using strategies B and D. We will check both of these separately. If uses strategies A and B, the matrix looks like: A B A (9, 3) (3, ) B (, 2) (6, ) Suppose that uses the mixed strategy ( p)a + pb. Then, s expected payoffs are: If chooses A, s expected value is: ( p)(3) + p(2) = 3 p If chooses B, s expected value is: ( p)() + p() = + 7p We set these equal: 3 p = + 7p Solving, we get that p = /. Thus, if uses the strategy 3A + B, will be indifferent between A and B (so that any mixed strategy involving A and B will be a best response for ). 0

11 To find the corresponding strategy for to use, we set the equations for the lines for A and B equal: 9 6p = 2p Solving, we get that p = /. Thus, if uses the strategy 3A + B, s best response will be A, B, or any combination of A and B (we can see this by looking at the previous graph). Thus, one mixed strategy Nash equilibrium is ( 3 A + B, 3A + B). Now, we consider using strategies B and D. If uses B and D, the matrix looks like: A B B (, 2) (6, ) D (, 5) (7, 9) Suppose that uses the mixed strategy ( p)b + pd. Then, s expected payoffs are: If chooses A, s expected value is: ( p)(2) + p(5) = 2 + 3p If chooses B, s expected value is: ( p)(5) + p(9) = 5 + p We set these equal: 2 + 3p = 5 + p Solving, we get that p = 3. Since 3 < 0, this does not correspond to a Nash equilibrium. Thus, there is not a Nash equilibrium corresponding to the second point in the graph. Thus, this game has three Nash equilibria: (A, A), (D, B), and ( 3 A + B, 3 A + B). 7. (a) Yes, this is a zero-sum game. Each entry sums to 9, so we can subtract 9 from all of s numbers to obtain a matrix where all the entries sum to zero. (b) No, this is not a zero-sum game. To see this, we try to create a linear function that will change s numbers into the negative of s numbers. This linear function would need to send to 2, 3 to, to, and to 0. Thus, we would need an equation for a line that contains the points (, 2), (3, ), (, ), and (, 0). We can find an equation for a line containing the first two points, and then we can check whether it also contains the last two points. The slope of a line through the first two points would be: m = y 2 y x 2 x = = 6 2 = 3 Thus, the line is of the form y = 3x + b. To solve for b, we plug in the values from the first point: 2 = 3() + b. Solving, we get b =. Thus, the line must

12 be y = 3x +. However, the fourth point (, 0), does not lie on this line, since 0 3()+. Thus, we cannot change s numbers into the negative of s numbers simply by multiplying by a constant and adding a constant. Thus, it is not a zero-sum game. (c) Yes, it is a zero-sum game. To see this, we create a linear function that will change s numbers into the negative of s numbers. This linear function would need to send 3 to and to 3 (and also work for all the other entries in the matrix). We can find the equation for the line containing the points ( 3, ) and (, 3). The slope of the line is: m = y 2 y x 2 x = = = 2 Thus, the line is of the form y = 2x + b. To solve for b, we plug in the values from the first point = 2( 3) + b. Solving, we get b = 5. Thus, the line is y = 2x 5. This equation does work for all of the other entries in the matrix, so it is a zero-sum game.. (a) i. If and both always choose A, then always receives 6, so her expected payoff is: ( ) ( ) 2 ( ) 3 ( ) = 6 7 = (b) Thus, s expected payoff is. ii. If and both always choose A, then always receives 7, so his expected payoff is: ( ) ( ) 2 ( ) 3 ( ) = = 56 Thus, s expected payoff is 56. i. On the first two rounds chooses A and chooses B, so receives 0 for the first two rounds. After that and both choose B, so receives 3 for the later rounds. Her expected payoff is: ( ) ( ) 2 ( ) 3 ( ) ( ) 2 7 = = 3 ( ) ( 7 ) = 7 =.375 Thus, s expected payoff is ( 7 ) +...

13 ii. On the first two rounds chooses A and chooses B, so receives 0 for the first two rounds. After that and both choose B, so receives 5 for the later rounds. His expected payoff is: ( ) [ = = ( 7 2 ) 7 ( ) ( ) Thus, s expected payoff is ( ) ( ) ( 7 ( ) ) ] +... = = 395 = (a) We create a table with all possible orderings of A, B, and C, and showing the marginal contributions of each of A, B, and C in the given order: Order A B C ABC ACB BAC BCA CAB CBA The average for A is 20/6 = 35, the average for B is 360/6 = 60, and the average for C is 270/6 = 5. Thus the Shapley allocation is: A B C (b) Yes, the allocation is rational. To see this we need to check that every coalition receives at least as much as it would on its own: A receives 35, which is greater than the 0 that A would receive on its own. BC receive = 05, which is greater than the 90 they receive on their own. AC receive = 0, which is greater than the 30 they would receive on their own. AB receive = 95, which is greater than the 60 they would receive on their own. Thus, the allocation is rational. 3