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Can One More Seat Make a State Lose Representation?

The largest-remainder method turns fair fractional shares into whole seats—and can punish a state when the legislature grows.

Difficulty
Intermediate
Area
Discrete mathematics and social choice
Reading time
10 minutes
Prerequisites
Fractions and percentages · Rounding down to a whole number · Comparing fractional remainders
Three cyan, magenta, and green population currents cross a dark maritime drafting grid toward two arrays of seat markers; the larger array contains one fewer green marker, illustrating a loss caused by adding a seat.

Three regions must divide ten seats in a council. Aster has 6,000 people, Brine has 6,000, and Cairn has 2,000. Their populations are in the ratio 6:6:26:6:2, so a proportional division gives the seats

4.2857ldots,qquad4.2857ldots,qquad1.4286ldots.4.2857ldots,qquad 4.2857ldots,qquad 1.4286ldots.

No person can occupy 0.42860.4286 of a council seat. Suppose the council rounds every share down, then gives the unassigned seat to the region with the largest fractional remainder. Aster receives four seats, Brine receives four, and Cairn receives two.

Now enlarge the council from ten seats to eleven. Every region’s proportional share increases. Yet the same rule awards five seats to Aster, five to Brine, and one to Cairn.

Cairn loses a representative because the council gained one.

This failure of ordinary expectations is called the Alabama paradox. The name comes from a calculation made after the United States census of 1880, but the paradox can occur whenever a fixed number of indivisible items must be divided proportionally.

A fair share meets a whole number

The rule in the opening puzzle is known as the largest-remainder method. In the history of United States congressional apportionment, the same rule is usually called Hamilton’s method.

Here is the rule in symbols. Let

  • nn be the number of regions;
  • pip_i be the population of region ii, where ii identifies one region from 11 through nn;
  • P=p1+p2+⋯+pnP=p_1+p_2+\cdots+p_n be the total population; and
  • HH be the positive whole number of seats to distribute.

Region ii has an exact proportional share, or standard quota, of

qi=HpiP.q_i=H\frac{p_i}{P}.

The floor symbol ⌊qi⌋\lfloor q_i\rfloor means “round qiq_i down to the nearest whole number.” Hamilton’s method proceeds in three steps:

  1. Give each region its lower quota ⌊qi⌋\lfloor q_i\rfloor.

  2. Count the seats still unassigned:

    L=H−∑i=1n⌊qi⌋.L=H-\sum_{i=1}^{n}\lfloor q_i\rfloor.
  3. Give one additional seat to each of the LL regions with the largest fractional remainders qi−⌊qi⌋q_i-\lfloor q_i\rfloor.

The letter LL names the number of leftover seats. If two regions tie for the last one, the governing rules need a separate tie-breaker. No tie-breaker is needed in our example: Aster and Brine tie when two leftover seats are available.

The interactive diagram below keeps the three populations fixed. Switch between ten and eleven seats, then compare both the seat markers and the remainder bars.

Interactive apportionment

Watch the leftover seat change hands

Total seats to distribute

Aster

6,000 people

Exact quota
4 + 2/7
Remainder
2/7

4 seats

No leftover seat

Brine

6,000 people

Exact quota
4 + 2/7
Remainder
2/7

4 seats

No leftover seat

Cairn

2,000 people

Exact quota
1 + 3/7
Remainder
3/7

2 seats

Receives the leftover seat

Fractional remainders

Aster
2/7
Brine
2/7
Cairn
3/7

Nine seats come from whole-number quotas. Cairn has the largest remainder, so Cairn receives the tenth seat.

What to notice: every exact quota rises when the total grows from 10 seats to 11. Cairn's remainder also rises, from 3/7 to 4/7. Aster's and Brine's remainders rise even faster and pass Cairn's, so the two new leftover seats go to them. Cairn falls from two seats to one.

Worked example: where Cairn’s seat goes

The total population is

P=6,000+6,000+2,000=14,000.P=6{,}000+6{,}000+2{,}000=14{,}000.

Because each population contains a factor of 1,0001{,}000, we can calculate with 66, 66, and 22 without changing any proportions.

With ten seats

Aster and Brine have the same quota:

qA=qB=10614=307=4+27.q_{\mathrm A}=q_{\mathrm B}=10\frac{6}{14}=\frac{30}{7}=4+\frac{2}{7}.

Cairn’s quota is

qC=10214=107=1+37.q_{\mathrm C}=10\frac{2}{14}=\frac{10}{7}=1+\frac{3}{7}.

Rounding down assigns 4+4+1=94+4+1=9 seats. One seat remains. Cairn’s remainder, 3/73/7, exceeds Aster’s and Brine’s remainders of 2/72/7, so Cairn receives the leftover seat:

(4,4,2).(4,4,2).

With eleven seats

Recalculate every quota because the total number of seats has changed:

qA=qB=11614=337=4+57,q_{\mathrm A}=q_{\mathrm B}=11\frac{6}{14}=\frac{33}{7}=4+\frac{5}{7},

and

qC=11214=117=1+47.q_{\mathrm C}=11\frac{2}{14}=\frac{11}{7}=1+\frac{4}{7}.

Rounding down still assigns only 4+4+1=94+4+1=9 seats, so two seats remain. The two largest remainders are now Aster’s and Brine’s, both 5/75/7. Those regions receive the two leftover seats:

(5,5,1).(5,5,1).

Cairn’s quota rose from 1+3/71+3/7 to 1+4/71+4/7. Cairn lost because the larger regions’ fractional parts rose faster, from 2/72/7 to 5/75/7, and overtook Cairn’s 4/74/7 remainder. Hamilton’s method reruns the remainder contest from scratch when HH changes; it does not preserve seats awarded in the previous contest.

Try another house size

Problem 1

Use Hamilton’s method to divide nine seats among Aster, Brine, and Cairn, whose populations remain in the ratio 6:6:26:6:2.

Reveal the solution

The quotas are

qA=qB=9614=277=3+67,q_{\mathrm A}=q_{\mathrm B}=9\frac{6}{14}=\frac{27}{7}=3+\frac{6}{7},

and

qC=9214=97=1+27.q_{\mathrm C}=9\frac{2}{14}=\frac{9}{7}=1+\frac{2}{7}.

The lower quotas assign 3+3+1=73+3+1=7 seats. The two leftover seats go to Aster and Brine because each has remainder 6/76/7, larger than Cairn’s 2/72/7. The result is

(4,4,1).(4,4,1).

Problem 2

Now divide twelve seats among the same regions. Does Cairn recover its second seat?

Reveal the solution

Aster’s and Brine’s quotas are

qA=qB=12614=367=5+17.q_{\mathrm A}=q_{\mathrm B}=12\frac{6}{14}=\frac{36}{7}=5+\frac{1}{7}.

Cairn’s quota is

qC=12214=127=1+57.q_{\mathrm C}=12\frac{2}{14}=\frac{12}{7}=1+\frac{5}{7}.

The lower quotas assign 5+5+1=115+5+1=11 seats. Cairn has the largest remainder, 5/75/7, so Cairn receives the final seat. The allocation is

(5,5,2).(5,5,2).

Yes. Cairn has one seat with nine seats available, two with ten, one with eleven, and two with twelve. More total seats do not produce a steady climb under Hamilton’s method.

The paradox came out of the census

Apportionment was never a bloodless rounding exercise in the United States. The Constitution tied representation in the House of Representatives to a census, so a choice among rounding rules changed the political weight of entire states.

The original constitutional formula was also built around slavery. Article I counted free persons and three-fifths of “all other Persons” when distributing representatives and direct taxes. That arrangement increased the representation of slaveholding states while denying enslaved people any political agency. Section 2 of the Fourteenth Amendment, ratified in 1868, replaced the original population rule with a count of whole persons and threatened reduced representation when a state denied voting rights to eligible male citizens. The National Archives’ constitutional texts and Fourteenth Amendment text preserve both formulas.

The largest-remainder method entered the dispute early. Alexander Hamilton supported an apportionment bill that used the approach after the 1790 census. On April 5, 1792, George Washington rejected that bill in the first presidential veto. Congress then passed a different calculation associated with Thomas Jefferson. The National Archives’ account of the first veto shows cabinet members arguing not only about numerical equality but also about what the Constitution permitted.

Congress later used the Vinton–Hamilton method for apportionments based on the censuses from 1850 through 1900, according to the Census Bureau’s historical survey of apportionment methods.

After the 1880 census, Census Office chief clerk C. W. Seaton calculated outcomes for House sizes from 275 through 350. The University of Wisconsin’s apportionment notes record Seaton’s finding: Alabama would receive eight seats in a 299-member House but only seven in a 300-member House. A Congressional Research Service history records Alabama’s quota rising from about 7.6467.646 to 7.6717.671 while the remainders of Illinois and Texas passed Alabama’s remainder. The arithmetic did not merely look peculiar: a rounding rule was deciding which state gained a vote in Congress.

The politics continued long after Alabama supplied the paradox’s name. Following the 1920 census, lawmakers from rural states that stood to lose influence prevented any reapportionment at all. The Census Bureau’s legislative history explains that the Reapportionment Act of 1929 later automated the process. In 1941 Congress adopted the method of equal proportions, also called the Huntington–Hill method, for future censuses.

Where the mathematics still matters

The United States no longer uses Hamilton’s method for the House. Under the present method, each state receives one seat first; priority values then assign seats 51 through 435. The Census Bureau provides the current equal-proportions formula and procedure.

Largest-remainder rules remain part of electoral mathematics elsewhere. Party-list proportional-representation systems convert votes into whole seats, and some use a Hare quota followed by largest remainders. The ACE Electoral Knowledge Network glossary defines both the quota and the allocation rule; International IDEA’s survey of open-list systems documents concrete variants.

The same arithmetic can appear outside elections. Imagine allocating a fixed number of indivisible clinic shifts among districts according to patient volume, or shared processors among research groups according to demand. Those examples are hypothetical, but the risk is real: if an organization uses largest remainders, expanding the resource pool can make one claimant lose a unit. Testing an allocation algorithm at several nearby totals can expose the behavior before policy turns it into a surprise.

Mathematicians Michel Balinski and H. Peyton Young later studied which fairness requirements can coexist. Their 1974 Proceedings of the National Academy of Sciences paper placed the congressional problem inside a broader theory of apportionment. One lesson from that theory is uncomfortable but useful: “proportional” does not identify a unique whole-number answer. A society must decide which failures it is prepared to tolerate, then encode that choice in a rule people can inspect.

One idea to remember

Hamilton’s method gives leftover seats to the largest fractional remainders. Adding a seat changes every remainder, so a previous winner can lose the new race—and lose a seat it already had.

Continue with Euler trails for another case where an everyday civic problem becomes discrete mathematics. Then compare Hamilton’s method with divisor methods, which assign new seats sequentially and avoid the Alabama paradox at the cost of different notions of proportional fairness.

Sources