Simon Fraser University
Spring 2016
Econ 302 D200 Midterm Exam Solution
Instructor: Songzi Du
Tuesday March 8, 2016, 8:30 – 10:20 AM
1. (20 points) Your club has 10 members, and Julie is interested in joining your club.
The procedure for admitting a new member is simple: each existing member (which includes
you) receives a ballot that has two options: (1) admit Julie and (2) do not admit Julie. Each
person can check one of those two options or abstain by not submitting a ballot. For Julie
to be admitted, she must receive at least six votes in favor of admittance (checking option
1). Letting m be the number of ballots submitted with option 1 checked, assume that your
payoff function is

1 if m ≥ 6 (Julie is admitted)
0
if m ≤ 5 (Julie is not admitted)
For this question, it does not matter what is the payoff function for the other members.
(i) Is checking option 1 your strictly dominant strategy? Why or why not.
No, because for example if only one other member checks option 1, then checking option
1 does not matter: Julie will not be admitted in any case.
(ii) Is checking option 1 your weakly dominant strategy? Why or why not. Recall that a
strategy X is weakly dominant if X weakly dominates every other strategy; that is, for
any other strategy Y , X is always (for any strategies chosen by other players) weakly
better than Y and is sometimes strictly better than Y.
Yes, checking option 1 either doesn’t matter, or is strictly better than checking option
2 or abstaining (when exactly five other members check option 1).
(iii) Explain why abstaining is a weakly dominated strategy for you. Recall that a strategy
X is weakly dominated if another strategy weakly dominates X.
Checking option 1 weakly dominates abstaining, as explained above.
(iv) Now suppose you are tired at the end of the day, so that it is costly for you to attend
the evening’s meeting to vote. By not showing up, you abstain from the vote. This is
1
reflected in your payoff function having the form:



1




1/2


0




−1/2
if m ≥ 6 and you abstained
if m ≥ 6 and you voted
if m ≤ 5 and you abstained
if m ≤ 5 and you voted
Argue that now abstaining is not a weakly dominated strategy for you.
Checking option 1 does not weakly dominates abstaining: for example if six other
members check option 1, then abstaining is strictly better than checking option 1.
Abstaining weakly dominates checking option 2, since they always result in the same
m but you have to show up to check option 2. Thus, abstaining is not weakly dominated
by any other strategy.
2. (20 points) Three firms are considering entering a new market. The payoff for each
, where n (1 ≤ n ≤ 3) is the number of firms that enter. The cost of
firm that enters is 150
n
entering is 62. (So the net payoff for a firm that enters is 150
− 62, and 0 for a firm that does
n
not enter.)
Part i: Find all the pure-strategy NE, and calculate the firms’ payoffs in the equilibrium.
Part ii: Find the symmetric mixed-strategy NE in which all three firms enter with the
same probability, and calculate the firms’ payoffs in the equilibrium.
Solution:
Part i:
Clearly, it cannot be an equilibrium for only one firm to enter, since 150/2 − 62 > 0.
Moreover, it cannot be an equilibrium for all three firms to enter, since 150/3 − 62 < 0.
Thus, the pure-strategy NE is for two firms to enter (each with payoff 150/2 − 62 = 13) and
one firm to not enter (with payoff 0). So there are three pure-strategy NE, in which firm i
(and only firm i) does not enter, for each i = 1, 2, 3.
Part ii:
In the symmetric mixed-strategy NE, suppose that each firm enters with probability
2
0 < p < 1. For both entering and not entering to be best responses, we must have:
p2 (150/3 − 62) + (1 − p)2 (150 − 62) + 2p(1 − p)(150/2 − 62) = 0,
i.e.,
50p2 − 150p + 88 = 0.
The solutions to the above quadratic equation are p = 4/5 and p = 11/5, so we take the
first solution since it must be a number between 0 and 1. Thus, in the symmetric mixedstrategy NE, each firm enter with probability 4/5. In this NE, each firm gets an expected
payoff of 0.
3. (15 points) Find all pure-strategy subgame perfect equilibria (if any exists) of the
following game. Explain your answer.
1
a
2
c
0
1
e
b
2
d
1
2
c
2
d
1
4
0
3
f
2
1
g
0
1
Solution:
Player 2 is indifferent between actions f and g in the node after e. Thus, both f and
g are admissible in a subgame perfect equilibrium. So we consider two possibilities. In the
first possibility, player 2 plays f . Converting the rest of the game into the normal form, we
have:
c, f d, f
a 0, 1 1, 2
b 1, 4 0, 3
e 2, 1 2, 1
Clearly, the pure-strategy equilibria here are (e; c, f ) and (e; d, f ), which are subgame
perfect equilibria for the whole game.
In the second possibility, player 2 plays g. Converting the rest of the game into the
normal form, we have:
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c, g d, g
a 0, 1 1, 2
b 1, 4 0, 3
e 0, 1 0, 1
Clearly, the pure-strategy equilibria here are (a; d, g) and (b; c, g), which are also subgame
perfect equilibria for the whole game.
4. (20 points) There are two players. Player 1 is either strong or weak (those are his
types), and chooses a meal: beer or quiche. Player 2 observes this food choice, and chooses
to fight or retreat.
Player 2 cannot tell if player 1 is strong or weak, and believes that player 1 is strong with
probability 0.1 and weak with probability 0.9. (So Nature chooses player 1 to be strong with
probability 0.1 and weak with probability 0.9.) Player 2 gets 1 if he fights the weak player,
−1 if he fights the strong player, and 0 if he retreats.
The strong player 1 prefers beer; the weak player 1 prefers quiche. Player 1, strong or
weak, gets 2 if Player 2 doesn’t fight him, 0 otherwise; player 1, strong or weak, gets an
additional 1 if he eats his preferred meal.
(i) Draw the game tree.
(ii) Find all pure-strategy subgame perfect equilibria (if any exists), and explain.
Solution:
Part i:
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Part ii:
We go through player 1’s four pure strategies:
• Suppose player 1 uses (S-Beer, W-Beer).
Strategy
Expected payoff
B-Fight Q-Fight
0.1 × (−1) + 0.9 × 1 = 0.8
Player 2: B-Fight Q-Retreat
0.1 × (−1) + 0.9 × 1 = 0.8
B-Retreat Q-Fight
0.1 × 0 + 0.9 × 0 = 0
B-Retreat Q-Retreat
0.1 × 0 + 0.9 × 0 = 0
So player 2’s best responses are (B-Fight, Q-Fight) and (B-Fight, Q-Retreat).
If player 2 uses (B-Fight, Q-Fight), then W-Beer is not a best response for the weak
type of player 1. If player 2 uses (B-Fight, Q-Retreat), then W-Beer is also not a best
response for the weak type of player 1.
Thus our conjecture of (S-Beer, W-Beer) cannot be a part of a SPE.
• Suppose player 1 uses (S-Beer, W-Quiche).
Strategy
B-Fight Q-Fight
Player 2: B-Fight Q-Retreat
B-Retreat Q-Fight
B-Retreat Q-Retreat
Expected payoff
0.1 × (−1) + 0.9 × 1 = 0.8
0.1 × (−1) + 0.9 × 0 = −0.1
0.1 × 0 + 0.9 × 1 = 0.9
0.1 × 0 + 0.9 × 0 = 0
So player 2’s best response is (B-Retreat, Q-Fight).
If player 2 uses (B-Retreat, Q-Fight), then W-Quiche is not a best response for the
weak type of player 1.
Thus our conjecture of (S-Beer, W-Quiche) cannot be a part of a SPE.
• Suppose player 1 uses (S-Quiche, W-Beer).
Strategy
B-Fight Q-Fight
Player 2: B-Fight Q-Retreat
B-Retreat Q-Fight
B-Retreat Q-Retreat
Expected payoff
0.1 × (−1) + 0.9 × 1 = 0.8
0.1 × 0 + 0.9 × 1 = 0.9
0.1 × (−1) + 0.9 × 0 = −0.1
0.1 × 0 + 0.9 × 0 = 0
So player 2’s best response is (B-Fight, Q-Retreat).
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If player 2 uses (B-Fight, Q-Retreat), then W-Beer is not a best response for the weak
type of player 1.
Thus our conjecture of (S-Quiche, W-Beer) cannot be a part of a SPE.
• Suppose player 1 uses (S-Quiche, W-Quiche).
Strategy
Expected payoff
B-Fight Q-Fight
0.1 × (−1) + 0.9 × 1 = 0.8
Player 2: B-Fight Q-Retreat
0.1 × 0 + 0.9 × 0 = 0
B-Retreat Q-Fight
0.1 × (−1) + 0.9 × 1 = 0.8
B-Retreat Q-Retreat
0.1 × 0 + 0.9 × 0 = 0
So player 2’s best responses are (B-Fight, Q-Fight) and (B-Retreat, Q-Fight).
If player 2 uses (B-Fight, Q-Fight), then S-Quiche is not a best response for the strong
type of player 1. If player 2 uses (B-Retreat, Q-Fight), then S-Quiche is also not a best
response for the strong type of player 1.
Thus our conjecture of (S-Quiche, W-Quiche) cannot be a part of a SPE.
We conclude that there is no pure-strategy SPE for this game.
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