[EM] Making any method DH3-proof

[Kristofer Munsterhjelm]

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