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This question is about rules for apportionment of parliament seats among parties. Many such rules have the property, that a party may gain a seat by splitting into two parties, or two parties may gain a seat by merging into one party.

As a simple example, consider a parliament with 10 seats, and four parties who win 21,22,22,35 votes respectively. The D'Hondt apportionment method would give the parties 2,2,2,4 seats respectively.

However, if the two 22 parties merge into a single party (with 44 votes), then the D'Hondt method gives the parties 2, 5, 3 seats respectively - so the two merged parties gained a seat: they now have 5 seats overall, instead of the 2+2 that they could have without merging.

My question is: is there an apportionment rule in which splitting / merging cannot change the outcome? I.e., if a party that splits into two parties, they get in total the same number of seats as the original one; and if two parties merge into one, it gets in total the same number of seats as the original two?

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  • I am not sure to understand the question. Parties do merge and split between elections, not while the election is taking place, so rules are applied before changes. At any system that I do know parties do not change the number of seats; in fact usually the seats belong to the candidate and not the party so there is no direct cause&effect. For example, if all MPs of both parties agree with the split then the merged party retains all the seats because all the MPs chose to, but if some decide to leave the party in disagreement then the merged party with "lose" those seats.
    – SJuan76
    Aug 2, 2021 at 11:20
  • @SJuan76 I'll try to clarify. Suppose you have a party, and you know - by opinion polls - that your party is going to get 17.4% of the votes. According to the apportionment rule in effect, this is going to give you 17 seats. But, you can split the party (before the elections) into two smaller parties, such that each small party will get approximately 8.7%. If the rule gives each small party 9 seats, then overall, your party has gained from the split - instead of 17 seats, they now have in total 18 seats. My question is: are there apportionment rules in which this CANNOT happen?
    – Erel Segal-Halevi
    Aug 2, 2021 at 12:01
  • I would say all of them. Opinion polls are not vinculant in any way; they could as well do not exist. The only thing that matters are the election results. So "the opinion poll ten days before gave us X seats" has no effect. And electoral rules benefit the big parties, not smaller ones. And the idea that a party that suddenly splits before an election would see an increase in vote is dubious.
    – SJuan76
    Aug 2, 2021 at 12:04
  • Voters typically dislike splits in parties - they are seen as signs of internal problems, and reduce trust. Even very popular politicians splitting off parts of larger parties often loose a significant portion of their former voters (or the "old" party collapses). So I suppose this is more a theoretical concern than a real exploit in most parliamentary systems. And as seats are not reallocated between elections, parties could only hope for an effect in the next election.
    – Hulk
    Aug 2, 2021 at 12:16
  • This doesn't 100% do what you want, but I think en.wikipedia.org/wiki/United_Kingdom_general_elections_overview comes closer to what you want to achieve than The D'Hondt apportionment method.
    – Thomas Koelle
    Aug 2, 2021 at 12:21

2 Answers 2

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If we add the condition that parties with the same number of votes must have the same number of seats, then this is impossible, unless you give out 1 seat per vote in which case your Parliament consists of the entire country.

Suppose there are, 100000 parties that received 1 vote. And suppose that another party got 100000 votes, and 1 seat.

If the 100000 parties merge together into a single party, they have 100000 votes. Either each party previously had 1 seat each, in which case they now have 100000 seats according to your rules, or each party previously had 0 seats each, in which case they now still have 0 seats. Neither case is going to match the other party which has 100000 votes and 1 seat.

You can also consider splitting the 100000-vote-1-seat party exactly down the middle into two parties that get 50000 votes each. Now they must have the same number of seats as each other, so the choices are 0 or 2. But they need to have 1. Not possible.

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    Interesting, thanks! But the condition "that parties with the same number of votes must have the same number of seats" is not possible even without my condition. For example, suppose two parties get exactly 50000 votes, but the parliament has an odd number of seats.
    – Erel Segal-Halevi
    Aug 2, 2021 at 12:45
  • @ErelSegal-Halevi It is common for parliaments using proportial voting systems to have some flexibility with how many members they have.
    – Arno
    Aug 2, 2021 at 16:23
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    They can try to minimize the amount of rounding by adjusting the number of seats but as long as there is rounding, there will be rounding.
    – user253751
    Aug 2, 2021 at 16:35
  • Some degree of rounding is almost always authorized in practice.
    – ohwilleke ♦
    Aug 2, 2021 at 21:32
  • Suppose that, instead of the condition "that parties with the same number of votes must have the same number of seats" (which cannot be guaranteed when the number of seats is fixed), we require only that each party gets at least its "lower quota" (proportional share rounded down). Can this be satisfied together with the condition I suggested?
    – Erel Segal-Halevi
    Aug 3, 2021 at 4:12
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There is no perfect mathematical formula that can guarantee a degree of proportionality where the number of votes you receive will always perfectly correspond to the same number of seats you win. (That is assuming the number of seats is fixed.)

When you get to that level of granularity, the only way to improve the proportionality is to tweak the mathematical formula here and there.

Here are some work-arounds:

  • Instead of using d'Hondt method (i.e. divisors of 1, 2, 3 ,4...), switch it to Sainte-Laguë method (divisors of 1, 3, 5, 7). This has been shown to favor smaller parties rather than bigger parties. I'm sure there are other variations out there.

  • You can straight up increase the number of seats. This might be the simplest way to brute force your way through the proportionality problem. This method does have limit though, I have yet to see a democratic legislature that can handle over 800 seats.

  • You can write the rules such that the legislature changes its own size to optimize proportionality. Consider the German Bundestag where the number of seats changes after each election. Each party is given as many supplementary seats as needed to ensure the balance of power corresponds to the result of last election.

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  • None of these aboslutely fix the problem, though. They just improve it.
    – user253751
    Aug 3, 2021 at 12:30

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