Fairness in Apportionment How do you decide whether a method for apportioning representatives is...
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Transcript of Fairness in Apportionment How do you decide whether a method for apportioning representatives is...
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Fairness in Apportionment
How do you decide whether a method for apportioning representatives is fair?
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Absolute Fairness
• If a state has a certain percentage of the total population, it should have that same percentage of seats in the House.
But, this usually results in a fractional number of seats, called the ideal quota or standard quota for the state. Thus, in most cases we have to go up or down to get an integer number or representatives.
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Upper and Lower Quota
Suppose a state has an ideal quota of 8.87 representatives.
• Its upper quota will be 9.
• Its lower quota will be 8.
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Satisfying Quota
• Any apportionment system which gives each state no fewer representatives than its lower quota and no more representatives than its upper quota is said to satisfy quota.
• An apportionment system which fails to do this for even one state is said to violate quota.
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An example
A state has an ideal quota of 23.71.
If, under a particular apportionment system, the state receives less than 23 representatives or more than 24 representatives, the system will have violated quota.
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What about our methods?
• Hamilton’s method never violates quota.
• Jefferson, Adams and Webster (and, in fact, any divisor method) can violate quota, although they do not always do so. We will see an historical example of this for Jefferson’s method on the next slide.
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Jefferson ♥ NY!
• New York’s population, 1820 census, was 1,368,775 while the total population of the US was 8,969,878.
• The house size was 213 so the standard divisor was 8,969,878 ÷ 213 = 42,112. Thus, NY’s ideal quota was 1,368,775 ÷ 42,112 = 32.503.
• Jefferson’s method, using a divisor of 39,900, gave NY int[1,368,775 ÷ 39,900] = 34 seats! This violated upper quota.
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Jefferson’s woes continued
• If Jefferson’s method had been used (it wasn’t) since 1850, it would have violated quota in every apportionment!
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Paradoxes
• Consider the following data for the Dale County schools.
Population
North High 9061
South High 7179
Valley High 5259
Meadow H. 3319
Ridge High 1182
Total 26,000
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Getting the Ideal Quotas
• We want to consider how to apportion Councils of size 26, 27 and 40 using Hamilton’s method.
26
Seat
27
seat
40
seat
Standard
Divisor
1000 963 650
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Table of Ideal Quotas
26 seats 27 seats 40 seats
North 9.061 9.4095 13.94 +1
South 7.179 7.455 +1 11.045
Valley 5.259 5.4613 +1 8.0908
Meadow 3.319 +1 3.4467 5.1062
Ridge 1.182 1.2275 1.8185 +1
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The Hamilton Apportionments
26 seats 27 seats 40 seats
North 9 9 14
South 7 8 11
Valley 5 6 8
Meadow 4 3 5
Ridge 1 1 2
26 27 40
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Alabama Paradox
• Notice what happened to Meadow High when Council size went from 26 to 27 seats! There was no change in total population or in any school population and yet Meadow lost a seat when the Council size increased by 1!!
• This is called the Alabama paradox. (Alabama would have lost a seat going from a 299 member house to a 300 member house after the 1880 census.) At this point Congress ditched the Hamilton method!
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Population Paradox
• This occurs when state X loses a seat to state Y even though X’s population grew at a higher rate than Y’s population!
• From 1900 to 1910 Virginia’s population grew by 19,767 while Maine grew by only 4648. Also, Virginia’s population grew at a 60% faster rate, yet, under Hamilton’s method, Virginia lost a seat to Maine!
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Population Paradox: A simple example
• We have 5 states: A,B,C,D,E with a total population of 900 and a 50 seat assembly, making the standard divisor d = 18. Using Hamilton, this gives:
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Population Paradox (cont.)
State Population Ideal Quota Apportion
A 150 8.33 8
B 78 4.33 4
C 173 9.61 +1 10
D 204 11.33 11
E 295 16.39 +1 17
Total 900 50
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Population grows:Only C and E grow
State Population Ideal Quota Apportion
A 150 8.25 8
B 78 4.29 +1 5
C 181 9.96 +1 10
D 204 11.22 11
E 296 16.28 16
Total 909 50
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Look what happened!
• State E, which gained population, lost a seat to state B, which did not gain population!!
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New States Paradox
• The addition of a new state, with its fair share of representatives can change the apportionments of other states.
• In 1907 Oklahoma joined the Union with 5 seats. Using Hamilton’s method to reapportion the House would have NY giving a seat to Maine, even though neither state (or any other) had a population change!
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Are there any apportionment methods that do not violate quota
and avoid all paradoxes??• (Balinsky and Young, 1980) There is
no perfect apportionment method! Any method which does not violate quota must produce paradoxes, and any method that does not produce paradoxes can violate quota.
• No apportionment method is completely fair!
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What about our methods?
• Hamilton satisfies quota but produces paradoxes (Alabama, population, new states)
• Jefferson, Adams, Webster (and all divisor methods) avoid all paradoxes, but can violate quota.