MS&E 246: Game Theory with Engineering Applications
Problem Set 1
1.
(a) (3pts) When
s

i
= 1, any strategy
s
i
yields a zero payoff.
So,
there are no strictly dominated strategies.
(b) (3pts)
s
i
= 0 is the only weakly dominated strategy. It is weakly
dominated by any strategy
s
i
∈
(0
,
1].
(c) (2+2pts) Any (
s
1
, s
2
) with
s
1
+
s
2
= 1 is a pure strategy NE. The
payoff of player
i
is
s
i
. Given the strategy
s

i
, player
i
has no
incentive to deviate from the strategy
s
i
= 1

s

i
. If he chooses
s
0
i
< s
i
his payoff will be only
s
0
i
. If he chooses
s
0
i
> s
i
, his payoff
will be zero. (1,1) is also a Nash equilibrium.
2.
(a) (6pts) Suppose player
i
has two weakly dominant strategies,
s
0
i
and
s
00
i
. Then
u
i
(
s
0
i
, s

i
)
≥
u
i
(
s
i
, s

i
) for all
s

i
and for all
s
i
. In
particular,
u
i
(
s
0
i
, s

i
)
≥
u
i
(
s
00
i
, s

i
) for all
s

i
. Since
s
00
i
is weakly
dominant, there is some
s

i
such that
u
i
(
s
00
i
, s

i
)
> u
i
(
s
0
i
, s

i
),
which is a contradiction.
Therefore, a player can not have two
weakly dominant strategies.
(b) (4pts) According to (a), there can be no such game.
3.
(a) (3pts) Playing LL, L or R is quite risky, since we do not know
what player 1 will be playing. The risk of obtaining a playoff of
49 is very large compared to the payoff of 1 if player 2 player L,
and the risk of obtaining a payoff of 100 is very large compared
to the payoff of 2 if player 2 played LL or R. Therefore, it seems
”reasonable” to play M.
(b) (4+6pts) The two pure NE of this game are (U,LL) and (D,R).
To check for mixed strategy NE, player 1 must mix between U
(with probability p) and D (with probability 1p). Player 2 has 11
possible mixing combinations:
{
LL,L
}
,
{
LL,M
}
,
{
LL,R
}
,
{
L,M
}
,
{
L,R
}
,
{
M,R
}
,
{
M,L,R
}
,
{
LL,M,R
}
,
{
LL,L,R
}
,
{
LL,L,M
}
and
{
LL,L,M,R
}
.
We will show that only the
{
LL,L
}
is part of a
mixed NE.
For player 2 to mix between LL and L (with probabilities q and
1q respectively), we must have
p
·
(2) + (1

p
)
·
(

100) =
p
·
(1) + (1

p
)
·
(

49) which yields
p
=
51
52
. The utility of player
2 from each strategy is then:
u
2
(
LL
) =
u
2
(
L
) =
1
26
,
u
2
(
M
) = 0
and
u
2
(
R
)
<
0. Then for player 1 to mix between U and D, we
1
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must have:
q
·
(100)+(1

q
)
·
(

100) =
q
·
(

100)+(1

q
)
·
(100)
which yields
q
=
1
2
and
u
1
(
U
) =
u
1
(
D
) = 0. Therefore,
p
=
50
51
and
q
=
1
2
is a mixed NE. We now show that no other mixing
combination of player 2 can be part of a mixed NE.
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 Winter '07
 JOHARI
 Game Theory, player, Highest Bid, price auction, First Price Auction

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