6.2: Apportionment Paradoxes
- Page ID
- 187017
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Describe and illustrate the Alabama paradox.
- Describe and illustrate the population paradox.
- Describe and illustrate the new-states paradox.
Controversy
There are certain characteristics that they would reasonably expect from a fair apportionment.
- If the house size is increased, the state quotas should all increase or remain the same, but never decrease.
- If one state’s population is growing more rapidly than another's, the faster-growing state should not lose a seat while the slower-growing state maintains or gains a seat.
- If there is a fixed number of seats, adding a new state should not cause an existing state to gain seats while others lose them.
However, apportionment methods are known to contradict these expectations. Before you decide on the right apportionment for Imaginarians, let’s explore the apportionment paradox, a situation that occurs when an apportionment method produces results that seem to contradict reasonable expectations of fairness.
There is a lot that the founders of Imaginaria can learn from U.S. history. The constitution of the United States requires that the seats in the House of Representatives be apportioned according to the results of the census that occurs every decade, but the number of seats and the apportionment method is not stipulated. Over the years, several different apportionment methods and house sizes have been used and scrutinized for fairness. This scrutiny has led to the discovery of several of these apportionment paradoxes.
After seeing Hamilton’s method, many people find that it makes sense, it’s not that difficult to use (or, at least, the difficulty comes from the numbers that are involved and the amount of computation that’s needed, not from the method), and they wonder why anyone would want another method. The problem is that Hamilton’s method is subject to several paradoxes. Three of them happened, on separate occasions, when Hamilton’s method was used to apportion the United States House of Representatives.
The Alabama Paradox
The Alabama Paradox is named for an incident that happened during the apportionment that took place after the \(1880\) census. (A similar incident happened ten years earlier involving the state of Rhode Island, but the paradox is named after Alabama.) The post-\(1880\) apportionment had been completed, using Hamilton’s method and the new population numbers from the census. Then it was decided that because of the country’s growing population, the House of Representatives should be made larger. That meant that the apportionment would need to be done again, still using Hamilton’s method and the same \(1880\) census numbers, but with more representatives. The assumption was that some states would gain another representative and others would stay with the same number they already had (since there weren’t enough new representatives being added to give one more to every state). The paradox is that Alabama ended up losing a representative in the process, even though no populations were changed and the total number of representatives increased.
The Alabama paradox occurs when an increase in the total number of items (quotas) to be apportioned results in a loss of an item for a group.
Example \(\PageIndex{1}\): Does Alabama Paradox Occur?
[1] Three states, A, B, and C, comprise a small country. The population of each state is given in the table below.
| State | A | B | C | Total |
|---|---|---|---|---|
| Population | \(11,251\) | \(18,245\) | \(16,468\) | \(45,964\) |
- According to the country's constitution, its congress shall have \(26\) seats, divided amongst the three states according to their respective populations. Use Hamilton's method to determine each state's apportionment of congressional seats.
- Use Hamilton's method to determine each state's apportionment of congressional seats if the number of seats in parliament is increased from \(26\) to \(27.\)
- Determine if the Alabama paradox occurred when the number of seats in Congress was increased from \(26\) to \(27.\)
- Answer
-
With \(26\) seats: Standard divisor = \(\frac{45,964}{26}=1767.84615\)
Table \(\PageIndex{1}:\) Apportionment with \(26\) seats State A B C Total Population \(11,251\) \(18,245\) \(16,468\) \(45,694\) Standard Quota \(6.36424\) \(10.32047\) \(9.31529\) Lower Quota \(6\) \(10\) \(9\) \(25\) Apportionment \(6+1=7\) \(10\) \(9\) \(26\) With \(27\) seats: Standard divisor = \(\frac{45,964}{27}=1702.37037\)
Table \(\PageIndex{2}:\) Apportionment with \(27\) seats State A B C Total Population \(11,251\) \(18,245\) \(16,468\) \(45,694\) Standard Quota \(6.60902\) \(10.71741\) \(9.67357\) Lower Quota \(6\) \(10\) \(9\) \(25\) Apportionment \(6\) \(10+1=11\) \(9+1=10\) \(27\) Yes, the Alabama occurred, since state A lost a seat in Congress.
Figure \(\PageIndex{2}\): Illustration of Alabama Paradox
The New State Paradox
The New States Paradox happened when Oklahoma became a state in \(1907.\) Oklahoma had enough population to qualify for five representatives in Congress. Those five representatives would need to come from somewhere, though, so five states, presumably, would lose one representative each. That happened, but another thing also happened: Maine gained a representative (from New York).
The new state paradox occurs when adding a new group changes the apportionment of other groups.
Example \(\PageIndex{2}\): Does New State Paradox Occur?
A small city is made up of three districts and governed by a committee with \(100\) members. District A has a population of \(5,310.\) District B has a population of \(1330,\) and District C has a population of \(3308.\) The city annexes a small area, District D, with a population of \(500.\) At the same time the number of committee members is increased by five. Hamilton’s method was used to find the apportionment before and after the annexation.
| State | A | B | C | Total |
|---|---|---|---|---|
| Population | \(5,310\) | \(1330\) | \(3,308\) | \(9,948\) |
| State | A | B | C | D | Total |
|---|---|---|---|---|---|
| Population | \(5,310\) | \(1330\) | \(3,308\) | \(500\) | \(10,448\) |
- Answer
-
Before annexation: Standard divisor = \(\frac{9948}{100}=99.48\)
Table \(\PageIndex{5}:\) Apportionment Before the Annexation State A B C Total Population \(5,310\) \(1330\) \(3,308\) \(9,948\) Standard Quota \(53.378\) \(13.370\) \(33.253\) \(100.000\) Lower Quota \(53\) \(13\) \(33\) \(99\) Final apportionment \(54\) \(13\) \(33\) 100 After annexation: Standard divisor = \(\frac{10448}{105}=99.505\)
Table \(\PageIndex{6}:\) Apportionment After the Annexation State A B C D Total Population \(5,310\) \(1330\) \(3,308\) \(500\) \(10,448\) Standard Quota \(53.364\) \(13.366\) \(33.245\) \(5.025\) \(105.000\) Lower Quota \(53\) \(13\) \(33\) \(5\) \(104\) Final apportionment \(53\) \(13+1=14\) \(33\) \(5\) \(105\) District D has a population of \(500,\) so it should get five seats. When District D is added with its five seats, District A loses a seat, and District B gains a seat. This is an example of the New-States Paradox.
The Population Paradox
The Population Paradox happened between the apportionments after the census of \(1900\) and of \(1910.\) In those ten years, Virginia’s population grew at an average annual rate of \(1.07\%\), while Maine’s grew at an average annual rate of \(0.67\%.\) Virginia started with more people, grew at a faster rate, grew by more people, and ended up with more people than Maine. By itself, that doesn’t mean that Virginia should gain representatives or Maine shouldn’t, because there are lots of other states involved. But Virginia ended up losing a representative to Maine.
The population paradox occurs when group A loses an item(s) to group B, even though group A's population grew faster than group B's population.
A mom decides to split \(11\) candy bars among three children based on the number of minutes they spend on chores this week. The time spent by each of her children is given in the table. Near the end of the week, Mom reminds the children of the deal, and they each do some extra work. Abby does an extra two minutes, Bobby an additional \(12\) minutes, and Charley an extra \(86\) minutes.
| Abby | Bobby | Charley | Total | |
|---|---|---|---|---|
| Population (Before) | \(54\) | \(243\) | \(703\) | \(1,000\) |
| Population (After) | \(56\) | \(255\) | \(789\) | \(1100\) |
- Use Hamilton’s method to apportion the candy bars before and after the extra work.
- Does the population paradox occur? Why or why not?
- Answer
-
Apportion using Hamilton before the extra work: Standard divisor = \(\frac{1000}{11}=90.90909\)
Table \(\PageIndex{8}:\) Apportion of Candy Bars Before the Extra Work Abby Bobby Charley Total Population \(54\) \(243\) \(703\) \(1,000\) Standard Quota \(0.594\) \(2.673\) \(7.733\) Lower Quota \(0\) \(2\) \(7\) \(9\) Apportionment \(0\) \(2+1= 3\) \(7+1=8\) \(11\) Apportion using Hamilton after extra work: Standard divisor = \(\frac{1100}{11}=100\)
Table \(\PageIndex{9}:\) Apportion of Candy Bars After the Extra Work Abby Bobby Charley Total Population \(56\) \(255\) \(789\) \(1,100\) Standard Quota \(0.560\) \(2.550\) \(7.890\) Lower Quota \(0\) \(2\) \(7\) \(9\) Final Apportionment \(0+1=1\) \(2\) \(7+1= 8\) \(11\) Abby time increase by: \(\frac{56-54}{54}=3.7\%\)
Bobby time increase by: \(\frac{255-243}{243}=4.9\%\)
Charley time increase by: \(\frac{789-703}{703}=12.2\%\)
Abby’s time increased by \(3.7\%\) while Bobby’s time increased by \(4.9\%,\) more than Abby's. However, Abby gains a candy bar while Bobby loses one. This is an example of the Population Paradox.
Example \(\PageIndex{4}\): Population Paradox
Suppose that \(18\) respirators are to be apportioned to three hospitals based on their capacities. The Hamilton method is used to allocate the respirators in \(2020,\) then to reallocate based on new capacities in \(2020.\) The results are shown in the table below. How does this demonstrate the population paradox?
| Hospital | Capacity in \(2020\) | Respirators in \(2020\) | Capacity in \(2021\) | Respirators in \(2021\) | Growth Rate |
|---|---|---|---|---|---|
| A | \(825\) | \(9\) | \( 882\) | \(9\) | \(\frac{882-825}{825}=6.01\%\) |
| B | \(613\) | \(7\) | \(626\) | \(6\) | \(\frac{626-613}{613}=2.12\%\) |
| C | \(239\) | \(2\) | \(242\) | \(3\) | \(\frac{242-239}{239}=1.26\%\) |
- Answer
-
Hospital B lost a respirator while hospital C gained one, even though hospital B had a higher growth rate (\(2.12\%\)) than hospital C (\(1.26\%.\))
Quota Rule Violation
A small college has three departments. Department A has \(98\) faculty, Department B has \(689\) faculty, and Department C has \(212\) faculty. The college has a faculty senate with \(100\) representatives. Use Jefferson’s method with a modified divisor of \(d = 9.83\) to apportion the \(100\) representatives among the departments.
Apportion using Hamilton before the extra work: Standard divisor = \(\frac{999}{100}=9.99\)
| State | A | B | C | Total |
|---|---|---|---|---|
| Population | \(98\) | \(689\) | \(212\) | \(999\) |
| Standard Quota | \(9.810\) | \(68.969\) | \(21.221\) | \(100\) |
| \(d = 9.83\) | \(9.969\) | \(70.092\) | \(21.567\) | |
| quota | \(9\) | \(70\) | \(21\) | \(100\) |
Department B has a standard quota of \(68.969\), so it should get either its lower quota \(68 \) or its upper quota \(69 \) seats. Using this method, District B received \(70\) seats, one more than its upper quota. This is a Quota Rule violation.
In \(1980,\) Michael Balinski (State University of New York at Stony Brook) and H. Peyton Young (Johns Hopkins University) proved that all apportionment methods violate the quota rule or suffer from one of the paradoxes. This means that it is impossible to find the “perfect” apportionment method. The methods and their potential flaws are listed in the following table.
| Paradoxes | ||||
|---|---|---|---|---|
| Method | Quota Rule | Alabama | Population | New-States |
| Hamilton | No violations | Yes | Yes | Yes |
| Jefferson | Upper-quota violations | No | No | No |
| Adams | Lower-quota violations | No | No | No |
| Webster | Lower- and upper-quota violations | No | No | No |
| Huntington-Hill | Lower- and upper-quota violations | No | No | No |