Game Theory
Exam June 2017
Name:
Group:
Grades:
I
II.1
II.2
II.3
II.4
Total
You have two and a half hours to complete the exam.
I. Short questions (5 points each)
I.1 When a player chooses a mixed strategy in a Nash equilibrium, the pure strategies that are part
of it may have different expected payoffs. Explain whether this statement is true or false.
I.2 Provide an example of a dynamic game that shows the following fact: if one player sees one
of her strategies eliminated, her utility in equilibrium increases.
I.3 There are 3 players and they vote (one round only) for A, B or C. Their preferences are:
Player 1: A better than B, B better than C,
Player 2: B better than A, A better than C,
Player 3: A better than C, C better than B.
The alternative with most votes gets selected. In case of a tie between the three options, no one gets
selected, which is regarded as the worst outcome for all players.
Is there any Nash equilibrium in non-weakly dominated strategies in which B is elected?
I.4. In the calculation of payoffs in an infinitely repeated game, show that discounting payoffs
using the rule of the discount rate is equivalent to consider that the game is repeated with some
probability.
I.1 False. If they have different payoff, the one that has the highest will be a better reply than the
mixed strategy. Then the mixed strategy cannot be part of a NE.
I.2
1
A
B
2
1
1
C
D
2
2
0
3
The SPNE is (A, D) with payoff 1 for Player 2. If D is eliminated, the new game is:
1
1
A
B
2
1
1
C
2
2
The SPNE is (A, C) with payoff 2 for Player 2.
I.3 (B, B, B) is such an equilibrium. Notice that Player 3 is not using a weakly dominated strategy,
as the outcome is B, better than the outcome after (A, C, B), which implies no option is selected.
Also (B, B, C) is an equilibrium that satisfies the requirements . Players 1 and 2 are the only ones
that can change the outcome by deviating, but the can only do it to elect C or not to elect any
option. Both are worse consequences than B.
I.4 Payoffs with discount 𝛿 are of the form:
𝑢 = 𝑢(𝑠1 ) + 𝛿𝑢(𝑠 2 ) + 𝛿 2 𝑢(𝑠 3 ) + 𝛿 3 𝑢(𝑠 4 ) + ⋯
Payoffs with probability p of repetition are of the form:
𝑢 = 𝑢(𝑠1 ) + 𝑝𝑢(𝑠 2 ) + 𝑝2 𝑢(𝑠 3 ) + 𝑝3 𝑢(𝑠 4 ) + ⋯
Both 𝛿 and p are numbers between 0 and 1, so the expressions are mathematically identical.
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II. Problems (20 points each)
II.1 Miren and Claudia, both students of game theory at UC3M, are fans of old board games. This
year the World-Board-Game-Day, a big fair, is held in Getafe and naturally both of them go to see
what is offered. They both end up at one stand where an original of John Nash's classic board game
“Hex” is offered in a second price auction (that is, the highest bid wins the auction, but the price
paid is the second highest bid). If they both bid the same, the winner is decided with the toss of a
coin (and the winner pays the bid, as the second bid coincides with the first). While both of them
are of course very interested, nobody else cares about the old game and participates in the auction.
Miren and Claudia know each other well. In particular, they both know the willingness to pay for
the game of the other. Miren is willing to pay up to 60 Euros, while Claudia’s maximum
willingness to pay is 70 Euros. The only admissible bids are 50, 60, and 70 Euros.
(a)
(b)
(c)
(d)
Represent the game in matrix form. (6 points)
Determine the set of rationalizable strategies for both players. (5 points)
Determine the set of pure strategy Nash equilibria. (5 points)
Is playing the maximum willingness to pay a weakly dominant strategy? (4 points)
(a)
50
50 5, 10
Miren 60 10, 0
70 10, 0
Claudia
60
0, 20
0, 5
0, 0
70
0, 20
0, 10
-5, 0
(b) There are no dominated strategies. The set {(50,50), (50,60), (50,70), (60,50), (60,60),
(60,70), (70,50), (70,60), (70,70)} is the set of razionalizable strategies.
(c)
50
50 5, 10
Miren 60 10, 0
70 10, 0
Claudia
60
0, 20
0, 5
0, 0
70
0, 20
0, 10
-5, 0
Nash equilibria in pure strategies: {(50,60), (50,70), (60,70), (70,50), (70, 60)}.
(d) For Miren, playing 60 (her maximum willingness to pay) is weakly dominant: it gives
higher or equal payoffs than playing 50 ((10, 0, 0) rather than (5, 0, 0)) or playing 70 ((10,
0, 0) rather than (10, 0, -5)). Similarly for Claudia.
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II.2 John and Peter are the only dock workers. They must unload a ship that just arrived, but for
that they need to cooperate. Cooperation implies that they agree on a way to share the total pay
established by the firm, which is a percentage of the cargo value. According to the labor norms,
each day John and Peter meet at the facilities, and John makes a proposal to divide the pay. If Peter
accepts they both unload the ship. If Peter does not accept, they wait until the next day and Peter
makes a counter-offer, and so on, alternating the player who makes the offer. The value of the
cargo that they will share is €4,000, but the cargo is perishable and every day that goes on without
unloading, a part is lost and the corresponding pay diminishes in €1,000. I.e., if they unload the first
day they share €4,000, a quantity that decreases by €1,000 every day after that.
(a) Represent a sketch of the extensive form game and indicate whether it is of perfect
information. (7 points)
(b) Find the subgame perfect Nash equilibrium. Note: as usual in these problems, assume
that in case of indifference, the offer is accepted. (13 points)
(a)
J
x1
P
A
R
P
x1
4,000-x1
x2
J
A
R
J
x2
3,000-x2
x3
P
A
R
P
x3
2,000-x3
x4
J
A
R
x4
1,000-x4
0
0
(b) Stage 4: J accepts any 𝑥 4 ≥ 0. Rejects otherwise. P offers 𝑥 4 = 0.
Stage 3: P accepts any offer such that 2,000 − 𝑥 3 ≥ 1,000 − 𝑥 4 with 𝑥 4 = 0; i.e., 𝑥 3 ≤ 1,000.
Rejects otherwise. J offers 𝑥 3 = 1,000.
Stage 2: J accepts any 𝑥 2 ≥ 𝑥 3 , with 𝑥 3 = 1,000; i.e, 𝑥 2 ≥ 1,000. Rejects otherwise. J offers
𝑥 2 = 1,000.
Stage 1: P accepts any offer such that 4,000 − 𝑥 1 ≥ 3,000 − 𝑥 2 with 𝑥 2 = 1,000; i.e., 𝑥 1 ≤
2,000. Rejects otherwise. J offers 𝑥 1 = 2,000.
In equilibrium, payoffs are (2,000, 2,000).
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II.3 Consider the normal form game between Player 1 and Player 2 represented by the following
matrix:
Player 2
A
B
C
A 6, 6 0, 9 0, 0
Player 1 B 9, 0 3, 3 0, 0
C 0, 0 0, 0 1, 1
(a) Find the Nash equilibria in pure strategies. (1 point)
(b) Find a symmetric (both players choose the same strategy) Nash equilibrium in mixed
strategies in which players assign positive probability to only two of their three pure
strategies. Note: remember to check that the pure strategy that is not used in the
equilibrium is not a better reply. (4 points)
Assume now that the game is played twice. After the first time, players observe what happened and
play again, so that they can condition the play in the second round to the actions in the first one.
Total payoffs are the sum of the payoffs in every round.
(c) Count the number of subgames in this repeated game, the number of information sets
for each player and the number of strategies for each player. (3 points)
(d) Can you find a subgame perfect Nash equilibrium in which (A, A) is played in the first
round? (6 points)
(e) Repeat (d) if the game is repeated 3 times. (6 points)
(a)
Player 2
A
B
C
A 6, 6 0, 9 0, 0
Player 1 B 9, 0 3, 3 0, 0
C 0, 0 0, 0 1, 1
Nash equilibria in pure strategies: {(B,B), (C,C)}.
(b) (𝑝[𝐴] + (1 − 𝑝)[𝐵], 𝑞[𝐴] + (1 − 𝑞)[𝐵]) cannot be a NE in mixed strategies: 𝑢(𝐴) =
𝑢(𝐵) implies 6𝑞 = 9𝑞 + 3(1 − 𝑞), which gives 0 = 3, a contradiction.
(𝑝[𝐴] + (1 − 𝑝)[𝐶], 𝑞[𝐴] + (1 − 𝑞)[𝐶]) cannot be a NE in mixed strategies: 𝑢(𝐴) =
1 1
6
1
9
𝑢(𝐶) implies 𝑝 = 𝑞 = 1/7. But, then, 𝑢𝑖 (7 , 7) = 7, while 𝑢𝑖 (𝐵, 7) = 7, which is higher.
(𝑝[𝐵] + (1 − 𝑝)[𝐶], 𝑞[𝐵] + (1 − 𝑞)[𝐶]) is a NE in mixed strategies: 𝑢(𝐵) = 𝑢(𝐶)
1 1
3
1
implies 𝑝 = 𝑞 = 1/4. Then 𝑢𝑖 (4 , 4) = 4, while 𝑢𝑖 (𝐴, 4) = 0, which is lower.
(c) There are 9 ways in which the first stage can finish. There are then 9+1=10 subgames.
Each player has 10 information sets. There are 3 ways to play in every information set, so
there are 310 strategies for each player.
(d) In the second stage only (B,B) or (C,C) can be played. Try play (A,A) in the first stage
and (B, B) in the second stage if (A,A) was played in the first stage and the mixed strategy
equilibrium in (b) otherwise, which is the best way to incentivize (A,A) in the first stage,
as gives the highest reward for doing it and also the highest punishment for not doing it.
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If players follow the strategy each one gets 6 + 3 = 9. If one deviates in the first stage
from A to B, he will get 9 + 3/4 = 9.75, and the strategy cannot be a subgame Nash
equilibrium.
(e) Play (A,A) in the first stage. Play (B,B) in the second stage if (A,A) was played in the
first stage and (C,C) in the second stage if (A,A) was not played in the first stage. Play
(B,B) in the third stage if (A,A) was played in the first stage and (B,B) was played in the
second stage, and play (C,C) in the third stage otherwise.
If players play according to the strategy each gets: 6 + 3 + 3 = 12. If one deviates in the
first stage from A to B, she gets 9 + 1 + 1 = 11, so it doesn’t pay. We do not need to
check for deviations in the second and third states as the strategy is already selecting NE in
those stages.
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II.4 Villages A and B have to decide simultaneously whether to contribute money to building a
road that connects them. Each can choose to contribute (paying a price of 10) or not to contribute.
The road will only be built if both of them contribute. Both villages know that having the road will
increase the utility of Village A by 20. The payoff of Village B from the road is either 10 or 20,
with probabilities 0.2 and 0.8, respectively. Only Village B knows its payoff from the road.
(a) Represent the situation as a Bayesian game. That is, show the set of players and the
players’ types, the beliefs about the types of other players, and the set of actions for each
type. Show also the players’ utilities in the usual matrix form. (6 points)
(b) List all the pure strategies for each player. (2 points)
(c) Find all pure-strategy Bayesian Nash equilibria of the game. (12 points)
(a) 𝑁 = {𝐴, 𝐵};
𝑇𝐴 = {𝐴}, 𝑇𝐵 = {𝐵10 , 𝐵20 };
(𝑝(𝐵10 |𝐴) = 0,2, 𝑝(𝐵20 |𝐴) = 0,8), (𝑝(𝐴|𝐵10 ) = 1), (𝑝(𝐴|𝐵20 ) = 1);
𝐶𝐴 = {0, 10}, 𝐶𝐵10 = {0, 10}, 𝐶𝐵20 = {0, 10}.
𝑝 = 0.2
𝐵10
0
10
0 0, 0 0, 0
A
10 0, 0 10, 0
1 − 𝑝 = 0.8
𝐵20
0
10
0, 0
0, 0
0, 0 10, 10
(b) 𝑆𝐴 = {0, 10}, 𝑆𝐵 = {(0, 0), (0, 10), (10, 0), (10, 10)}.
(c) 𝐵𝑁𝐸 = {(0, (0,0)), (10, (0,10)), (10, (10,10))}.
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