14.12 Game Theory
Fall 2004
Answers for Midterm 1, Fall 2004
1. Consider the following game:
A
B
C
a
3
,
0
0
,
3
0
, x
b
0
,
3
3
,
0
0
, x
c
x,
0
x,
0
x, x
(a) The question does not ask to
fi
nd all Nash equilibria. One way is to
fi
nd all the purestrategy Nash
equilibria and then use the symmetry of the game to guess a symmetric mixedstrategy Nash equilibria
and check that it is indeed a Nash equilibria. The Nash equilibria of that game are
(
c, C
)
(
1
3
a
+
1
3
b
+
1
3
c
;
1
3
A
+
1
3
B
+
1
3
C
)
(
1
2
a
+
1
2
b
;
1
2
A
+
1
2
B
)
In the solutions, we
fi
nd all the equilibria starting with the purestrategy ones. The best responses of
player 1 are
BR
1
(
A
)
=
{
a
}
BR
1
(
B
)
=
{
b
}
BR
1
(
C
)
=
{
c
}
The best responses of player 2 are
BR
2
(
a
)
=
{
B
}
BR
2
(
b
)
=
{
A
}
BR
2
(
c
)
=
{
C
}
Hence the only purestrategy Nash equilibrium is
(
c, C
)
In a mixedstrategy equilibrium, player 2 is playing
A
with probability
p
,
B
with probability
q
and
C
with probability
1
−
p
−
q
. The expected payo
ff
s of player 1 are
EU
1
(
a
)
=
3
p
EU
1
(
b
)
=
3
q
EU
1
(
c
)
=
1
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Hence his best responses are
p
>
1
3
and
p > q
=
⇒
a
q
>
1
3
and
q > p
=
⇒
b
1
3
>
p
and
1
3
> q
=
⇒
c
p
=
q >
1
3
=
⇒
{
a, b
}
p
=
1
3
> q
=
⇒
{
a, c
}
q
=
1
3
> p
=
⇒
{
b, c
}
p
=
q
=
1
3
=
⇒
{
a, b, c
}
By symmetry, if player 1 is playing
a
with probability
r
,
b
with probability
s
and
c
with probability
1
−
r
−
s
, the best responses of player 2 are
r
>
1
3
and
r > s
=
⇒
B
s
>
1
3
and
r > s
=
⇒
A
1
3
>
r
and
1
3
> s
=
⇒
C
r
=
s >
1
3
=
⇒
{
A, B
}
r
=
1
3
> s
=
⇒
{
B, C
}
s
=
1
3
> r
=
⇒
{
A, C
}
r
=
s
=
1
3
=
⇒
{
A, B, C
}
 Let’s
fi
rst look at an equilibrium where player 2 is mixing among
{
A, B, C
}
with positive probabili
ties. In order for this strategy to be a best response it must be the case that player 1’s probabilities
are
r
=
s
=
1
3
. Player 1 is playing his three strategies with a positive probability which is a best
response to
p
=
q
=
1
3
. We conclude that
(
1
3
a
+
1
3
b
+
1
3
c
;
1
3
A
+
1
3
B
+
1
3
C
)
is a Nash equilibrium.
 Then look at an equilibrium where player 2 is mixing among
{
A, B
}
. This means that
p
+
q
= 1
and the best responses of player 1 are
p
>
1
2
=
⇒
1
a
=
⇒
2
B
cannot be an equilibrium
p
=
1
2
=
⇒
1
{
a, b
}
p
<
1
2
=
⇒
1
b
=
⇒
2
A
cannot be an equilibrium
Player 1 is thus mixing among
{
a, b
}
.And using the symmetry, this can be the case in equilibrium
if and only if
r
=
s
=
1
2
. We conclude that
(
1
2
a
+
1
2
b
;
1
2
A
+
1
2
B
)
is a Nash equilibrium.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '05
 MuhammadYildiz
 Equilibrium, Game Theory, player

Click to edit the document details