Simon Fraser University
Fall 2014
Econ 302 D100 Final Exam Solution
Instructor: Songzi Du
Monday December 8, 2014, 12 – 3 PM
This brief solution guide may not have the explanations necessary for full marks.
NE = Nash equilibrium, SPE = subgame perfect equilibrium, PBE = perfect
Bayesian equilibrium
1. (10 points) Consider the following simultaneous-move game (Table 1).
W
X
Y
Z
A
9, 2
4, 3
1, 2
4, 2
B
3, 2 3, 6
2, 5
7, 5
C
4, 0
7, 1
3, 0
3, 2
D
4, 3
4, 4
9, 3
3, 3
Table 1
Part i: Find the strategies that survive iterative deletion of strictly dominated strategies
(ISD). For each strategy that you delete, write down the strategy that strictly dominates it.
Part ii: Find all NE (pure and mixed) in this game, and for each NE that you find,
calculate the two players’ expected payoffs in the equilibrium.
Solution:
Part i, ISD:
Round 1: both W and Y are strictly dominated by X. Delete W and Y .
Round 2: both A and D are strictly dominated by 1/2B + 1/2C. Delete A and D.
The strategies that survive: B, C for player 1, and X, Z for player 2.
Part ii, NE:
For NE we can look at the smaller game given by ISD.
Clearly, there is no pure-strategy NE.
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Suppose player 1 uses mixed strategy pB + (1 − p)C and player 2 uses mixed strategy
qX + (1 − q)Z.
For both B and C to be player 1’s best responses, we must have: 3q + 7(1 − q) =
7q + 3(1 − q), i.e., q = 1/2.
For both X and Z to be player 2’s best responses, we must have: 6p+(1−p) = 5p+2(1−p),
i.e., p = 1/2.
Thus, the unique mixed-strategy NE is (1/2B + 1/2C, 1/2X + 1/2Z). In this equilibrium
player 1 gets an expected payoff of 5, and player 2 gets an expected payoff of 3.5.
2. (10 points) Three firms are considering entering a new market. The payoff for each
, where n is the number of firms that enter. The cost of entering is 62.
firm that enters is 150
n
(So the net payoff for a firm that enters is 150
− 62, and 0 for a firm that does not enter.)
n
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
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 mixed2
Figure 1
strategy NE, each firm enter with probability 4/5. In this NE, each firm gets an expected
payoff of 0.
3. (10 points) Find and report all pure-strategy SPE in Figure 1. Explain your answers.
Solution:
Clearly, in any SPE player 2 must use b and c, and player 1 must use D. In the subgame
following B, there are two pure-strategy NE: (E, e) and (F, f ).
Thus there are two pure-strategy SPE:
SPE #1: player 1 plays (A, D, E) and player 2 plays (b, c, e).
SPE #2: player 1 plays (A, D, F ) and player 2 plays (b, c, f ).
4. (10 points) Two players must choose among three alternatives, a, b, and c. Their
payoffs from each alternative are as follows:
a
b
c
player 1 10
8
-1
player 2
3
1
2
The rules are that player 1 moves first and vetoes one of the three alternatives. Then
player 2 chooses one of the remaining two alternatives.
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Figure 2
Part i: Draw the game tree.
Part ii: Find and report the pure-strategy SPE. Be sure to report a complete strategy
for each player.
Solution:
Part i: see Figure 2.
Part ii: the SPE is player 2 chooses b if a is vetoed, chooses a if b is vetoed, and
chooses b if c is vetoed; and player 1 chooses to veto b. (You may also write the SPE
by clearly marking the strategies on the tree.)
5. (10 points) There are two players. Player 1 is either a high (H) or low (L) type
worker, with probability 0.3 and 0.7 respectively. Player 1 knows his type and chooses to
get an MBA degree (D) or be content with his undergraduate degree (U). Player 2 who is
an employer observes player 1’s degree but not his type, and must decide to assign player 1
to be a manager (M) or a blue-colar worker (B).
Getting an MBA degree has a cost of cH = 3 for a high type worker and cL = 5 for a
low type worker. The market wage for a manager is wM = 10, and the market wage for
a blue-collar worker is wB = 6. The payoff of a high type worker who gets a blue-collar
assignment is wB − cH if he has gotten an MBA and is wB if he has not, and likewise in the
other cases. The employer does not care about the MBA degree per se. The employer’s net
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payoff (after paying the wage) depends on the worker’s type and assignment and is given by
the following table:
M
B
H
10
5
L
0
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Part i: Draw the game tree.
Part ii: Find all pure-strategy PBE (if there is any), and explain. Include conditional
beliefs (such as P(H | D) = 1 and P(H | U ) ≤ 1/2.) in your description of PBE.
Solution:
Part i: see Figure 3.
Figure 3
Part ii:
After observing MBA degree D, player 2 gets 10P(H | D) from assigning manager job
(D-M) and 5P(H | D)+4(1−P(H | D)) = 4+P(H | D) from assigning blue collar job (D-B);
so D-M is a best response if P(H | D) ≥ 4/9 and D-B is a best response if P(H | D) ≤ 4/9.
Likewise, after observing undergraduate degree U , U-M is a best response if P(H | U ) ≥
4/9 and U-B is a best response if P(H | U ) ≤ 4/9.
We now check four cases:
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1. If both types of player 1 get MBA (H-D, L-D), then player 2 would assign blue collar
job given MBA degree (D-B), since P(H | D) = 3/10 < 4/9. This means a payoff of 3
for the high type of player 1, but he can get at least 6 from not getting MBA (H-U).
So (H-D, L-D) cannot be a part of a PBE.
2. If both types of player 1 are do not get MBA (H-U, L-U), then player 2 would assign
blue collar job given undergraduate degree (U-B), since P(H | U ) = 3/10 < 4/9. For
both types of player 1 to not pursue MBA, player 2 must also assign blue collar job
given MBA degree (D-B), which is possible because P(H | D) is arbitrary (D happens
with probability 0). Therefore, it is a PBE that player 1 plays (H-U, L-U), and player
2 plays (U-B, D-B) with beliefs P(H | U ) = 3/10 and P(H | D) ≤ 4/9.
3. If high type of player 1 gets MBA and low type does not (H-D, L-U), then player 2 would
assign manager job given MBA degree and assign blue collar job given undergraduate
degree (D-M, U-B), since P(H | D) = 1 and P(H | U ) = 0. Given (D-M, U-B), it is
easy to check that (H-D, L-U) are best responses for player 1 of both types. Therefore,
it is a PBE that player 1 plays (H-D, L-U), and player 2 plays (D-M, U-B) with beliefs
P(H | D) = 1 and P(H | U ) = 0.
4. If high type of player 1 does not get MBA and low type gets MBA (H-U, L-D), then
player 2 would assign manager job given undergraduate degree and assign blue collar
job given MBA degree (D-B, U-M), since P(H | D) = 0 and P(H | U ) = 1. Given
(D-B, U-M), it is easy to check that (H-U, L-D) is not a best response for player 1 of
either type. So (H-U, L-D) cannot be a part of a PBE.
6. (15 points) Players 1 and 2 put a dollar each in a pot, and player 1 is dealt a card
which is either a king (K) or an ace (A) (with equal probability). Player 1 observes his card
and then decides whether to fold, forfeiting his dollar to player 2, or to bid, proceeding with
the game. If player 1 bids, then without knowing the card of player 1 player 2 can fold and
forfeit his dollar to player 1, or bid, in which case each player must add another dollar to
the pot. After bidding by both players, if player 1 has a king then player 2 wins the pot,
while if player 1 has an ace then player 1 wins the pot. See the game tree in Figure 4.
Part i: Find all pure-strategy PBE, if there is any. Explain your answers.
Part ii: Find a mixed-strategy PBE, with the following steps:
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Figure 4
(0) Notice that if player 1 has an ace, then he never folds (i.e., A-Bid); suppose player 1 with
a king bids with probability p, and player 2 bids with probability q, where 0 < p < 1
and 0 < q < 1. We need to calculate p and q that make a PBE.
(1) For player 1 with a king to mix between bidding and folding, he must be indifferent;
what probability q makes player 1 with a king indifferent?
(2) For player 2 to mix between bidding and folding, he must be indifferent; what belief
about player 1’s card (conditional on player 1 bidding) makes player 2 indifferent?
(3) What probability p gives (via Bayes’ rule) the conditional belief that makes player 2
indifferent?
(4) Steps 0 – 3 determine the mixed-strategy PBE. What are players 1 and 2’s expected
payoffs in this equilibrium? (For player 1, calculate his expected payoff before he is dealt
the card.) Is the game rigged in a player’s favor?
Solution:
Part i:
Clearly, player 1 with an ace will bid (A-Bid) in any PBE.
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1. If player 1 with a king bids (K-Bid), then player 2 believes that P(A | Bid) = 1/2 so
his best response is to bid if player 1 bids. If player 2 bids, then player 1 with a king
prefers to fold.
2. If player 1 with a king folds (K-Fold), then player 2 believes that P(A | Bid) = 1 so
his best response is to fold if player 1 bids. If player 2 folds, then player 1 with a king
prefers to bid.
Therefore, there is no pure-strategy PBE.
Part ii:
(1) −2q + 1(1 − q) = −1, i.e., q = 2/3.
(2) −2P(A | Bid) + 2(1 − P(A | Bid)) = −1, i.e., P(A | Bid) = 3/4.
(3)
1/2
1/2+p/2
= P(A | Bid) = 3/4, i.e., p = 1/3.
(4) In this equilibrium, player 1 with an ace gets 2 × 2/3 + 1 × 1/3 = 5/3, and player 1 with
a king gets −1 (since he is indifferent between bidding and folding). So overall (before
player 1 getting his card), player 1 gets 5/3 × 1/2 − 1 × 1/2 = 1/3.
If player 1 bids, then player 2 gets −1 (since he is indifferent between bidding and
folding), and if player 1 does not bid, then player 2 gets 1. Since player 1 bids with
probability 1/2 + 1/2 × 1/3 = 2/3, player 2 gets 2/3 × (−1) + 1/3 × 1 = −1/3.
The game is rigged in player 1’s favor, since his expected payoff is positive. If the game
were fair, then the expected payoffs of both players would be zero.
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