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WEIGHTED VOTING
Suppose a county commission consists of three members, one representing each of the three cities in the
county.
Voting power on the commission is proportional to the population of the cities with the commissioner from
city A getting 49 votes, the commissioner from city B getting 40 votes, and the commissioner from city C getting 11
votes.
Since there are 100 total votes, 51 votes are required to carry any measure. In this scenario the representative
from city C has as much power as the representative from city A since no vote can be won without support from two
commissioners.
In fact all three commissioners have equal power.
This is counterintuitive since one might have
expected A to have more than 4 times the power of C.
This is an example of a weighted voting system.
The number of votes required to win a vote, in this case 51,
is called the
quota
.
The number of votes each voter casts is called the voter’s
weight
.
The notation we use to describe
this voting system is [51: 49, 40, 11].
Let’s consider the set of all commissioners: {A, B, C}. In the study of weighted voting, each subset is called a
coalition
.
The weight of the coalition is defined to be the sum of the weight of its voters.
If the weight of the
coalition is greater than or equal to the quota, we call that coalition a
winning coalition
.
Otherwise, it is a
losing
coalition.
Coalition
Weight
Type
{A}
49
losing
{B}
40
losing
{C}
11
losing
{A, B}
89
winning
{A, C}
60
winning
{B, C}
51
winning
{A, B, C}
100
winning
{
}
0
losing
Example:
What is wrong with the following weighted voting system? [21: 6, 5, 5, 3]
Solution:
The quota is too high.
Even if every voter voted for a proposition, it couldn’t pass.
The quota must always
be less than or equal to
the sum of the weights.
Example:
What is wrong with the following weighted voting system? [8: 6, 5, 5, 3]
Solution:
The quota is too low.
If the first and last voters voted yes and the middle two voters voted no, both
coalitions would win.
We can’t allow a proposition to both pass and fail.
The quota must always be greater than
half
the total number of votes.
Example:
A weighted voting system has 5 voters with weights 1, 1, 2, 2, and 3.
Find all possible values of the quota
q.
Solution:
The sum of the weights is 9. q
≤
9 and q > 9/2.
Hence q must be 5, 6, 7, 8, or 9.
Example:
A corporation has four partners with weights 10, 8, 7 and 1.
The bylaws of the corporation require that
twothirds of the votes are needed to pass a motion.
What is the quota?
Solution:
10 + 8 + 7 + 1 = 26.
Twothirds of 26 is 52/3 = 17 1/3. Since 17 is less than twothirds, the quota must be
18.
Example:
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 Spring '08
 Storfer

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