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,
a iaiai a ai
≤ 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
ia
≤ 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 <
ia ia ia ia ia ia ia ia ia ia i a ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia
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
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 ia
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
ia ia ia ia ia ia ia ia ia ia ia ia ia ia i a i a ia ia ia ia i a ia i a i a ia ia i a ia ia ia ia
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
ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ai ia ia
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
ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia i a
e point of this exercise is to formulate a precise statement to be proved, and then to prove it.
ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia
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
ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia
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
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
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
ia
+ 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
ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ai ia i a ia ia ia ia
rove this formally. ia ia
First we show that if the positions of the form 1, m, n with m n are losing when
ia
≤ 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
ia ia ia ia ia ia ia ia ia ia ia ia ia i a ia i a i a i a ia
. Namely, if m = n then a winning
i a
≤ 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,
+
ia ia ia i a i a ia ia ia ia ia ia ia ia ia ia i a i a ia ia ia ia ia ia ia ia ia
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).
– −
ia ia ia ia ia ia ai ai ia ia ia ia ia ia ia ia
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-
ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia
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
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 ia ia ia ia ia ia
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-
ia ia ia ia ia ia ia ia ia ia ia i a ia i a ia ia ia ia ia ia ia ia ia
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
ia ia ia ia ia ia ia i a ia ia ia ia ia ia i a ia ia ia ia ia ia ia ia ia i a ia ia ia ia ia ia
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
ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia i a i
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
ia ai ia ia ia ai ia ia ia ia ia ia ia ai ia ia ia ia ia ia ia ia
ust be the same powers of 2, except for 1 = 20, which occurs in only m or n, where we have assumed that n
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
is the larger number, so 1 appearsin the representation of n: We have m = 2a 2b 2c
ia ia ia ia ia ia ai ia ia ia ia i a ia ia i a ia
iaiaiaiaiaia iaiaiaiaiaia
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
+ + + ··· + +
ia ia ia ia i a ai ia ai ai ai i a i a ia ia ia iaiaiai a i a
∗1 +m∗ +n ∗ 0.≡The ∗ following is an example using the bit representation where
i a ia ia i a
iaiaiaiaiai a iaiaiaiai a iaiaiaiaiai a i a ia ia ia ia ia ia ia ia ia
i a ia ia ia
m = 12 (which determines the bit pattern 1100, which of course depends on m):
ia ia ia ia ia ia ia ia ia ia ia ia ia ia
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-
ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia
sum of the binary representations 01, 10, 11 is 00. Examples show that any other position iswinni
ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ai
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
ia ia ia ia ia ia i a ia i a i a ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia
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
+ +
ia ia ia i a ia ia ia ia ia ia ia ia ia ia ia ia ia ia i a ia ia ia ia iaiaia ia i a iaiaia iaiaia ia ia
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).
( + )+
ia ia
i a i a ai
4
Game Theory Basics 1st Edition
By Bernhard von Stengel. Chapters 1 - 12
1
,TABLE OF CONTENTS ia ia ia
1 - Nim and Combinatorial Games
ia ia ia ia ia
2 - Congestion Games
ia ia ia
3 - Games in Strategic Form
ia ia ia ia ia
4 - Game Trees with Perfect Information
ia ia ia ia ia ia
5 - Expected Utility
ia ia ia
6 - Mixed Equilibrium
ia ia ia
7 - Brouwer’s Fixed-Point Theorem
ia ia ia ia
8 - Zero-Sum Games
ia ia ia
9 - Geometry of Equilibria in Bimatrix Games
ia ia ia ia ia ia ia
10 - Game Trees with Imperfect Information
ia ia ia ia ia ia
11 - Bargaining
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12 - Correlated Equilibrium
ia ia ia
2
,Game Theory Basics ia ia
Solutions to Exercises i a i a
© Bernhard von Stengel 2022
i a ia ia ia
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
ia ia ia ia ia i a ia ia ia ia ia ia ia ia ia ia ia i a ia ia ia ia ia ia ia ia ia ia ia ia
≤ 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
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 ia ia ia ia ia ia ia
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
ia ia ia ia ia ia ia ia ia ia ia ia ia ia iaiaia ia ia ia ia ia ia ia ia ia ia ia ia
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,
a iaiai a ai
≤ hence x < y, as required.
ia ia ia ai ai ai ai ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia
(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
ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia iaiaiaia ia ia iaiaiaia ia ia ia ia ia ia ia ia
x y and y ≤x and hence x = y by≤
iaiaiaiaia antisymmetry of , which contradicts
ia
≤ x ≠ y, so
≤ we have x z and x ≠ z
ia iaiaiaiaia ia ia ia ia ia ia ia ia iaiaiaia ia ia ia ia ia ia ia ia ia iaiaiai a ia ia ia ia
, that is,x < z by (1.6), as required.
≤ ≤
ia ia ai ia ia ia ia ia ia
Also, < is irreflexive, because x < x would by definition mean x x and x ≠ x, but ≤
ia ia the latter is not true.
ia ia ia ia ia ia ia ia ia ia iaiaia ia ia ia ia ia ia ia ia ia ia
Finally, we show (1.7), so we have to show that x ≤ y implies x < y or x = y and vice versa, given that <
ia ia ia ia ia ia ia ia ia ia i a ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia
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
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 ia
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
ia ia ia ia ia ia ia ia ia ia ia ia ia ia i a i a ia ia ia ia i a ia i a i a ia ia i a ia ia ia ia
d if x = y then x ≤ y because ≤is reflexive. This completes the proof.
ia ia ia ia ia ia ia ia ia ia ai ia i a ia ia ia
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
ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ai ia ia
les, and then use mathematical induction to prove what we conjecture to be the losing positions. A losin
ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia
g position is one where every move is to a winning position, because then the opponent will win. Th
ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia i a
e point of this exercise is to formulate a precise statement to be proved, and then to prove it.
ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia
First, if there are only two heaps recall that they are losing if and only if the heaps are of equal size. I
ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia i a
f they are of unequal size, then the winning move is to reduce thelarger heap so that both heaps have
ia ia ia ia ia ia ia ia ia ia ia ia ia ai ia ia ia ia ia ia
equal size.
ia ia
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
ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia
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
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
o 1, 3, 2. For 1, 4, 5, reducing any heap produces a winning position, so this is losing.
≥ ≥
ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia
The general pattern for the losing positions thus seems to be: 1, m, m 1, for even numbers
ia
+ 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
ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ai ia i a ia ia ia ia
rove this formally. ia ia
First we show that if the positions of the form 1, m, n with m n are losing when
ia
≤ m is even and n = m
ia ia ia ia ia ia ia ia ia ia ia ia ia iaiaiaiaiaia ia ia ia ia ia ia ia ia ia ia ia
1, then these are the only losing
ia
+ positions because any other position 1, m, n with m n is winning
ia ia ia ia ia ia ia ia ia ia ia ia ia i a ia i a i a i a ia
. Namely, if m = n then a winning
i a
≤ 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,
+
ia ia ia i a i a ia ia ia ia ia ia ia ia ia ia i a i a ia ia ia ia ia ia ia ia ia
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
+ +
ia i a i 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
so be a winning move from 1, m, m so there the winning move is not unique).
– −
ia ia ia ia ia ia ai ai ia ia ia ia ia ia ia ia
Second, we show that any move from 1, m, m + 1 with even m is to a winning position,using as inductive
ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ai ia ia i
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-
ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia
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
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 ia ia ia ia ia ia
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-
ia ia ia ia ia ia ia ia ia ia ia i a ia i a ia ia ia ia ia ia ia ia ia
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
ia ia ia ia ia ia ia i a ia ia ia ia ia ia i a ia ia ia ia ia ia ia ia ia i a ia ia ia ia ia ia
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
ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia i a i
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
ia ai ia ia ia ai ia ia ia ia ia ia ia ai ia ia ia ia ia ia ia ia
ust be the same powers of 2, except for 1 = 20, which occurs in only m or n, where we have assumed that n
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
is the larger number, so 1 appearsin the representation of n: We have m = 2a 2b 2c
ia ia ia ia ia ia ai ia ia ia ia i a ia ia i a ia
iaiaiaiaiaia iaiaiaiaiaia
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
+ + + ··· + +
ia ia ia ia i a ai ia ai ai ai i a i a ia ia ia iaiaiai a i a
∗1 +m∗ +n ∗ 0.≡The ∗ following is an example using the bit representation where
i a ia ia i a
iaiaiaiaiai a iaiaiaiai a iaiaiaiaiai a i a ia ia ia ia ia ia ia ia ia
i a ia ia ia
m = 12 (which determines the bit pattern 1100, which of course depends on m):
ia ia ia ia ia ia ia ia ia ia ia ia ia ia
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-
ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia
sum of the binary representations 01, 10, 11 is 00. Examples show that any other position iswinni
ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ai
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
ia ia ia ia ia ia i a ia i a i a ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia
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
+ +
ia ia ia i a ia ia ia ia ia ia ia ia ia ia ia ia ia ia i a ia ia ia ia iaiaia ia i a iaiaia iaiaia ia ia
by the same argument, a winning move is to reduce the Nim heap of size n to 1 (which only works if
+ +
ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia ia
n > 1).
( + )+
ia ia
i a i a ai
4
