Game Theory
Steven Heilman
Please provide complete and wellwritten solutions to the following exercises.
Due March 1st, in the discussion section.
Homework 7
Exercise 1.
Let
n
be a positive integer. Let
v
: 2
{
1
,...,n
}
→ {
0
,
1
}
be a characteristic function
that only takes values 0 and 1. Assume also that
v
is monotonic. That is, if
S, T
⊆ {
1
, . . . , n
}
with
S
⊆
T
, then
v
(
S
)
≤
v
(
T
). The
ShapleyShubik power index
of each player is defined
to be their Shapley value.
By monotonicity of
v
, we have
v
(
S
∪ {
i
}
)
≥
v
(
S
) for all
S
⊆ {
1
, . . . , n
}
and for all
i
∈
{
1
, . . . , n
}
. Also, since
v
only takes values 0 and 1, we have
v
(
S
∪ {
i
}
)

v
(
S
) =
(
1
when
v
(
S
∪ {
i
}
)
> v
(
S
)
0
when
v
(
S
∪ {
i
}
) =
v
(
S
)
.
Consequently, we have the following simplified formula for the ShapleyShubik power index
of player
i
∈ {
1
, . . . , n
}
:
φ
i
(
v
) =
X
S
⊆{
1
,...,n
}
:
v
(
S
∪{
i
}
)=1 and
v
(
S
)=0

S

!(
n
 
S
 
1)!
n
!
.
Compute the ShapleyShubik power indices for all players on the UN security council, with
pre1965 and post1965 structure. Which structure is better for nonpermanent members?
In pre1965 rules, the UN security council had five permanent members, and six nonperma
nent members. A resolution passes only if all five permanent members want it to pass, and
at least two nonpermanent members want it to pass. So, we can model this voting method,
by letting
{
1
,
2
, . . . ,
11
}
denote the council, and letting
{
1
,
2
,
3
,
4
,
5
}
denote the permanent
members.
Then we use the characteristic function
v
: 2
{
1
,...,
11
}
→ {
0
,
1
}
so that, for any
S
⊆ {
1
, . . . ,
11
}
,
v
(
S
) = 1 if
{
1
,
2
,
3
,
4
,
5
} ⊆
S
and if

S
 ≥
7. And
v
(
S
) = 0 otherwise.
This voting method was called unfair, so it was restructured in 1965. After the restructuring,
the council had the following form (which is still used today). The UN security council has
five permanent members, and now ten nonpermanent members. A resolution passes only if