1. The setup: divisor and quota
Apportionment problems have three inputs: a fixed total number of seats $S$ to hand out, a list of groups (states, departments, parties) with populations $p_1, p_2, \dots, p_n$, and the constraint that each group ends up with a whole number of seats summing to $S$.
Two derived quantities do almost all the work.
The total population divided by the total number of seats:
$$ d = \frac{p_1 + p_2 + \cdots + p_n}{S} $$It's the "ideal" population per seat — how many people one seat ought to represent if everything were continuous.
For group $i$:
$$ q_i = \frac{p_i}{d} $$The exact (usually fractional) number of seats group $i$ deserves under perfect proportionality. The lower quota is $\lfloor q_i \rfloor$ — round down. The upper quota is $\lceil q_i \rceil$ — round up.
A method "satisfies quota" if every group always ends up with either its lower or its upper quota — no group gets fewer than $\lfloor q_i \rfloor$ or more than $\lceil q_i \rceil$ seats. This sounds like the bare minimum of fairness. Most methods violate it sometimes.
Throughout we'll use a running 3-state example:
| State | Population | Standard quota | Lower | Upper |
|---|---|---|---|---|
| $A$ | $15{,}475$ | |||
| $B$ | $ 6{,}210$ | |||
| $C$ | $ 3{,}315$ | |||
| Total | $25{,}000$ |
With $S = 25$ seats, the standard divisor is $d = 25{,}000 / 25 = 1{,}000$. The standard quotas:
| State | $p_i$ | $q_i = p_i/d$ | $\lfloor q_i \rfloor$ | $\lceil q_i \rceil$ |
|---|---|---|---|---|
| $A$ | $15{,}475$ | $15.475$ | $15$ | $16$ |
| $B$ | $ 6{,}210$ | $ 6.210$ | $ 6$ | $ 7$ |
| $C$ | $ 3{,}315$ | $ 3.315$ | $ 3$ | $ 4$ |
| Total | $25{,}000$ | $25.000$ | $24$ | $27$ |
Lower quotas sum to 24 — one seat short. Upper quotas sum to 27 — two seats too many. Every method below is a different rule for closing that one-seat gap.
2. Hamilton's method
Also called the method of largest remainders. The simplest, the most intuitive — and the one with the most baroque pathologies.
Procedure:
- Compute every group's standard quota $q_i$.
- Give each group its lower quota $\lfloor q_i \rfloor$ to start.
- If seats remain, hand them out one at a time to the groups with the largest fractional parts of $q_i$, until you reach $S$.
On the running example: lower quotas $A: 15$, $B: 6$, $C: 3$ sum to 24. One seat remains. Fractional parts: $A: .475$, $B: .210$, $C: .315$. $A$ has the largest, so $A$ gets the extra seat. Final apportionment: $A: 16$, $B: 6$, $C: 3$. Sum $= 25$. ✓
Hamilton always satisfies quota — every group ends up at its lower or upper quota by construction. That's its great virtue. Its great vice is that it suffers three separate paradoxes; we'll meet them all in §7.
The US used Hamilton from 1852 to 1900 before the Alabama paradox forced a switch.
3. Jefferson's method
The first divisor method: instead of patching up standard quotas, find a modified divisor $d'$ such that when you round each $p_i / d'$ down, the rounded values sum to exactly $S$.
Procedure:
- Try $d' = d$ (the standard divisor). Round each $p_i / d'$ down.
- If the sum is too low (typical), decrease $d'$ so each quota grows. Repeat.
- If the sum is too high, increase $d'$.
- Stop when the rounded-down values sum to $S$.
On the example, $d = 1000$ gives $\lfloor 15.475 \rfloor + \lfloor 6.210 \rfloor + \lfloor 3.315 \rfloor = 24$ — one short. Try $d' = 970$:
$$ A: \lfloor 15{,}475 / 970 \rfloor = \lfloor 15.954 \rfloor = 15 $$ $$ B: \lfloor 6{,}210 / 970 \rfloor = \lfloor 6.402 \rfloor = 6 $$ $$ C: \lfloor 3{,}315 / 970 \rfloor = \lfloor 3.418 \rfloor = 3 $$Still 24. Try $d' = 950$:
$$ A: \lfloor 16.289 \rfloor = 16, \quad B: \lfloor 6.537 \rfloor = 6, \quad C: \lfloor 3.489 \rfloor = 3 $$Sum 25 ✓. Jefferson gives $A: 16$, $B: 6$, $C: 3$ — same as Hamilton here. In general it doesn't agree.
Jefferson systematically favors large states. Because every group rounds down, the bigger you are, the more raw seats you collect before the rounding kicks in — and to fix the shortfall, the divisor has to shrink, which proportionally helps the giants more than the small fry. Used by the US from 1791 to 1842 until it gave New York 40 seats when its quota was 38.59 — a clear quota violation.
4. Webster's method
The same divisor-search idea as Jefferson, but with standard rounding — round to nearest, with $.5$ going up — instead of always rounding down.
Procedure:
- Find a modified divisor $d'$ such that $\sum_i \text{round}(p_i / d')$ equals $S$.
- If the rounded values are too low, decrease $d'$; if too high, increase $d'$.
With the running data and $d = 1000$, standard rounding gives $\text{round}(15.475) + \text{round}(6.210) + \text{round}(3.315) = 15 + 6 + 3 = 24$. Decrease the divisor slightly — try $d' = 985$:
$$ A: \text{round}(15.711) = 16, \quad B: \text{round}(6.305) = 6, \quad C: \text{round}(3.365) = 3 $$Sum 25 ✓. Same answer again on this example — but Webster's statistical bias is much smaller than Jefferson's. Studies after the 1980 US census found Webster has nearly zero population bias on realistic data, which is why it's the academic default. It was used by the US House from 1842-1852 and 1901-1941.
5. Adams's method
The mirror image of Jefferson: always round up. Find a divisor $d'$ such that $\sum \lceil p_i / d' \rceil = S$.
With ceiling rounding, the standard divisor over-counts dramatically (every group gets pushed up to its upper quota). To bring the total down, $d'$ must be increased. On the running data, try $d' = 1100$:
$$ A: \lceil 14.068 \rceil = 15, \quad B: \lceil 5.645 \rceil = 6, \quad C: \lceil 3.014 \rceil = 4 $$Sum 25 ✓. Adams gives $A: 15$, $B: 6$, $C: 4$ — a different apportionment from the previous three. $C$ gains a seat at $A$'s expense.
By the same logic that makes Jefferson favor giants, Adams favors midgets — the smallest group is guaranteed at least one seat (it can never round down to 0), while the largest absorbs the deficit. Adams was proposed by John Quincy Adams in 1832 as a counterweight to Jefferson; it was never adopted.
6. Huntington-Hill (the US method)
A divisor method like the others, but with a clever rounding rule: round each $q'_i = p_i / d'$ at the geometric mean of $\lfloor q'_i \rfloor$ and $\lceil q'_i \rceil$, rather than at the arithmetic mean (which is what Webster uses).
$$ \text{geometric mean of } n \text{ and } n+1 = \sqrt{n(n+1)} $$If $q'_i$ falls below $\sqrt{n(n+1)}$ it rounds down to $n$; if above, up to $n+1$. The geometric mean is always slightly less than the arithmetic mean ($(2n+1)/2$), so Huntington-Hill rounds up a bit more aggressively than Webster — favoring small states slightly.
The US has used Huntington-Hill since 1941 to apportion the 435 seats of the House of Representatives. The method satisfies the constraint that no state can be given zero seats (every state gets at least one), and it minimizes a particular measure of relative population per seat across all states.
| $n$ (lower) | $n+1$ | Webster cutoff $(n + .5)$ | Huntington-Hill cutoff $\sqrt{n(n+1)}$ |
|---|---|---|---|
| $1$ | $2$ | $1.5$ | $1.414$ |
| $2$ | $3$ | $2.5$ | $2.449$ |
| $5$ | $6$ | $5.5$ | $5.477$ |
| $15$ | $16$ | $15.5$ | $15.492$ |
Huntington-Hill's cutoff is always slightly below Webster's — a quota of $5.48$ rounds up under HH but down under Webster. The size of this gap shrinks as $n$ grows, which is why for the largest states the two methods almost always agree.
7. Apportionment paradoxes
Three counterintuitive things that can happen with Hamilton's method, and that the divisor methods avoid.
The Alabama paradox
Increasing the total number of seats — with no change in any group's population — can cause a group to lose a seat. Discovered in 1880: when the House considered going from 299 to 300 seats, Alabama's apportionment dropped from 8 to 7 under Hamilton. (See Example 2 below for a constructed numerical case.)
The population paradox
One group's population grows faster than another's — yet under Hamilton, the faster-growing group loses a seat to the slower-growing one. Real US census data triggered this around 1900.
The new-states paradox
Adding a new group (with appropriately rounded extra seats) changes the apportionment between the original groups. Triggered in 1907 when Oklahoma was admitted to the Union — it received 5 seats as expected, but New York lost one and Maine gained one in the reshuffle.
All three paradoxes share a common cause: Hamilton uses fractional remainders as its tie-breaker, and remainders aren't stable under changes elsewhere in the table. Divisor methods are immune because they only ever look at one group at a time relative to a single divisor.
8. The Balinski-Young theorem
Mathematicians Michel Balinski and H. Peyton Young proved in 1980 that the apportionment problem has its own impossibility theorem — directly analogous to Arrow's.
No apportionment method can simultaneously (i) satisfy quota — every group gets either $\lfloor q_i \rfloor$ or $\lceil q_i \rceil$ — and (ii) be free of the Alabama, population, and new-states paradoxes.
You can satisfy quota (Hamilton does) but then you suffer the paradoxes. Or you can dodge the paradoxes (any divisor method does) but then you sometimes violate quota — Jefferson, Webster, Adams, and Huntington-Hill all occasionally give a state one more (or one fewer) seat than its lower/upper quota allows.
Like Arrow, this isn't a verdict on any one method. It's a statement that "fully fair apportionment" is a logical impossibility — and the choice of method is the choice of which kind of unfairness to accept. The US settled on Huntington-Hill in 1941 after 150 years of switching back and forth in the wake of each new fairness crisis.
Methods used by the US House of Representatives, in historical order: Jefferson (1791-1842), Webster (1842-1852), Hamilton (1852-1900), Webster (1901-1941), Huntington-Hill (1941-present).
9. Common pitfalls
The standard quota uses the standard divisor $d = (\text{total pop}) / S$. The modified quota uses a search-tuned $d'$. Only the modified quota is the thing that gets rounded in Jefferson/Webster/Adams/HH; the standard quota is just the starting point and a sanity check.
There isn't one. You find $d'$ by trial: try a value, round, check the sum, nudge. With practice it converges in 2-3 tries; with software it's a one-line binary search. Don't try to solve for it algebraically — for many populations the valid $d'$ is an entire interval, not a single number.
Jefferson rounds down. Adams rounds up. Webster rounds to nearest (cutoff at the arithmetic mean). Huntington-Hill rounds at the geometric mean. Picking the wrong rounding rule will give you "the right method's answer for the wrong method."
It doesn't. Hamilton's method is what apportionment is usually introduced with — because it's the most intuitive — but the US House has used Huntington-Hill since 1941 (signed by FDR after the 1940 census). Don't carry over a textbook example to claim something about real US politics.
10. Worked examples
Example 1 · Apportion 25 seats among 3 states by each method
Populations: $A = 15{,}475$, $B = 6{,}210$, $C = 3{,}315$; total $25{,}000$; seats $S = 25$. Standard divisor $d = 1{,}000$. Standard quotas $15.475, 6.210, 3.315$.
| Method | $A$ | $B$ | $C$ | Sum |
|---|---|---|---|---|
| Hamilton | 16 | 6 | 3 | 25 |
| Jefferson ($d' = 950$) | 16 | 6 | 3 | 25 |
| Webster ($d' = 985$) | 16 | 6 | 3 | 25 |
| Adams ($d' = 1100$) | 15 | 6 | 4 | 25 |
| Huntington-Hill | 15 | 6 | 4 | 25 |
Hamilton/Jefferson/Webster agree; Adams and Huntington-Hill agree differently. Same populations, two different "fair" answers. Notice how Adams & HH transfer a seat from the largest state to the smallest — the small-group bias of upward-leaning rounding.
For Huntington-Hill, the cutoff between 15 and 16 is $\sqrt{15 \cdot 16} = 15.492$; for 3 vs 4 it's $\sqrt{12} = 3.464$. With $d' = 1{,}100$: $A$'s quota is $14.068$, well below $15.492$, so it rounds down to 15. $C$'s quota is $3.014$, below $3.464$, so it rounds down to 3 — sum is only $24$, so the divisor must shrink. After more iteration, the converged HH apportionment turns out to be $A: 15, B: 6, C: 4$.
Example 2 · The Alabama paradox in action
Three states $X$, $Y$, $Z$ with populations $940$, $9{,}030$, $10{,}030$ (total $20{,}000$). Apportion $S = 10$ seats, then $S = 11$, using Hamilton.
With $S = 10$: standard divisor $d = 2{,}000$. Quotas $0.470$, $4.515$, $5.015$. Lower quotas $0, 4, 5$ sum to $9$; one seat remains. Largest fractional part is $X$ ($.470$). Apportionment: $X: 1$, $Y: 4$, $Z: 5$.
With $S = 11$: standard divisor $d = 20{,}000 / 11 \approx 1{,}818.18$. Quotas $0.517$, $4.967$, $5.517$. Lower quotas $0, 4, 5$ sum to $9$; two seats remain. Largest fractional parts: $Y$ ($.967$) and $Z$ ($.517$). Apportionment: $X: 0$, $Y: 5$, $Z: 6$.
Adding a seat made $X$ lose its seat — going from 1 to 0. Nothing about $X$ changed; the math of the remainders simply shifted. This is the Alabama paradox in miniature.
Example 3 · Huntington-Hill on a small case
Apportion 5 seats among 3 groups with populations $5{,}300$, $3{,}500$, $1{,}200$ (total $10{,}000$). Standard divisor $d = 2{,}000$; quotas $2.65, 1.75, 0.60$.
HH cutoffs: between $0$ and $1$ → $\sqrt{0 \cdot 1} = 0$ (so any positive quota rounds up to 1 — guaranteeing every group at least 1 seat). Between $1$ and $2$ → $\sqrt{2} \approx 1.414$. Between $2$ and $3$ → $\sqrt{6} \approx 2.449$.
At $d' = 2{,}000$: group 1's quota $2.65 > 2.449$ → round to 3. Group 2's $1.75 > 1.414$ → round to 2. Group 3's $0.60 > 0$ → round to 1. Sum $= 6$, one too many. Increase $d'$.
At $d' = 2{,}200$: quotas $2.409, 1.591, 0.545$. Group 1: $2.409 < 2.449$ → round to 2. Group 2: $1.591 > 1.414$ → round to 2. Group 3: round to 1. Sum $= 5$ ✓.
Apportionment: $2, 2, 1$. Notice that group 3, despite having a quota of only $0.60$, walks away with 1 seat — characteristic of HH's small-state guarantee.
Example 4 · Jefferson violates quota
Populations $100$, $100$, $100$, $700$; total $1{,}000$; $S = 10$. Standard divisor $d = 100$; quotas $1, 1, 1, 7$. Lower quotas sum to $10$ already — Jefferson, Webster, and Hamilton all give $1, 1, 1, 7$. ✓ This is the easy case.
Now move 50 from the big state to make $150, 100, 100, 650$. Standard quotas with $d = 100$: $1.5, 1.0, 1.0, 6.5$. Lower quotas sum to $9$; one short.
Jefferson: shrink $d'$ until floors sum to $10$. At $d' = 92$: $\lfloor 1.630 \rfloor + \lfloor 1.087 \rfloor + \lfloor 1.087 \rfloor + \lfloor 7.065 \rfloor = 1 + 1 + 1 + 7 = 10$ ✓. So Jefferson gives the big state 7 — but its upper quota is $\lceil 6.5 \rceil = 7$, so we're at the boundary, not violating yet.
Construct a more lopsided case (e.g., $1{,}000$ small states of population $10$ each plus 1 huge state of population $9{,}000$, with $S = 100$): Jefferson awards the huge state significantly more than its upper quota, while the small states absorb the loss. The bigger and more skewed the populations, the worse Jefferson's quota violations get — the historical reason for switching away from it after 1842.