# Chapter 13 Weighted Voting

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```					Chapter 13: Weighted Voting

• Banzhaf Power Index
• Shapley-Shubik Power Index
• Equivalent Systems
• Examples
Chapter 13 - Lecture Part 1
•   Vocabulary
•   Assumptions
•   Notation
•   Examples of Weighted Voting
•   The idea of power within a voting system
Weighted Voting Systems – An Example

• The United States Electoral College is an example of a weighted
voting system.

• We elect the President of the United States, not by a direct “one
person-one vote” method, but by the Electoral College. When you
vote for President, you are voting to have your state’s electoral votes
cast in favor of your choice for President.

• This system provides different weights to each of the states (and the
District of Columbia) based on population.

• Population size is assessed every 10 years through the census and
electoral votes are re-distributed among the states. This gives the
states differing voting weights.

• For example, California, with the highest population, currently has
55 votes in the Electoral College, while Florida has 27. Some states
(and the District of Columbia) have only 3 electoral votes. The
number of electoral votes a state has is the same as the total
representatives that state has in Congress (House and Senate).
The Electoral College – A Weighted Voting System

• There are a total of 538 votes divided among the states and a
simple majority of the electoral votes (270 votes) is required to win
the Presidency.

• A state’s electoral votes are awarded to the candidate with a
plurality of votes in that state.

• There are some exceptions: Maine and Nebraska divide their
electoral votes proportionally (according to the plurality winner of
each district).

• Colorado recently rejected a proposal (in 2004) to divide electoral
Weighted Voting Systems – Vocabulary & Notation

• In the language of voting theory, we say that the number of electoral
votes given to a state is that state’s voting weight.

• The number of electoral votes needed to elect a President, 270 in the
U.S. Electoral College, is called the quota of the weighted voting system.

• In general, with n voters, we use the notation

[ q : w1, w2, w3, w4, …, wn ]

to represent a weighted voting system with quota q and
weights w1, w2, w3, w4, …, wn .

• In the example of the Electoral College, q = 270, n = 51 (50 states +
D.C.) and wi is the weight (number of electoral votes) for state i.
Weighted Voting Systems - Vocabulary

• Some vocabulary:

– A coalition is a subset of the set of all voters within a voting
system. A coalition may consist of some, all or none of the
voters.
– Weighted voting systems decide measures.
– A coalition’s weight is the sum of the voting weights of its
members.
– A winning coalition consists of a set of voters with enough votes
to pass a measure. A losing coalition does not have enough
– A coalition is winning if it has a weight that is greater than or
equal to the quota of the system.
– A blocking coalition has enough votes to prevent a measure from
passing.
Weighted Voting Systems - Assumptions

• Some assumptions:

– A weighted voting system must be able to pass a measure.
Therefore, the quota can not be greater than the total weight of
all voters.

– We can not have opposing coalitions both win. Therefore, the
quota must be more than half the total weight of all voters.

Symbolically, we have

1
w1  w2  w3  ...  wn   q  w1  w2  w3  ...  wn 
2
Weighted Voting Systems - Assumptions

• This statement represents a fundamental assumption about
weighted voting systems:

1
w1  w2  w3  ...  wn   q  w1  w2  w3  ...  wn 
2
• Let w represent the total weight of all voters …

w  w1  w2  w3  ...  wn
1
• Then the above statement simply becomes:             wqw
2
• This translates into the statement: The quota of any weighted
voting system must less than or equal to the total weight of the
system and more than half the weight of the system.
Weighted Voting Systems - Assumptions

• What does the assumption about the quota of a weighted system

1
wqw
2
imply about the number of votes needed to block a measure ?

• To answer this question, suppose we have a voting system with
weight w and quota q satisfying the condition stated above.

• By definition, any coalition will need q votes to pass a measure.
Also, any coalition can block passage of a measure if it can
prevent any other coalition from collecting q votes.

• We’ll define the blocking quota of a system to be w – q + 1.
Weighted Voting Systems - Assumptions

• We have w = total weight of the system
q = quota
w – q + 1 = blocking quota

• Suppose coalition X has q votes. Then any opposing coalition
could collect at most the remaining w – q votes.

Coalition X            Coalition Y
q                     w-q

• Assuming q is more than half the total weight w, then w – q is less
than half the total weight.

• Coalition Y could block coalition X if it had more than w – q votes.
In that way, coalition X could no longer reach the quota q. That is,
Y can block if and only if it has the blocking quota w – q + 1.
Two Fundamental Questions

Given any weighted voting system

–   Can the blocking quota equal the quota ?

–   Can the blocking quota be greater than the quota ?
Two Fundamental Questions

1.   Can the blocking quota equal the quota ?
The answer to this question is yes.

 Suppose w – q + 1 = q. Then w + 1 = 2q and thus
q = (w + 1)/2.

 Note that q is an integer whenever w is odd.

 Note that q = (w +1)/2 will satisfy w/2 < q < w.

The blocking quota equals the quota when w is odd and
q = ( w + 1)/2.

It is possible that the quota and blocking quota are equal in a weighted
voting system.
Two Fundamental Questions

2.   Can the blocking quota be greater than the quota ?

The answer to this question is no.

   Suppose w – q + 1 > q. Then w + 1 > 2q and thus
q < (w + 1)/2.
   As in any weighted voting system, it must still be true that
w/2 < q.
   Consequently we have, w/2 < q < (w+1)/2.
   This implies w < 2q < w+1 which can never be true for integer
values of w and q.

We have shown deductively that w – q + 1 > q is impossible in a
weighted voting system using a proof by contradiction.
Weighted Voting Systems – An Example

Consider the weighted voting system [ 16 : 9, 9, 7, 3, 1, 1 ]. This
weighted system this could represent shareholders in a company.
Each shareholder has a different proportion of the vote when
measures are considered.
The total weight of the system is 30.
Thus 16 votes is a majority.

Because of the assumption
w/2 < q < w , we deduce the quota
must be more than 15 and less than
or equal to 30. Thus, in this case
16 < q < 30.

If q = 16, a majority of votes is
required to pass a measure. If
q = 30 then passage of a measure
requires unanimous support.
Note that any coalition of voters with 15 or more votes is a blocking
coalition in this system. Can you form a coalition that can block passage of
any measure but is unable to pass a measure ?
Weighted Voting Systems – An Example

Consider the weighted voting system [ 16 : 9, 9, 7, 3, 1, 1 ].

Here are some basic questions -

•   How many voters do we have ?

•   What is the quota ?

•   What is the total weight of all
voters combined ?

•   What is the blocking quota ?

5. What is the weight of the coalition of Dr.’s Mansfield, Ide, Lambert and
Edwards ?

6. Does the coalition mentioned above have sufficient votes to pass a
measure ? Do they have sufficient votes to block a measure ?
Weighted Voting Systems - Vocabulary

• More vocabulary:

– A dictator is a voter who can pass a measure even when all
others oppose the measure.

– A voter has veto power when that voter can block a measure
even when all others support it.

– A dummy voter is a voter who is never needed to win or block a
measure.
An Example of a Dummy

• In the weighted voting system [ 51: 26, 26, 26, 22 ] it seems as
though voting weights are at least near to being equally divided.

• Let’s name these voters A, B, C, and D, in that order.
That is, A, B, C each have 26 votes and D has 22 votes.

• We might consider this fair – perhaps D is a new partner in a
corporation, or is a new board member, perhaps D holds less stock in
some company, or represents a state with a slightly smaller
population.

• In this particular voting system, D is a dummy.

• We will find that D is never critical to any blocking or winning
coalitions. That is, any blocking or winning coalition will remain as
such with or without the support of D.
An Example of a Dummy

• In the weighted voting system [ 51: 26, 26, 26, 22 ] it seems as
though voting weights are at least close to equally divided.

• We have named these voters A, B, C, and D, in that order.

• Note that the total weight of this system is 100 and that the quota is
51. Also, note that the blocking quota is 50.

• Consider any coalition that includes D, the voter with 22 votes. D
alone cannot pass or block a measure. If D joins with exactly one
other voter, that coalition will still have insufficient votes (48 together)
to pass or block a measure.

• A third voter is still required to pass or block a measure … and with a
third voter, D is not needed. D is a dummy because D is never
required by any coalition to pass or block a measure.
An Example of a Dictator

• Consider the voting system [ 51: 60, 40 ].

• Suppose A is the voter with a voting weight of 60 and that B has
weight equal to 40.

• In this example, A is a dictator. A can pass or block a measure with
or without the support of B. The voting weight of A alone exceeds
both the quota and blocking quota.

• In a system with a dictator, all other voters are dummies.
An Example of Veto Power

• Consider the system [ 3 : 2, 1, 1 ] with voters A, B, and C,
respectively.

• In this example, voter A has veto power because A can block any
measure.

• The quota is 3 and the blocking quota is w – q + 1 = 4 – 3 + 1 = 2.

• Voter A ( a coalition of one ) has enough votes to block any measure
alone and therefore has veto power in this system.

• Note that A is not a dictator and can not pass a measure alone
because A does not have sufficient weight to pass a measure that all
others oppose.
Measuring Power

• Consider again the voting system [ 51: 26, 26, 26, 22 ] for voters A,
B, C and D, respectively.

• How can we conceive of the power of individual voters within this
system?

• The total weight w of all voters is 26 + 26 + 26 + 22 = 100.

• It seems reasonable to say each voter has some measure of power
corresponding to that voter’s share of the total votes.

• Perhaps one might suggest that individual voter power be measured
by the ratio

wi            where wi is the weight of voter i and w is the
w             total sum of the weights of all of the voters.
Measuring Power

• In the voting system [ 51: 26, 26, 26, 22 ] with voters A, B, C and D,
respectively, we consider the voting power of individual voters …

wi
• We will call the ratio        the nominal power for voter i.
w
• Does the nominal power truly represent voter D’s power?

• In this case 22/100 = .22 meaning we would say D has a nominal
power of 22%. ( That is, D has 22% of voting weight of this system.)

• Remember that, in another sense, D really has no power in this
particular voting system because we have already discovered that D
is a dummy voter.
Measuring Power

• In this chapter, we study two other methods for measuring power in a
weighted voting system:

– The Banzhaf Power Index

– The Shapley-Shubik Power Index

• The goal is to provide a more reasonable measure of power for
voters within a weighted voting system.

• We will consider when these methods are appropriate and when they
may not be appropriate.

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