14.12  Fall 05
Solutions for Problem Set #2
Question 1 (30)
By inspection, it can easily be seen that there is no pure strategy Nash Equilibrium of the game. Whenever
a cell is considered, it is always possible to
f
nd a deviation by at least one of the players. Thus we conclude
that there is no pure strategy Nash Equilibrium.
In order to
f
nd all the possible mixed strategy Nash equilibria, we have to clear the game from the
strictly dominated strategies since we know from the lectures and recitations that in equilibrium, none of
the strictly dominated strategies are given a positive probability.
Let’s try to see if there is any strictly dominated strategies:
For player 1
B
is a bestresponse (BR) to
L
. Similarly
C
is a BR to
M
and
A
is a BR to
R
. Therefore
there is no strictly dominated strategy for player 1 at the
f
rst round
of reasoning.
For player 2 playing
L
is a BR to
A
. Similarly
M
is a BR to
B
and
L
is a BR to
C
. Therefore the only
possible candidate for strictly dominated strategy is
R
. Let’s see if there is any possible combination of
L
and
M
which dominates
R
. Calling the probability assigned to
L
as
p,
the combination of
L
and
M
has to
satisfy
all
of the following inequalities.
p
∗
1+(1
−
p
)
∗
0
>
0
p
∗
1+(1
−
p
)
∗
2
>
1
p
∗
2+(1
−
p
)
∗
1
>
1
These inequalities can be reduced to:
p>
0
,
1
>p, p>
0
respectively. As a result, any combination of
L
and
M
with
0
<p<
1
is doing better than playing
R
.
Therefore
R
is a strictly dominated strategy for player 2 at the
f
rst round of reasoning (Do not think that
they are playing sequentially. They are just doing the reasoning before playing the
simultaneous
game).
Since player 1 knows that player 2 will not play
R
, he eliminates his own strategy
A
since in the reduced
game (which excludes the strategy R of player 2)
A
is strictly dominated by both
B
or
C
.Noww
ea
r
el
e
f
t
with the following game
LM
B
2,1
1,2
C
1,2
2,1
Let’s say player 1 plays
B
with probability
µ
and player 2 plays
L
with probability
λ.
Then the mixed
equilibrium should satisfy the following equations
µ
∗
1+(1
−
µ
)
∗
2=
µ
∗
2+(1
−
µ
)
∗
1
λ
∗
2+(1
−
λ
)
∗
1=
λ
∗
1+(1
−
λ
)
∗
2
Therefore the solution is
µ
=
λ
=1
/
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View Full DocumentThis implies that the only Nash equilibrium is
NE
=
∙
1
2
B
+
1
2
C,
1
2
L
+
1
2
M
¸
Remember that a Nash equilibrium is the equilibrium strategies 
NOT
the payo
f
s.
About grading
: It was enough to give just one example for
p.
The details here were only for instructive
purposes.
Question 2 (30)
a
d
c
b
R
L
D
C
a
1
a
3
a
2
z
y
x
n
m
h
g
f
3
1
2
3
1
2
2
5
1
0
0
0
1
1
1
1
6
0
1
3
3
3
0
5
0
3
0
0
2
2
2
0
0
0
0
2
1
2
1
0
1
1
1
SG 1
SG 2
SG 3
SG 4
SG 5
SG 6
SG 7
Starting from the smallest subgame (SG 1), we are solving the game backwards. All the proper subgames
are taken into a circle in the above picture. Once we solve a SG, that particular subgame is replaced with
the sequentially rational decision’s payo
f
. Just to illustrate the substitution process, after solving the
f
rst
2
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 Spring '05
 MuhammadYildiz
 Game Theory, vi, player

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