[EM] Participation criterion and Condorcet rules
John
john.r.moser at gmail.com
Thu Aug 9 12:28:54 PDT 2018
On Thu, Aug 9, 2018 at 2:29 PM Kristofer Munsterhjelm <km_elmet at t-online.de>
wrote:
> On 2018-08-07 18:05, John wrote:
> > Current theory suggests Condorcet methods are incompatible with the
> > Participation criterion: a set of ballots can exist such that a
> > Condorcet method elects candidate X, and a single additional ballot
> > ranking X ahead of Y will change the winner from X to Y.
> >
> > https://en.wikipedia.org/wiki/Participation_criterion
> >
> > This criterion seems ill-fitted, and I feel needs clarification.
> >
> > First, so-called Condorcet methods are simply Smith-efficient (some are
> > Schwartz-efficient, which is a subset): they elect a candidate from the
> > Smith set. If the Smith set is one candidate, that is the Condorcet
> > candidate, and all methods elect that candidate.
>
> Not all Condorcet methods are Smith-efficient. For instance, Minmax is not.
>
>
True. Most methods attempt to resolve a Condorcet cycle, but must be
Smith-efficient for the above to be true. Most methods people talk about
(Schulze, Ranked Pairs) when advocating Condorcet over IRV in public
discourse are Smith-efficient.
> > From that standpoint, each Condorcet method represents an arbitrary
> > selection of a candidate from a pool of identified suitable candidates.
> > Ranked Pairs elects the candidate with the strongest rankings; Schulze
> > elects a more-suitable candidate with less voter regret (eliminates
> > candidates with relatively large pairwise losses); Tideman's Alternative
> > methods resist tactical voting and elect some candidate or another.
>
> I think that's more true of methods that go "If the CW exists, elect
> him, otherwise...".
Not really. If a method provably always elects from a particular subset
(Smith, Schwartz) which can be identified by some algorithm, then that
method essentially elects from a pool of suitable candidates and excludes
other candidates identified as not-suitable. The decision to use such a
method inherently assumes that this subset is suitable and those outside
this subset are non-suitable.
> > Given that Tideman's Alternative methods resist tactical voting, one
> > might suggest a bona fide Condorcet candidate is automatically resistant
> > to tactical voting and thus unlikely to be impacted by the no-show
> paradox.
>
> James Green-Armytage's paper on strategy resistance,
> http://jamesgreenarmytage.com/strategy-utility.pdf , gives some proofs
> as to when "Condorcetifying" a method only improves its strategic
> resistance. If I recall correctly, making a method Condorcet-compliant
> usually doesn't alter its susceptibility to burial while it improves its
> resistance to compromising.
>
>
I intended that a set of ballots which produces a Condorcet cycle is
more-vulnerable to manipulation than a set of ballots which produces a
single Condorcet winner.
If the winner is just A (Condorcet winner), you have to defeat A before the
outcome can change. If B defeats A and is undefeated by all other
candidates, B becomes the Condorcet winner. If B defeats A but C (also
undefeated by all except A) defeats B, you have a Condorcet cycle: the
Smith Set is A B C.
In that situation, you want to rank B C A, because you need B to defeat C
or C to defeat A to change the winner from A, and you want B to defeat C.
If you rank B A C, you strengthen A's win over C.
So exploiting Participation failure probably isn't a very viable
> strategy, and for a Condorcet-compliant analog of some other method,
> it's not more viable than doing it in the other method. The exception
> would be methods that automatically pass Participation (DAC, DSC,
> Plurality).
>
>
Burying a Condorcet candidate seems unlikely in real life, too. Think
about a two-nominee system using STV to pick a spectrum from each party
(two Democrats, two Greens, two Libertarians, two Republicans). Who is
getting buried? This is an even more interesting question when you add
Electoral Fusion, and so the Democrats and Greens end up nominating three
candidates instead of four. (Double-nomination avoids the problem of the
politically-active subset being more-extreme and reducing voter choice in
general elections in our two-plus system; Fusion is another approach).
In practice, ranked voting controls marginal risk: maybe your candidate of
choice is Ben Jealous, but you think Ian Schlakman would have also made an
excellent Democratic candidate. Do you bury Ian, or do you rank him second
to push back against Hogan? If you think Ben and Ian are going to split
the vote—that the Greens are going to suck away 5% of the votes and make
Hogan the winner—ranked voting gives you a way to say hey, I like Ian, too,
but my vote is for Ben. If Hogan BEATS Ben by 1% and most Democrats ranked
Ian #2 while Republicans either bullet-voted or tried to bury Ben, guess
what? Ian and Ben have a mutual majority vote over Hogan.
In a Condorcet cycle, Ian would still probably fall away, and Greens who
voted Ben #2 would likewise push Ben above Hogan. That's actually the
likely outcome.
Hogan voters casting Hogan-Ian to try and bury Ben are only likely to push
Ian above Ben and end up sending Ben's votes to Ian (Hogan beats Ben, Ian
beats Ben, Ian wins).
You start adding in a second middle-right candidate (the Libertarian?) and
it gets more-complex.
Alternative Smith seems to me as if a burying strategy would only work if
you elected a less-desirable candidate. That is: if you like the
Republican (Hogan) more than the Libertarian, and you like the Libertarian
more than the Democrat (Ben), any attempt to bury potential winner Ben will
only succeed if you rank the Green (Ian) FIRST, then Hogan; and Ben is
between Ian and Hogan. Why?
You need Ian to defeat Ben and knock him out of the Condorcet Cycle, so you
need to rank Ian above Ben.
If there is a Condorcet cycle, then a third candidate must be ranked above
Hogan—likely Ian.
In such a case, if Ben is in the cycle, Ian probably brings Hogan into the
cycle by losing to Ben but defeating Hogan, while Hogan beats Ben.
Alternative Smith removes the plurality loser; Ian has the fewest
first-ranked votes and goes first. You want to remove Ben? You rank Ian
above Hogan. Ian's votes and Hogan's strategic votes would have to total
together to beat Ben but not defeat Hogan; and Ben's second votes would
have to be placed on Ian so infrequently as to not close the gap and elect
Hogan.
Good luck with that.
> But I imagine Participation is more a paradox-avoidance criterion than
> it is a strategic criterion, similar to monotonicity. (Again in my
> opinion,) IRV's monotonicity failure isn't something that can be
> exploited in strategy as much as it is evidence of the method "getting
> it wrong". You have two ballot sets where going from the first to the
> second only improves candidate A's situation, but A wins according to
> the first ballot set yet loses in the second.
>
>
Yes. Voters need confidence that their vote does what they want. I think
the best we can do is say it usually does what they want.
IRV's failure is that the candidate elected seems to not be one favored by
anyone: you can have a Condorcet Winner, a Plurality Winner, and an IRV
Winner in a race between three candidates, and each candidate can be the
winner of each of these three methods. If the Condorcet Winner beats the
other two head-to-head and the Plurality winner just bluntly gets the most
votes when everyone is asked to pick one of the three, what in the heck is
the IRV winner?
With a Smith-efficient method, every candidate not in the Smith set would
lose one-on-one to any and all candidates in the Smith set. We can tell
the voters, with absolute authority, that those candidates excluded are
candidates who cannot win against any of the candidates we might elect.
You have that confidence that we'll elect someone meaningful to the
expressed will of the voters.
As for what the method does besides that, well, it may do something
strange. It follows a mathematical rule and will not give anyone the power
to dictate the winner; the precise outcome may have a confusing
relationship with the ballots cast, even though the group from which we
elect a winner has a very clear relationship.
If nothing else, it's less about ground game and more about the voters as a
whole finding the candidate acceptable. Everyone's vote matters. Under
plurality (and with sufficiently-few candidates), your vote doesn't matter
if your district is 60% Democrat or 60% Republican.
> > I ask if the following hold true in Condorcet methods where tied
> > rankings are disallowed:
> >
> > 1. In methods independent of Smith-dominated alternatives (ISDA),
> > ranking X above Y will not change the winner from X to Y /unless/ Y
> > is already in the Smith Set prior to casting the ballot.
> > 2. In ISDA methods, ranking X above Y will not change the winner from X
> > to Y /unless/ some candidate Z both precedes X and is in the Smith
> > set prior to casting the ballot.
> > 3. In ISDA methods, ranking X above Y will not change the winner from X
> > /unless/ some candidate Z both precedes X and is in the Smith set
> > /after/ casting the ballot.
> > 4. In ISDA methods, ranking X above Y and ranking Z above X will either
> > not change the winner from X /or/ will change the winner from X to Z
> > if Z is not in the Smith Set prior to casting the ballot and is in
> > the Smith Set after casting the ballot.
> > 5. in ISDA methods, ranking X above Y will not change the winner from X
> > to Y /unless/ Y precedes Z in a cycle after casting the ballot
> > /and/ Z precedes X on the ballot.
> >
> > I have not validated these mathematically.
>
> Markus Schulze replied to this more succinctly than I could, but to
> restate: Suppose X is the CW. Then by 1., adding a ballot ranking X
> first should not deprive X of the victory. Hence every Condorcet method
> should pass mono-add-top if the starting scenario is one where the
> winning candidate is the CW.
>
> I don't know if that's true, but at a first glance, it seems to be too
> strong. Suppose we have an election with an A>B>C>A cycle, and all of
> these candidates beat candidate D pairwise, so that D is the Condorcet
> loser and the Smith set is {A,B,C}. Suppose that the method being used
> elects A. Also suppose the B>D pairwise victory is very weak, so that
> adding two ADBC ballots reverses it to D>B. Then that could admit D into
> the Smith set, and the internal logic of the method could make D win.
> Yet D was ranked below A, the winner, on those ADBC ballots.
>
> E.g. for Smith//Plurality:
>
> 34: D>A>B>C
> 33: D>B>C>A
> 33: D>C>A>B
> 51: A>B>C>D
> 50: B>C>A>D
>
> The Smith set is {A, B, C}. Eliminating the non-Smith member D gives the
> election:
>
> 85: A>B>C
> 83: B>C>A
> 33: C>A>B
>
> where A has the most first preference votes and thus wins.
>
> Adding two A>D>B>C ballots gives
>
> 34: D>A>B>C
> 33: D>B>C>A
> 33: D>C>A>B
> 51: A>B>C>D
> 50: B>C>A>D
> 2: A>D>B>C
>
> where the Smith set is {A, B, C, D}, and thus D wins with 100 first
> preference votes.
>
> Since Smith//Plurality passes ISDA, that should answer your five
> questions in the negative.
>
>
Interesting. I wonder if close victories are a solvable election problem
or if that's an error that simply will never go away. We just had an
election here where a candidate won by about 20 votes, and three candidates
all received nearly the same number of votes. Two hundred votes change and
the third-place winner is the first-place winner—out of 80,000 votes.
If fifteen voters had voted Brochin instead of Olszewski, the results would
have reversed.
Now you tell me: what's the difference between that and the election you
describe above? I've been approaching this from a technical standpoint;
I'm starting to think there's a philosophical problem here—one that can't
be solved. Condorcet tries—it achieves something akin to mutual majority
consensus—but if the needle only has to move a few fractions of a
millimeter to change the winner, have we elected someone or did they simply
win a contest?
> The only Condorcet method I know of that passes both Condorcet and
> mono-add-top is Minmax (margins), but that method fails Smith. As
> Schulze said, the question of whether Smith and mono-add-top are
> compatible is open.
>
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