4.2 Hamilton’s Method

Alexander Hamilton proposed the method that now bears his name. His method was approved by Congress in 1791, but was vetoed by President Washington. It was later adopted in 1852 and used through 1911. He begins by determining, to several decimal places, how many things each group should get. Since he was interested in the question of Congressional representation, we’ll use the language of states and representatives, so he determines how many representatives each state should get. He follows these steps:

Note

Hamilton’s Method

Determine how many people each representative should represent. Do this by dividing the total population of all the states by the total number of representatives. This answer is called the divisor.
Divide each state’s population by the divisor to determine how many representatives it should have. Record this answer to several decimal places. This answer is called the quota. Since we can only allocate whole representatives, Hamilton resolves the whole number problem, as follows:
Cut off all the decimal parts of all the quotas (but don’t forget what the decimals were). These are called the lower quotas. Add up the remaining whole numbers. This answer will always be less than or equal to the total number of representatives (and the “or equal to” part happens only in very specific circumstances that are incredibly unlikely to turn up).
Assuming that the total from Step 3 was less than the total number of representatives, assign the remaining representatives, one each, to the states whose decimal parts of the quota were largest, until the desired total is reached.

Make sure that each state ends up with at least one representative!

Note on rounding: Today we have technological advantages that Hamilton (and the others) couldn’t even have imagined. Take advantage of them, and keep several decimal places.

Example 1

The state of Delaware has three counties: Kent, New Castle, and Sussex. The Delaware state House of Representatives has 41 members. If Delaware wants to divide this representation along county lines (which is not required, but let’s pretend they do), let’s use Hamilton’s method to apportion them. The populations of the counties are as follows (from the 2010 Census):

# Math in Society 4.2 -- Hamilton's Method, step by step
# Divide the population by the seats to get the DIVISOR, divide each population by
# the divisor to get its QUOTA, hand out the whole numbers, then give the leftover
# seats to the largest decimal remainders.
def hamilton(pops, seats, label):
    total = sum(pops.values())
    divisor = total / seats
    quota = {k: v / divisor for k, v in pops.items()}
    lower = {k: int(q) for k, q in quota.items()}
    leftover = seats - sum(lower.values())
    ranked = sorted(pops, key=lambda k: (-(quota[k] - lower[k]), k))
    final = dict(lower)
    for k in ranked[:leftover]:
        final[k] += 1

    print(f"{label}: {seats} seats")
    print(f"   divisor = {total:,} / {seats} = {divisor:,.5f}"
          f"   (one seat per {divisor:,.0f} people)")
    print(f"   {'':<12}{'population':>12}{'quota':>11}{'lower':>7}"
          f"{'remainder':>11}{'final':>7}")
    for k in sorted(pops, key=lambda k: -pops[k]):
        got = " <-- leftover seat" if k in ranked[:leftover] else ""
        print(f"   {k:<12}{pops[k]:>12,}{quota[k]:>11.4f}{lower[k]:>7}"
              f"{quota[k]-lower[k]:>11.4f}{final[k]:>7}{got}")
    print(f"   {'total':<12}{total:>12,}{seats:>11.4f}{sum(lower.values()):>7}"
          f"{'':>11}{sum(final.values()):>7}")
    broke = [k for k in pops if not (quota[k] - 1 < final[k] < quota[k] + 1)]
    print("   quota rule: " + ("every county is within 1 of its quota"
                               if not broke else f"VIOLATED by {broke}"))
    print()
    return final

delaware = {"Kent": 162310, "New Castle": 538479, "Sussex": 197145}
rhode_island = {"Bristol": 49875, "Kent": 166158, "Newport": 82888,
                "Providence": 626667, "Washington": 126979}

hamilton(delaware, 41, "Delaware")            # EDIT: change 41 to any house size
hamilton(rhode_island, 75, "Rhode Island")

print("Bristol's remainder is over .5 and it still gains nothing: Hamilton does not")
print("round, it RANKS. Three counties had bigger remainders and only three seats")
print("were left over.")
print("Try it: run hamilton(rhode_island, 74, 'Rhode Island') and then 76. Watch how")
print("a single extra seat can reshuffle which counties win the leftovers.")

County	Population
Kent	162,310
New Castle	538,479
Sussex	197,145
Total	$897, 934$

Show solution

1. First, we determine the divisor: $897, 934 / 41 equals 21, 900.82927$

2. Now we determine each county’s quota by dividing the county’s population by the divisor:

County	Population	Quota
Kent	162,310	7.4111
New Castle	538,479	24.5872
Sussex	197,145	9.0017
Total	$897, 934$

3. Removing the decimal parts of the quotas gives:

County	Population	Quota	Initial
Kent	162,310	7.4111	7
New Castle	538,479	24.5872	24
Sussex	197,145	9.0017	9
Total	$897, 934$		$40$

4. We need 41 representatives and this only gives 40. The remaining one goes to the county with the largest decimal part, which is New Castle:

County	Population	Quota	Initial	Final
Kent	162,310	7.4111	7	7
New Castle	538,479	24.5872	24	25
Sussex	197,145	9.0017	9	9
Total	$897, 934$		$40$	$41$

Example 2

Use Hamilton’s method to apportion the 75 seats of Rhode Island’s House of Representatives among its five counties.

County	Population
Bristol	49,875
Kent	166,158
Newport	82,888
Providence	626,667
Washington	126,979
Total	$1, 052, 567$

Show solution

1. The divisor is $1, 052, 567 / 75 equals 14, 034.22667$

2. Determine each county’s quota by dividing its population by the divisor:

County	Population	Quota
Bristol	49,875	3.5538
Kent	166,158	11.8395
Newport	82,888	5.9061
Providence	626,667	44.6528
Washington	126,979	9.0478
Total	$1, 052, 567$

3. Remove the decimal part of each quota:

County	Population	Quota	Initial
Bristol	49,875	3.5538	3
Kent	166,158	11.8395	11
Newport	82,888	5.9061	5
Providence	626,667	44.6528	44
Washington	126,979	9.0478	9
Total	$1, 052, 567$		$72$

4. We need 75 representatives and we only have 72, so we assign the remaining three, one each, to the three counties with the largest decimal parts, which are Newport, Kent, and Providence:

County	Population	Quota	Initial	Final
Bristol	49,875	3.5538	3	3
Kent	166,158	11.8395	11	12
Newport	82,888	5.9061	5	6
Providence	626,667	44.6528	44	45
Washington	126,979	9.0478	9	9
Total	$1, 052, 567$		$72$	$75$

Note that even though Bristol County’s decimal part is greater than .5, it isn’t big enough to get an additional representative, because three other counties have greater decimal parts.

Hamilton’s method obeys something called the Quota Rule. The Quota Rule isn’t a law of any sort, but just an idea that some people, including Hamilton, think is a good one.

Controversy

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 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.

# Math in Society 4.2 -- The Alabama Paradox, found by brute force
# Nobody moves, nobody is born, nobody dies. Only the size of the house changes.
def hamilton(pops, seats):
    divisor = sum(pops.values()) / seats
    quota = {k: v / divisor for k, v in pops.items()}
    seats_for = {k: int(q) for k, q in quota.items()}
    leftover = seats - sum(seats_for.values())
    for k in sorted(pops, key=lambda k: (-(quota[k] - seats_for[k]), k))[:leftover]:
        seats_for[k] += 1
    return seats_for

pops = {"A": 6000, "B": 6000, "C": 2000}      # EDIT: any population table
LOW, HIGH = 6, 26                              # EDIT: range of house sizes to sweep

names = list(pops)
W = max(7, max(len(k) for k in names) + 2)
print("populations: " + ", ".join(f"{k} {v:,}" for k, v in pops.items())
      + f"   (total {sum(pops.values()):,})\n")
print("  house" + "".join(f"{k:>{W}}" for k in names))
paradoxes = 0
previous = None
for seats in range(LOW, HIGH + 1):
    now = hamilton(pops, seats)
    note = ""
    if previous is not None:
        lost = [k for k in names if now[k] < previous[k]]
        if lost:
            paradoxes += 1
            note = "   <-- " + ", ".join(lost) + " LOSES a seat as the house GROWS"
    print(f"{seats:>7}" + "".join(f"{now[k]:>{W}}" for k in names) + note)
    previous = now

print(f"\n{paradoxes} Alabama paradox(es) in house sizes {LOW}-{HIGH}.")
print("Adding a seat gives every quota a slightly bigger fractional part, but not")
print("by the same amount: a big state's remainder can climb past a small state's")
print("and steal the leftover seat the small state was holding.")
print("\nTry it: use the tutoring numbers from the exercises,")
print("   pops = {'Math': 330, 'English': 265, 'Chemistry': 130, 'Biology': 70}")
print("with LOW, HIGH = 10, 45. Biology loses a tutor when the college hires its")
print("40th -- with not one student changing subject.")
print("\n(Note: this is plain Hamilton. The rule that every group gets at least one")
print("seat is not enforced here, so a group whose quota is below 1 can show a 0.)")

The New States Paradox happened when Oklahoma became a state in 1907. Oklahoma had enough population to quality for five representatives in Congress, so they raised the total number of representatives by five. When the apportionment was recalculated, unsurprisingly Oklahoma received five representatives. However, despite their populations not changing, Maine gained a representative and New York lost one.

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.

Adapted from Math in Society by David Lippman, hosted on LibreTexts (math.libretexts.org) and licensed under CC BY-SA 3.0. Changes were made. License: CC-BY-SA-3.0.