Gerrymandering can keep a relatively unpopular party in power for longer than it otherwise would be. Assuming the unpopular party becomes even more unpopular, just how unpopular would they need to be before the stratagem of gerrymandering itself is insufficient to support them?

For example, in 2016 gerrymandering enabled Republican candidates who received 50% of the popular vote to win 9 out of 12 available Congressional seats in North Carolina. That is, half the votes, but 3/4 of the seats and power. And suppose that as long as they hold 3/4 of the power, they'll continue to gerrymander.

Now suppose in the future, the Republicans become more unpopular, and receive only 25% or 10% or even 1% of the North Carolina vote, can gerrymandering alone sustain their hold on the majority of seats, or is there some necessary numeric limit that would eventually be reached?

Note: for the purposes of this question, please ignore other concomitant stratagems by which unpopular parties might squeak by, and also suppose there are only two parties contending. For US voting systems Reynolds v. Sims, 377 U.S. 533 (1964) et al would apply, so that the gerrymandered districts must have approximately equal populations.

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    In the United States, Congressmen do not perform gerrymandering. In most states that have gerrymandered districts, it is the state legislature the determines the districts.
    – Jasper
    Commented Jun 27, 2019 at 20:14
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    @Jasper, Re " it is the state legislature that determines...": True, but Republicans also hold majorities in the NC state legislature. Anyway it's just an example; the Q. is about gerrymandering in the abstract rather than NC in particular.
    – agc
    Commented Jun 27, 2019 at 20:27

8 Answers 8


This is a rather simple mathematical exercise. If you allow me total freedom to draw districts within the current requirements, I can place anyone in any district I want provided they have equal population in the end. Connectedness in two dimensions is not enough of a barrier to stop this. In that case, my best bet is to fill as many districts as possible with 50%+1 of my supporters, and fill in the rest with my opponents. Since at least half the population of each district I win must consist of my supporters, we can easily see there is an upper limit on how many districts I can win:

A party with x% of the vote cannot win more than 2x% of the districts no matter how they are drawn.

In practice there's no way I can reach this limit--no court in the land would allow me to draw the kind of tortured districts needed for arbitrary assignment. 50%+1 is also nowhere close to safe--I'd need a small buffer in my districts so that changing demographics and opinions don't turn it into a gerrymander against me. I'd guess that overall this limits one's practical chances to 1.5x%, but this just an estimate, not a mathematical truth.

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    If there's an odd number of voters, you only need 50% plus 0.5. This is a plea to use the simpler and more accurate phrase "more than 50%" to describe a majority.
    – phoog
    Commented Jun 28, 2019 at 2:56
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    So if I understand this correctly, given only two parties, about 25.5% of voter support is the lower limit, below which gerrymandering would fail.
    – agc
    Commented Jun 28, 2019 at 3:26
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    This answer relies on a fact that is largely overlooked in the rest of the Q&A but still doesn't seem prominently noted enough: it is SCOTUS precedent that Congressional and State legislature districts within a state must have equal populations (give or take some minor bit of fudging due to non-divisibility and such). Commented Jun 28, 2019 at 6:17
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    Is the requirement to win a district 50%+1 of the votes or just a simple majority (more than any other candidate) ? If a simple majority suffices, the limit can be lower, if the opposition is splintered.
    – Chieron
    Commented Jun 28, 2019 at 10:38
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    @DavidRice, This is a simplified question, so abstracting out non-voters is OK. But, please post an answer incorporating various additional variables if that seems like it would make for a better answer.
    – agc
    Commented Jun 28, 2019 at 14:32

The problem here is that "one person one vote" can be based on residents rather than citizens or voters. So if you could assign people purely arbitrarily, you could fill up your districts with non-voters such that only a small number of voters are required. In theory, one voter and eight hundred thousand non-voters per district would work. Pack all the other side's actual voters into one district (no non-voters and none of your voters).

However, you then run into a few problems.

  1. It's not the number of federal House seats that determines your ability to gerrymander. It's the number of state legislative seats. So you need more than just twelve federal seats. You need a larger number of state seats.
  2. You can't gerrymander statewide races. So to make this work, you have to disempower the governor's ability to veto a districting plan.
  3. Most states don't have enough non-voters to make this work.
  4. You can't actually assign people arbitrarily per district.

Realistically, this would never work. But it is theoretically possible with just a few hundred voters given sufficient non-voters.

In actuality, Republicans in North Carolina received slightly more votes than Democrats. But instead of winning seven to six, they won ten to three in North Carolina. But this didn't overcome the Democratic-favoring gerrymander in California, where Republicans won about 33% of the vote but only 13% of the seats. That's a surplus of eleven for the Democrats. Even netting out the three for the Republicans from North Carolina, that's still an advantage of eight seats for the Democrats from those two states.

Apparently in the last House election, the gerrymanders almost exactly cancelled out. The Democrats won about as many House seats (54%) as their share of the vote (53.4%). Which is surprising because Democrats face a natural handicap in that their voters are more concentrated than Republican voters.

Source: Wikipedia from Johnson, Cheryl L. (February 28, 2019). "Statistics of the Congressional Election of November 6, 2018". Clerk of the U.S. House of Representatives.

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    It's a bit more difficult to see clear gerrymandering in 33% of the vote not translating into 33% of the seats than a 50/50 split.
    – Jontia
    Commented Jun 28, 2019 at 8:14
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    FWIW, California has an officially nonpartisan redistricting committee, so I don't think that the Democratic advantage in the state can be attributed to a gerrymander. It's not out of the question (perhaps in practice the committee is biased), but more plausible to me is Peter Schneider's explanation.
    – A. Howells
    Commented Jun 28, 2019 at 19:05
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    @A.Howells Which is my point. Independent commissions produce non-proportional results--just like partisan legislatures.
    – Brythan
    Commented Jun 28, 2019 at 19:09
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    I see. I didn't really get that from the answer. Perhaps it would be good to incorporate (part of) Peter's comment? +1 regardless
    – A. Howells
    Commented Jun 28, 2019 at 19:24
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    @Brythan You expressly call it a "Democratic gerrymander", which is decidedly not "even districts created by an independent commission can produce disproportionate party representation", and very clearly not your point. Commented Jun 28, 2019 at 22:22

One of the problems is that people look at Gerrymandering from the lens of Republicans vs Democrats. It's not - it's a terrible example of the two parties working together.

Imagine you belong to Team Yellow, and you've been elected - but it was kinda close. There's a heavy Pro-Purple area along the northern edge of your district that nearly cost you your election. Meanwhile, Team Purple elected in a guy who has a section of Pro-Yellow people that was giving him grief. So when it comes to redistricting, what do you do? You 'trade' your pro-purple area for his pro-yellow area - entrenching both of you into your spots. It's not Yellow-Vs-Purple at all - it's about both reps working to make their reelection easier.

This helps highlight why the premise of the question is wrong. While the Republican party is probably happy with the current state of things, the Democrat representatives are probably much happier than their Republican counterparts. Because they don't have to worry about reelection. The Republican reps are likely thinking about doing the opposite as what the question would imply. If they could do a redistrict right now, they'd probably "sacrifice" a district in order to more solidly claim the other 8 - several of which they came close to losing (better to cleanly have 8 than gamble for 9.) Which the Democrats reps would probably frown at - it'd make their reelection efforts harder. The "party" doesn't mean much if you're thrown out of office, after all - no matter which party you're part of.

  • In a nutshell "parties hate competition", or to quote Walter Karp: "The left and right wings of the party establishment--two great pinions of an ancient bird of prey." Makes one wonder what districting schemes exist which could be used maximize competition.
    – agc
    Commented Jul 1, 2019 at 13:52

In theory, if the stars align perfectly, one can gerrymander a 50-50 vote to win 11 out of 12 districts. This can be done by drawing the district boundaries so that one district has 61% of voters supporting your opponents, and the other 11 have only 49% supporting your opponents.

In practice, it will come down to how tightly packed, geographically, your opponents are. If they are evenly spread, then it becomes much more difficult than if they are densely packed in a few geographic areas. Thanks to a combination of voter registrations (where the state knows voter affiliations in each house), and cheaper computing power allowing for more complex modelling, a willing party can find a way to optimise the boundaries for their benefit.

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    This example shows how a party with 40% of the vote in a 12-district state can engineer a majority in its congressional delegation, but it doesn't answer the question, which is asking for the smallest proportion of the vote a party could have while still being able to gerrymander a majority of congressional districts in its favor.
    – phoog
    Commented Jun 28, 2019 at 3:05
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    > "In theory, if the stars align perfectly" -- but not in practice. As State Representative David Lewis, senior chair of the House Select Committee on Redistricting points out: "I propose that we draw the maps to give a partisan advantage to 10 Republicans and 3 Democrats because I do not believe it’s possible to draw a map with 11 Republicans and 2 Democrats."
    – BurnsBA
    Commented Jun 28, 2019 at 16:48
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    "where the state knows voter affiliations in each house" It's worth noting that this is not true in every case. Several states (including my own state of Tennessee) do not require voters to state any party affiliation. You can only vote in one party's primary, but you can choose that at the time of the election, not based on any party registration. In lots of municipalities, most of the local offices effectively only have candidates from one party, so it makes sense to vote in that party's primary, even if that's not necessarily the party with whose platform you might most agree.
    – reirab
    Commented Jun 28, 2019 at 20:49

Actually, the key to successful gerrymandering is not spreading out the votes of the controlling party but concentrating the votes of the minority.

Say that voters have equal inclination to vote for each party, (that is the electorate is split 50-50 but with guaranteed loyalty to their party), but that one party completely controls redistricting. In that case with perfect knowledge of the electorate it would be possible for the majority to win all but one seat. The majority makes it so the minority wins that one seat with a 100% margin while all others go to the 'majority' party with much narrower margins.

Of course reality does not feature such perfect knowledge but it should give a basis for thinking about whatever terms you choose for modeling reality.

  • You misunderstood the answer. Say there's 10 districts and 1 million people each with a 50/50 split (5million each). Now you have 1 district that's 100% blue (1 million) and 9 districts that are 55/45 (5 million red, 4 million blue) red.
    – xyious
    Commented Jun 28, 2019 at 19:22

In gerrymandering we have 2 divisions to manipulate; blue vs red voters, and voters vs non-voters.

10% of the vote, 83% of the seats

Consider 12 seats, a 29% turnout and a 90% blue (10% red) vote. If we pack the (26.1%) blue voters into 3 seats, reds win 9/12 seats. However, what if we pack 2 blue seats? That would absorb 16.7% out of the 26.1% of blue voters, leaving 9.43% blue voters vs 10% red voters. Distributing the remainder evenly would yield a win in all the remaining seats, for a total of 10/12.

This strategy clearly depends on the geographical distribution of voters vs non-voters, and red vs blue. In a world where they were all evenly distributed it would be impossible to gerrymander an unfair victory. But the more uneven the distribution, the greater the opportunity. So enacting policies which simply increase the clustering of similar groups (say, localised taxes which can align with whichever is the ruling local group, or tax breaks for family groups) increases the degree to which this approach can work.

Another limitation would be the safety margin you wish to retain - in the above example, 10 out of 12 seats are won on a margin of 1.5% (51.5/48.5). If the weather makes a difference of 2% this would be a risky strategy.

It is clear, however, that gerrymandering can have surprisingly large effects as long as there is an uneven distribution and low turnout. Raising turnout 1% (without changing the ratio of red/blue) would yield an additional seat for blue.

Can 1% win a majority? Yes

Let's change our scenario to 11 seats to make the majority apparent, we'll make it 1% of votes red and 99% blue, leaving only the turnout as a variable, say T. The requirement is to win 6/11 seats which means the maximum blue vote is 55% packed into the 5 seats lost, plus T * 1% in the remaining seats (which we will match with the T * 1% blue). So (0.55 + 0.01T)/(0.01T) = 99 -> 0.55 + 0.01T = 0.99T, T = 0.561. So turnout of 56.1% or less can permit a red vote of 1% to win a majority over 99% (in theory).

  • And of course this assumes that the turnout rate is essentially identical across all voters. If you had 100% red turnout and 1% blue turnout you could make things more severe. That's a severe disparity, but even a small difference in turnout rates can translate into a major effect (especially when they are already low by default, as in the US), and is a major reason why drives to increase voter turnout are both common and potentially significant. Commented Jul 2, 2019 at 0:46
  • @zibadawatimmy: 1. Thanks, I've fixed the red/blue. 2. I'm not assuming anything about the turnout of different parties, only that a proportion of voters will not turn out and therefore they can be used for gerrymandering, if that makes sense. The model here implicitly assumes perfect knowledge of who will vote and which way, so in reality there is a degree of party turnout and a degree of randomness which isn't modelled here. Simply put, higher turnout overall reduces the opportunities to gerrymander.
    – Phil H
    Commented Jul 2, 2019 at 8:24

While it's not in the USA, there was a period where some parliamentary seats in the UK were voted on by 7 voters. And about 1% of the population voted on 37% of the seats (152/406).


Summarizing the information in the link, towns in the UK had a set number of seats assigned to them. If the number of people in the town shrank (closed industries, the town literally falling into the sea, and so on), then the few remaining people in the town effectively controlled the seat. In some cases, one person controlled the seats due to being the landlord of over half the houses in the town, and the ballot not being secret...

The gerrymandering here is not correcting the district boundaries when the population changes. In theory, you could gerrymander by altering existing boundaries so that almost all districts contain one voter, with the rest of the country being put together in another district. This means that in the extreme case you could have two districts with one voter each, and the rest of the population in the other district. Those two voters would control which party/faction got into power.

  • Yes, this may be possible in the U.K., but it's not possible in the U.S., due to the Constitutional requirement that districts contain even portions of the population. The Constitution requires a census every 10 years and districts must be apportioned based on it, so the party in power can't just decide to not redistrict if it would result in them losing a seat (or more.)
    – reirab
    Commented Jun 28, 2019 at 20:52

100% turnout, Identical sized districts, Infinite population = 25% minimum.

E.g. 1200 voters in 12 districts:

  • 49 blue votes, 51 red votes in 7 districts
  • 100 blue votes in 5 districts
  • Total 843 blue voters, 357 red votes. (29.75% to stay in power)

1.2 million voters in 1200 districts:

  • 499 blue votes, 501 red votes, in 601 districts
  • 1000 blue votes in 599 districts
  • total 898899 blue votes. 301101 red votes (25.09% to stay in power)

As you increase the population size, you can get an asymptote to 25%

100% turnout, district sizes allowed to vary by up to a factor of 2. = 1/6 minimum

If you divide a state into 1 division per N people, and then round to the nearest integer, the division sizes will vary by a factor of up to 2. (Montana has 994k people per district, Rhode Island 1st has 526k.)

Eg 1200 districts:

  • 249 blue voters, 251 red voters, in 601 districts of 500 people.
  • 1000 blue votes in 599 districts of 1000 people.
  • 150851 red voters, 748649 blue. Red wins with 16.77%

Increasing the number of votes or districts to infinity and the ratio will go to 1/6

<100% Turnout - no lower bound (a single vote can win)

Uniform turnout won't affect the answer (if every voter has a 50% chance of not voting - then the vote counts half evenly, same result). So you need to weight the turnout.

Turnout at extremes can vary from 0% to 100%, which can effect the vote entirely - 1 vote can win an election if the other sides turnout is 0%

A 1% turnout for blue, and 100% turnout for red:

  • 249 blue votes from 24900 supporters, 251 red voters from 251 supporters, in 601 districts of 25151 people.
  • 503 blue votes from 50302 supporters, in 599 districts
  • Total 45095798 blue supports, 150851 red supporters. Red wins with 0.33% of the vote.

Increasing the number of votes or districts to infinity and the ratio will go to 1/300 (with a 100:1 turnout discrepancy)

Uneven district sizes - no lower bound (2 votes can win).

With 100% turnout, and 3 districts:

  • 1 red vote and 0 blue votes in 2 districts
  • infinite blue votes in 1 district
  • Red wins 2 to 1 with ~0% of the vote.
  • On the non-voters -- that would fall under "other concomitant stratagems by which unpopular parties might squeak by", (i.e. voter suppression, et al), which the Q. noted should be ignored. Presume for any given projection that there was some constant fixed set of voters who invariably vote, and gerrymander accordingly.
    – agc
    Commented Dec 16, 2020 at 23:19

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