[EM] STAR cloneproof variant based on Score Chain Climbing

Tue May 31 11:29:14 PDT 2022

I've fleshed out a cloneproof STAR proposal using Score Sorted Margins.

   - Score ballots, 0-5
   - Aggregate total scores for all candidates.
   - We will find 3 primary winners. The first two are the candidates with
   top two score totals, Score winner (SW) and Score Runner-up (SRU)
   - The third primary winner, X, is found by reweighting each ballot
   according to its scores for SW and RU:
      - weight = 1 / (1 + (Ballot-Score[A] + Ballot-Score[B]/MaxScore)
      - and then re-summing total scores for reweighted ballots.

The next round is decided using Score Sorted Margins. Using either original
scores, normalized scores, or scores in a separate runoff election, sort
the candidates SW, SRU and X in descending order of score. If using the
original scores, the ordering will be SW, SRU, X, but for other cases,
possibly not. Let's call that sorted ordering A, B, C for now. The Score
Sorted Margins winner for a 3 candidate case can be found as follows:

   - Find the three candidate score totals, sort them in descending order
   of score. We call the seeded order A, B, C.
   - Find the pairwise counts for A vs B, B vs C, and C vs A.
   - If there is a beats-all winner, that candidate is the winner.
   - Otherwise, if A>B, A wins
      - Why? Because there 8 possible cases (excluding ties) for the 3
      pairwise contests. 2 of those lead to A winning as CW, 2 lead to B as CW,
      and 2 to C as CW. The remaining cases are
      - A > B > C (> A), and
      - A < B < C (< A)
      - In the first cycle, the seeded ordering is already in pairwise
      sorted order, so the cycle is broken below C. So after filtering out CW
      cases, A>B determines we have the first cycle case.
   - Otherwise if margin(A,B) > margin(B,C), A wins
      - This is where a sorted margins iteration would actually take place
      -- both AvsB and BvsC are out of order pairwise. If the AB margin is
      greater, then the BC pair is swapped first, leading to a pairwise-sorted
      ordering of A > C > B.
   - Otherwise, B wins

If the original Score order of candidates is preserved, then the third
primary candidate can only win if it's the Condorcet Winner. And the runner
up can win only if it's CW, or its margin with A is less than its margin
with C. So, overall, there is a bias toward the score winner.

Personally, I would prefer a separate primary and runoff if there are 4 or
more candidates -- the primary has the effect of winnowing the field and
enabling closer scrutiny of candidates for the general election. If there
are only 2 or 3 candidates, a primary is unnecessary and you can go
straight to sorted margins for the general.

On Sun, May 29, 2022 at 11:56 AM Ted Stern <dodecatheon at gmail.com> wrote:

> Reviving an old topic.
>
> I had another thought for clone proofing STAR or Top Two Approval runoff,
> based on SPAV or RRV. It is not summable.
>
> Round one:
>
> Approval or Score ballots.
>
> Advance the top two approved or top two total score candidates.
>
> Advance a third candidate using SPAV with Approval ballots or RRV if score
> ballots, as if the two already advanced candidates were chosen by that
> method. That is, if using approval, a ballot's weight for the third count
> has weight 1 if it approved no previous winners, 1/2 if it approved one of
> the first two winners, or 1/3 if it approved both the previous winners.
> Similarly for score ballots using RRV.
>
> If round one uses Approval ballots, a score ballot runoff will be held. If
> using Score, the round one ballots can be recounted to find pairwise
> preferences, using ratings to infer rankings. Or a separate score runoff
> could also be held with the three winners, which I prefer.
>
> In round two, use a cloneproof and burial resistant Condorcet method.
> Let's say, score sorted margins or score chain climbing.
>
> The difference from my first proposal is that there are always 3
> candidates, representing either 2 factions, if top two are clones, or 3
> factions, if not cloned.
>
> My overall preference is for an approval first round, to eliminate most
> candidates, then the cloneproof third candidate will generally represent a
> different perspective to be debated before the runoff.
>
>
> On Sat, Mar 12, 2022, 00:17 Forest Simmons <forest.simmons21 at gmail.com>
> wrote:
>
>> Thanks, Kevin. It was a comment of yours that made me realize that burial
>> punishment (via chain climbing) was not enough ... but of course, looking
>> away and pretending the burier was probably sincere ... that is no good
>> either.
>>
>> So a sincerity check is natural ... if the sincere ballots contradict the
>> strategic ballots, then in this case you have both detection and correction.
>>
>> Perhaps we could forget Chain Climbing and just use the sincerity check
>> on the weakest defeat that was critical in determining the winner.
>>
>> El vie., 11 de mar. de 2022 11:42 p. m., Kevin Venzke <stepjak at yahoo.fr>
>> escribió:
>>
>>> Hi Forest,
>>>
>>> Le vendredi 11 mars 2022, 23:03:30 UTC−6, Forest Simmons <
>>> forest.simmons21 at gmail.com> a écrit :
>>> > SCC:
>>> >
>>> > Initialize a variable X as the (name of) the lowest score candidate.
>>> Then ...
>>> >
>>> > While more than one candidate remains, eliminate all of the candidates
>>> pairwise
>>> > defeated by X, before storing a new name into X, the name of the
>>> lowest score
>>> > remaining candidate.
>>> > EndWhile
>>> >
>>> > The last value of X (the SCC winner Xf) is one of the finalists.
>>> >
>>> > The other finalist is the second to the last value of X, which we
>>> designate Xf'.
>>>
>>> For the case that the initial value of X is the CW, should an
>>> elimination order
>>> be specified?
>>>
>>> > But doesn't the last X defeat all of the previous X's?
>>> >
>>> > Yes, according to the ballots. But there is a good chance that the
>>> only reason
>>> > Xf defeats Xf' on the ballots is that Xf' was insincerely buried under
>>> Xf.
>>>
>>> In my terminology, that would mean Xf' is the sincere CW and Xf is the
>>> "pawn."
>>> The strategists' own candidate (the "rival") has been eliminated, so
>>> their
>>> strategy failed (and would be a backfire, if the last X simply won).
>>>
>>> This probably implies that the sincere CW was unexpectedly the Score
>>> loser.
>>>
>>> > So how do we vindicate (or expose as fraudulent) the finalist Xf?
>>> >
>>> > We could take another trip to the polls for a runoff between between
>>> Xf and Xf'.
>>> >
>>> > Otherwise, we can require voters to submit two ballots ... one to
>>> determine the
>>> > two finalists, and the other to choose between them.
>>> >
>>> > Sincere voters simply duplicate their first ballot to produce their
>>> second one.
>>> > The strategy burdened voters adjust their insincerities to produce
>>> their second
>>> > ballot.
>>> >
>>> > It is crucial that the second ballot be used exclusively for choosing
>>> the winner
>>> > between the two finalists.
>>> >
>>> > However, once the final winner has been certified , these ballots can
>>> be used
>>> > for forensics.
>>>
>>> All true. It seems like the effect of this is to make "backfired
>>> strategy"
>>> outcomes impossible. Is that the goal? It seems like that might risk
>>> encouraging
>>> voters to *try* burial strategies, unless it's sufficient to "name and
>>> shame"
>>> strategists through the forensics performed afterwards.
>>>
>>> It seems like this proposal could even prevent a backfire when *both* of
>>> two
>>> major factions are ranking the same pawn insincerely high, so that the
>>> pawn
>>> becomes the voted CW.
>>>
>>> Kevin
>>>
>>
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