[EM] Making any method DH3-proof

[Kristofer Munsterhjelm]

As either I or Forest once mentioned, his honest runoff mechanism 
prevents the election of an honest Condorcet loser.

It can therefore be used to stop any method from electing an universally 
despised dark horse if the runoff candidates are picked in such a way 
that at least one of them cannot be the/a dark horse.

So here's a proof of concept:

Let the voters rank the candidates twice. One of these rankings is 
marked "honest" and will only be used to resolve a runoff. On the other 
(say the one marked "ordinary"), they may strategize as much as they desire.

Choose, for the virtual runoff, the winner of some base method using the 
O (ordinary) ballots, and the Plurality (or IRV) winner using the same 
ballots. If this elects the same candidate, pick the second place 
Plurality (or IRV, etc) finisher as the other finalist, instead.

Finally, elect the one that beats the other pairwise according to the H 
ballots.


Now, an objection of mine is that this kind of ballot format may seem 
ridiculous or bizarre. "Why do I have to fill in my ballot twice? And 
how do I know that I can be honest on the H ballot?". If so, just do an 
ordinary runoff instead of using Forest's virtual runoff. (Sophisticated 
game theory arguments don't help if the voters don't believe them.)

If that ordinary runoff has a debate period where the voters can more 
thoroughly scrutinize the candidates (through the media), then it would 
also help distinguish weak centrists from strong winners, if that's 
deemed to be a problem with the base method. (E.g. people who don't like 
Condorcet often say its problem is that it elects weak candidates too 
often.)


So in short: if you use a Forest or real runoff between the winner of 
any method and the highest placed different winner of a method that 
isn't susceptible to DH3, then the combined method isn't either.

-km