[EM] Two round Condorcet (or maybe 1.01 round)

[Forest Simmons]

How about a runoff between the candidate with minimal max pairwise
opposition and the candidate with maximal minimum pairwise (inclusive)
support.

With complete rankings they're equivalent. But equal rankings and
truncations arising from strategy or indifference can make the runoff more
interesting.

El jue., 23 de sep. de 2021 8:11 a. m., Kristofer Munsterhjelm <
km_elmet at t-online.de> escribió:

> On 23.09.2021 09:39, Juho Laatu wrote:
> > What do you think of this rather simple method?
> >
> > (1) Arrange a Condorcet election
> > (2) If there is a a Condorcet winner / no top loop / no tie, elect the
> first round winner
> > (3) Otherwise arrange a second round, using the same Condorcet variant,
> and elect the second round winner
> >
> > The idea is to encourage sincere voting at the first round. And to
> > reduce the need for defensive strategic voting at the first round,
> > against strategies that make use of creating a top loop. And to
> > discourage people that are planning to win by voting strategically.
> >
> > Rule (2) could be also such that any first round winner will be
> > elected, unless there is a formal request to arrange the second round.
> > The request should maybe be based on a claim that there indeed was
> > insincere voting that may have influenced the outcome.
> >
> > In rule (2) a tie / no Condorcet winner could mean also having
> > multiple equally strong winners. Maybe we would go automatically to the
> > second round in this case (instead of picking one of the winners at
> > random), even if there was no insincere voting.
> >
> > Round two uses the same (sub)method as in the first round to remove
> > any interest of trying to go to the second round because of better
> > chances of winning with that (sub)method.
>
> If you use your first suggestion, then every Condorcet method is equal
> in the eyes of point 2, so "using the same submethod" doesn't really
> make sense. I suppose you're referring to the case where step one is
> e.g. "Arrange a minmax election" and then round two only happens if
> enough people complain about it.
>
> From a utility perspective, such a method *could* work, because the
> voters' willingness to petition for a second round is what enables it to
> distinguish an okay winner from a truly horrible one. However, if I were
> to argue against it, I would probably say that voting isn't narrowly
> utilitarian to begin with (the well-known paradox of voting), and that
> the mechanism is complex enough that it might lead to too much of a
> status quo bias.
>
> Or to put it differently, the idea is that there's enough of a
> barrier/threshold to petitions, then a second round would only happen
> when the first-round winner is a bad one. But it's not obvious how this
> threshold should be adjusted. In one extreme case, you always hold a
> runoff whenever the Smith set is larger than one. In another, it takes a
> supermajority to hold the runoff. I would be more in favor of the former
> than the latter, particularly if there's some time between the two
> rounds so that the Smith set candidates have to defend themselves,
> participate in debates, etc.
>
> Another option I suggested to Forest goes like this:
>
> 1. Arrange a Condorcet election with a strategy-resistant method and a
> method wth good performance under honesty, e.g. Smith,IRV and Ranked
> Pairs (probably something even better can be constructed).
> 2. If the winners are the same (which happens whenever there's a CW, but
> also in some other cases), then that winner wins outright.
> 3. Otherwise arrange a two candidate runoff between the two winners.
>
> Simple, it's not, but the methods could cover each other's weaknesses.
> Taking this reasoning further, it should be possible to design "minmax
> robust methods" that return two candidates so that the VSE after a
> honest two-candidate runoff is the best possible, no matter the level of
> strategy.
>
> Well, in theory. Actually designing such a method with desirable
> properties (monotonicity, etc) would be much harder :-)
>
> -km
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