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Days 21 and 22
August 10 and 11, 2009
Last time
We finished our discussion of voting theory by proving Arrow’s theorem. This theorem stated that it is impossible to have a voting system that produces a single winner from among three (or more) candidates and that satisfies the Pareto condition, IIA, and monotonicity. This was in stark contrast to the two- candidate case, in which majority rule works as well as we’d like it to: it treats all voters fairly, it treats all candidates equally, and it satsifies the monotonicity condition. May’s theorem, remember, stated that majority rule is the only such social choice procedure for two-candidate elections that always elects a single winner and which satisfies these three properties. There is more to voting theory than what we’ve seen. I would recommend looking at Taylor’s “Mathe- matics and Politics” for more on the applications of mathematics to voting, weighted voting (which occurs when voters have varying degrees of power), game-theoretic problems such as arms races and a general theory of escalation, and others. In the last two days of class, we will examine the fundamental problem that arises in apportionment theory. There is no main result we plan to prove. The goal is to simply start with an innocent sounding problem, work out some example solutions, run into paradoxes, try to find ways around them, and hopefully learn along the way something about how mathematical thinking can be applied to real-world problems.
Apportionment theory: Introduction
The characters in this story are based on those involved every ten years in the U.S., when the U.S. Census comes out and the number of seats in the U.S. House of Representatives awarded to each state is recalcu- lated. First, we have n states (here, n = 50, but let’s let n be arbitrary) numbered 1-n with populations p 1 , p 2 ,... , pn. Second, we let p = p 1 + p 2 +... + pn
denote the total population. State i then has a fraction pi/p of the population; expressed as a percentage, we say that state i has pi p
× 100%
of the country’s population. Now we supposed that the House of Representatives has h seats (here, h = 435, but again we can state the problem for an arbitrary h). Since it’s written in the U.S. Constitution that states should receive House seats proportionately (“according to their respective Numbers”), we conclude that state i should receive qi =
pi p × h
seats in the House. The variables qi denote apportionment quotas. If these numbers of seats were integers, then we would assign each state pi × h/p, or pih/p seats for a total of
q 1 + q 2 +... + qn =
(p 1 + p 2 +... + pn) p
× h = h
seats, as expected. An obvious problem arises here: the numbers pih/p are seldom integers. The first issue is therefore that we must round the (fractional) number of each seats each state deserves to whole numbers (integers). Suppose we tried to most obvious rounding method that we learned in grade school: Round a number up to the nearest integer if it’s fractional part is .5 or higher, and round if not. (Quick review question for the final: Which way would we round 27.49?) Some states will get fewer seats than they deserve, and others more, but at least every state would agree that the relative advantages and disadvantages would cancel over time. In fact, the 1790 apportionment bill submitted to President George Washington (the date on the bill refers to the first year in which it would apply, not to the year it was submitted) declared that this very method should be used to apportion seats in the House of Representatives. Washington vetoed the bill, however, citing his (or his advisor’s, I’m not sure) observation that rounding could, using other possible data sets,
result in fewer or more than 105 congressmen and women being seated. (The original size of the House was 105 seats.) As an easy example of how this can happen, suppose there are n = 3 states and h = 10 seats to be apportioned, and suppose the states have equal populations, so that pi/p = 1/3 in each case. Then each state deserves approximately 3.33 seats. If we naively round these numbers (down in each case), we see that each state gets 3, for a total of 9 seats. Who gets the extra seat? You might object to this scenario saying that these percentages are contrived. Consider the more generic scenario where the states have 31.1%, 34.4%, and 34.5% of the population, so the exact numbers of seats they deserve are 3.11, 3.44, and 3.45 seats, respectively. If we round these percentages, then we see again that each state gets three seats, leaving us one short of a full House (sorry for that). Perhaps the third state should get the tenth seat, but can we see why the second state wouldn’t be too happy about this?
Apportionment theory: The notation
We recall and in other cases define the following notation, listed here for convenient reference:
- n is the number of states
- h is the number of seats in the House
- pi is the population of State i
- p = p 1 + p 2 +... + pn is the total population
- s = p/h is the standard divisor, which isn’t necessary but which is convenient notation
- qi = pi/s is the quota for State i (this is the number of seats a state would receive in the event that all of these numbers are whole numbers and no rounding is needed)
- An apportionment is a sequence (a 1 , a 2 ,... , an) of nonnegative integers, where ai represents the number of seats given, or apportioned, to State i
- For a nonnegative real number x, bxc represents the greatest integer less than or equal to x. This is the number you obtain by simply ignoring the decimal part of x. For example, b 3 c = 3, b 3. 145 c = 3, bπc = 3, b 2. 99 c = 2, and
= 3. (Make sure you understand the last one for the final!)
- Similarly, define dxe for nonnegative real numbers x to be the least integer greater than or equal to x. This is the number you obtain when you “round up”, whereas bxc is the number you get by “rounding down”. For example, d 3. 34 e = 4 and d 4 e = 4.
- When rounding “as usual”, denote the result by 〈x〉. So 〈 3. 45 〉 = 3 and 〈 3. 5 〉 = 4.
With the notation set up as listed, we can state the general apportionment problem:
Suppose States numbered 1 through n have populations p 1 through pn, and that there are h seats to be apportioned to these states, find a method of apportioning the seats to the states according to their respective populations in the “fairest way possible”.
More concisely, given an h and a sequence (p 1 , p 2 ,... , pn) of state populations, compute the sequence (q 1 , q 2 ,... , qn) of state quota and round in “the fairest way possible” to obtain a sequence (a 1 , a 2 ,... , an) of apportionments. We, of course, have to say what we mean by “fair”. For example, one might expect that ai, the number of seats State i receives, is never below what’s called its lower quota, bqic, and is never above its upper quota, dqie. Also, we might require that each ai is nonzero (so that every state is represented); this can certainly be forgotten by computerized procedures which compute and round long lists of state quotas to obtain apportionments. Other methods of defining “fair” will be discussed later; for now, we consider some methods for computing apportionments.
population. (This isn’t the population paradox, but it is similar.) Also, consider the example with populations 94, 147, and 160. Let h = 15. Add three to State A’s population and 5 to State B’s population. State A grew relative to State B, but it lost a seat relative to B.
- New states paradox: This event occurs when a new state joins the country, its approximate quota of seats is added to the House, and as a result, one of the original states loses a seat to another one of the original states. As an example, use cut-the-knot.org again with state populations 275, 604, and 335 and House size 37. Suppose we add State D, which has population 300. Since the old standard divisor was (population of States A, B, and C) divided by (size of House), or 32.8, we divide 32.8 into 300 to get 9. Hence the new House size is 46(= 37 + 9). The old House had 8, 19, and 10 seats for A, B, and C, respectively. The new house has 9, 18, 10, and 9 seats for States A, B, C, and D, respectively. Hence State B has lost a seat to State A. Nothing was supposed to change except that we added State D’s 9 deserved seats to the House!
In the U.S. House, the number of states and seats has been fixed, so the new states and new seats paradoxes aren’t a problem. The population paradox still is, however. In the 1970s, mathematicians Balinski and Young, like their voting theory analogues Kenneth (May) and Kenneth (Arrow), proved that if an apportionment method is such that it avoids the population paradox (and the others, though they only need to assume it avoids one), then it must be a divisor method. We define this type of method next. As we’ll see however, divisor methods have their own problems: every one fails the quota rule, which states that no state’s apportionment should be below its lower quota bqic or above its upper quota dqie. Combining these results, we have the following impossibility theorem:
Theorem 1. There is no apportionment method (as defined in these notes) that avoids that population paradox and satisfies the quota rule.
As we’ve emphasized before, our failure to find a suitable method isn’t a matter of not yet being clever enough; no such method exists! From a political standpoint, this means we must make compromises when deciding how to apportion House seats.
Apportionment theory: Divisor methods
Recall that the standard divisor s is the total population p divided by the size of the House h. Each state’s quota qi is then computed by dividing s into its population pi. (Hence a state’s quota qi = pi/h, which is the same as computing its fraction pi/p of the population and then computing the number of seats (pi/p) × h it should have. This is a matter of rearranging fractions.) In Hamilton-like methods, one first rounds these quotas qi down then assigns the left over seats in some way. (An equivalent Hamilton-like method is to round these quotas up and then take away seats according to which states had their quotas rounded up by the “most”, according to whether one considers relative or absolute differences between qi and dqie.) In divisor methods, one wishes to avoid the problem of having to assign (or take away) seats after the initial apportionment. Hence what one does is to fix a method of rounding (say, “always round down”), use trial and error to determine a suitable divisor d that is close to but different from the standard divisor s, then compute modified state quotas by dividing d into each pi. Since the modified divisor was specially chosen, the result by definition will be that the rounded modified quotas will sum to h. Here are the examples:
- Jefferson method. In this method, one always rounds down. Given a sequence of state populations p 1 , p 2 ,... , pn, one uses trial and error to find a modified divisor d. Then each states modified quota pi/d is computed. If d has been chosen correctly, the apportionments bp 1 /dc , bp 2 /dc ,... , bpn/dc will sum to h, the number of House seats. To determine a suitable d, one starts with the standard divisor s and computes the standard quotas
bp 1 /sc , bp 2 /sc ,... , bpn/sc.
If the sum of these is h, then one uses d = s. If the sum is less than h, one chooses a smaller d so as to increase the modified quotas pi/d. One then adds
bp 1 /dc , bp 2 /dc ,... , bpn/dc.
If the sum is h, then d is our “suitable” modified divisor. If the sum is less than h, then one again decreases d and repeats; if the sum is more than h, then one increases d so as to decrease the sum. (In the last case, one wouldn’t increase d past s, because we already knew that s was too large. One could proceed most quickly by averaging s and d and trying the result.) One continues until one finds a modified divisor d such that
bp 1 /dc + bp 2 /dc +... + bpn/dc = h.
Then one sets ai = bpi/dc; that is, one apportions bpi/dc seats to State i. As an example, we compute how many seats the Jefferson method would have apportioned to Virginia and Delaware after the 1790 census: Jefferson had used trial and error to come up with a modified divisor of d = 33, 000. (The standard divisor s = p/h was 34,437. Using this divisor, assigning bpi/sc seats to State i would have resulted in too small of a House size. To increase apportionments, we have decrease the divisor. After some trial and error, Jefferson found that d = 33, 000 was sufficiently small so that the apportionments were sufficiently increased such that the sum of the states’ apportionments bpi/dc was 105, the size of the House at the time.) The states of Virginia, Pennsylvania, and Delaware had populations of 630,560, 432,879, and 55,540, respectively, so their apportionments (according to the Jefferson method) were, respectively, ⌊ 630 , 560 33 , 000
= b 19. 108 c = 19, ⌊ 432 , 879 33 , 000
= b 13. 118 c = 13, and ⌊ 55 , 540 33 , 000
= b 1. 683 c = 1.
At least in this case, we see that Virginia, the biggest state at the time, gets a seat that Delaware had from Hamilton’s method. Another example (also from the Chapter 15 reading) is when, in 1820, New York had a population of 1,368,775, the total U.S. population was 8,969,878, and the House size was 213. The standard divisor, then, was s =
p h
New York’s quota was therefore
qNew York =
pNew York s
With the other data in hand, we would see that Hamilton’s method gives New York 33 seats. However, using the Jefferson method, one first obtains a suitable divisor of d = 39, 900, and then one computes the modified quota for New York: pNew York d
When one rounds this number down to the nearest integer, one obtains that the Jefferson method apportions 34 seats to New York. Notice that this is more than 1.4 seats more than it deserves. This is an example of a quota rule violation. The quota rule states that if a state has standard quota qi = pi/s, then its apportioned number ai of seats should be between bqic and dqie. This example shows that the Jefferson method fails to satisfy the quota rule. We won’t show this in general, but it turns out that every divisor method fails to satisfy this rule.
- Adams method. This method is identical to Jefferson’s method, only one rounds up where Jefferson’s method rounds down. (The “J” in Jefferson might be turned into a down arrow, and the “A” in Adams might be turned into an up arrow. This helps me remember which way each method rounds.)
Apportionment theory: Properties of, and fairness criteria for, apportionment
methods
In the course of giving examples, we have seen that the Hamilton-like methods can produce paradoxes, any such method that does not produce paradoxes is a divisor method, and that the divisor methods fails to satisfy the quota rule. Moving on, then, we might at least consider which methods treat big and small states equally. We claim the following (try to understand for yourself before continuing):
- The Jefferson method favors big states, and the Adams method favors small states. (To see this, consider the rule that gives every state its lower quota (resp. its upper quota). The large states would like it if all of the tiny states got their lower quotas (of 1 or 2, etc.). Losing its fractional part of a seat wouldn’t hurt California or Texas that much, but it would hurt Wyoming and Delaware. Since the Jefferson method is based on rounding down, it favors large states. Similarly, rounding up boosts small states more than it does big states, so the Adams method favors small states.)
- The Hamilton and Webster methods do not favor big states or small states. (The fractional part of a number has equal likelyhood of being below or above 1/2, whether the integer part is big or small. Hence big states and small obtain extra seats in a pretty random manner. People typically agree that randomness is fair.)
- The Lowndes and Hill-Huntington methods favor small states. Both methods seek to minimize relative differences in the states’ representative share. However, the geometric mean between two numbers n and n + 1 is always less than n plus a half. Hence, when rounding according to geometric means, numbers are more often rounded up than down. As discussed two items above, rounding up more often than down favors small states.
If you like hard data, let’s look back to the 2000 U.S. Presidential election. According to the method we use, the Hill-Huntington method, Bush won with 271 electoral votes to Gore’s 267. (Actually Gore got 266, because an D.C. elector didn’t vote because she wanted to protest D.C.’s lack of representation. In this discussion, I will assume she’d represent her district faithfully and vote for Gore in the event that her vote would be the tying or winning vote.) See the Excel spreadsheet. Notice that Gore wins the election using the Jefferson method, Bush wins when using the Lowndes, Adams, or Hill-Huntington methods, and they tie when using the Hamilton or Webster methods. I’ve uploaded this file to the Lecture Notes page, so enjoy!
