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I corrected what i thought as undue criticism of Hare quota, and disregarding its use outside STV. Quite a few sources claim it, along Sainte-Lague, gives most proportional results (though Im personally sceptical that Sainte-Lague could give equally perfect proportionality to Hare quota), and thats hardly an unimportant advantage in comparison to Droop. It also seems to me to be equivalent to the most obvious definition of proportionality, i.e. that percentage of seats a party gets is equal, within the rounding margin of error, to the percentage of votes (since (Vp/Vt)*S=Vp/(Vt/S) Vp being votes a party won, Vt the total num of votes, and S the num of available seats). Seeing how precise STV methods get sofisticated and computationally intensive anyways, finding a quota that allocates as much places as possible prior to fractions and transfers doesnt seem like an important saving. Btw does anyone know more about QPQ method and its Swedish predecessor, and its (incredibly small?) computational intensity, and has any data on its proportionality?--Aryah 03:03, 19 July 2006 (UTC)Reply[reply]

In general, the 3 unbiased methods in common use today are Hamilton's method, Webster's method, and Hill's method. All 3 of them will, in the long run, tend to favor neither large nor small parties, so long as every party exceeds the Droop quota threshold. In all 3 methods, it's easy to manipulate .

Hamilton, Webster, and Hill are all dirt-cheap to calculate. Hamilton is slightly easier to calculate if you want to do it by hand, but Webster/Hill will converge in at most a few steps--the maximum number of steps for exact convergence is equal to half the number of parties, and in most cases it's half the square root of the number of parties. In practice, this means 1-2 extra steps, which isn't exactly hard if you have a calculator on hand. STV is also dirt-cheap.

The only computationally-intensive voting method I'm aware of is Kemeny-Young. Proportional approval voting is expensive in theory, but modern mixed-integer linear programming algorithms can find approximate solutions of +/-1 seat in seconds. Closed Limelike Curves (talk) 20:28, 11 February 2024 (UTC)Reply[reply]

I dont understand why such a fuss is made out of the fact that Hare quota can give a minority of seats to a majority of votes (btw it has a significantly smaller problem with this than Sainte-Lague) - giving more than 1/2 of the seats to a party of more than 1/2 of the votes is not mathematically more important (thus making it not a technical but a political flaw) than giving more than 1/3 of seats to a party of more than 1/3 of votes, or more than 1% of the seats to a party of more than 1% of the votes, and all of this cannot be simultaneously satisfied with the system of allocating the seats. It is not related to proportionality, but is a political demand - so it seems quite appropriate for it to be satisfied at the end of the calculation, by giving some premium seats to the majority party - an ad-hoc sollution to an ad-hoc problem. It certanly doesnt seem to be a sufficient reason to sacrifice superior proportionality throuought the calculation, that Hare ensures. Particulary not on such a way, as with Hagenbach-Bischoff quota needed for this majority rule ensurance, as to open the possibility of aditional bias of giving more seats to some constituancies than to others --Aryah 05:50, 19 July 2006 (UTC)Reply[reply]

I think the issue is not giving less than majority to a party with more than half the votes but the constant related inverse - giving a majority of seats to a party with less than half the votes. Not all mis-representation is viewed the same - a variance of ten percent from a party's due proportion of seats that wrongly give majority control is more important than difference between 30 percent of seats and 40 percent of seats -- it actually gives a party a majority of seats in a chamber and that is where power lies. 2604:3D09:887C:7B70:0:0:0:6A17 (talk) 19:31, 20 November 2023 (UTC)Reply[reply]

Hello,

in the example there are 100 votes and 2 seats. The latter divides the first, making the example unrealistically simplistic. It would be more interesting to see how they deal with non integer fractions (it can be of importance!) Evilbu 13:04, 22 October 2006 (UTC)Reply[reply]

After this recent Judd Gregg and US Census flap, I've been considering the hot potato that apportionment of US Representatives will be after 2010. One problem is that of counting (who to count and who not to), but that's a different problem than what is concerning me at the moment. So, assuming we have undisputed census figures for each state, the (hopefully blind and objective) mathematical method for determining how many Representatives each state gets sure seems different than what is depicted at United_States_Congressional_Apportionment#The_Equal_Proportions_Method.

The constraints applied to this problem is that the total number of Representatives is fixed and determined in advance by law; 435, and that each state, even the least populous, must get at least one Representative.

Let

${\displaystyle K\ }$ be the number of states (currently 50).

${\displaystyle 1\leq k\leq K\ }$ be the index of k^th state. It doesn't matter how they're ordered.

${\displaystyle P_{k}\ }$ be the agreed census population for the k^th state.

${\displaystyle P=\sum _{k=1}^{K}P_{k}\ }$ is the total population of all K states (excluding DC and the territories).

${\displaystyle N_{k}\ }$ is the number of Representatives for the k^th state that we are trying to determine.

${\displaystyle N=\sum _{k=1}^{K}N_{k}\ }$ is the total number of Representatives in the House for all K states (excluding DC and the territories) which is currently 435.

${\displaystyle q>0\ }$ is the nationwide constant of proportionality or quota ratio for proportionately allocating Representatives or a state as a function of its population.

So, if we could actually have fractional numbers of persons as Representatives,

${\displaystyle N_{k}\ =\ q\ P_{k}\quad \quad \forall \ 1\leq k\leq K}$

${\displaystyle \sum _{k=1}^{K}N_{k}\ =\sum _{k=1}^{K}\ q\ P_{k}=q\sum _{k=1}^{K}P_{k))$

or

${\displaystyle \ N=qP\quad }$

${\displaystyle \ q={\frac {P}{N))\quad }$

But, of course, we cannot divide Congressional Representatives into fractions even if we might like to tear them apart on occasion. Each states House delegation must be an integer number of people at least as big as one. Wouldn't this mean:

${\displaystyle N_{k}\ =\lceil q\ P_{k}\rceil \quad \quad \forall \ 1\leq k\leq K}$ ?

where

${\displaystyle \lceil x\rceil =\min \,\{n\in \mathbb {Z} \mid n\geq x\))$ is the ceiling function (which means always round up).

Now if q>0 was arbitrarily small (but positive), then each state would get 1 Representative. It wouldn't be particularly well apportioned and it wouldn't add up to N=435. We want

${\displaystyle N=\sum _{k=1}^{K}N_{k}=\sum _{k=1}^{K}\lceil q\ P_{k}\rceil \ }$

Then couldn't q be increased monotonically, thus increasing some of the N_k and ${\displaystyle \sum _{k=1}^{K}N_{k}=\sum _{k=1}^{K}\lceil q\ P_{k}\rceil \ }$ until it reaches the legislated N=435 value? Would that not be the meaning of proportional representation with the constraints that N_k must be an integer at least as large as 1? How is there any paradox in this method (assuming that, as q increases we don't have two states simultaneously increasing their integer N_k and the total jumping from 434 to 436) and where the heck does that Huntington-Hill method that is depicted at United_States_Congressional_Apportionment#The_Equal_Proportions_Method come from? How does that possibly have anything to do with true proportional allocation of a fixed number of seats in the House?

Can someone explain this? 96.237.148.44 (talk) 01:57, 14 February 2009 (UTC)Reply[reply]

Answered at Talk: Huntington–Hill method #Can the Huntington-Hill method, sometimes called "method of equal proportions" be clarified or justified?. --84.151.17.20 (talk) 23:11, 17 November 2009 (UTC)Reply[reply]

a simple way to allocate seats (perhaps too simple?)

is to divide total pop by 435

this gives you pop per member call it A

any state with less than A gets a seat.

subtract population in these small states from total pop to get B

subtract number of small states (X) from 435 to get C

divide B by C to get new pop per member -- call it M

apply M to each state by dividing state population by M, indicating fractional M

first allocate seats to whole Ms. if that allocation does not equal C, give seats one at a time to states with largest fraction of an M left over until the number of allocated seats comes up to C (C plus X = 435). (largest remainder method) 2604:3D09:887C:7B70:0:0:0:6A17 (talk) 23:21, 18 November 2023 (UTC)Reply[reply]

correct algorithm for counting ranked choice votes (counts each vote exactly once, doesn't lose voter preference information when not neccessary, results in minimal bayesian regret):

1. if no candidate has quota on first choice:
* take the candidate with the fewest first choice votes, and:
  temprarily ignore that first choice, thus the second choice becomes first choice, third becomes second, etc.
  if the first choice was already ignored on this ballot, ignore the second choice in addition to the first, etc.
2. repeat step 1 as neccessary until one candidate has quota or there is only 1 candidate remaining 
3. the candidate with the most votes now wins a seat (regardless of whether quota is reached, or multiple candidates reached quota)
4.a if that is at or under the quota, just eliminate these votes
4.b if that is above the quota, each ballot included in that quota should now count as ((count - quota) / count) ballots.
5. eliminate the winning candidate from all ballots, all choice positions, since they can only be elected once.
6. un-ignore all the choices you ignored - that was just temporary for that round.
7. repeat from step 1 until all seats are filled

Kevin Baas^talk 20:28, 29 March 2016 (UTC)Reply[reply]

You might be providing the method of doing Multiple Majority Winner voting but not STV.

in STV, once a person is counted as elected or eliminated, he or she is out of the running and never comes back in, and many mnore differences. 2604:3D09:887C:7B70:0:0:0:6A17 (talk) 23:25, 18 November 2023 (UTC)Reply[reply]

Firstly, the description of the Brazilian electoral system doesn't belong in the introduction. I suggest that a new section "Examples of Use" be added, and this paragraph moved there. (As it may grow into a long tabulation over time, I suggest making it the last paragraph in the article.)

Secondly, the description of the Hong Kong electoral system largely repeats what's in the Criticisms section. The introduction should include just a one-sentence summary of the comparisons, and the other details should be incorporated into the Criticisms section (or move to the aforementioned "Examples of Use".

Possibly the Comparison with the Droop quota heading should be made subordinate to a new "Comparisons with other Methods" heading, leaving room for the Brazilian example as a "Mixed" system. (It would be helpful to add some explanation of the effect that the mixed system has in Brazil.) Martin Kealey (talk) 06:50, 16 June 2019 (UTC)Reply[reply]

This wikipedia page should just be on the Hare quota, versus say the Droop quota. The flaws of the operation of STV or largest remainder systems should not be blamed on the use of the Hare quota.

this page had this paragraph:

The difference between the two quotas comes down to what the quota implies. In the Hare system, winners elected under a Hare system represent that proportion of the electorate; winners under a Droop system are elected by that proportion of the electorate.^{[clarification needed]}

but I took it out because there is no difference like that. Those elected under both are elected by the votes that elected them and the voters behind them. It seems to me that no winner is elected by a proportion but by votes, in fact by the quota in question.

As it is, the example is perhaps too STV-specific, which possibly makes it too complex. For example, the D'Hondt method#Jefferson_and_D'Hondt mentions that it can be used to distribute the available national seats to sub-national multi-member constituencies. I think what is effectively a Hare-quota system is used for apportionment in Finland at least, where the number of seats in each constituency is the population in the constituency divided by the national population times 199, rounded down to an integer value. Reorganizing that, the part with 199 divided by the national population is a Hare quota, if I've understood it correctly, and the constituency population are the "votes" of each constituency, which translate to their share of the seats. Parliament has 200 seats (1 is always reserved for autonomous Åland, outside the normal math), and any remaining seats are distributed in descending order of the previously discarded decimal values.

128.214.185.52 (talk) 09:08, 28 February 2020 (UTC)Reply[reply]

Hare quota can be used in systems other than STV.

The quota in Finland's case is based on population, not based on valid votes so is not Hare.

Finland's quota is like Hare but is not. It is like a uniform quota (like as used in New York City's STV 1937-1945) but (apparently) sets out a set percentage, not a set number of votes. 2604:3D09:887C:7B70:0:0:0:6A17 (talk) 19:58, 20 November 2023 (UTC)Reply[reply]

I made some minor steps in this direction of a more-neutral article w.r.t. STV. I've made much more extensive edits over at the Droop quota article. I think the best way to make these changes is to focus first on the Hare/Hamilton quota as a mathematical quantity, i.e. explain what it means. Then, explain how it can be used (or misused) in other cases--an appropriate use being in apportionment, and a typically-inappropriate use being in STV. Closed Limelike Curves (talk) 20:31, 11 February 2024 (UTC)Reply[reply]

This article has ended up as an article about Hong Kong politics for reasons I don't quite understand. (It also doesn't seem to make sense. Splintering isn't a typical result of the Hare quota--Germany has used Hare for most of its history within a stable 4-party system. It definitely doesn't incentivize splitting for parties that can win more than 1 seat, for which it's effectively unbiased.) Closed Limelike Curves (talk) 20:50, 11 February 2024 (UTC)Reply[reply]