Transcript Chapter 13 – Weighted Voting

Chapter 13 – Weighted Voting
Part 3:
The Shapley-Shubik Power Index
Shapley-Shubik Power Index
• List all permutations of all voters within a weighted voting system.
• Add weights of individual voters in each permutation, consecutively,
from left to right.
• A voter is critical ( or pivotal ) in a particular permutation if that voter’s
weight changes the corresponding coalition from losing to winning.
• An individual voter’s Shapley-Shubik power index is the fraction of
times that voter is critical out of the total number of permutations of
all voters.
Shapley-Shubik Power Index – Example 1
• Consider the weighted voting system [ 7: 5, 3, 2 ].
• We will calculate the Shapley-Shubik power index for this system.
We must list all permutations of the three voters – let’s name them A, B,
and C, respectively. There are 3! = 6 permutations of 3 voters.
Permutations
Sums, cumulative, left to right
Critical
A
B
C
5
8
10
B
A
C
B
5
7
10
C
B
A
C
3
8
10
A
B
C
A
3
5
10
A
C
A
B
2
7
10
A
C
B
A
2
5
10
A
Shapley-Shubik Power Index – Example 1
Permutations
Sums, cumulative, left to right
Critical
A
B
C
5
8
10
B
A
C
B
5
7
10
C
B
A
C
3
8
10
A
B
C
A
3
5
10
A
C
A
B
2
7
10
A
C
B
A
2
5
10
A
Conclusion: The Shapley-Shubik power index for this weighted voting
system is given by ( 2/3, 1/6, 1/6 ).
This is because voter A is critical 4 out of 6 times, hence has power
index 4/6 = 2/3. Likewise, voters B and C are each critical in 1 out of 6
possible permutations and thus have power indices of 1/6 each.
Comparing Results
• We may now compare the results of the Shapley-Shubik and the
Banzhaf analysis of power.
• For the system [ 7: 5, 3, 2 ] we find the Banzhaf power index to be
( 6, 2, 2 ) as shown below …
Coalition
Weight
Extra votes
Critical voters
A, B, C
10
3
A
A, B
8
1
A, B
A, C
7
0
A, C
B, C
5
A
5
B
3
C
2
Empty
0
Comparing Results
Given the weighted voting system [ 7: 5, 3, 2 ]
– We found the Banzhaf power index to be ( 6, 2, 2 ). Writing this index in
the form where each value corresponds to the fraction of the total power
each voter has as measured by the Banzhaf approach we have ( 6/10,
2/10, 2/10 ) or ( 3/5, 1/5, 1/5 ).
According to the Banzhaf approach, the first voter has 60% of the power
while the other two voters each have 20% of the power.
– According to the Shapley-Shubik approach, we found the power index
to be ( 2/3, 1/6, 1/6 ). That is, the first voter has 67% of the power, while
the other two voters each have approximately 17% of the power.
– If we measure the nominal power of these voters we find the nominal
power index for this system is (5/10, 3/10, 2/10) or (1/2, 3/10, 1/5). That
is, the analysis of nominal power shows the first voter has 50% of the
power while the second voter has 30% and the last voter has 20% of
the power.
Shapley-Shubik Power Index – Example 2
• We need not always list all permutations of the given voters. For
example, if all voters except one have the same weight, it is possible
to count permutations without listing them.
• For example, consider the weighted voting system
[ 6 : 5, 1, 1, 1, 1, 1 ].
• To calculate the Shapley-Shubik power index directly requires listing
6! = 720 permutations of voters. To avoid this, we note that only one
voter has weight 5 while the others have weight 1.
• We will determine the Shapley-Shubik power for the voter with weight
5 and first, and then for each voter with weight 1.
• We can show all of the voters with weight 1 will share power equally
within this system.
Shapley-Shubik Power Index – Example 2
• We are considering the voting system [ 6 : 5, 1, 1, 1, 1, 1 ] which
has 6 voters.
• Considering the permutations of these voters, we note that there are
6 possible positions for the voter with weight 5.
__ __ __ __ __ __
• The voter with weight 5 will be critical, for example, when in the
second position in the list.
__ 5 __ __ __ __
• How many ways can this happen ? We must count all of the
rearrangements (permutations) of the other 5 voters, which is 5! =
120.
Shapley-Shubik Power Index – Example 2
• We are considering the voting system [ 6 : 5, 1, 1, 1, 1, 1 ] which
has 6 voters.
• The voter with weight 5 will also be critical when in the third place in
permutations of these voters.
__ __ 5 __ __ __
• How many ways can this happen ? Again, the answer is 5! = 120 by
counting rearrangements of the other 5 voters.
• In fact, the voter with weight 5 will be critical in any position in the
sequences of voters except when in the first position. That is, the
voter with weight 5 will be critical 5*120 = 600 out of 720 times.
Shapley-Shubik Power Index – Example 2
• We are considering the voting system [ 6 : 5, 1, 1, 1, 1, 1 ] which
has 6 voters.
• Summary: The voter with weight 5 will be critical in each of these
cases: For each case we count the ways of rearranging (permuting)
the remaining 5 voters.
__
__
__
__
__
5 __ __ __ __
__ 5 __ __ __
__ __ 5 __ __
__ __ __ 5 __
__ __ __ __ 5
which can happen 5! ways
can happen 5! ways
can happen 5! ways
can happen 5! ways
can happen 5! ways
• The voter with weight 5 is critical 5*5! = 600 times out of 6! = 720
total permutations.
• So the Shapley-Shubik power for this voter is 600/720 = 5/6.
Shapley-Shubik Power Index – Example 2
• We are considering the voting system [ 6 : 5, 1, 1, 1, 1, 1 ] which
has 6 voters.
• We have determined the voter with weight 5 has a Shapley-Shubik
power of 5/6. We know that the other voters will share the
remaining 1/6 of the power. Thus, each of the other voters has
1
6  1 th of the power
5
30
• Alternatively, we can deduce the power of the remaining voters by
noting that each is critical only when in the second position in the
permutations of all voters and only when following the voter with
weight 5. This can happen for each of these voters 4! = 24 ways.
Thus, we find each of these voters has power 24/720 = 1/30.
Shapley-Shubik Power Index – Example 2
• We are considering the voting system [ 6 : 5, 1, 1, 1, 1, 1 ] which
has 6 voters.
• Result: The Shapley-Shubik power index for this system is
5 1 1 1 1 1 
 , , , , , .
 6 30 30 30 30 30 
• Can you determine the Banzhaf power index for this system ?
United Nations Security Council – Shapley Shubik Power Index
• We will determine the power of member nations in the United
Nations Security Council using the Shapley – Shubik power index.
• The United Nations Security Council is made of 15 member nations:
5 of these are considered permanent members while 10 other
member nations are nonpermanent members elected to serve a twoyear term.
• To pass a measure, a coalition of nations in the U.N. Security Council
must have the support of all of the permanent members and at least
4 of the nonpermanent members.
• These conditions mean that the U.N. Security Council is a weighted
voting system with the following configuration:
[ 39 : 7, 7, 7, 7, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 ].
United Nations Security Council – Shapley Shubik Power Index
• It would be extraordinarily tedious to list all 15! = 1.3 trillion
permutations of these members so we take another approach…
• Considering the system [ 39 : 7, 7, 7, 7, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 ]
we note that a nonpermanent member x would be critical if and only
if that member were in the 9th position in a permutation of all voters.
__ __ __ __ __ __ __ __ x __ __ __ __ __ __
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
• This is because, for a nonpermanent member to be critical, the votes
of the 5 permanent members must be secured as well as the votes of
three other nonpermanent members.
United Nations Security Council – Shapley Shubik Power Index
• A winning coalition in which a nonpermanent member is critical could
look something like this
P P P P P NP NP NP x NP NP NP NP NP NP
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
• We must count all of the ways in which x could be the ninth member
nation to join such a coalition and therefore be pivotal to it’s winning.
• First, there are 15! possible permutations total (about 1.3 trillion)
• There are 9C6 = 84 ways to select the remaining 6 nations that follow
x and then 6! = 720 ways to permute just those 6 members.
• There are 8! = 40320 ways to permute the 8 nations before x. Five
of them must be permanent members and therefore three of them
must be nonpermanent members.
United Nations Security Council – Shapley Shubik Power Index
• A winning coalition in which a nonpermanent member is critical could
look something like this
P P P P P NP NP NP x NP NP NP NP NP NP
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
• Thus, the Shapley-Shubik power for a nonpermanent member will be
C 6!8! 
9
6
15!

84  6  5  4  3  2 1
84  4


15 14 13 12 1110  9 14 13 1110  9
22
4

13 11 5  3 2145
United Nations Security Council – Shapley Shubik Power Index
• We have determined the Shapley-Shubik power for an individual
nonpermanent member of the U.N. Security Council to be 4/2145.
• Because there are 10 such members, this makes up 10 times 4/2145
of the power which leaves (2145 – 40)/2145 = 2105/2145 of the
power for each of the 5 permanent members.
• Thus the permanent members each have 1/5 of 2105/2145 of the
power which is 421/2145.
• Conclusion: According to the Shapley-Shubik analysis of power in
the United Nations Security Council each of the permanent member
nations has 421/2145 (or about 19.6%) of the power while each of
the nonpermanent members each has about 4/2145 (or about
0.19%) of the power.
United Nations Security Council – Banzhaf Power Index
• We proceed now to determine the Banzhaf power index for the
United Nations Security Council.
• We are going to find the Banzhaf power index for the weighted voting
system [ 39: 7, 7, 7, 7, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 ].
• As before, we will avoid a direct approach, which in this case would
require listing 210=1024 different coalitions (subsets) of the member
nations.
• Instead, we first recognize that, as stated previously, any winning
coalition must include all 5 of the permanent members and at least 4
nonpermanent members. Also, because we are to determine the
Banzhaf power index, we do not count different permutations of the
voters, only combinations.
United Nations Security Council – Banzhaf Power Index
• We recognize that any winning coalition must include all 5 of the
permanent members and at least 4 nonpermanent members.
• The total number of distinct winning coalitions, without regard to the
order in which they are formed, is given by
10C4
+
10C5
+
10C6
+
10C7
+
10C8
+
10C9
+
10C10 =
= 210 + 252 + 210 + 120 + 45 + 10 + 1 = 848
• The five permanent members of the Security Council are critical in
each of these winning coalitions because any winning coalition must
include all five permanent members. Thus, the Banzhaf power for
each permanent member is 2*848 = 1696, where we multiply by 2 to
account for the number of blocking coalitions to which each
permanent member is critical.
United Nations Security Council – Banzhaf Power Index
• We have a total of 848 distinct winning coalitions possible in the U.N.
Security Council with each of the permanent members being critical
in each of these winning coalitions.
• Of all 848 winning coalitions, a nonpermanent member would be
critical when joined by the five permanent members and exactly
three other non-permanent members.
• For each nonpermanent member, there are 9C3 = 84 ways to choose
3 nonpermanent members from the remaining 9 nonpermanent
members. Thus each nonpermanent member is critical in 84 winning
coalitions and therefore also critical to 84 blocking coalitions.
• The Banzhaf power for each nonpermanent member of the U.N.
Security Council is 84*2 = 168.
United Nations Security Council – Comparing Measures of Power
• For the weighted voting system [ 39: 7,7,7,7,7,1,1,1,1,1,1,1,1,1,1 ]
which represents the United Nations Security Council
• The Banzhaf power index for this system is
– 1696 for each permanent member and 168 for each
nonpermanent member.
– Note that the total number of time all voters are critical is
1696(5)+168(10) = 10160
– This means that according to the Banzhaf model, each
permanent member has 1696/10160 = 16.7% of the power while
each nonpermanent member has 1.6% of the power.
United Nations Security Council – Comparing Measures of Power
• For the weighted voting system [ 39: 7,7,7,7,7,1,1,1,1,1,1,1,1,1,1 ]
which represents the United Nations Security Council
• As stated previously, the Shapley-Shubik power index for this
system is
– 421/2145 for each permanent member and 4/2145 for each
nonpermanent member.
– This means that each permanent member has approximately
19.6% of the power, while each nonpermanent member has
approximately 0.196% of the power.
United Nations Security Council – Comparing Measures of Power
• For the weighted voting system [ 39: 7,7,7,7,7,1,1,1,1,1,1,1,1,1,1 ]
which represents the United Nations Security Council
• Measuring the nominal power of member nations we find
– each permanent member has nominal power 7/45 = 15.6%
– each nonpermanent member has nominal power 1/45 = 2.2%