[EM] STAR cloneproof variant based on Score Chain Climbing

[Ted Stern]

One more comment, below.

On Tue, Feb 15, 2022, 12:51 Ted Stern <dodecatheon at gmail.com> wrote:

> 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.
>

And if both A and B defeat X, you get regular STAR, and the winner of A vs.
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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