Making any method DH3-proof
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
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?
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.
Would more than 50 percent implicit approval for the common winner of the
two methods be enough for such a runoff exemption?
El lun., 25 de jul. de 2022 3:26 a. m., Kristofer Munsterhjelm <
km_elmet@t-online.de> escribió:
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
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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
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@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
On 7/26/22 8:27 PM, Forest Simmons wrote:
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.
There's probably a connection between UD methods that try to deter
burial and this kind of construction. The UD methods attempt to
automatically perform the runoff while this construction makes sure
(when there is a risk of strategy) that at least the runoff result will
be honest. They're probably not entirely the same, since the implied
honesty of the second round would make some runoff combinations possible
that would otherwise be too risky, but it wouldn't be surprising if they
were broadly similar.
-km