[EM] What are some simple methods that accomplish the following conditions?
C.Benham
cbenham at adam.com.au
Sun Jun 2 15:34:24 PDT 2019
Forest (and interested others),
I (or maybe we) discovered/decided some years ago that Approval Sorted
Margins is definitely better than DMC.
Regarding alternatives to IRV that lay claim to be at least as good, two
of my strong standards are that the method must meet
one of Condorcet and FBC and that it must meet one of (at least in the
case of methods that don't allow an explicit approval
cut-off in the rankings) Minimal Defense (and maybe some yet-to-coined
irrelevant-ballot independent version of it) and Chicken
Dilemma. (I say "one of" in both cases because we know that two of
isn't possible).
The main other ways I categorise methods is according to types of
ballots used and relatively simple (to explain and sell and use)
versus less so.
I am negative on both "Asset" voting and weakening the Plurality
criterion.
Chris Benham
On 2/06/2019 8:01 am, Forest Simmons wrote:
> just one other tweak to MDDA: Kevin's official version of MDDA says
> that in the case of two of more candidates surviving the
> disqualification test, in that case the most approved candidate should
> be elected. My tweak is to elect from among the undisqualified
> candidates the one furthest from being disqualified; no need to switch
> horses in this case.
>
> On Sat, Jun 1, 2019 at 2:41 PM Forest Simmons <fsimmons at pcc.edu
> <mailto:fsimmons at pcc.edu>> wrote:
>
> Kevin, Chris, and Kristofer:
>
> Great ideas!.
>
> DMC (explicit approval version) also comes to mind as fairly
> simple to describe:
>
> *Remove candidates from the bottom of the approval list until
> there is a ballot pairwise beats-all candidate among the remaining
> candidates (to elect).
> *
>
> And analogous to Chris's equivalence between Approval Sorted
> Margins and Condorcet(approval margins), we have the equivalence
> between DMC and Condorcet(winning approval).
>
> Of course we are talking explicit approval every time we say
> approval in this context.
>
> We know that none of these Condorcet efficient methods satisfies
> the FBC, but how about MDDA with explicit approval instead of its
> implicit approval default?
>
> This has the problem that Chris has pointed out in MJ and other
> Bucklin versions where addition of irrelevant ballots can change a
> majority of ballots into a minority.
>
> So here's my idea: do MDDA with explicit approval AND symmetric
> completion of all of the ballots except for the Top Rank. By
> excusing the top rank from symmetric completion, we preserve the
> FBC, but we lose the Condorcet Criterion.
>
> [This shows how close we can get to both the CC and the FBC, if we
> do a full symmetric completion including the top rank, then this
> version of MDDA satisfies the CC but not the FBC. We can toggle
> back and forth.
> Suppose that we balance on the edge (of symmetric completion in
> the top rank or not) with asset voting so the proxies could
> finesse this difference. The unsophisticated voters could just
> vote their favorites, etc. Would this come even closer to
> satisfying both the CC and the FBC? In other words mightn't this
> expedient externalize the strategizing enough to satisfy both FBC
> and CC for all practical purposes?]
>
> I think that the FBC is more important in this context, so suggest
> that we exempt at least the top rank from symmetric completion if
> not all of the approved ranks.
>
> Now what about the irrelevant ballots problem? I think that
> symmetric completion, if only for the truncated candidates, would
> suffice. If X and Y are both truncated on N new ballots then N/2
> of them count for X>Y and N/2 of them count for Y>X so a majority
> defeat between X and Y is preserved.
>
> Note that symmetric completion among only the truncated candidates
> is the same as "half power truncation."
>
> This brings me to another question. Suppose that voters knew that
> candidate B became a plurality loser because of insincere
> truncation. Would they feel so bad it B won anyway?
>
> This leads to a weak form of Plurality: If a candidate fails
> Plurality, even after the ballots have been symmetrically
> completed, then that candidate must not be elected.
>
> Would this weak form be adequate?
>
> Thanks for your great ideas and interest in these questions!
>
> Forest
>
>
>
>
>
>
> On Sat, Jun 1, 2019 at 12:49 PM Kevin Venzke <stepjak at yahoo.fr
> <mailto:stepjak at yahoo.fr>> wrote:
>
> Hi Forest,
>
> I had two ideas.
>
> Idea 1:
> 1. If there is a CW using all rankings, elect the CW.
> 2. Otherwise flatten/discard all disapproved rankings.
> 3. Use any method that would elect C in scenario 2. (Approval,
> Bucklin, MinMax(WV).)
>
> So scenario 1 has no CW. The disapproved C>A rankings are
> dropped. A wins any method.
> In scenario 2 there is no CW but nothing is dropped, so use a
> method that picks C.
> In both versions of scenario 3 there is a CW, B.
>
> If step 3 is Approval then of course step 2 is unnecessary.
>
> In place of step 1 you could find and apply the
> majority-strength solid coalitions (using all rankings)
> to disqualify A, instead of acting based on B being a CW. I'm
> not sure if there's another elegant way
> to identify the majority coalition.
>
> Idea 2:
> 1. Using all rankings, find the strength of everyone's worst
> WV defeat. (A CW scores 0.)
> 2. Say that candidate X has a "double beatpath" to Y if X has
> a standard beatpath to Y regardless
> of whether the disapproved rankings are counted. (I don't know
> if it needs to be the *same* beatpath,
> but it shouldn't come into play with these scenarios.)
> 3. Disqualify from winning any candidate who is not in the
> Schwartz set calculated using double
> beatpaths. In other words, for every candidate Y where there
> exists a candidate X such that X has a
> double beatpath to Y and Y does not have a double beatpath to
> X, then Y is disqualified.
> 4. Elect the remaining candidate with the mildest WV defeat
> calculated earlier.
>
> So in scenario 1, A always has a beatpath to the other
> candidates, no matter whether disapproved
> rankings are counted. The other candidates only have a
> beatpath to A when the C>A win exists. So
> A has a double beatpath to B and C, and they have no path
> back. This leaves A as the only candidate
> not disqualified.
>
> In scenario 2, the defeat scores from weakest to strongest are
> B>C, A>B, C>A. Every candidate has
> a beatpath to every other candidate no matter whether the
> (nonexistent) disapproved rankings are
> counted. So no candidate is disqualified. C has the best
> defeat score and wins.
>
> In scenario 3, the first version: B has no losses. C's loss to
> B is weaker than both of A's losses. B
> beats C pairwise no matter what, so B has a double beatpath to
> C. However C has no such beatpath
> to A, nor has A one to B, nor has B one to A. The resulting
> Schwartz set disqualifies only C. (C needs
> to return B's double beatpath but can't, and neither A nor B
> has a double beatpath to the other.)
> Between A and B, B's score (as CW) is 0, so he wins.
>
> Scenario 3, second version: B again has no losses, and also
> has double beatpaths to both of A and
> C, neither of whom have double beatpaths back. So A and C are
> disqualified and B wins.
>
> I must note that this is actually a Condorcet method, since a
> CW could never get disqualified and
> would always have the best worst defeat. That observation
> would simplify the explanation of
> scenario 3.
>
> I needed the defeat strength rule because I had no way to give
> the win to B over A in scenario 3
> version 1. But I guess if it's a Condorcet rule in any case,
> we can just add that as a rule, and greatly
> simplify it to the point where it's going to look very much
> like idea 1. I guess all my ideas lead me to
> the same place with this question.
>
> Oh well, I think the ideas are interesting enough to post.
>
> Kevin
>
>
>
> >Le jeudi 30 mai 2019 à 17:32:42 UTC−5, Forest Simmons
> <fsimmons at pcc.edu <mailto:fsimmons at pcc.edu>> a écrit :
> >
> >In the example profiles below 100 = P+Q+R, and 50>P>Q>R>0.
> One consequence of these constraints is that in all three
> profiles below the cycle >A>B>C>A will obtain.
> >
> >I am interested in simple methods that always ...
> >
> >(1) elect candidate A given the following profile:
> >
> >P: A
> >Q: B>>C
> >R: C,
> >and
> >(2) elect candidate C given
> >P: A
> >Q: B>C>>
> >R: C,
> >and
> >(3) elect candidate B given
>
> >
> >P: A
> >Q: B>>C (or B>C)
> >R: C>>B. (or C>B)
> >
> >I have two such methods in mind, and I'll tell you one of
> them below, but I don't want to prejudice your creative
> efforts with too many ideas.
> >
> >Here's the rationale for the requirements:
> >
> >Condition (1) is needed so that when the sincere preferences are
>
> >
> >P: A
> >Q: B>C
> >R: C>B,
> >the B faction (by merely disapproving C without truncation)
> can defend itself against a "chicken" attack (truncation of B)
> from the C faction.
> >
> >Condition (3) is needed so that when the C faction realizes
> that the game of Chicken is not going to work for them, the
> sincere CW is elected.
> >
> >Condition (2) is needed so that when sincere preferences are
>
> >
> >P: A>C
> >Q: B>C
> >R: C>A,
> >then the C faction (by proactively truncating A) can defend
> the CW against the A faction's potential truncation attack.
> >
> >Like I said, I have a couple of fairly simple methods in
> mind. The most obvious one is Smith\\Approval where the voters
> have
> >control over their own approval cutoffs (as opposed to
> implicit approval) with default approval as top rank only. The
> other
> >method I have in mind is not quite as
> >simple, but it has the added advantage of satisfying the FBC,
> while almost always electing from Smith.
>
>
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