[EM] STAR cloneproof variant based on Score Chain Climbing

[Ted Stern]

Replies inline below:

On Tue, Feb 15, 2022 at 12:09 PM Kevin Venzke <stepjak at yahoo.fr> wrote:

> Hi Ted,
>
> Ted Stern wrote:
> > Here's a proposal for a STAR variant that handles clones:
> >
> > Top two score winners (A and B), plus the score winner when you exclude
> the
> > top score winner ballots (by score weight), that's the clone-proof part.
> Call
> > that the exclusive winner, X. In other words, if a ballot gives a score
> of 5
> > out of 10 to the top score winner, remove half that ballot's weight.
>
> Is it allowed for B and X to be the same candidate? (I'm not sure it's
> totally
> ruled out that even A and X could be the same.)
>

In the absence of clones, X could very easily be B.


>
> If a ballot gave A a 10/10, I assume all weight is removed when
> determining X.
>

Exactly. A ballot voting for A gets A-score/max-score removed.


>
> > Eliminate any candidates defeated by X.  If more than one remains, the
> winner
> > is the one who defeats the other.
>
> When you say "if more than one remains" I assume you mean "of A and B." Or
> possibly "of A, B, and X."
>

Assuming A, B and X are distinct:

If X defeats both A and B, only X remains and X wins.

If X is defeated by A, but defeats B, B is removed, and A wins.

If X is defeated by B, but defeats A, A is removed and B wins.


>
> When you say "the winner is the one who defeats the other," this is a
> second
> round of voting, correct? So in all cases of this second round it will be
> between two of {X,A,B}?
>

See above


>
> > This follows the logic of Forest Simmons' Score Chain Climbing to resolve
> > cycles.
> >
> > The clone problem is that A and B could be clones. Removing A's ballot
> > contributors finds the non-clone while avoiding pushover incentive. If X
> > defeats both A and B, it's likely the CW. Otherwise, whichever of A or B
> is
> > defeated by X is "weaker" (low probability, but possible in cycles).
> > Using a lower-scoring candidate as an eliminator reduces burial
> incentive.
>
> Not sure I follow. The claim is counter-intuitive because usually a
> candidate to
> be given artificial preferences is a weaker one (lower-scoring) because the
> insincere voters believe this candidate can't win, and don't wish for him
> to win.
>
> For example in your case if either A or B (doesn't matter) believes they
> will
> beat X pairwise then they are able to try to use X to knock the other of
> A/B
> out. Just rate X a 1/10, and the other 0/10. No?
>
> As I say, I think it's not chain climbing generally, but TACC(implicit)
> specifically, that has anti-burial value.
>

I'd love to see an example of what you're suggesting. Can you think of
one?  I don't necessarily need my proposal to win, I'm just looking for the
best way to cloneproof STAR while still selecting the best candidates.


>
> > Why do I propose finding X that way instead of by a Hare or Droop quota?
> > Well, for one thing, it's a summable process. Next, if A has >50%
> approval,
> > A and B are probably the two candidates to choose from anyway. If A has
> > <50% approval, X is being found with something like a Hare quota, moving
> > more toward Droop as A's approval decreases.
> >
> > Your thoughts?
>
> I think the goal of decloning the election is a bit different from the
> goal of
> selecting the "best" two finalists for a runoff (if this method actually
> has a
> runoff, wasn't sure there). So I'm not so worried if you don't use some
> more
> "proper" decloning method like a quota.
>
> Kevin
>
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