[EM] Participation criterion and Condorcet rules
Kristofer Munsterhjelm
km_elmet at t-online.de
Thu Aug 9 11:29:17 PDT 2018
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.
> 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...". Consider the Ranked Pairs method, for instance. The
RP method consists of sorting the pairwise victories in order of
magnitude (and tiebreaking by random voter hierarchy if necessary). It
then goes down the list, affirming pairwise victories unless there's a
contradiction with previously affirmed victories.
The procedure makes no explicit use of the Smith set, but always elects
from the Smith set because of how it works - if A is in Smith, and B is
not, then the affirming procedure will reach A>B before it reaches B>A,
so A will be ranked before B in the final ordering.
Such a proof is implicit, and thus is similar to say, a proof that IRV
passes mutual majority. (If a majority of the voters rank k candidates
above everybody else but not necessarily in the same order, then after
at least k eliminations, one of these candidates must be the only one
left from that group, and since Plurality meets Majority, the remaining
candidate can't be eliminated.)
Since one wouldn't usually say "IRV is a method that selects an
arbitrary candidate from a pool of identified suitable candidates - the
smallest mutual majority set", I don't think the description works for
implicitly Smith methods like Ranked Pairs either. Mathematically, both
are true (Ranked Pairs elects from Smith and IRV elects from the
smallest mutual majority set), but neither methods' logic go "first
identify the set, then do something to pick someone from it".
> 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.
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).
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.
> 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.
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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