[EM] Making any method DH3-proof

[Forest Simmons]

Here's another possibility: when both methods yield the same finalist,
enlist the most likely victim of burial as the other finalist.

For example, the Smith member with the least implicit approval is usually a
likely victim.

With that choice, the winner of the honest runoff (besides not being the
HCL) would be a member of the strategic ballot Smith set.

El mar., 26 de jul. de 2022 2:35 a. m., Kristofer Munsterhjelm <
km_elmet at t-online.de> escribió:

> On 7/26/22 1:29 AM, Forest Simmons wrote:
> > I see ... the only reason you cannot dispense with the runoff step
> > whenever the two methods have the same winner, is to make sure the
> > sincere Condorcet loser never gets elected ... right?
>
> That's right. If you just want to prevent DH3 and nothing more, then
> there's no problem with electing without a runoff if the two methods
> agree about the winner and at least one of them is immune to DH3.
>
> Let's call the criterion that the method never elects the honest
> Condorcet loser, the honest Condorcet loser criterion.
>
> To pass honest Condorcet loser, you can only dispense with the runoff if
> the election is so that, for at least one of the methods, the honest
> Condorcet loser is not elected in a strategic equilibrium. Sort of like
> Monroe's "nonelection of irrelevant alternatives".
>
> > But perhaps the runoff could be dispensed with if both methods yielded
> > the same winner AND this common winner had more than fifty percent first
> > place votes.
>
> I would guess so, because if that weren't the case, then that would mean
> that every method that passes majority can elect the honest Condorcet
> loser in a strategic equilibrium, which seems implausible? Even if I
> don't know if it's false.
>
> Another possibility is that not every method can elect the HCL in an
> equilibrium, but both of the methods have some kind of arms race dynamic
> so that the HCL could win in an equilibrium *and* have majority support.
> That, too, sounds implausible, but it's perhaps less implausible than
> the above.
>
> > Would more than 50 percent implicit approval for the common winner of
> > the two methods be enough for such a runoff exemption?
>
> I'm less certain of that. Suppose everybody fully ranks. Then (if I
> understand implicit approval correctly) that means that the
> Antiplurality winner can never be the HCL... which doesn't seem right.
> If everybody tries to exaggerate by pushing their hated candidate to
> last place, then DH3 might happen. Because the dark horse doesn't have
> any last preferences (everybody is burying), then he has unanimous
> implicit approval (I think?)
>
> There are of course caveats that could make 50% IA work. If one of the
> base methods is so that DH3 isn't rewarded, then the escalation towards
> the situation where the dark horse has >50% IA doesn't happen, and then
> it *should* be safe. But to say the same about the Honest Condorcet
> Loser criterion, we'd have to know that for one of the method, there's
> no equilibrium where the HCL has >50% implicit approval. Again, I don't
> know for sure, but intuitively, it seems that implicit approval has a
> lot more room for strategic shenanigans than does first-place only.
>
> We really lack tools to figure out what the strategic equilibria (e.g.
> Myerson-Weber) are for given methods, and if they can elect honest
> Condorcet losers, universally despised candidates, Pareto-dominated
> candidates, etc.
>
> I think Warren once commented that even if (say) burial is not
> strategically rewarding, a number of voters would do so anyway because
> it's such an obvious thing to do. That would kind of ruin everything
> that isn't cardinal, Plurality or IRV, though, and seems very pessimistic.
>
> -km
>
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