[EM] "inconclusively-dominated sets", IRV-completed Condorcet

[James Green-Armytage]

Alex small wrote:
>I think I prefer IRV-completed Condorcet to this method.  Although from
>the description it seems to use information on the voters' full preference
>rankings, in the end it's equivalent to a method using far less
>information to break cycles.  IRV would break a cycle using 2 pieces of
>information:  (1) how many first place votes each candidate has and (2)
>the results of a pairwise contest.  This method really only needs the
>first piece of information when there's a cycle.  So IRV-completed
>Condorcet seems preferable.

	This paragraph confused me a bit. You are arguing that IRV-completed
Condorcet doesn't use full preference rankings, and then you use that
argument to conclude that it's a better method than the
lowest-two-elimination-runoff (L2ER) / orphan method. I might agree with
you that IRV-completed Condorcet is better, but I still don't follow this
line of argument. But oh well.
	I wanted to talk about Condorcet completed by IRV (CCIRV) a bit more
here, beef up the proposal a bit. 
	First of all, if there is a top cycle in the pairwise result, I'd like to
eliminate the non-members of the Schwartz set (union of minimal
undominated sets) before moving on to the IRV tally.
	Second of all, I think that cycles where all defeats are by a majority
(of the valid vote) should be treated differently from cycles where one or
more defeats are by a minority. Again, this is for the same
anti-truncating-strategy reasons that brought us winning votes instead of
margins. 
	So, in trying to define the rule here more precisely, I made up a set
called the"union of minimal
inconclusively-dominated sets". (I'm using the phrase
"inconclusively-dominated" as a sort of euphemism for being
beaten/dominated by a minority instead of a majority. Sorry if it doesn't
seem to fit for some people.) If you'd like to abbreviate it, I guess it
would be the UMID set.
	My proposal for IRV-completed Condorcet is to eliminate every candidate
outside the UMID set before proceeding to the IRV tally.
	Okay, so the UMID set is like the Schwartz set, but a little different,
because it takes into account whether a candidate is beaten by a majority
or a minority. If there is a set such that no candidate within the set is
majority-beaten by any candidate outside the set, then it's an
inconclusively-dominated set. If it doesn't contain other
inconclusively-dominated sets, then it's a minimal
inconclusively-dominated set. So the union of minimal
inconclusively-dominated sets consists of all the candidates who belong to
a minimal inconclusively-dominated set.
	For example, let's say that pairwise, A beats B beats C beats A, but the
C-->A defeat is by a minority, while the other two defeats are by a
majority. So, {A}, by himself, constitutes a minimal
inconclusively-dominated set, because he is not majority-beaten by any
other candidate. {A, B, C} all together constitute an
inconclusively-dominated set, but not a MINIMAL inconclusively dominated
set... so the only minimal inconclusively dominated set is {A}. Hence my
CCIRV version eliminates all candidates besides A, rendering the IRV tally
pointless.
	Another example. A minority-beats B majority-beats C minority-beats A.
{A} constitutes a MID set; no one majority-beats A. {B} also constitutes a
MID set. {C}, however, doesn't constitute a MID set, because B
majority-beats C. {A, B, C} constitutes an ID set but not a MID set, so no
go. Hence, the union of MID sets, the UMID set, is {A, B}. So, my method
would eliminate C, and then proceeds to the IRV tally, where A, of course,
would win.
	Does it seem arbitrary to you that A should win, and not B? Not me.
Because look, those who put B in first place are the ones responsible for
the fact that the C-->A defeat is by a minority. As it is, we don't know
if the A:C contest is a sincere A-->C majority beat or a sincere C-->A
majority beat. If it was a sincere C-->A majority beat, than it's the B
voters' own fault for not voting it out to a majority. If it was a sincere
A-->C beat, then A is a sincere Condorcet winner, and the B voters
shouldn't be rewarded for their evil truncation scheme.
	So, yup, that's my proposal for CCIRV. It's CC-UMID-IRV! Yesssss!
Actually, it's not such a bad method, maybe. It doesn't share the same
truncation incentives as regular CC-IRV, which is nice... you have to do
order-reversal if you want to mess things up. Of course, you know me, my
primary single-winner proposals are still weighted pairwise and iterative
pairwise, but I'm just saying that this one is not-too-bad.
	By the way, I'm pretty sure that winning votes Condorcet always chooses
from the UMID set. As does weighted pairwise, with the majority/minority
defeat strength provision.
 
my best,
James Green-Armytage