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Chapter 15: Apportionment Part 7: Which Method is Best? Paradoxes of Apportionment and Balinski & Young’s Impossibility Theorem

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Page 1: Chapter 15: Apportionment Part 7: Which Method is Best? Paradoxes of Apportionment and Balinski & Young’s Impossibility Theorem

Chapter 15: Apportionment

Part 7: Which Method is Best?

Paradoxes of Apportionment and

Balinski & Young’s Impossibility Theorem

Page 2: Chapter 15: Apportionment Part 7: Which Method is Best? Paradoxes of Apportionment and Balinski & Young’s Impossibility Theorem

Which Method is Best?

• Why does the Huntington-Hill method have to be so complicated?

• Is it really worth the trouble?

• And how do we decide which method is better?

• To compare the Hamilton, Jefferson, Webster and Huntington-Hill methods of apportionment, we will need to understand certain terms…

Page 3: Chapter 15: Apportionment Part 7: Which Method is Best? Paradoxes of Apportionment and Balinski & Young’s Impossibility Theorem

Which Method is Best?

• We define representative share for state i to be ai / pi where pi and ai are the population and final apportionment of state i.

• In the case of the apportionment in the U.S., we can think of representative share as being the fraction of a Congressional seat given to each individual in a particular state.

• For example, if there are 100,000 people in a certain state and that state has 5 seats, then representative share for that state is 5/100,000 = .00005 seats.

• For convenience we define 1 microseat = 1/1,000,000 seat. Then .00005 seats is 50 microseats.

Page 4: Chapter 15: Apportionment Part 7: Which Method is Best? Paradoxes of Apportionment and Balinski & Young’s Impossibility Theorem

Which Method is Best?

• Another term we will use is district population. It is the number of people per seat within a given state. That is, district population is pi/ai.

• Notice that if r = “representative share” and d = “district population” then r = 1/d and d = 1/r.

• Finally, we need to define the relative difference between two numbers:

• Given two real numbers A and B, suppose that A > B, then

A – B is the absolute difference between these two numbers and

{(A-B)/B}*100 is the relative difference between these numbers, expressed as a percentage.

Page 5: Chapter 15: Apportionment Part 7: Which Method is Best? Paradoxes of Apportionment and Balinski & Young’s Impossibility Theorem

Which Method is Best?

• It can be shown that the Webster Method will minimize absolute differences in representative share.

• Also, it can be shown that the Huntington-Hill method will minimize relative differences in both representative share and district populations.

• For example, consider the 1941 apportionment, when the Huntington-Hill method was first used. There were two competing methods of apportionment being considered at the time: Webster’s and Huntington-Hill.

Page 6: Chapter 15: Apportionment Part 7: Which Method is Best? Paradoxes of Apportionment and Balinski & Young’s Impossibility Theorem

Which Method is Best?

• Two states stood to either gain or lose a seat depending on which method of apportionment was used.

• Based on the 1940 census, the populations of those states were as shown in the following table.

State Population

Michigan 5,256,106

Arkansas 1,949,387

Page 7: Chapter 15: Apportionment Part 7: Which Method is Best? Paradoxes of Apportionment and Balinski & Young’s Impossibility Theorem

Which Method is Best?

Using the Huntington-Hill method

State Population Seats Representative share

Michigan 5,256,106 17 17/5,256,106 = 3.234 microseats

Arkansas 1,949,387 7 7/1,949,387 = 3.591 microseats

Using Webster’s method

State Population Seats Representative share

Michigan 5,256,106 18 18/5,256,106 = 3.425 microseats

Arkansas 1,949,387 6 6/1,949,387 = 3.078 microseats

Page 8: Chapter 15: Apportionment Part 7: Which Method is Best? Paradoxes of Apportionment and Balinski & Young’s Impossibility Theorem

Which Method is Best?

Using the Huntington-Hill method

State Population Seats Representative share

Michigan 5,256,106 17 17/5,256,106 = 3.234 microseats

Arkansas 1,949,387 7 7/1,949,387 = 3.591 microseats

Absolute difference

3,306,719 10 0.357 microseats

Using Webster’s method

State Population Seats Representative share

Michigan 5,256,106 18 18/5,256,106 = 3.425 microseats

Arkansas 1,949,387 6 6/1,949,387 = 3.078 microseats

Absolute difference

3,306,719 12 0.347 microseats

Page 9: Chapter 15: Apportionment Part 7: Which Method is Best? Paradoxes of Apportionment and Balinski & Young’s Impossibility Theorem

Which Method is Best?

Using the Huntington-Hill method

State Population Seats Representative share

Michigan 5,256,106 17 17/5,256,106 = 3.234 microseats

Arkansas 1,949,387 7 7/1,949,387 = 3.591 microseats

Absolute difference

3,306,719 10 0.357 microseats

Using Webster’s method

State Population Seats Representative share

Michigan 5,256,106 18 18/5,256,106 = 3.425 microseats

Arkansas 1,949,387 6 6/1,949,387 = 3.078 microseats

Absolute difference

3,306,719 12 0.347 microseats

Absolute difference in representative share between any two states is always

smaller with Webster’smethod.

Page 10: Chapter 15: Apportionment Part 7: Which Method is Best? Paradoxes of Apportionment and Balinski & Young’s Impossibility Theorem

Which Method is Best?

Using the Huntington-Hill method

State Population Seats Representative

share

District population

Michigan 5,256,106 17 3.234 microseats 309,183

Arkansas 1,949,387 7 3.591 microseats 278,484

Absolute difference

3,306,719 10 0.357 microseats 30,699

Using Webster’s method

State Population Seats Representative share

District population

Michigan 5,256,106 18 3.425 microseats 292,006

Arkansas 1,949,387 6 3.078 microseats 324,898

Absolute difference

3,306,719 12 0.347 microseats 32,892

Page 11: Chapter 15: Apportionment Part 7: Which Method is Best? Paradoxes of Apportionment and Balinski & Young’s Impossibility Theorem

Which Method is Best?

Using the Huntington-Hill method

State Population Seats Representative

share

District population

Michigan 5,256,106 17 3.234 microseats 309,183

Arkansas 1,949,387 7 3.591 microseats 278,484

Absolute difference

3,306,719 10 0.357 microseats 30,699

Relative difference

170% 143% Approx. 11.0% Approx. 11.0%

Relative difference in representative share and relative difference in district population are always the same. Any difference is due to rounding in the previous values.

Page 12: Chapter 15: Apportionment Part 7: Which Method is Best? Paradoxes of Apportionment and Balinski & Young’s Impossibility Theorem

Which Method is Best?

Using Webster’s Method …

State Population Seats Representative

share

District population

Michigan 5,256,106 18 3.425 microseats 292,006

Arkansas 1,949,387 6 3.078 microseats 324,898

Absolute difference

3,306,719 12 0.347 microseats 32,892

Relative difference

170% 200% Approx. 11.3% Approx. 11.3%

Again, these really are always the same. Any difference is due to rounding in the previous values.

Page 13: Chapter 15: Apportionment Part 7: Which Method is Best? Paradoxes of Apportionment and Balinski & Young’s Impossibility Theorem

Which Method is Best?

Using Webster’s Method …

State Population Seats Representative

share

District population

Michigan 5,256,106 18 3.425 microseats 292,006

Arkansas 1,949,387 6 3.078 microseats 324,898

Relative difference

170% 200% Approx. 11.3% Approx. 11.3%

Using the Huntington-Hill method

State Population Seats Representative

Share

District population

Michigan 5,256,106 17 3.234 microseats 309,183

Arkansas 1,949,387 7 3.591 microseats 278,484

Relative difference

170% 143% Approx. 11.0% Approx. 11.0%

Page 14: Chapter 15: Apportionment Part 7: Which Method is Best? Paradoxes of Apportionment and Balinski & Young’s Impossibility Theorem

Which Method is Best?

Using Webster’s Method …

State Population Seats Representative

share

District population

Michigan 5,256,106 18 3.425 microseats 292,006

Arkansas 1,949,387 6 3.078 microseats 324,898

Relative difference

170% 200% 11.3% 11.3%

Using the Huntington-Hill method

State Population Seats Representative

share

District population

Michigan 5,256,106 17 3.234 microseats 309,183

Arkansas 1,949,387 7 3.591 microseats 278,484

Relative difference

170% 143% 11.0% 11.0%

Relative differences in both representative share and district population are always less with

the Huntington-Hill method.

Page 15: Chapter 15: Apportionment Part 7: Which Method is Best? Paradoxes of Apportionment and Balinski & Young’s Impossibility Theorem

Which Method is Best?

So which method of apportionment is best ?

While we don’t have a definitive mathematical answer to the question which method is better, we can use mathematics to make some useful comparisons.

And there are still other ways to mathematically compare methods of apportionment that we will not consider here. For example, do some methods benefit small states, do other methods benefit large states?

We looked at only two particular states from the 1940 apportionment of the U.S. House and compared results from just two methods. At the very least, we have observed part of an illustration of the following facts.

Webster’s method is better at minimizing absolute differences in representative share between any two states. The Huntington-Hill method is better at minimizing relative differences in both representative share and in district population between any two states.

Page 16: Chapter 15: Apportionment Part 7: Which Method is Best? Paradoxes of Apportionment and Balinski & Young’s Impossibility Theorem

Balinski and Young’s Impossibility Theorem

Not only is there no mathematical answer to the question which method is best? - there is a mathematical theorem demonstrating that there never will be a “perfect” method of apportionment.

In the 1970s, two mathematicians by the names of Michel Balinski and Peyton Young studied apportionment in an effort to determine which method is best.

In 1980, the result of their research was a theorem, called Balinski and Young’s Impossibility Theorem, that states that no apportionment method can satisfy certain reasonable criteria for fairness.

Page 17: Chapter 15: Apportionment Part 7: Which Method is Best? Paradoxes of Apportionment and Balinski & Young’s Impossibility Theorem

Balinski and Young’s Impossibility Theorem

Balinski and Young’s Impossibility Theorem states the following:

It is mathematically impossible to have a perfect method of apportionment: Any method of apportionment that satisfies the quota rule will produce paradoxes and any method of apportionment that avoids paradox will violate the quota rule.

This result is similar to Arrow’s Impossibility Theorem (1952) which states no voting method can satisfy a certain set of reasonable criteria of fairness.

Page 18: Chapter 15: Apportionment Part 7: Which Method is Best? Paradoxes of Apportionment and Balinski & Young’s Impossibility Theorem

Balinski and Young’s Impossibility Theorem

To fully understand Balinski and Young’s Impossibility Theorem, we need to understand the quota rule and the types of paradoxes that can occur in apportionment. We have seen one example of a paradox (the Alabama paradox) and will consider others, but first we consider the quota rule…

• When we calculate the quota for a particular state, we usually get a real number value that is not an integer. Depending on the method we use, we will round this number up or down to an integer value. After the initial apportionment, again depending on the method we use, the initial apportionment may decrease or increase.

• It is reasonable to expect that, for example, if a state has a quota of, say 16.78, that by the end of the apportionment process the state will have either 16 or 17 seats.

Page 19: Chapter 15: Apportionment Part 7: Which Method is Best? Paradoxes of Apportionment and Balinski & Young’s Impossibility Theorem

The quota rule

• It might seem unfair to some other state if a state with a quota of 16.78 ended up with an apportionment of 18 seats. Or it certainly would be understandable if a state with a quota of 16.78 considered it to be unfair if they ended up with an apportionment of only 15 seats.

• We say that an apportionment method satisfies the quota condition (or quota rule) if every state’s final apportionment is equal to either it’s upper or lower quota.

Page 20: Chapter 15: Apportionment Part 7: Which Method is Best? Paradoxes of Apportionment and Balinski & Young’s Impossibility Theorem

Paradoxes of Apportionment

• The Alabama Paradox - An increase in the total number of seats to be apportioned causes a state to lose a seat.

• The Alabama Paradox first surfaced after the 1870 census. With 270 members in the House of Representatives, Rhode Island got 2 representatives but when the House size was increased to 280, Rhode Island lost a seat.

• After the 1880 census,C. W. Seaton (chief clerk of U. S. Census Office) computed apportionments for all House sizes between 275 and 350 members. He then wrote a letter to Congress pointing out that if the House of Representatives had 299 seats, Alabama would get 8 seats but if the House of Representatives had 300 seats, Alabama would only get 7 seats.

Page 21: Chapter 15: Apportionment Part 7: Which Method is Best? Paradoxes of Apportionment and Balinski & Young’s Impossibility Theorem

Paradoxes of Apportionment

• The Population Paradox - An increase in a state’s population can cause it to lose a seat.

• The Population Paradox was discovered around 1900, when it was shown that a state could lose seats in the House of Representatives as a result of an increase in its population. (Virginia was growing much faster than Maine--about 60% faster--but Virginia lost a seat in the House while Maine gained a seat.)

Page 22: Chapter 15: Apportionment Part 7: Which Method is Best? Paradoxes of Apportionment and Balinski & Young’s Impossibility Theorem

Paradoxes of Apportionment

• The New States Paradox - Adding a new state with its fair share of seats can affect the number of seats due other states.

• The New States Paradox was discovered in 1907 when Oklahoma became a state. Before Oklahoma became a state, the House of Representatives had 386 seats. Comparing Oklahoma's population to other states, it was clear that Oklahoma should have 5 seats so the House size was increased by five to 391 seats. The intent was to leave the number of seats unchanged for the other states. However, when the apportionment was mathematically recalculated, Maine gained a seat (4 instead of 3) and New York lost a seat (from 38 to 37).

Page 23: Chapter 15: Apportionment Part 7: Which Method is Best? Paradoxes of Apportionment and Balinski & Young’s Impossibility Theorem

Comparison of Methods

Hamilton Jefferson Webster Huntington-Hill

Violates quota rule

No Yes Yes Yes

Alabama paradox

Yes No No No

Population paradox

Yes No No No

New states paradox

Yes No No No

These apportionment methods confirm Balinski and Young’s Impossibility Theorem: Any apportionment method will violate the quota rule or exhibit paradoxes.