Talk:Copeland's method

Latest comment: 3 years ago by Colin.champion in topic Ties (again)

Ties

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How does this method handle ties? Presumably a candidate with 2 wins, 1 tie and 2 losses would finish ahead of a candidate with 2 wins and 3 losses? Grover cleveland (talk) 19:08, 18 October 2010 (UTC)Reply

The term "Copeland's method" refers to all those methods that always choose the winner from the candidates with the most pairwise victories. The term "Copeland's method" doesn't refer to a concrete tie-breaking rule. Markus Schulze 21:04, 18 October 2010 (UTC)Reply
Perhaps the article should be edited to make this clearer. Right now, the article gives the impression that "Copeland's method" is a specific algorithm. Cheers. Grover cleveland (talk) 22:18, 19 October 2010 (UTC)Reply
I've found a source saying that Copeland's takes the number of victories and subtracts the number of defeats. This means that the number of ties is relevant (since a tie would otherwise be a defeat). I'll amend the article accordingly. Grover cleveland (talk) 22:20, 19 October 2010 (UTC)Reply

tags

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I tagged two statements, below. I didn't delete them because I understand that it's difficult to obtain sources for such obscure subjects, even on the most basic of matters. However, I don't understand them.

When there is no Condorcet winner (i.e. when there are multiple members of the Smith set), this method often leads to ties. For example, if there is a three-candidate majority rule cycle, each candidate will have exactly one loss, and there will be an unresolved tie between the three.[citation needed]

I don't understand this at all. What is a three candidate-majority rule cycle? Surely with any reasonable number of voters a tie is highly unlikely, in this or any other voting system.

Critics argue that it also puts too much emphasis on the quantity of pairwise victories and defeats rather than their magnitudes. [citation needed]

I understand this to mean the system ignores how much people prefer one candidate over another. Wouldn't this criticism apply to any system which does not employ range voting? Most voting systems do not account for the intensity of support for a candidate, only its extent. BillMasen (talk) 22:52, 19 October 2010 (UTC)Reply

I think that a "three-candidate majority rule cycle" would be a situation where, A pairwise defeats B, B defeats C and C defeats A. This would result in a tie between A B and C (assuming that A, B and C are preferred equally to any remaining candidates). I'm not sure that this situation would be particularly improbable. Grover cleveland (talk) 05:11, 20 October 2010 (UTC)Reply

Why would it be more probable than a tie in a public SMDP election? Which has never happened? BillMasen (talk) 11:31, 20 October 2010 (UTC)Reply

See: Resolvability criterion. Markus Schulze 22:20, 20 October 2010 (UTC)Reply

Suppose there are 3 candidates. Then the probability, that the plurality winner is not unique, goes to zero when the number of voters increases towards infinity. However, the probability, that there is a circular tie, goes to about 5% when the number of voters increases towards infinity. Markus Schulze 22:33, 20 October 2010 (UTC)Reply

I appreciate the attempts to clear this up, but I'm sorry to say I'm still in the dark. The page you link just describes the criterion, not why this particular method would violate it. What kind of tie are we talking about? where three candidates have exactly the same number of pair victories? I don't see how that can be more likely than three or two candidates gaining the same number of plurality votes.

Thanks! :) BillMasen (talk) 22:52, 21 October 2010 (UTC)Reply

See e.g. this paper. Markus Schulze 10:36, 22 October 2010 (UTC)Reply

Criticism

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The unsourced criticism that this method relies too heavily on the pairwise outcomes (rather than magnitudes) is largely rebutted by the reality (for which I presently have no source) that votes do not reflect natural preference, but devolve upon strategic considerations. I happen to hate bipartisan politics, so I vote to give credence to a third voice, whether or not I think the 3rd voice is fit to govern, which deprives my vote of a say in the magnitude of the winning margin by the top candidates. Many voters presume the outcome is a foregone conclusion (and come up with any number of peculiar rationales to feel less depressed or cynical about their democratic privilege), making magnitude a dubious statistic for margin of preference. — MaxEnt 00:08, 5 August 2011 (UTC)Reply

Voting on a spectrum

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Copeland’s method has an attractive feature, which is that in the one-dimensional analogue of the Tennessee capital example it will produce a winner close to the median voter. This must be well known – is it worth mentioning it in the article?

(Suppose that everyone has a position on the spectrum represented by a real number; assume that no two people are coincident and that the number of voters is odd. Assume that each voter gives his first preference to the candidate closest to him on the spectrum, his second preference to the next closest, and so forth. Then there is a median voter. We can define the left median candidate as the rightmost candidate to the left of the median voter, and the right median candidate as the leftmost candidate to the right of the median voter. The left median candidate will outscore everyone to his left, and the right median candidate will outscore everyone to his right, so the winner must be one of them. Thus the Copeland method extends the Median voter theorem to multiway choices.) Colin.champion (talk) 14:47, 6 January 2021 (UTC)Reply

Tied votes

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The article says that victories score 1 and defeats score –1. Isn’t it more natural (and equivalent) to let victories score 1 and ties score ½? This was the system adopted in the English Football League 63 years before Copeland’s paper. Many people would take it for granted that a tie was half way to a win. I wonder whether the presentation here makes the method seem less natural than it really is, breaks the analogy with sporting tournaments, and gives Copeland undeserved credit for filling in an obvious detail. Colin.champion (talk) 16:02, 6 January 2021 (UTC)Reply

I have expanded the description of the system. I hope what I have written is acceptable. I tilt the article somewhat towards ‘1/½/0’ scoring. Colin.champion (talk) 14:03, 15 January 2021 (UTC)Reply
I deleted the section on the second order Copeland method. It seeks to resolve ties in a way which has the beneficial feature of posing an NP-hard problem to malicious parties; but so far as I can see it has the drawback of not being likely to resolve ties in the first place. In particular, if a tie arises in a 3-way election, then a second Copeland round never resolves it. Colin.champion (talk) 09:52, 22 January 2021 (UTC)Reply

Pairwise voting methods

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Requested move 5 February 2021

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The following is a closed discussion of a requested move. Please do not modify it. Subsequent comments should be made in a new section on the talk page. Editors desiring to contest the closing decision should consider a move review after discussing it on the closer's talk page. No further edits should be made to this discussion.

The result of the move request was: Not moved. Sorry, but this proposal is a nonstarter. It's generally a bad idea to change the topic scope of articles by moving them to a different title. I suggest that the plan start with drafting a new article at Draft:Pairwise voting methods and then moving it to article space when ready. This article seems to be a subtopic, which may then either be retained as a WP:Summary style subtopic or merged into the new methods article. – wbm1058 (talk) 22:53, 2 March 2021 (UTC)Reply


Copeland's methodPairwise voting methods – See discussion on Talk:Condorcet method. RobLa (talk) 02:03, 5 February 2021 (UTC)Reply

@Colin.champion: At first I didn't like your changes to this article, but now I think I see a plan forward without reverting your changes:

  1. . Rename this article to Pairwise voting methods - this would clear up the confusion about whether pairwise voting methods should be called "Llull's method", "Condorcet's method" or "Copeland's method"
  2. . Transform this article into a history article - since Ramon Llull came up with the idea of using pairwise comparisons, we can have sections for #Ramon_Llull, Copeland's method#Marquis_de_Condorcert, and Copeland's method#Arthur Herbert Copeland (and possibly merge the "Arthur Herbert Copeland" article into this one).
  3. . Create a new Copeland's method article - that version should be based on oldid=1000095080 of the current "Copeland's method" article.

If we don't do something like my plan listed here, then I'm not sure what to do next (perhaps revert to oldid=1000095080?). We need need to figure out a way to keep from conflating the many Condorcet criterion-compliant articles, and this article actually confuses more than it helps now. -- RobLa (talk) 01:59, 5 February 2021 (UTC)Reply

I'm reformatting this section to make it a little less confusing after I inserted {{subst:move}} and the template inserted a comment on my behalf. See also the discussion on Talk:Condorcet method (particularly Talk:Condorcet method#Scope of this article). -- RobLa (talk) 03:12, 5 February 2021 (UTC)Reply

Comments

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Oppose. See User_talk:RobLa/Condorcet#Resuming_discussion. Colin.champion (talk) 08:30, 5 February 2021 (UTC)Reply
Support - as the person who hopes to make the set of changes described in the comments that I authored above. I also described much of my rationale in my response to User:Colin.champion in the "Resuming discussion" section of "User_talk:RobLaCondorcet" (here: User_talk:RobLa/Condorcet#Resuming_discussion) -- RobLa (talk) 10:14, 5 February 2021 (UTC)Reply
Oppose Moving this article to "Pairwise voting methods" will make the readers mistakenly believe that Copeland's method was the only pairwise voting method. Markus Schulze 08:35, 9 February 2021 (UTC)Reply
The discussion above is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

Ties (again)

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RobLa pours cold water on the possiblity of ties. Maybe he knows best, but I’m not persuaded by his arguments. He says that “there has not been a published example of a large, publicly-administered, ranked ballot election that had a circular preference”, which may be true – but has there been a large, publicly-administered, ranked ballot election at all under Copeland’s method? It seems to be easier to induce spurious ties by tactical voting under Copeland’s method than to obtain an outright win, so perhaps voting practices would change if it was used. At any rate, it’s dangerous to assume that people would vote in exactly the same way under a different system.

Also, isn’t the wording rather tendentious?– “Some people claim... theoreticians... people who work on game theory... Regardless... ” But I admit that I have no knowledge of my own as to whether ties are likely. Colin.champion (talk) 11:17, 5 February 2021 (UTC)Reply

The results of many ranked ballot elections have been published. It would be possible for anyone to take the results of any ranked ballot election (e.g. the mayoral elections in Burlington, Vermont in 2006 and 2009, or any of the annual Debian Project Leader elections over the years), and tally the election using Copeland's method. It may be possible to find something pretty quickly; I've been looking for such an example for years, and I haven't seen one, but I've only been doing it as a hobby. Certainly, one of the professionals reading this article would be able to find one fairly quickly and make note of it. Also, could we please assume good faith and not come out of the gate accusing me of tendentious editing? -- RobLa (talk) 11:48, 5 February 2021 (UTC)Reply
Yes, but were these elections conducted under Copeland’s system? My point is that voters may adapt their behaviour to the voting system used. There’s no accusation of bad faith in saying that wording is tendentious. Colin.champion (talk) 13:10, 5 February 2021 (UTC)Reply
Incidentally I’m reassured by your statement that cycles don’t arise in practice – the opposite observation would have made Copeland’s method much less attractive. I don’t doubt that this fact is worth stating. My only worry is that I don’t think Wikipedia should imply that ties can therefore be dismissed as an academic nicety. They’re important even if they don’t occur because the system should make sure it doesn’t create circumstances in which they start to occur. Colin.champion (talk) 17:26, 5 February 2021 (UTC)Reply
As I've informally analyzed the data from several real-world ranked ballot elections, I haven't found a case of a tie, but I haven't looked at that many election results. I have looked at a few elections really closely: e.g. the 2009 Burlington mayoral election, the 2018 San Francisco mayoral special election, and the 2003 Debian Project Leader election.[1] It only seems to be in artificial simulations that ties seem to be common (where "common" is still defined as a pretty low percentage). There are very few "ranked ballot" elections in the world, and most that have published data use "ranked choice voting" or "RCV" (which is really "instant-runoff voting" or "IRV" or "alternative vote" as they call it in Australia). I've read/skimmed many papers over the years, and every plausible "tie" that I've seen with Copeland looks like it would be a close (and contentious) election using any method. We can chat elsewhere about our respective preferences in voting systems, but generally, I think that "Copeland's method" makes an excellent method when paired with the right tie-breaking procedure. I don't want to leave the reader thinking that tie-breaking isn't important (because I'm pretty sure that a Monte carlo Simulation would show that Copeland's method has way more ties than the Schulze method or with Ranked Pairs or almost any of the other Condorcet methods (monotonic or not). It would depend on the model being used and many other factors, though. Any system being set up for municipal elections should have a detailed tie-breaking procedure, and Copeland is no exception.
Folks that oppose Condorcet methods love to say "but what if there's a circular tie? Huh? Checkmate!!!!" They then promote methods that have real-world flaws that have caused them to get repealed (e.g. 2009 Burlington mayoral election). Hence why I'm trying to tone down the emphasis on ties in this article. -- RobLa (talk) 09:56, 7 February 2021 (UTC) (and p.s. - what strategy would mainstream voters use to induce a tie in a Copeland-based election?)Reply
Rob, I don’t think you should use a Wikipedia article to push an agenda, just state the facts and let readers decide.
As an example, suppose that the electorate is split equally between Trump, Romney and AOC; Trump supporters view Romney as second best (or second worst), Romney supporters view Trump as second best/worst, and AOC supporters view Romney as second best/worst. Romney wins if everyone votes sincerely under Copeland, and this is the right result. Trump voters can do nothing to force a win, but if they give their second preference votes tactically to AOC there’s a Copeland tie. If the tie-breaker is bad enough, Trump may come out winning. If the tie-breaker is not so bad, Romney will still win. So the moral is to do the best you can with the tie-break, thereby depriving people of a motive to vote tactically in the first round.
It’s not an entirely convincing example, and in the end there’s nothing you can do if there is too much tactical voting; but the onus is to prove that ties are not a legitimate concern, not to prove that they are.
There’s another reason to treat historical evidence with scepticism. If Condorcet cycles don’t occur this isn’t by chance: it’s because some underlying factors impose a structure on voters’ preferences. In the past century politics has been largely class-based, leading to views lying along a spectrum and the median voter theory determining the result. But these things can change at the drop of a hat. A ranked-preference election in England in the 1820s might have been full of chaotic cycles. A change in voters’ interests may lead to the same thing happening again. If you assume that the future will resemble the recent past you leave a lot of hostages to fortune. Colin.champion (talk) 10:19, 7 February 2021 (UTC)Reply
I rewrote the sections on Copeland ties, including a reference which indicates that they are unlikely to arise under honest voting (but also giving an example of how they may be induced tactically). Colin.champion (talk) 13:27, 3 June 2021 (UTC)Reply

References

Non-equivalence of 1/0/-1 versus 1/(1/2)/0

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Currently the article seems to claim two methods of Copeland scoring are equivalent: "An alternative (and equivalent) way to construct the results matrix is by ..". I'm pretty sure they're not equivalent, so I'm going to delete "equivalent." dfrankow (talk) 17:12, 6 April 2021 (UTC)Reply

Let xij be the (i,j) entry in the results matrix under 1/0/–1 and let yij be the corresponding entry under 1/½/0; then xij = 2yij–1; the row sum ∑xij = 2 ∑yijn, so “1/0/–1 score” = 2ד1/½/0 score” – n. Ignore the subtraction of n, which is a fixed offset, and one score is simply double the other. Colin.champion (talk) 06:55, 7 April 2021 (UTC)Reply