© 2010 Pearson Prentice Hall. All rights reserved. CHAPTER 14 Voting and Apportionment

© 2010 Pearson Prentice Hall. All rights reserved Flaws of Apportionment Methods

© 2010 Pearson Prentice Hall. All rights reserved. 3 Objectives 1.Understand and illustrate the Alabama paradox. 2.Understand and illustrate the population paradox. 3.Understand and illustrate the new-states paradox. 4.Understand Balinski and Young’s Impossibility Theorem.

© 2010 Pearson Prentice Hall. All rights reserved. 4 The Alabama Paradox An increase in the total number of items to be apportioned results in the loss of an item for a group.

© 2010 Pearson Prentice Hall. All rights reserved. 5 Example 1: Illustrating the Alabama Paradox A small country with a population of 10,000 is composed of three states. According to the country’s constitution, the congress will have 200 seats. A table of the country’s population distribution is given below. Use Hamilton’s method and show that the Alabama paradox occurs if the number of seats is increased to 201. StateABCTotal Population ,000

© 2010 Pearson Prentice Hall. All rights reserved. 6 Solution: We begin with 200 seats and compute the standard divisor. Using this value, we obtain the apportionment for each state using Hamilton’s method. Example 1: Illustrating the Alabama Paradox

© 2010 Pearson Prentice Hall. All rights reserved. 7 Next, we find each state’s apportionment with 201 congressional seats. Again, we compute the standard divisor. Example 1: Illustrating the Alabama Paradox

© 2010 Pearson Prentice Hall. All rights reserved. 8 The final apportionments are summarized below. When the number of seats increased from 200 to 201, state C’s apportionment decreased. This is an example of Alabama’s Paradox. –States A and B benefit at state C’s expense. Example 1: Illustrating the Alabama Paradox

© 2010 Pearson Prentice Hall. All rights reserved. 9 The Population Paradox Group A loses items to group B, even though the population of group A grew at a faster rate than that of group B.

© 2010 Pearson Prentice Hall. All rights reserved. 10 Example 2: Illustrating the Population Paradox A small country with a population 10,000 is composed of three states. There are 11 seats in congress, divided among the three states according to their respective populations. Using Hamilton’s method, the apportionment of congressional seats for each state is shown. a. Find the percent of increase in the population of each state. b. Use Hamilton’s method to show that the population paradox occurs. StateABCTotal Original Population ,000 New Population ,000

© 2010 Pearson Prentice Hall. All rights reserved. 11 Solution: a. From the percentages, state C is increasing at a faster rate than state A and B. From the percentages, state B is increasing at a faster rate than state A. Example 2: Illustrating the Population Paradox

© 2010 Pearson Prentice Hall. All rights reserved. 12 b.We use the Hamilton’s method to find the apportionment for each state with its new population. First we compute the standard divisor. Example 2: Illustrating the Population Paradox

© 2010 Pearson Prentice Hall. All rights reserved. 13 The final apportionments are shown below. Even though state A now has a congressional seat, state B has lost a seat to state A, but recall state B’s population grew faster than state A’s. This is an example of population paradox. Example 2: Illustrating the Population Paradox

© 2010 Pearson Prentice Hall. All rights reserved. 14 New-States Paradox The addition of a new group changes the apportionments of other groups.

© 2010 Pearson Prentice Hall. All rights reserved. 15 Example 3: Illustrating the New-States Paradox A school district has two high schools: East High and West High. The school district has a counseling staff of 48 counselors. The standard divisor is SchoolPopulation Standard Quota Lower Quota Hamilton’s Apportionment East High West High Total

© 2010 Pearson Prentice Hall. All rights reserved. 16 Suppose a new high school, North High, is added to the district with 1448 more students. Using the standard divisor of 200, the district hires 7 counselors.Show that the new-states paradox occurs when the counselors are reapportioned. Solution: The new standard divisor is Example 3: Illustrating the New-States Paradox

© 2010 Pearson Prentice Hall. All rights reserved. 17 Before North High, East High had 8 counselors and West High had 40. After we added North High, East High added 1 counselor, West High lost 1 counselor. Hence, the addition of another high school changed the apportionments of the other high schools. This is an example of new-states paradox. Example 3: Illustrating the New-States Paradox

© 2010 Pearson Prentice Hall. All rights reserved. 18 Balinski and Young’s Impossibility Theorem Is there an ideal apportionment method? There is no perfect apportionment method. Any apportionment method that does not violate the quota rule must produce paradoxes. Any apportionment method that does not produce paradoxes must violate the quota rule.