practicemidterm_solutions.pdf - Game Theory Homework#3 All the problems are worth 4 points each and will be graded on a 0\/1\/2\/3\/4 scale Due on Sunday

practicemidterm_solutions.pdf - Game Theory Homework#3 All...

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Unformatted text preview: Game Theory Homework #3, 09/22/2018 All the problems are worth 4 points each and will be graded on a 0/1/2/3/4 scale. Due on Sunday 09/30/2018 before 11:59 pm to be uploaded via Gradescope. 1. Find the values of the zero-sum game given by the following payo matrix, and determine all the optimal strategiesfor both players in the games: (a) 20 2 10 5 3 1 4 3 1 3 8 4 4 8 7 6 9 6 1 2 2 7 9 1 5 Solution 1. The 4th row is strictly dominated by 3rd. So we can eliminate it, without losing optimal strategies: 20 2 10 ∗ 3 20 2 10 ∗ 3 20 ∗ 10 ∗ 3 ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ 1 4 3 1 3 8 4 4 8 7 6 9 ∗ ∗ ∗ ∗ 7 9 1 5 1 ∗ 3 1 3 ∗ 4 4 8 ∗ 6 9 ∗ ∗ ∗ ∗ 7 ∗ 1 5 1 ∗ 3 1 ∗ ∗ ∗ ∗ 8 ∗ 6 9 ∗ ∗ ∗ ∗ 7 ∗ 1 5 1 ∗ 3 1 ∗ ∗ ∗ ∗ 8 ∗ 6 9 ∗ ∗ ∗ ∗ 7 ∗ 1 5 ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ 8 ∗ 6 9 ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ 6 ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗  The 3rd column is strictly dominated by the 4th,  The 2nd row is strictly dominated by the 3rd,  The 1st column is strictly dominated by the 4th,  The 1st and last rows are strictly dominated by the 3rd,  The 2nd and last columns are strictly dominated by the 4th,  The only pair of safety strategies is (e3 , e4 ), the value of the game is 6. Solution 2. Note that (e , e ) is a saddle point. Thus, the strategies (e , e ) are optimal. So, any optimal 3 4 3 4 strategy of the rst player should be an optimal response to y = e4 , but x = e3 is the only optimal response to y = e4 , so e3 is the only optimal strategy of PI. Analogously, e4 is the only optimal strategy of PII. (b)  1 6 2 4 1 3 5 7 2 1  Solution. This is a 2 × 5 problem, so we can solve it graphically. The strategy of the rst player is (1 − p, p): 7 6 5 4 3 2 1 V = 5/3 1 0 1 p p∗ = 1/3 The value of the game is 5/3, the unique strategy of PI is (2/3, 1/3)T . From picture we can see that PII can only use 3rd and 5th strategies (since other strategies give 1st player more than V when she plays (2/3, 1/3)T .) Thus, to nd all optimal strategies of the 2nd player we need to solve ( y3 + 2(1 − y3 ) ≤ 5/3, 3y3 + (1 − y3 ) ≤ 5/3. ⇐⇒ 1/3 ≤ y3 ≤ 1/3 ⇐⇒ y3 = 1/3. Thus, the only optimal strategy of PII is (0, 0, 1/3, 0, 2/3)T . 2. (a) Find the Nim sum of 5 , 6 and 10. In the game of Nim, is this position in N or in P? Solution. 5 = 01012 6 = 01102 10 = 10102 5 ⊕ 6 ⊕ 10 = 10012 = 9 Since the Nim sum is not zero, the position is in N. (b) In a subtraction game players may remove 1, 2 or 3 chips. Find the P positions and determine which player wins if the game starts with a pile of 100 chips. Solution. In part (c) we will show that the SG function for this subtraction game is g(x) = x (mod 4). Thus, the P-positions are {x : g(x) = 0} = {x = 4k : x ∈ Z, x ≥ 0}. Since 100 is divisible by 4, this is a P-position. (c) Solution. Now we are going to prove that the SG function for a subtraction game with 1 pile and subtraction set {1, 2, . . . , k} is g(x) = x (mod k + 1). We just need to check that this function satises the denition: g(x) = mex{g(y) : x → y is a legal move.}. It means that we need to show two things: • We cannot go to the position with the same residual by mod k + 1. This is obvious, since we always subtract the number which is less than k + 1, so it is not divisible by k + 1, so the residual must change. • From any position x and for any naonnegative integer l < (x (mod k + 1)) we can go from x to the position with residual l. We indeed can do that since we can just subtract (x (mod k + 1)) − l from x. It makes the residual exactly l, and it is a legal move since (x (mod k + 1)) − l ≤ (x (mod k + 1)) < k + 1 and x ≥ x (mod k + 1) ≥ (x (mod k + 1)) − l. Now we can complete the solution by using the SG theorem: the SG function for the game with 3 piles is g(x1 , x2 , x3 ) =(x1 =(x1 (mod (k1 + 1))) ⊕ (x2 (mod 4)) ⊕ (x2 (mod (k2 + 1))) ⊕ (x3 (mod 6)) ⊕ (x3 (mod 12)). g(100, 25, 32) = 0 ⊕ 1 ⊕ 8 = 9 6= 0, so this position is in N. (mod (k3 + 1))) 3. Consider the general sum game given by (1, 1) (0, 0) (0, 0) (0, 0) (2, 2) (0, 0) . (0, 0) (0, 0) (3, 3) Answer the following: (a) Find safety strategies for both players. Solution. First, let's nd the safety strategies for the rst player: in the zero-sum game with matrix A 1 0 0 0 2 0 we can nd a fully mixed equalizing strategy of the rst player: 0 0 = 3 x1 = 2x2 = 3x3 =⇒ x = 1 (6, 3, 2)T . 11 Since the matrix A is symmetric, the same strategy is also equalizing for the second player in that zero-sum game. Since a pair of fully equalizing strategies in a zero-sum game is always a pair of safety strategies, 111 (6, 3, 2)T is a safety strategy for the rst player, and the value of the zero-sum game is 6/11. Now let's prove that we found all the safety strategies of PI. Note that the rst player must use all the strategies: if she does not use some strategy ei , then the PII can just use ei , which results in zero payment, which is less than the value of the game. Thus the safety strategy of the rst player must be fully mixed, so the strategy of the second player must be fully equalizing. We found that the only fully equalizing strategy is also fully mixed, so the strategy of PI should also be fully equalizing, which is exactly what we found. So there are no other safety strategies of PI. Second, nd the safety strategy for the second player. Since B T = A, the game is symmetric for both players, so the same strategy 111 (6, 3, 2)T is also the unique safety strategy for the second player. Thus,   is the pair of safety strategies. 1 1 (6, 3, 2)T , (6, 3, 2)T 11 11 (b) Find all Pure Nash Equilibrium, and determine which of them are strict Nash equilibrium. Solution. A cell of the payo matrix corresponds to a pure Nash equilibrium i the payo to the second player is maximal in the row and the payo to the rst  in the column. Let's underline those maximal elements: (1, 1) (0, 0) (0, 0) (0, 0) (2, 2) (0, 0) . (0, 0) (0, 0) (3, 3) One can see that there are three pure Nash equilibria in this game: (1, 1), (2, 2), and (3, 3), where (a, b) means that the rst player plays strategy number a, and the second  b. Now let's check if those equilibria are strict. Note that in each of those equilibria both players have strictly positive payo. However, if one of the players unilaterally deviates, the cell with the payo moves from the diagonal of the matrix, and both players get zero payo. Thus, it is strictly worse for any of the players to deviate than to play according to the equilibrium (the payo of the player who deviates decreases from the positive number to zero.) Thus, all those pure equilibria are all strict. (c) Find all symmetric mixed Nash equilibrium (not necessarily fully mixed). Solution. We are looking for the symmetric mixed Nash equilibria. Symmetric means that both players play the same strategy. Nash equilibrium means that each player plays optimally in response to another player. So we need to nd all strategies z which are optimal response to itself. (Note that since the game is symmetric: (A = B ) it is enough to only have that z is an optimal response of the rst player, when the second player also plays z). Use equalizing principle: x is an optimal response to y if for all i holds xi > 0 =⇒ (Ay)i = max(Ay)j , j which is same as (Ay)i 6= max(Ay)j =⇒ xi = 0. j In words, the response is optimal if and only if it mixes only the best options. If the option is not one of the best, the optimal response always has zero probability of playing that option. Now, for our z = (z1 , z2 , z3 )T there are 4 cases: • z1 > 0, z2 > 0, z3 > 0. According to the equalizing principle, such z is an optimal responce to itself i all the elements of Az = (z1 , 2z2 , 3z3 ) are equal. Thus, z1 = 111 (6, 3, 2)T gives our rst simmetric mixed NE (z1 , z1 ). • z1 > 0, z2 > 0, z3 = 0. According to the equalizing principle, such z is an optimal responce to itself i the rst two elements of Az = (z1 , 2z2 , 3z3 ) are equal and maximal (not less than the third element). Thus, such z should have   z3 = 0 z1 = 2z2 ≥ 3z3 = 0 ⇐⇒ z = z1 + z2 + z3 = 1 So for z2 = 2 1 , ,0 3 3 T . , (z2 , z2 ) is our second NE. • z1 > 0, z2 = 0, z3 > 0 and z1 = 0, z2 > 0, z3 > 0  analogously to the second case we get symmetric Nash   equilibria with strategies z3 = 34 , 0, 14 T and z4 = 0, 35 , 25 T . Overall, we considered all possible cases of what elements of z are zero, and in each case showed that there is unique mixed NE. So there are no other symmetric mixed NE in this game and the answer is (z1 , z1 ), (z2 , z2 ), (z3 , z3 ), (z4 , z4 ), where z1 , . . . , z4 are given above. T 2 1 3, 3, 0 4. (a) Give an example of a two-person zero-sum game where there are no pure Nash equilibria and all the entries of the matrix are dierent. Solution.   4 . 2 1 3 (b) Give an example of a zero sum game represented by a 3 × 3 matrix where there is a Nash equilibrium which is pure for one player and fully mixed for the other player. Solution. 2 1 1 2 1 1 2 1 . 1 Note that the pair (e1 , (1/3, 1/3, 1/3)T ) is a NE. 5. Three neighboring colleges have n students each, that hit the nightclubs on weekends. Each of the two clubs, C1 and C2, chooses a price in [0, 1, 2, . . . , 100]. College A students go to C1, College C students go to C2, and College B students choose whichever of C1 or C2 has the lower cover charge that weekend, breaking ties in favor of C1. Find all the pure equilibria (if any). Solution. Suppose C2 sets the price b. Let's nd the optimal response a of C1: • If a ≤ b, then 2n students go to C1, and pay 2an in total. This is maximized by a = b (among a ≤ b). • If a > b, then n students fo to C1, and pay an. This is maximized by a = 100 (among a > b, if they exist). Thus, the optimal response to b is either b (which gives 2bn) or 100 (which gives 100n if b < 100 and 2bn if b = 100.) More precisely, • If b < 50, the optimal a is a = 100, • If b > 50, the optimal a is a = b, • If b = 50, the both a = 100 and a = 50 are optimal. However, note that if a = 100, the optimal response of C2 is to set b = 99, so the rst case, and the rst part of the last case never give a NE. If b ≥ 50 and a = b then C2 can change b to b = a − 1, which will give more prot. Thus the second case, and the second part of the last case also do not give NE. Thus, there are no pure NE in this game. ...
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