Econ 302
Assignment 4 — Solution
These three games are sequential games of perfect information. We solve them by backward
induction: start at the nodes/information sets one level above the terminal nodes, and solve
for the best responses; after solving those best responses, move up one more level and solve
for the best responses there; and so on.
1. In the subgame (in stage 2) in which firm 1 had produced q1 units, firm 2’s best response
is to solve:
max(a − b(q1 + q2 ))q2 − cq2 .
q2
The first-order condition is:
a − bq1 − 2bq2 − c = 0,
or
a − c − bq1
a−c 1
=
− q1 .
(1)
2b
2b
2
Equation (1) is the equilibrium strategy of firm 2; it specify firm 2’s best-responding
action after every contingency (which is a quantity of firm 1, q1 ).
Given that player 2 is playing according to the strategy (1), firm 1 in the beginning of
the game (stage 1) solves:
q2∗ (q1 ) =
a−c 1
max a − b q1 +
− q1
q1 − 2cq1 .
q1
2b
2
The first order condition is:
a − bq1 −
a−c
− 2c = 0,
2
or
a − 3c
.
(2)
2b
Equation (2) is the equilibrium strategy of firm 1, and together with the strategy of firm
2 in (1) they form the subgame-perfect equilibrium of this game.
In this equilibrium firm 1 produces q1∗ = a−3c
units, and firm 2 produces q2∗ (q1∗ ) = a+c
2b
4b
a+5c
a+c
units. The market price in this equilibrium is P = a − b a−3c
+
=
.
2b
4b
4
q1∗ =
1
2. In the subgame (in stage 3) in which firm 1 had produced q1 units and firm 2 had
produced q2 units, firm 3’s best response is to solve:
max(a − b(q1 + q2 + q3 ))q3 − cq3 .
q3
The first-order condition is:
a − b(q1 + q2 ) − 2bq3 − c = 0,
or
a−c 1
a − c − b(q1 + q2 )
=
− (q1 + q2 ).
(3)
2b
2b
2
Equation (3) is the equilibrium strategy of firm 3; it specify firm 3’s best-responding
action after every contingency (which is a quantity of firm 1, q1 , plus a quantity of firm 2,
q2 ).
In the subgame (in stage 2) in which firm 1 had produced q1 units, and anticipating the
strategy (3) of firm 3 in stage 3, firm 2’s best response is to solve:
q3∗ (q1 , q2 ) =
a−c 1
− (q1 + q2 )
q2 − cq2 .
max a − b q1 + q2 +
q2
2b
2
The first order condition is
b
a−c
a − q1 − bq2 −
− c = 0,
2
2
or
a−c 1
− q1 .
(4)
2b
2
Equation (4) is the equilibrium strategy of firm 2; it specify firm 2’s best-responding
action after every contingency (which is a quantity of firm 1, q1 ).
Finally, in the beginning of the game (stage 1), firm 1’s best response, anticipating the
strategies (4) and (3), is to solve:
q2∗ (q1 ) =
a−c 1
a−c 1
a−c 1
max a − b q1 +
− q1 +
−
q1 +
− q1
q1 − cq1
q1
2b
2
2b
2
2b
2
3(a − c) 1
+ q1
q1 − cq1
= max a − b
q1
4b
4
2
The first order condition is
a−
3(a − c) b
− q1 − c = 0,
4
2
or
a−c
.
(5)
2b
Strategies (5), (4) and (3) form a subgame perfect equilibrium.
In this equilibrium firm 1 produces q1∗ = a−c
units, firm 2 produces q2∗ (q1∗ ) = a−c
, and firm
2b
4b
3 produces q3∗ (q1∗ , q2∗ ) = a−c
units. Notice that firm 1 and 2 produce the same quantities as if
8b
there are only two firms. The market price in this equilibrium is P = a − 78 (a − c) = 81 a + 78 c.
q1∗ =
3.
Given the huge negative payoff (−1000000000) of death, each player wants to
minimize his probability of death. Conditional on being alive, the player wants to kill as
many opponents as possible: a payoff of 2 for two dead, 1 for one dead, 0 for zero dead.
Here is one subgame perfect equilibrium:
In the (last) subgame where Z (assuming that he is alive) is shooting, Z shoots randomly
at one of the alive players. That is, if only one other player is alive, Z shoots him; if both X
and Y are alive, Z plays the mixed strategy of shooting X with probability 1/2 and shooting
Y with probability 1/2. This is a best-response for Z, because he hates X and Y equally.
In the subgame where Y (assuming that he is alive) is shooting, Y shoots Z if Z is alive,
shoots X if Z is dead. This is a best response for Y, because if Z is alive he will be shot at
by Z.
In the subgame where X is shooting (the beginning of the game), X shoots shoots into
the air. The probability that X dies when he shoots into air is 0.4 × 0.5 = 0.2: Y misses
in his shot of Z (probability 0.4), and Z chooses to shoot X (probability 0.5) in the final
round. When X shoots Y, the probability that X dies is 0.3 × 1 + 0.7 × 0.2 = 0.44: if X’s
bullet hits Y, X dies for sure (shot by Z, who never misses, in the final round); if X’s bullet
misses Y, we are back to the scenario of shooting into the air. Likewise, when X shoots Z
the probability that X dies is 0.3 × 0.6 + 0.7 × 0.2 = 0.32. Therefore, shooting into the air
is the best response.
This completes the description of a subgame perfect equilibrium. The game tree in the
next page is produced by Haiyun Kevin Chen (notice that D = −1000000000).
3
Air
   
D
1
D D
2
1
X
Z
chance
   
D
D
D  1 
2
1
Y
Air
[.6] Hit
Z
[.3] Hit
X
Y
Z
Miss [.4]
Air
X
Z
 
1
1
D
[.6] Hit
chance
Miss [.7]
X
Y
Z
Miss [.4]
Air
     
D
1
0
 1   D  0
1
1
0
Y
Air
Y
Air
     
D
1
0
 1   D  0
1
1
0
X
Z
Miss [.4]
Y
Z
Air
[.6] Hit
chance
   
D
D
D  1 
2
1
 
1
1
D
Air
Y
Air
     
D
1
0
 1   D  0
1
1
0
X
Z
Miss [.4]
X
Z
 
1
1
D
[.6] Hit
chance
Miss [.7]
X
Y
Z
Miss [.4]
Air
     
D
1
0
 1   D  0
1
1
0
Y
Air
Y
Z
Air
Air
     
D
1
0
 1   D  0
1
1
0
X
Figure 1: Extensive Form Representation of the Three-Way Duel in Assignment 4
   
D
1
2 1
D
D
[.6] Hit
chance
X
Y
[.3] Hit
chance
Z
Air
   
D
D
D  1 
2
1
Y
Z
[.6] Hit
chance
Y
Air
     
D
1
0
 1   D  0
1
1
0
X
Z
Miss [.4]
X
 
1
1
D
[.6] Hit
chance
Z
X
Y
Z
Miss [.4]
Air
     
D
1
0
 1   D  0
1
1
0
Y
The black letters {X, Y, Z} denote the players who are shooting and the red letters {X, Y, Z, Air} denote the targets at which a player shoots. The blue letters represent chance and its
“moves”. Letter D in the payoff vector indicates a player is dead.
     
D
1
0
 1   D  0
1
1
0
chance
Y
X
Air
Y
Air
     
D
1
0
 1   D  0
1
1
0
X
Z