Solution Manual for Game Theory Basics 1st Edition By Bernhard von Stengel, ISBN: 9781108843300, All 12 Chapters Covered

SOLUTION MANUAL
Game Theory Basics 1st Edition
By Bernhard von Stengel. Chapters 1 - 12




1

,TABLE OF CONTENTS ia ia ia




1 - Nim and Combinatorial Games
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2 - Congestion Games
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3 - Games in Strategic Form
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4 - Game Trees with Perfect Information
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5 - Expected Utility
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6 - Mixed Equilibrium
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7 - Brouwer’s Fixed-Point Theorem
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8 - Zero-Sum Games
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9 - Geometry of Equilibria in Bimatrix Games
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10 - Game Trees with Imperfect Information
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11 - Bargaining
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12 - Correlated Equilibrium
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2

,Game Theory Basics ia ia




Solutions to Exercises i a i a




© Bernhard von Stengel 2022
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Solution to Exercise 1.1 ia ia ia




(a) Let ≤ be defined by (1.7). To show that ≤ is transitive, consider x, y, z with x ≤ y and y ≤ z. If x = y then x
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≤ z, and if y = z then also x ≤ z. So the only case left is x < y and y < z, which implies x < z because < is tr
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ansitive, and hence x ≤ z. ia ia ia ia ia




Clearly, ≤is reflexive because x = x and therefore x ≤ x. ia ai ia ia ia ia ia ia ia ia ia ia




To show that is≤antisymmetric, consider x and y with x y and y x. If ≤
ia we had x ≠ y then
≤ x < y and y <
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x, and by transitivity x < x which contradicts (1.38). Hence x = y, as required. This shows that ≤ is a
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partial order. ia




Finally, we show (1.6), so we have to show that x < y implies x y and x ≠ y and vice
≤ versa. Let x < y, whi
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ch implies x y by (1.7). If we had x = y then x < x, contradicting
ia
≤ (1.38),sowealsohave x ≠ y. Conversely, ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ai ai ai ai ai ia ia ia ia i




x y and x ≠ y implyby(1.7)x < y or x = y where the second case is excluded,
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≤ hence x < y, as required.
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(b) Consider a partial order and assume ≤ (1.6) as a definition of <. To show that < is transitive, suppose ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia




x < y, that is, x y and x ≠ y, and y < z, that is, y≤z and y ≠ z. Because is transitive, x z. If we
ia ia ia
≤ had x = z then
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x y and y ≤x and hence x = y by≤
iaiaiaiaia antisymmetry of , which contradicts
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≤ x ≠ y, so
≤ we have x z and x ≠ z
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, that is,x < z by (1.6), as required.
≤ ≤
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Also, < is irreflexive, because x < x would by definition mean x x and x ≠ x, but ≤
ia ia the latter is not true.
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Finally, we show (1.7), so we have to show that x ≤ y implies x < y or x = y and vice versa, given that <
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is defined by (1.6). Let x ≤ y. Then if x = y, we are done, otherwise x ≠ y and then by definition x < y. H
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ence, x ≤ y implies x < y or x = y. Conversely, suppose x < y or x = y. If x < y then x ≤ y by (1.6), an
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d if x = y then x ≤ y because ≤is reflexive. This completes the proof.
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Solution to Exercise 1.2 ia ia ia




(a) In analysing the games of three Nim heaps where one heap has size one, we first lookat some examp
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les, and then use mathematical induction to prove what we conjecture to be the losing positions. A losin
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g position is one where every move is to a winning position, because then the opponent will win. Th
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e point of this exercise is to formulate a precise statement to be proved, and then to prove it.
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First, if there are only two heaps recall that they are losing if and only if the heaps are of equal size. I
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f they are of unequal size, then the winning move is to reduce thelarger heap so that both heaps have
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equal size.
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3

, Consider three heaps of sizes 1, m, n, where 1 m n. We≤observe ≤ the following: 1, 1, m is winning, b ia ia ia ia ia ia ia ia ia iaiaiaiaia iaiaiaiaia ia ia ia ia ia ia ia ia ia ia




y moving to 1, 1, 0. Similarly, 1, m, m is winning, by moving to 0, m, m. Next, 1, 2, 3 is losing (observe
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d earlier in the lecture), and hence 1, 2, n for n 4 is winning. 1, 3, n is winning for any n 3 by moving t
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o 1, 3, 2. For 1, 4, 5, reducing any heap produces a winning position, so this is losing.
≥ ≥
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The general pattern for the losing positions thus seems to be: 1, m, m 1, for even numbers
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+ m. This i ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia




ncludes also the case m = 0, which we can take as the base case foran induction. We now proceed to p
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rove this formally. ia ia




First we show that if the positions of the form 1, m, n with m n are losing when
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≤ m is even and n = m
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1, then these are the only losing
ia
+ positions because any other position 1, m, n with m n is winning
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. Namely, if m = n then a winning
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≤ move from1, m, m is to 0, m, m, so we can assume m < n. If m is even ia ia i a ia i a ia ia ia ia ai ia ia ia ia ia ia ia ia ia ia ia ia ia ia i a ia ia ia ia




then n > m 1 (otherwise we would be in the position 1, m, m 1) and so the winning move is to 1,
+
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m, m 1. If m is odd then the winning move is to 1, m, m 1, the same as position 1, m 1, m (this would al
+ +
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so be a winning move from 1, m, m so there the winning move is not unique).
– −
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Second, we show that any move from 1, m, m + 1 with even m is to a winning position,using as inductive
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hypothesis that 1, mJ, mJ + 1 for even mJ and mJ < m is a losing position. The move to 0, m, m + 1 produ
a ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia




ces a winning position with counter- ia ia ia ia ia




move to 0, m, m. A move to 1, mJ, m + 1 for mJ < m is to a winning position with the counter-
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move to 1, mJ, mJ + 1 if mJ is even and to 1, mJ, mJ − 1 if mJ is odd. A move to 1, m, m is to a winning positi
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on with counter- ia ia




move to 0, m, m. A move to 1, m, mJ with mJ < m is also to a winning position with the counter-
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move to 1, mJ − 1, mJ if mJ is odd, and to 1, mJ 1, mJ if mJ is even (in which case mJ 1 < m because m is e
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ven). This concludes the induction proof. ia ia ia ia ia




+ +
This result is in agreement with the theorem on Nim heap sizes represented as sums of powers of 2: 1
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a m n is losing
i
∗ +∗ a only if, exceptfor 20, the powers of 2 making upm and n come in pairs. So these m
ifand+∗ i a i a ia ia
i a
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ust be the same powers of 2, except for 1 = 20, which occurs in only m or n, where we have assumed that n
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is the larger number, so 1 appearsin the representation of n: We have m = 2a 2b 2c
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for a > b > c > 1,so m is
+ + + ··· ··· ≥
i a i a i a i a i a i a iaiaiaiaiaiaiai a ai i a ia ia




even, and, with the same a,b, c, .. ., n = 2 a 2 b 2 c 1 = m 1. Then i a i a i a i a i a i a
i a ia ia ia ia ia




+ + + ··· + +
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∗1 +m∗ +n ∗ 0.≡The ∗ following is an example using the bit representation where
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m = 12 (which determines the bit pattern 1100, which of course depends on m):
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1 = 0001
12 = 1100
13 = 1101
Nim-sum 0 = 0000

(b) We use (a). Clearly, 1, 2, 3 is losing as shown in (1.2), and because the Nim-
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sum of the binary representations 01, 10, 11 is 00. Examples show that any other position iswinni
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ng. The three numbers are n, n 1, n 2. If n+is even+then reducing the heap of size n 2 to 1 creates the
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position n, n 1, 1 which is losing as shown in (a). If n is odd, then n 1 is even and n 2 = n 1 1 so
+ +
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by the same argument, a winning move is to reduce the Nim heap of size n to 1 (which only works if
+ +
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n > 1).
( + )+
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i a i a ai




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