10 November 2008

A quirk of electoral apportionment

I was curious: how will the electoral vote apportionment change between now and 2012? (Reapportionment is done after each census, and censuses take place in years divisible by 10; the apportionment takes effect the year after the census. Thus the 2004 and 2008 presidential elections were done under one apportionment, and the 2012, 2016, and 2020 elections will be done under another one.)

I don't know (my first attempt at programming the apportionment gave some really strange-looking results) but I wanted to share an amusing fact.

Each of the 50 states receives a number of electoral votes equal to its number of Representatives, plus two. So the question is really one of determining the number of Representatives that each state gets. The way this works is as follows. First, each state receives one seat. Then, let the populations of the states be P1, P2, ..., P50; let

Qi,j = Pi / (j(j-1))1/2

for 1 ≤ i ≤ 50 and all positive integers j. Sort these numbers, and take the 385 largest of these numbers. Now state i (the state with population Pi) gets r = ri representatives, where r is the unique integer such that Qi,r is one of the 385 largest Q's, and Qi,r+1 is not. (385 is 435-50; 435 is the number of seats in the House of Representatives, and 50 seats were already assigned in the previous step, one for each state.) Essentially, this assigns the seats in the House "in sequence", so we can speak of the 51st seat, 52nd seat, ..., 435th seat.

So what if there's a tie for 385th place in that ordering? This can occur, of course, if two states have the same population, and I bet some tiebreaker is written into the law. But what if two states have different populations, but after dividing by the square root factor, two of the Qi,j are the same? Surprisingly, this can happen. Let P1 = 6P2. Then it's not hard to see Q1,9 = Q2,2; that is, state 1 gets its ninth seat "simultaneously with" state 2 getting its second seat. More generally, if

P1 / (m(m-1))1/2 = P2 / (n(n-1))1/2

then state 1 gets its mth seat simultaneously with state 2 getting its nth seat. Note that P1/P2 is rational. So a tie can only occur when (m(m-1)/n(n-1))1/2 is rational; when does this happen?

When n = 2, this amounts to asking when (m(m-1)/2) is a square; this happens for m = 2, 9, 50, ... (the indices of the square-triangular numbers in the sequence of triangular numbers) So one state can receive its second seat at the same time another one gets its 9th seat, its 50th seat, ... if the larger state has 6, 35, ... times the population of the smaller one.

Somehow I doubt the law covering apportionment has a provision for this. I suspect the provision taken would be similar to what happens if there's a tie in an election; I know there are some jurisdictions that just flip a coin in that case.

Edit, 10:53 pm: Boris points out in the comments that somebody's done the projection. Texas gains 4, Florida and Arizona each gain 2; the Carolinas, Georgia, Utah, Nevada, and Oregon each gain 1. New York and Ohio each lose 2; Massachusetts, New Jersey, Pennsylvania, Michigan, Illinois, Minnesota, Iowa, Missouri, Louisiana, and California each lose 1. At first glance this shift seems like it would favor the Republicans in the presidential race; nine of the seats created are in states that voted for McCain in '08, and only two of the seats destroyed are. But I'm not sure about this analysis; states are made of people, so as a state's population grows or shrinks its political makeup changes as well. Maybe Nate Silver will have something to say about this?

9 comments:

Jonah said...

Question: if Washington, DC gets big enough, do they get a second non-voting representative?

Michael Lugo said...

Jonah,

I don't know. I know that they do not get a fourth electoral vote; the 23rd Amendment (which gives electoral votes to DC) says that DC gets the number of electoral votes it would get if it were a state with that population, except it doesn't get more than the least populous state. I think the non-voting representatives are not mentioned in the Constitution, so I suppose it would be up to whoever decides the internal workings of Congress.

Unknown said...

Actually, the most likely outcome would just be expanding the house to 436, which would just require a change of law (the states in question would lobby hard for it, and I can't see anyone fighting them). There was talk of doing this in 2000, since Utah and NC were so close (the difference was probably a miscount in a dorm at UNC).

Michael Lugo said...

Ben,

I heard something similar, although I recall that they were considering expanding the House to 437 -- for partisan balance, basically, because DC would almost certainly elect a Democrat -- and that the plan was to give the extra seat to Utah because Utah was close to getting another seat and had argued that their population was being undercounted because of the large number of Mormon missionaries who are Utah residents but who were elsewere at the time of the census.

Unknown said...

Re what will change: Wikipedia has projections for 2010.

Buddha Buck said...

The real tragedy is that the House stopped growing nearly a century ago, and the population hasn't. A simple calculation (300M/435) shows that each Rep represents, on average, 690K people, which is way too many for a single person to represent. It also means that, despite population growth, representation in the House can, and will, fall for many states.

I'd like to see the laws changed to allow a dynamic number of House members. One method would be to allocate using the existing "Equal Proportions" method until all states had at least two representatives. Another would be to continue allocating until the average constituency size was a fixed number significantly smaller than the current size -- say a measly quarter million people.

Unknown said...

Re ties: Huntington, one of the originators of the "method of equal proportions" (Congress seems to often call it the "Hill method"; I've always heard "Huntington-Hill method") actually commented on the issues of ties in his 1928 paper "The Apportionment of Representatives in Congress" in the Transactions of the AMS. See page 15 of 26 of his article (page 99 in the journal) -- hopefully my deep link works if you have access to JSTOR.

He says that there are very few possibilities for ties under his method, and actually considers this a "theoretical argument in favor of [his] method".

Incidentally, it appears that your intuition as to which tie-breaking is written into the law might be incorrect. The actual law for the matter simply says to use the method of equal proportions, which isn't helpful at all with regards to specifics. As far as I can tell, there is no actual description of the method anywhere in the U.S. Code; thus, in some sense, neither tie-breaking scenario is considered.

However, Huntington mentions that in the case of a tie like the one you're talking about (different n and m in your notation), "convention .. provides that in case of a tie preference shall be given to the state having the larger population".

Michael Lugo said...

Boris,

I agree that there's a very small probability of ties -- but in any case the numbers involved are so large that one really shouldn't worry about the probability of ties.

The link seems to work; I'm not on campus right now, and I don't feel like dealing with my library's web site, but I'll probably take a look at the paper tomorrow.

Anonymous said...

Sorry if this is familiar to you (or if you wrote it about it, though a search of your blog suggested not), but you might enjoy this while on this topic of apportionment: http://en.wikipedia.org/wiki/Apportionment_paradox#Alabama_Paradox