chapter 13 – weighted voting part 3: the shapley-shubik power index

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


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.


• 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.


• 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.



• 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.



• 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 __ __ __ __ which can happen 5! ways

__ __ 5 __ __ __ can happen 5! ways

__ __ __ 5 __ __ can happen 5! ways

__ __ __ __ 5 __ can happen 5! ways

__ __ __ __ __ 5 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.



• 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

th of the power

• 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.

30

1

561



• Result: The Shapley-Shubik power index for this system is

• Can you determine the Banzhaf power index for this system ?

.30

1,

30

1,

30

1,

30

1,

30

1,6

5

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 two-year 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 ].


• 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.


• 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.


• 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

2145

4

351113

22

910111314

484

9101112131415

12345684

!15

!8!69

6

C


• 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.


• 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.


• 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.



• 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.



• 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%

chapter 13 – weighted voting part 3: the shapley-shubik power index

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