Fair shares: apportionment and Arrow's theorem
Tonight's idea
The US Constitution shares House seats among the states in proportion to population, but a share comes out as 7.8 seats and you cannot send 0.8 of a person to Washington. Rounding fairly is apportionment.
Hamilton's method: compute each state's exact share, its quota; give every state the whole-number part; hand the leftover seats, one each, to the states with the largest fractional parts. Washington vetoed it (the first veto ever); Congress adopted Jefferson's method: pick a divisor \(D\), give each state \(\lfloor \text{population}/D \rfloor\) seats (\(\lfloor x \rfloor\) means round \(x\) down), and adjust \(D\) until the seats add up. Rounding down favours big states, and Jefferson was from the biggest one.
Hamilton's method returned, and in 1880 a clerk found that Alabama would get 8 seats in a house of 299 but 7 in a house of 300: the Alabama paradox. Hamilton stays within quota (the share rounded down or up) but can give a state fewer seats when the house grows. Jefferson avoids that house-size paradox but can break quota. Methods doing both exist; other fairness demands still clash (stretch).
Voting is worse. Arrow (1951) asked for a rule turning everyone's rankings into one group ranking such that:
- it always gives an answer, whatever the rankings;
- unanimity: if everyone prefers A to B, so does the group;
- independence of irrelevant alternatives: the group's order of A against B depends only on how voters rank A against B;
- no dictator: no one voter's ranking is automatically the group's.
Arrow's theorem: with three or more options, no rule satisfies all four.
Watch one: Hamilton with three states
Populations 7, 6 and 2 million; 10 seats. Total 15 million, so one seat per 1.5 million. Quotas: \(7/1.5 = 4.667\), \(6/1.5 = 4\), \(2/1.5 = 1.333\). Whole parts 4, 4, 1: nine seats. The largest fraction, 0.667, gives A the spare seat: 5, 4, 1. Jefferson with \(D = 1.4\) million gives \(\lfloor 5 \rfloor, \lfloor 4.29 \rfloor, \lfloor 1.43 \rfloor = 5, 4, 1\) too.
Your turn
- The Alabama paradox. Populations 6, 6 and 2 million. Apportion 10 seats by Hamilton's method. Predict: can an eleventh seat cost anyone a seat? Now apportion 11. What happens to the small state, and exactly why?
- Jefferson. Five states with populations 650, 100, 100, 100 and 50 thousand share 12 seats. Compute the quotas. Show that \(D = 72\) thousand works for Jefferson and write down the result. Which state beats its quota rounded up? Which gets nothing?
- Twelve voters, three candidates. Their rankings:
Who wins by plurality? By a runoff between the top two? Is there a Condorcet winner, who beats each other candidate head to head?5 voters 4 voters 3 voters 1st A B C 2nd B C B 3rd C A A - The spoiler. Suppose C withdraws from question 3 and C's voters switch to their second choice. Who wins by plurality now? Which of Arrow's conditions does plurality violate, and what is the everyday name for this?
Stretch: the theorem that says you must choose
(a) Why does Hamilton always stay within quota? (b) Check that a dictatorship satisfies (1), (2) and (3): those three alone force a dictator. (c) Why does a Condorcet cycle (A beats B, B beats C, C beats A) make pairwise majority rule fail (1)?
Answers
- Ten seats: quotas 4.286, 4.286, 1.429; whole parts 4, 4, 1; the largest fraction (0.429) is the small state's, so 4, 4, 2. Eleven seats: quotas 4.714, 4.714, 1.571; whole parts 4, 4, 1; the two largest fractions are now the big states' 0.714, so 5, 5, 1. The small state loses a seat as the house grows: multiplying every quota by 11/10 grew the big states' fractional parts by 0.429 each and the small state's by only 0.143.
- Quotas 7.8, 1.2, 1.2, 1.2, 0.6. With \(D = 72\): \(650/72 = 9.03\), \(100/72 = 1.39\), \(50/72 = 0.69\); rounding down gives 9, 1, 1, 1, 0, which is 12 seats. The big state gets 9 against a quota of 7.8, more than its quota rounded up. The smallest state gets nothing despite a quota of 0.6.
- Plurality: A 5, B 4, C 3, A wins. Runoff A against B: C's voters prefer B, so B wins 7 to 5. Head to head B beats A 7–5 and C 9–3, so B is the Condorcet winner, and plurality failed to elect it.
- Without C, its three voters go to B: B 7, A 5, B wins. The A-versus-B result changed because of C, who was not going to win either way: a violation of independence of irrelevant alternatives (3). Everyday name: the spoiler effect.
Stretch. (a) Hamilton starts every state at \(\lfloor q \rfloor\) and adds at most one seat, so each ends at \(\lfloor q \rfloor\) or \(\lfloor q \rfloor + 1\). The clash Balinski and Young proved (1982) is with population, not house size: for four or more states, no method stays within quota and also never lets a state that grew faster than another lose a seat to it. (b) Dictatorship: it gives a ranking for any input (1); if everyone, including the dictator, ranks A above B so does the output (2); the output's A-versus-B order is the dictator's, which depends on nothing else (3). It fails only (4). (c) A Condorcet cycle makes pairwise majority rule output A > B > C > A, which is not a ranking, so the rule has no answer for that profile and fails (1).