GAME THEORY
EXERCISES—PART I
(FROM THE TEXTBOOK OF C. HOLT)
Gabriele Camera
Fall 2016 - University of Basel
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PROBLEM 1
What is the Nash Equilibrium for the Traveler’s Dilemma game where choices
must be in the set A = {−50, −49, · · · , −1, 0, 1, · · · , 49, 50} and there is a
penalty/reward rate R = 5?
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ANSWER TO PROBLEM 1
Action set: A1 = A2 = {−50, −49, · · · , −1, 0, 1, · · · , 49, 50}
Payoff matrix:
-50
-50 (-50,-50)
-49 (-55,-45)
-48 (-55,-45)
..
..
49 (-55,-45)
50 (-55,-45)
-49
(-45,-55)
(-49,-49)
(-54,-44)
..
(-54,-44)
(-54,-44)
-48
(-45,-55)
(-44,-54)
(-48,-48)
..
(-53,-43)
(-53,-43)
...
...
...
...
...
...
...
49
50
(-45,-55) (-45,-55)
(-44,-54) (-44,-54)
(-43,-53) (-43,-53)
..
..
(49,49)
(54,44)
(44,54)
(50,50)
(−50, −50) is the unique Nash Equilibrium
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NASH EQUILIBRIUM: A PROOF
Step 1: Existence (direct proof)
Conjecture(-50,-50) is a NE. This conjecture follows from what we saw in class.
π1(−50, −50) = −50 > π1(a1, −50) = −55
for all a1 > −50.
π2(−50, −50) = −50 > π2(−50, a2) = −55
for all a2 > −50.
Hence, (−50, −50) is a NE
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NASH EQUILIBRIUM: A PROOF
Step 2: Uniqueness (indirect proof)
By contradiction, suppose that (â1, â2) 6= (−50, −50) is a NE
Case 1: â1 < â2 (the case â2 < â1 is equivalent)
π1(â1, â2) = â1 + R
π2(â1, â2) = aˆ1 − R
But player 2 can do better, choosing â1:
π2(â1, â1) = â1 > π2(â1, â2) = â1 − R
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(contradiction)
NASH EQUILIBRIUM: A PROOF
Case 2: â1 = â2
Player 1’s payoff is
π1(â1, â1) = â1
But player 1 can do better, by cutting â1 by 1:
π1(â1 − 1, â1) = â1 − 1 + R = â1 + 4 > â1 = π(â1, â1) (contradiction)
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PROBLEM 2
Consider a Traveler’s Dilemma with choices in A = {80, 81, . . . 200}, and R = 10.
Two players play four rounds with claims (80, 195), (98, 117), (96, 117), (130, 100)
respectively. Calculate the earnings of the first player. What would his earnings
be if the first player chose 200 in each round?
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ANSWER TO PROBLEM 2
Action set for each player: A = {80, 81, . . . , 199, 200}
Penalty/reward rate: R = 10
Player 1’s payoff in each round: π1(a1, a2) = min(a1, a2) + θ1(a1, a2) × R, where


 1 if a1 < a2
θ1(a1, a2) :=
0 if a1 = a2

 −1 if a > a
1
2
So, the payoff of player 1 is
π1 = (80 + 10) + (98 + 10) + (96 + 10) + (100 − 10) = 394
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TAKING MORE RISK
However, if the first player took more risk by choosing 200 in each round, his
earnings would be
π̄1 = (195 − 10) + (117 − 10) + (117 − 10) + (100 − 10) = 489
which is more profitable than π1
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PROBLEM 3
Consider a Traveler’s Dilemma with R=$5 and N players and choices in the set
A = {80, 81, . . . 200}. If all choices are not equal, there is a $5 penalty if one’s
choice is not the lowest and a $5 reward rate for the person with the lowest choice.
Speculate on what the effect on average claims of increasing N from 2 to 4 players
is in a setup with repeated random matchings
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ANSWER TO PROBLEM 3
Case N = 2: unique NE = (80, 80)
Since R = 5 is small compared with any claim in {80, 81, . . . , 199, 200}, individuals
might be tempted to choose high claims
In addition, in a repeated game setup the two players may learn from round to
round what to expect.
Consequently, they might coordinate on higher claims over time
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FAR FROM THE EQUILIBRIUM PATH
Effort Choice Frequencies with High Effort Cost (light bars), and Low Effort Cost
(dark bars)
See figure 0
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MULTIPLE PLAYERS
Case N = 4: Players are matched together
Players select choices a1, a2, a3, a4 ∈ {80, 81, . . . , 199, 200}
Player i’s payoff:
πi(a1, a2, a3, a4) = min(a1, a2, a3, a4) + θi(a1, a2, a3, a4) × R


 1
θi(a1, a2, a3, a4) :=
0

 −1
if ai = min(a1, a2, a3, a4) 6= aj for some j 6= i
if a1 = a2 = a3 = a4
if ai 6= min(a1, a2, a3, a4) = aj for some j =
6 i
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NASH EQUILIBRIUM
Example:
Suppose that a1 is the lowest choice with ai > a1 for each i = 2, 3, 4, then
• Player 1 gets a1 + R
• Players 2, 3 and 4 get a1 − R
By symmetry, (80,80,80,80) is the unique NE (see Problem 2)
Again, individuals may increase their earnings from round to round by coordinating
on higher choices
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CONVERGENCE?
However, anticipating others’ intentions becomes tougher
On average, one might expect:
• Choices converge to higher values more slowly
• Choices should be smaller than choices expected in the 2-player case: cutting
the expected others’ choice only by 1 makes, now, the chances of success more
unlikely
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PROBLEM 4
Consider a Matching Pennies Game with the following payoff matrix
Left
Right
Top
30, 40
40, 80
Bottom
40, 80
80, 40
(i) Does there exist a Nash Equilibrium in pure strategies?
(ii) Does there exist a Nash Equilibrium in mixed strategies?
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ANSWER TO PROBLEM 4 (QUESTION I): CASE 1
Row player chooses Top and Column player chooses Right
Left
Right
Top
30, 40
40, 80
Bottom
40, 80
80, 40
{Top,Right} is not a Pure NE, since Row player can do better by choosing Bottom
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CASE 2
Row player chooses Bottom and Column player chooses Right
Left
Right
Top
30, 40
40, 80
Bottom
40, 80
80, 40
{Bottom,Right} is not a Pure NE, since Column player can do better by choosing
Left
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CASE 3
Row player chooses Top and Column player chooses Left
Left
Right
Top
30, 40
40, 80
Bottom
40, 80
80, 40
{Top,Left} is not a Pure NE, since either Column player can do better by choosing
Right or Row player can do better by choosing Bottom
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CASE 4
Row player chooses Bottom and Column player chooses Left
Left
Right
Top
30, 40
40, 80
Bottom
40, 80
80, 40
{Bottom,Left} is a Pure NE, since both players cannot do better by changing
action
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ANSWER TO PROBLEM 4 (QUESTION II)
Let {q, 1 − q} be the set of probability distribution over {Top, Bottom}, and
{p, 1 − p} be the set of probability distribution over {Left, Right}
Row player: choose Top with probability q to make column player indifferent
column player chooses Left
column player chooses Right
z
}|
{
q40 + (1 − q)80 =
z
}|
{
q80 + (1 − q)40
⇒
1
q= ,
2
Column player: choose Left with probability p to make row player indifferent:
row player chooses Top
row player chooses Bottom
z
}|
{
z
}|
{
30p + 40(1 − p) = 40p + 80(1 − p)
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⇒
p=
4
>1
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MIXED STRATEGY NASH EQUILIBRIUM
Every finite strategic game has a mixed strategy Nash Equilibrium
Here, the mixed strategy Nash Equilibrium is the profile ({0, 1}, {1, 0})
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