Talk:Quota method

changes

apologies if I made a wrong change out of ignorance, but the claim that Hare could ellect more seats than there are available made no sense, and is contradicted by: Droop quota - "This gives the Droop quota the special property that it is the smallest integral quota which guarantees that the number of candidates able to reach this quota cannot exceed the number of seats. " - by implication a larger quota would certanly guarantee this also, and seems only logical, since Hare wouldnt give a seat automatically unless it had at least the number of votes equal to the votes/seats, and sum(bottom((Vpn*S)/Vt))<=sum(Vpn)*(S/Vt)=S since sum(Vpn)=Vt by def (Vpn being votes of the candidate n, S - total seats, Vt - total). And it was claimed that Hagenbach-Bischoff quota would not suffer from this problem, which is explicitly denied by the article. Also I wonder what measure of proportionality considers Droop more proportional? Ive found quite a few articles stating contrary, none supporting this claim.--Aryah 13:11, 19 July 2006 (UTC)[reply]

Originally the claim about electing too many candidates was about the Imperali quota only. As to proportionality, most STV supporters think Droop is better (i.e. more proportional) than Hare because it means more candidates get elected with full quotas. I believe an intuitive test helps identify the underlying issue: consider an election for two seats where one party gets 70% and the other 30%. Many people would think (before doing calculations predicated on a particular approach) that a 2-0 split was fairer than a 1-1 split.--Henrygb 16:44, 19 July 2006 (UTC)[reply]

Well, the claim seems to have expanded over time :) I did notice that the discussion of Largest remainder methods on wiki is so far limited to their use in STV. Whatever their merits might be in that use, largest remainder methods are used in party list systems too (typically Hare or a Hare/some highes averages method combination). And Hare was used in calculating the num of seats each country of the U.S. got in the congress (or in some other body, not sure) - but was replaced with highest averages methods (interestingly, currently uses a system significantly more generous to small states than would applying Hare quota be, and this was defended by some famous mathematicas of the time as even more proportional) due to problems with monotonicity that largest remainder methods have. I think Droop is sometimes preffered not on the basis of its better proportionality - something Id consider difficult to argue, and would in any case require statistical research, and an argument about the applied criteria, not only examples, cuz there are allways counterexamples - but because it allocates more seats before taking fractions into account, and because it makes majority rule more likely. But these are unrelated to proportionality of the results. In any case, I would not be among those that would consider 2:0 an _obviously_ fairer split than 1:1 if two places were ellected and the votes were 70%:30%. The choice would indeed seem quite ambiguous before choosing a specific method. But, however thats divided, it would be horribly disproportionate, since its impossible to speak of proportionality with such miniature districts. Works I quoted considered Hare to be neutral on the basis of its application to various actuall ellection result data - along with Sainte-Lague (Webster) method, the only method thats truly neutral to both large and small parties. Dont get me wrong, I think STV as a method is incredibly versatile, powerfull and promising, but am not necessarery enthustiatical about the specific 'tweaks' its common real life applications pick. But the articles should in any case not speak of these methods only in the context of their application to STV, nor reflect only the common use (and Id think essentially provisional) prefferences of particular methods in only that context. And outside STV, I think that Droop, unlike Hare, is quite a marginal choice - whereas the articles on wiki are strong on claiming 'technical inferiority' etc of Hare in comparison to Droop --Aryah 08:50, 20 July 2006 (UTC)[reply]

The Hare Quota can result in parties with very small percentages of the vote winning a single seat due to a split of the remainder between many parties. Say, a ten-seat district with parties getting 42%, 22%, 21%, 12% and 3% of the vote respectively. The quotas add up: A) 4.2 = 4 seats. B) 2.2 = 2 seats. C) 2.1 = 2 seats. D) 1.2 = 1 seat. E) 0.3 = 1 seat. Despite getting only a quarter of the vote of C, D received the same number of seats, one. Note this could have happened with any party, A could have gotten 43% of the vote and won the last seat. But the Hare Quota's very nature of being "unbiased" in allocating the last seats helps the smaller parties, for it is a proportionately bigger boon to them. A on 43% with 5 seats is overrepresented, but because it has more votes anyway, it is less so per vote cast than a small party. It receives a seat for every 8.6% of the vote, as opposed to every 3% as D would. This will occur with other methods but it is more pronounced with the Hare Quota. Oh, and while strictly proportionally speaking, a 70/30 2-0 or 1-1 split might seem like a tossup, I believe it is more desirable that a party with an obvious large majority of the vote receive a majority of the seats than a smaller be represented. -Nichlemn 09:36, 5 August 2006 (UTC)[reply]

Im also tempted to do something about this sentence: "The Sainte-Laguë method avoids these paradoxes but is less easy for the average voter to understand." . It is true, however it seems to have proven irrelevant; Sainte-Lague and the essentially identical and equally 'complex' d'Hondt algorithm are use in a huge number of countries, including most of europe. People seem to be not overly concearned by the fact that most dont know how their votes beget parliament seats. I come from such a country, and have known how the allocation works only after learning it in a small constitutional law class at the uni, by pure coincidence. When reading up on the issues around introducing STV, i was in sheer wonder what big a deal was made from the fact that most people would not know the algorithm of allocation, and Im guessing its from this cultural experience that this note is comming from. Again, it is a true fact, and it is possibly fair for it to be mentioned. But leaving it at that does not, imo, take into account that so much of the world is absolutely indifferent to this 'problem', and sees no problem with not knowing the algorithm (and certaly not finding it the least bit 'undemocratic' or untrustworthy because of it). In any case, even the U.S.A uses essentially a similar, and even more complex algorithm in determining the number of seats each state gets in their congress, so can be also added to the countries that use it and gives its complexity not a second thought, giving additional strenght to claiming this 'complexity' to be largly irrelevant. --Aryah 12:43, 20 July 2006 (UTC)[reply]

Hare v. Droop

Here is another example, stolen from Droop_quota#Advantage_over_the_Hare_quota. Suppose there are five seats and 120 votes and the distribution is:

Blue 63

Pink 18

Red 19

Scarlet 20

Then with Droop Blue will win three seats, while with Hare Blue will only win two seats, despite having more than half the total votes and more than three times as many votes as any other party. --Henrygb 21:57, 22 August 2006 (UTC)[reply]

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Independence of Irrelevant Alternatives

Some largest-remainder methods fail independence of irrelevant alternatives when looking at the seat shares of each party.

Under LR-Hagenbach-Bischoff, in a 2-winner election, if Party A gets 12 votes and B 6 votes, then the HB quota is 18/3=6 votes, so A gets 1 seat and then there's a tie for the second seat, meaning B has a 50/50 probability (under a random tiebreaker) to get 0 or 1 seats. But if Party C enters the race and 3 new voters pick them, then the quota is now 21/3=7 votes, so Party A will now have only 5 votes remaining after winning their first seat, so that B will win the second seat with 100% probability. (This all occurs despite C not winning any seats in either scenario.) GreekApple123 (talk) 22:03, 1 September 2020 (UTC)[reply]

All of them do, actually! See new states paradox. :) – Closed Limelike Curves (talk) 02:51, 8 October 2024 (UTC)[reply]