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Every ten years, the census reapportions Congressional representation. There are 385 seats in the House that are "up for grabs" based on each state's population. Since a majority in the House is 218 seats, it's theoretically possible for one state to be so populated that it is apportioned enough Representatives that its delegation automatically is the majority.

The Senate is specifically not subject to this issue, but the Electoral College is. A big enough state would have 270 electors and would be able to single-handedly decide the next President. This would take even more population, since the state would need to be apportioned 268 Representatives.

Assuming the current estimate of roughly 324,424,000 people in the country, how many of them would need to live in one state to reach these thresholds? And assuming that state was California, since it's already the most populated, how densely populated would it have to be to accommodate that many people?

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    This seems like a math problem rather than an actual political question. – user1530 Jan 24 '17 at 18:18
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    This kind of unbalance was a major structural problem for the Soviet Union. If the Russian SFSR had caucused before each Supreme Soviet vote, the Russian SFSR would have won every vote, and the other republics would not have had any say. To prevent this, the Communist Party was set up so that each SSR had its own Communist Party, but the Russian SFSR did not. Russia only had the overall Communist Party of the Soviet Union. This meant that (even as other ethnic groups complained about the power of the "Great Russian" majority), Russians complained about not having a Party of their own. – Jasper Jan 24 '17 at 19:02
  • @Jasper - Interesting. I hadn't known that, but I can see how both sides would be upset about that. – Bobson Jan 24 '17 at 19:41
  • Having seen the answers it doesn't look like CA is a specially interesting case of "how many people would need to move to make one state control the electoral college?" – user9389 Jan 24 '17 at 20:10
  • @notstoreboughtdirt - Agreed. I picked it because it's the most populous to begin with and because it tries to lead the country in regulation. I expected densities more on the order of a major city across the entire state, though. – Bobson Jan 24 '17 at 20:14
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California needs around 200 million people to control the presidency.

The Math

There are 538 Electoral College votes and a candidate needs to get 270 votes to win.

California gets 2 Electoral College votes from their Senate seats so they need 268 votes from their House of Representatives seats.

There are 435 total seats in the House so California needs about 268/435 of the population, roughly 61.6%.

268/435 * 324,424,000 = 199,875,017 (rounded up)

The Reality

The United States first gives each state one seat then uses the Huntington-Hill method to apportion the rest. This method assigns seats one at a time to whichever state has the worst representation, as decided by their formula.

There is no single correct number because the seats California gets is dependent on the exact population for each of the other 49 states. The math above is an estimate of how many would be necessary but in practice the number could be a little higher or lower.

Some other answers have calculated the absolute minimum needed in the best case scenario for California, I recommend looking at those as well.


Population Density

California is 163,696 square miles so there would be 1,221 people per square mile.

For comparison New Jersey is the most densely populated state with 1,210 people per square mile.

Sources:

https://simple.wikipedia.org/wiki/List_of_U.S._states_by_population_density

http://www.ipl.org/div/stateknow/popchart.html

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    Given that no serious scholar or reporter would much care about this question, I think that this is one of the few times that independent analysis should be encouraged and rewarded, especially since the math is straightforward. Well done – K Dog Jan 24 '17 at 17:12
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    I doubt it's that simple. The apportionment math seems more complicated than that. This seems like a good back-of-the-napkin estimate, though. – Bobson Jan 24 '17 at 17:13
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    Absolute population density might not be the best way to look at it, 35M -> 200M is nearly 6x – user9389 Jan 24 '17 at 18:10
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    For anyone interested I threw together a little program to calculate the number of seats for California following the Huntington-Hill method. I don't feel comfortable updating my answer with the results because it's completely unproven and I assumed giving all other states equal populations is best for California. I prefilled it with the nation population in the question and the lowest population I could give California and get 268 seats. jsfiddle.net/kr59zez3 – JonK Jan 24 '17 at 19:55
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    This answer has a few problems. Even if some states do not have 1/435 of the population, they are guaranteed 1 house seat each. That's where the number 385 comes from, but that's not the end of it, some states will have more than 1/385, and even be close to 2/385, so they may be awarded an elector at the expense of a more populous state. Every time the census reapportions electors, several lawsuits are opened because one state didn't get enough and wants to steal an elector from CA, and CA didn't quite get enough for its share of the population either. This answer doesn't reflect any of that. – Nathan L Jan 25 '17 at 16:46
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I'll take a stab at this one. There are some factors to consider prior to estimating the population required for a single state to control both the House and the Electoral Collage. First, the total population (assumed in the original question to be 324,424,000 ppl) is not the population used in determining the number of representatives a particular state has. To get that figure, the populations of all US territories and DC must be removed. For simplicity I'm going to use 2016 population estimates from here. This means that the US population subjected to congressional representation is 322,446,343 ppl.

Another thing on the top that is important, there are 435 members of the House, not the 385 the OP suggests. This means that there are 538 total electors in the Electoral College (435 from the House, 100 from the Senate, and 3 from DC). Our theoretical majority state is still going to need 268 representatives though.

Next, we need to make sure we understand how congressional apportionment works currently. Taking data from the 2010 census, California had a population of 37,254,503 ppl, which lead to 53 seats in the House. Looking at the fraction of the total population living in California,

(37,254,503/308,156,338)*435 = 52.59

This tells us that congressional apportionment rounds up for representation, though there appears to be some considerations required to make total number of voting representatives equal to 435. Here we will consider standard rounding and no funny business to get to 268.

Now let's try to find out the population required for an Electoral College majority. Starting with the total population and the fact that we only need greater than 267.5 effective representatives we can directly calculate the required population:

322446343*(267.5/435) = 198,285,970 ppl (rounded up to the nearest whole person)

This represents approximately 61.5% of the total US population.

The next question you asked was concerning population density. Assuming total state land areas found here, the corresponding population density for California would be 1272.87 pp/mi^2. The lowest possible population density for a state to have a majority of electors in the Electoral College would be Alaska with a population density of 347.48 pp/mi^2.


EDIT

Following the method of equal proportions referred by @AlonzoMuncy in another answer (which falls under the category of the more you know), it is possible to calculate the population bounds required to a majority of the Electoral College.

In particular, using the priority formula

A = P/Sqrt[n*(n+1)]

it is possible to calculate the upper population bound for our fictitious state directly using the following relationship:

P/Sqrt[n*(n+1)] = (Ptot - P)/Sqrt[(ntot - n)*((ntot - n) + 1)],

where P is the population in our state, Ptot is the total US population, ntot is the total number of representatives (ntot = 435), and n is the number of representatives in our state. Substituting in n = 268 and Ptot = 322,446,343 we get that the maximum state population is

P =  198,570,949 ppl

This assumes that all the population is contained in the next populated state. The lower bound can be calculated assuming all other 49 states are equally populated

P/Sqrt[n*(n+1)] = ((Ptot - P)/49)/Sqrt[(ntot - n)/49*((ntot - n)/49 + 1)],

yielding a state population

P = 188,856,488.

This means that our state has to have a population between 188,856,488 and 198,570,949 assuming my understanding of the method is correct.

  • This is ~1.5M less than the other answer. No idea which one is correct though :-) Still, it's in the same ballpark: about 200M. – user11249 Jan 24 '17 at 17:15
  • Nice answer. And very good point about removing the non-represented population. I hadn't considered that. As for the 385 number, that's what's left after removing 50 representatives to account for each state's minimum. – Bobson Jan 24 '17 at 17:18
  • @Carpetsmoker: The total effective population between the two answers is slightly different, as well as some other assumptions in calculating the total. – Marchi Jan 24 '17 at 17:26
  • @Bobson: You are absolutely correct that every state gets a representative. This answers the question what is the minimum population required by a state to have a majority in the Electoral College. Obviously, state population distributions will effect the required population in practice, especially now that the states are likely to have drastically different ratios of population to representatives. – Marchi Jan 24 '17 at 17:26
  • Same goes for this answer, self-analysis should be allowed in this instance – K Dog Jan 24 '17 at 17:26
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TL;DR: 163,006,794 for the House; 201,521,490 for the Presidency.


I wrote the code to actually run these calculations, based on the formula on the Wikipedia page (and in Alonzo Muncy's answer). I took population data from this apportionment calculator, which has 2016 numbers for the population that has representatives (as Marchi's answer pointed out). I was able to successfully reproduce the results that that calculator output. Then I added a "Population Moved" multiplier which moved X% of the population from each state to California, to preserve the initial 322,446,343 number (this is slightly lower than I asked about) and fiddled with that percentage by hand.

The upshot is that it would require 43.7% of the population of every other state to move to CA in order to give it a controlling share of the House. That would give CA a new population of 163,006,794, for 218 seats. (Texas has the next highest with 21 seats.)

For control of the Electoral College, it's 57.3% of the population moving to California, giving it a new population of 201,521,490, for 268 seats. (Texas has the next highest with 16 seats).


Based on CA's 155,779 sq. mi. (per Marachi's answer again), that would be a population density of 1046 pp/sq.mi. or 1293 pp/sq.mi. So, about New Jersey's density, over a much bigger area.

  • sigh. and that's why I hate HNQs. best asnwer is lowest voted. – user4012 Jan 25 '17 at 17:05
  • @user4012 - I'm probably going to accept this one once I can, if no one else comes up with a better one. – Bobson Jan 25 '17 at 17:09
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The current method to assign house seats is the method of equal proportions.

Step 1: Each of the 50 states is given its one guaranteed seat in the House of Representatives, leaving 385 seats to assign. (All states have 1 house seat at this step)

Step two: Assign priority numbers according to a priority below

Priority = population / (Square root of (n * (n + 1)) where n is the current number of seats that the state has.

I can't do the calculation right now. However, Bobson did so in a different answer and got 201,521,490 to give California 268 seats in the house and control over the electoral college.

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No matter how many people California has, it would never be able to control the presidency.

If you're speaking from a strictly mathematical sense, the above answers serve that purpose quite well. But since this is a site for the discussion of politics, and not math, I feel a need to point out the obvious - the United Sates as a whole would never allow one singular state to control the presidency. If California's population exploded to the extents listed above, the Electoral College would be redesigned to keep the votes reasonably distributed across the 50 states.

  • Why the downvote? While the question may be mathematical in nature, the political side is that it could never happen, regardless of population. I feel that needs to be pointed out. – vavskjuta Jan 24 '17 at 19:43
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    Do you have a reference? It seems like anything like that would require an amendment which would require 3/4 of congress which would would be like 60% from CA if an apportionment like this happened. Consider if 200M people decided to be in CA during the 2020 census. – user9389 Jan 24 '17 at 19:56
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    According to the National Archives, "The Constitution provides that an amendment may be proposed... by a constitutional convention called for by two-thirds of the State legislatures." So even if California had enough votes to block an amendment in the House and Senate, they wouldn't be able to prevent the rest of the states from calling a constitutional convention. There they could propose an amendment altering EC which could then be ratified by the states. Since ratification only needs 3/4 of the states, CA could be bypassed again. – vavskjuta Jan 24 '17 at 20:09
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    This answer, is, however, speculation. It used to be common knowledge that if a president were elected with a minority of the popular vote, the electoral college would be eliminated or reformed. Well, that has happened twice in the last two decades, and there are no credible proposals for reform under discussion. – phoog Jan 25 '17 at 0:22
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    One other factor is to consider how California came to hold such a high percentage of the population. It's unlikely that California could support ~200M people due to water constraints so it probably couldn't happen with the current population. But if some disaster (like an east coast megatsunami or large asteroid strike) caused loss of population along with mass migration to California, then California could end up with a majority of population and representatives quickly, without a chance for other states to block it. (not to mention the social and political turmoil caused by the disaster) – Johnny Jan 25 '17 at 5:51

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