- 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.

- 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

For my final (?) cloneproof STAR proposal, I think the virtual three candidate primary needs modification. If the first two candidates are the top two score winners, which third candidate would maximize equity in the That candidate can be found by, for each ballot, subtracting (ballot_score(score winner) + ballot_score(score runner up)/maxscore from the ballot's voting strength [or reducing voting strength to zero if that quantity is greater than 1], then finding the score winner over all reweighted ballots. This approach also minimizes pushover incentive. The goal here is to have at least 2 distinct factions represented in the extreme case where the top two scoresum candidates are clones. To test this, does anyone have any non-Smith STAR examples? On Tue, May 31, 2022, 11:29 Ted Stern <dodecatheon@gmail.com> wrote: > 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@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@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@yahoo.fr> >>> escribió: >>> >>>> Hi Forest, >>>> >>>> Le vendredi 11 mars 2022, 23:03:30 UTC−6, Forest Simmons < >>>> forest.simmons21@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 >>>> >>>

- 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.

- 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

I was pointed to this voting simulator developed by Kyle Brockman that contains a version of STAR3: https://www.chocolatepi.net/voteapp/ In this model, STAR3 picks the top three scoring candidates, then the winner is chosen by pairing S1 against the winner of S2 vs S3. Which is the same as the Smith//Score winner for three candidates. This method has most of the features I look for: * More resistance to clones. Obviously, a crowding attack could still work, but it's much less likely. * Some resistance to chicken dilemma. Not perfect, but enough to make the strategy unpredictable. * Bias toward score winner. * Simplicity. As easy to describe as Vote321. On Wed, Jun 15, 2022, 12:56 Ted Stern <dodecatheon@gmail.com> wrote: > For my final (?) cloneproof STAR proposal, I think the virtual three > candidate primary needs modification. > > If the first two candidates are the top two score winners, which third > candidate would maximize equity in the > > That candidate can be found by, for each ballot, subtracting > (ballot_score(score winner) + ballot_score(score runner up)/maxscore from > the ballot's voting strength [or reducing voting strength to zero if that > quantity is greater than 1], then finding the score winner over all > reweighted ballots. > > > This approach also minimizes pushover incentive. > > The goal here is to have at least 2 distinct factions represented in the > extreme case where the top two scoresum candidates are clones. > > To test this, does anyone have any non-Smith STAR examples? > > On Tue, May 31, 2022, 11:29 Ted Stern <dodecatheon@gmail.com> wrote: > >> 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@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@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@yahoo.fr> >>>> escribió: >>>> >>>>> Hi Forest, >>>>> >>>>> Le vendredi 11 mars 2022, 23:03:30 UTC−6, Forest Simmons < >>>>> forest.simmons21@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 >>>>> >>>> ---- > Election-Methods mailing list - see https://electorama.com/em for list > info >

- 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.

- 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

Might as well go all the way: while there are two or more uneliminated candidates, from among these remaining candidates eliminate the pairwise loser between the two with the lowest (remaining) scores. The entire finish order (if desired) can be obtained by bubble sorting the score order until no adjacent candidates are out-of-order pairwise. Rectification priority goes to the out-of-order adjacent pair whose pairwise loser has the lowest score. In fact, this method is identical to Ranked Pairs with lowest losing score corresponding to greatest defeat strength. The resulting full clone independence is well worth the tiny tweak of always starting at the bottom score candidate rather than the strangely arbitrary rule of starting at the third candidate from the top. El lun., 25 de jul. de 2022 9:28 p. m., Ted Stern <dodecatheon@gmail.com> escribió: > I was pointed to this voting simulator developed by Kyle Brockman that > contains a version of STAR3: > > https://www.chocolatepi.net/voteapp/ > > In this model, STAR3 picks the top three scoring candidates, then the > winner is chosen by pairing S1 against the winner of S2 vs S3. Which is the > same as the Smith//Score winner for three candidates. > > This method has most of the features I look for: > > * More resistance to clones. Obviously, a crowding attack could still > work, but it's much less likely. > > * Some resistance to chicken dilemma. Not perfect, but enough to make the > strategy unpredictable. > > * Bias toward score winner. > > * Simplicity. As easy to describe as Vote321. > > On Wed, Jun 15, 2022, 12:56 Ted Stern <dodecatheon@gmail.com> wrote: > >> For my final (?) cloneproof STAR proposal, I think the virtual three >> candidate primary needs modification. >> >> If the first two candidates are the top two score winners, which third >> candidate would maximize equity in the >> >> That candidate can be found by, for each ballot, subtracting >> (ballot_score(score winner) + ballot_score(score runner up)/maxscore from >> the ballot's voting strength [or reducing voting strength to zero if that >> quantity is greater than 1], then finding the score winner over all >> reweighted ballots. >> >> >> This approach also minimizes pushover incentive. >> >> The goal here is to have at least 2 distinct factions represented in the >> extreme case where the top two scoresum candidates are clones. >> >> To test this, does anyone have any non-Smith STAR examples? >> >> On Tue, May 31, 2022, 11:29 Ted Stern <dodecatheon@gmail.com> wrote: >> >>> 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@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@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@yahoo.fr> escribió: >>>>> >>>>>> Hi Forest, >>>>>> >>>>>> Le vendredi 11 mars 2022, 23:03:30 UTC−6, Forest Simmons < >>>>>> forest.simmons21@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 >>>>>> >>>>> ---- >> Election-Methods mailing list - see https://electorama.com/em for list >> info >> >

- 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.

- 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

Hi Forest, We discussed your proposed method back in 2004. It's basically the same as Definitive Majority Choice, using score instead of Approval. DMC(score) has a simple basic procedure, as you note, which is identical to Smith//Score in the 3 candidate case, but is less satisfying for larger cycles. Which is why I prefer Score Sorted Margins in general. Back at the start of the thread, I mentioned that I like STAR except for the clone problem. The basic idea is that the score winner is good in general, but a coarse discrete scale lacks resolution, so the runoff increases satisfaction. STAR3 is a minor modification that still makes the case for score, but reduces the incentive for crowding strategy. And by restricting the number of pairwise comparisons, it remains human countable and summable on the second round (the pairwise part). On Tue, Jul 26, 2022, 17:30 Forest Simmons <forest.simmons21@gmail.com> wrote: > Might as well go all the way: while there are two or more uneliminated > candidates, from among these remaining candidates eliminate the pairwise > loser between the two with the lowest (remaining) scores. > > The entire finish order (if desired) can be obtained by bubble sorting the > score order until no adjacent candidates are out-of-order pairwise. > Rectification priority goes to the out-of-order adjacent pair whose > pairwise loser has the lowest score. > > In fact, this method is identical to Ranked Pairs with lowest losing score > corresponding to greatest defeat strength. > > The resulting full clone independence is well worth the tiny tweak of > always starting at the bottom score candidate rather than the strangely > arbitrary rule of starting at the third candidate from the top. > > El lun., 25 de jul. de 2022 9:28 p. m., Ted Stern <dodecatheon@gmail.com> > escribió: > >> I was pointed to this voting simulator developed by Kyle Brockman that >> contains a version of STAR3: >> >> https://www.chocolatepi.net/voteapp/ >> >> In this model, STAR3 picks the top three scoring candidates, then the >> winner is chosen by pairing S1 against the winner of S2 vs S3. Which is the >> same as the Smith//Score winner for three candidates. >> >> This method has most of the features I look for: >> >> * More resistance to clones. Obviously, a crowding attack could still >> work, but it's much less likely. >> >> * Some resistance to chicken dilemma. Not perfect, but enough to make the >> strategy unpredictable. >> >> * Bias toward score winner. >> >> * Simplicity. As easy to describe as Vote321. >> >> On Wed, Jun 15, 2022, 12:56 Ted Stern <dodecatheon@gmail.com> wrote: >> >>> For my final (?) cloneproof STAR proposal, I think the virtual three >>> candidate primary needs modification. >>> >>> If the first two candidates are the top two score winners, which third >>> candidate would maximize equity in the >>> >>> That candidate can be found by, for each ballot, subtracting >>> (ballot_score(score winner) + ballot_score(score runner up)/maxscore from >>> the ballot's voting strength [or reducing voting strength to zero if that >>> quantity is greater than 1], then finding the score winner over all >>> reweighted ballots. >>> >>> >>> This approach also minimizes pushover incentive. >>> >>> The goal here is to have at least 2 distinct factions represented in the >>> extreme case where the top two scoresum candidates are clones. >>> >>> To test this, does anyone have any non-Smith STAR examples? >>> >>> On Tue, May 31, 2022, 11:29 Ted Stern <dodecatheon@gmail.com> wrote: >>> >>>> 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@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@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@yahoo.fr> escribió: >>>>>> >>>>>>> Hi Forest, >>>>>>> >>>>>>> Le vendredi 11 mars 2022, 23:03:30 UTC−6, Forest Simmons < >>>>>>> forest.simmons21@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 >>>>>>> >>>>>> ---- >>> Election-Methods mailing list - see https://electorama.com/em for list >>> info >>> >>

- 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.

- 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

El mié., 27 de jul. de 2022 1:52 p. m., Ted Stern <dodecatheon@gmail.com> escribió: > Hi Forest, > > We discussed your proposed method back in 2004. It's basically the same as > Definitive Majority Choice, using score instead of Approval. > Actually, DMC sorts from the favorable to the unfavorable, whereas SPE sorts from the unfavorable to the favorable, like your top three proposal, only SPE starts from the very bottom of the score list instead of from the third one down. > > DMC(score) has a simple basic procedure, as you note, which is identical > to Smith//Score in the 3 candidate case, but is less satisfying for larger > cycles. Which is why I prefer Score Sorted Margins in general. > > Back at the start of the thread, I mentioned that I like STAR except for > the clone problem. The basic idea is that the score winner is good in > general, but a coarse discrete scale lacks resolution, so the runoff > increases satisfaction. > > STAR3 is a minor modification that still makes the case for score, but > reduces the incentive for crowding strategy. And by restricting the number > of pairwise comparisons, it remains human countable and summable on the > second round (the pairwise part). > > On Tue, Jul 26, 2022, 17:30 Forest Simmons <forest.simmons21@gmail.com> > wrote: > >> Might as well go all the way: while there are two or more uneliminated >> candidates, from among these remaining candidates eliminate the pairwise >> loser between the two with the lowest (remaining) scores. >> >> The entire finish order (if desired) can be obtained by bubble sorting >> the score order until no adjacent candidates are out-of-order pairwise. >> Rectification priority goes to the out-of-order adjacent pair whose >> pairwise loser has the lowest score. >> >> In fact, this method is identical to Ranked Pairs with lowest losing >> score corresponding to greatest defeat strength. >> >> The resulting full clone independence is well worth the tiny tweak of >> always starting at the bottom score candidate rather than the strangely >> arbitrary rule of starting at the third candidate from the top. >> >> El lun., 25 de jul. de 2022 9:28 p. m., Ted Stern <dodecatheon@gmail.com> >> escribió: >> >>> I was pointed to this voting simulator developed by Kyle Brockman that >>> contains a version of STAR3: >>> >>> https://www.chocolatepi.net/voteapp/ >>> >>> In this model, STAR3 picks the top three scoring candidates, then the >>> winner is chosen by pairing S1 against the winner of S2 vs S3. Which is the >>> same as the Smith//Score winner for three candidates. >>> >>> This method has most of the features I look for: >>> >>> * More resistance to clones. Obviously, a crowding attack could still >>> work, but it's much less likely. >>> >>> * Some resistance to chicken dilemma. Not perfect, but enough to make >>> the strategy unpredictable. >>> >>> * Bias toward score winner. >>> >>> * Simplicity. As easy to describe as Vote321. >>> >>> On Wed, Jun 15, 2022, 12:56 Ted Stern <dodecatheon@gmail.com> wrote: >>> >>>> For my final (?) cloneproof STAR proposal, I think the virtual three >>>> candidate primary needs modification. >>>> >>>> If the first two candidates are the top two score winners, which third >>>> candidate would maximize equity in the >>>> >>>> That candidate can be found by, for each ballot, subtracting >>>> (ballot_score(score winner) + ballot_score(score runner up)/maxscore from >>>> the ballot's voting strength [or reducing voting strength to zero if that >>>> quantity is greater than 1], then finding the score winner over all >>>> reweighted ballots. >>>> >>>> >>>> This approach also minimizes pushover incentive. >>>> >>>> The goal here is to have at least 2 distinct factions represented in >>>> the extreme case where the top two scoresum candidates are clones. >>>> >>>> To test this, does anyone have any non-Smith STAR examples? >>>> >>>> On Tue, May 31, 2022, 11:29 Ted Stern <dodecatheon@gmail.com> wrote: >>>> >>>>> 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@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@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@yahoo.fr> escribió: >>>>>>> >>>>>>>> Hi Forest, >>>>>>>> >>>>>>>> Le vendredi 11 mars 2022, 23:03:30 UTC−6, Forest Simmons < >>>>>>>> forest.simmons21@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 >>>>>>>> >>>>>>> ---- >>>> Election-Methods mailing list - see https://electorama.com/em for list >>>> info >>>> >>>

- 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.

- 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

Read the page again. https://electowiki.org/wiki/Definite_Majority_Choice On Wed, Jul 27, 2022, 14:40 Forest Simmons <forest.simmons21@gmail.com> wrote: > > > El mié., 27 de jul. de 2022 1:52 p. m., Ted Stern <dodecatheon@gmail.com> > escribió: > >> Hi Forest, >> >> We discussed your proposed method back in 2004. It's basically the same >> as Definitive Majority Choice, using score instead of Approval. >> > > Actually, DMC sorts from the favorable to the unfavorable, whereas SPE > sorts from the unfavorable to the favorable, like your top three proposal, > only SPE starts from the very bottom of the score list instead of from the > third one down. > >> >> DMC(score) has a simple basic procedure, as you note, which is identical >> to Smith//Score in the 3 candidate case, but is less satisfying for larger >> cycles. Which is why I prefer Score Sorted Margins in general. >> >> Back at the start of the thread, I mentioned that I like STAR except for >> the clone problem. The basic idea is that the score winner is good in >> general, but a coarse discrete scale lacks resolution, so the runoff >> increases satisfaction. >> >> STAR3 is a minor modification that still makes the case for score, but >> reduces the incentive for crowding strategy. And by restricting the number >> of pairwise comparisons, it remains human countable and summable on the >> second round (the pairwise part). >> >> On Tue, Jul 26, 2022, 17:30 Forest Simmons <forest.simmons21@gmail.com> >> wrote: >> >>> Might as well go all the way: while there are two or more uneliminated >>> candidates, from among these remaining candidates eliminate the pairwise >>> loser between the two with the lowest (remaining) scores. >>> >>> The entire finish order (if desired) can be obtained by bubble sorting >>> the score order until no adjacent candidates are out-of-order pairwise. >>> Rectification priority goes to the out-of-order adjacent pair whose >>> pairwise loser has the lowest score. >>> >>> In fact, this method is identical to Ranked Pairs with lowest losing >>> score corresponding to greatest defeat strength. >>> >>> The resulting full clone independence is well worth the tiny tweak of >>> always starting at the bottom score candidate rather than the strangely >>> arbitrary rule of starting at the third candidate from the top. >>> >>> El lun., 25 de jul. de 2022 9:28 p. m., Ted Stern <dodecatheon@gmail.com> >>> escribió: >>> >>>> I was pointed to this voting simulator developed by Kyle Brockman that >>>> contains a version of STAR3: >>>> >>>> https://www.chocolatepi.net/voteapp/ >>>> >>>> In this model, STAR3 picks the top three scoring candidates, then the >>>> winner is chosen by pairing S1 against the winner of S2 vs S3. Which is the >>>> same as the Smith//Score winner for three candidates. >>>> >>>> This method has most of the features I look for: >>>> >>>> * More resistance to clones. Obviously, a crowding attack could still >>>> work, but it's much less likely. >>>> >>>> * Some resistance to chicken dilemma. Not perfect, but enough to make >>>> the strategy unpredictable. >>>> >>>> * Bias toward score winner. >>>> >>>> * Simplicity. As easy to describe as Vote321. >>>> >>>> On Wed, Jun 15, 2022, 12:56 Ted Stern <dodecatheon@gmail.com> wrote: >>>> >>>>> For my final (?) cloneproof STAR proposal, I think the virtual three >>>>> candidate primary needs modification. >>>>> >>>>> If the first two candidates are the top two score winners, which third >>>>> candidate would maximize equity in the >>>>> >>>>> That candidate can be found by, for each ballot, subtracting >>>>> (ballot_score(score winner) + ballot_score(score runner up)/maxscore from >>>>> the ballot's voting strength [or reducing voting strength to zero if that >>>>> quantity is greater than 1], then finding the score winner over all >>>>> reweighted ballots. >>>>> >>>>> >>>>> This approach also minimizes pushover incentive. >>>>> >>>>> The goal here is to have at least 2 distinct factions represented in >>>>> the extreme case where the top two scoresum candidates are clones. >>>>> >>>>> To test this, does anyone have any non-Smith STAR examples? >>>>> >>>>> On Tue, May 31, 2022, 11:29 Ted Stern <dodecatheon@gmail.com> wrote: >>>>> >>>>>> 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@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@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@yahoo.fr> escribió: >>>>>>>> >>>>>>>>> Hi Forest, >>>>>>>>> >>>>>>>>> Le vendredi 11 mars 2022, 23:03:30 UTC−6, Forest Simmons < >>>>>>>>> forest.simmons21@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 >>>>>>>>> >>>>>>>> ---- >>>>> Election-Methods mailing list - see https://electorama.com/em for >>>>> list info >>>>> >>>>

- 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.

- 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

That's the Benham formulation of DMC where (unlike in SPE) the eliminations are not done pairwise: *While no undefeated candidates exist, eliminate the least-approved candidate.* Another formularon of DMC is simply ... elect the least approved candidate that defeats every candidate with greater approval. Another equivalente formulation of DMC is .. elect the Ranked Pairs Winner where defeat strength is measured by winning approval, i.e. how close the winner of the pairwise defeat is to the top of the approval list. Compare that with the Ranked Pairs formulation of Sequential Pairwise Elimination: elect the Ranked Pairs winner where defeat strength is gauged by how close the defeated loser of the pairwise contest is to the bottom of the approval list. Compare that with ASM: Bubble sort the approval list by rectifying the out-of-order adjacent pairs giving priority to pairs with smallest approval margin between loser and winner of the pairwise defeat. Corresponding formulation of the DMC finish order: Bubble sort the approval list by rectifying the out-of-order adjacent pairs giving priority to pairs nearest the high approval end of the list. Corresponding formulation of the SPE finish order: Bubble sort the approval list by rectifying the out-of-order adjacent pairs ...while giving priority to pairs nearest the low approval end of the list. It's easy to get confused because of the similarities :-) -Forest El mié., 27 de jul. de 2022 7:57 p. m., Ted Stern <dodecatheon@gmail.com> escribió: > Read the page again. > > https://electowiki.org/wiki/Definite_Majority_Choice > > On Wed, Jul 27, 2022, 14:40 Forest Simmons <forest.simmons21@gmail.com> > wrote: > >> >> >> El mié., 27 de jul. de 2022 1:52 p. m., Ted Stern <dodecatheon@gmail.com> >> escribió: >> >>> Hi Forest, >>> >>> We discussed your proposed method back in 2004. It's basically the same >>> as Definitive Majority Choice, using score instead of Approval. >>> >> >> Actually, DMC sorts from the favorable to the unfavorable, whereas SPE >> sorts from the unfavorable to the favorable, like your top three proposal, >> only SPE starts from the very bottom of the score list instead of from the >> third one down. >> >>> >>> DMC(score) has a simple basic procedure, as you note, which is identical >>> to Smith//Score in the 3 candidate case, but is less satisfying for larger >>> cycles. Which is why I prefer Score Sorted Margins in general. >>> >>> Back at the start of the thread, I mentioned that I like STAR except for >>> the clone problem. The basic idea is that the score winner is good in >>> general, but a coarse discrete scale lacks resolution, so the runoff >>> increases satisfaction. >>> >>> STAR3 is a minor modification that still makes the case for score, but >>> reduces the incentive for crowding strategy. And by restricting the number >>> of pairwise comparisons, it remains human countable and summable on the >>> second round (the pairwise part). >>> >>> On Tue, Jul 26, 2022, 17:30 Forest Simmons <forest.simmons21@gmail.com> >>> wrote: >>> >>>> Might as well go all the way: while there are two or more uneliminated >>>> candidates, from among these remaining candidates eliminate the pairwise >>>> loser between the two with the lowest (remaining) scores. >>>> >>>> The entire finish order (if desired) can be obtained by bubble sorting >>>> the score order until no adjacent candidates are out-of-order pairwise. >>>> Rectification priority goes to the out-of-order adjacent pair whose >>>> pairwise loser has the lowest score. >>>> >>>> In fact, this method is identical to Ranked Pairs with lowest losing >>>> score corresponding to greatest defeat strength. >>>> >>>> The resulting full clone independence is well worth the tiny tweak of >>>> always starting at the bottom score candidate rather than the strangely >>>> arbitrary rule of starting at the third candidate from the top. >>>> >>>> El lun., 25 de jul. de 2022 9:28 p. m., Ted Stern < >>>> dodecatheon@gmail.com> escribió: >>>> >>>>> I was pointed to this voting simulator developed by Kyle Brockman that >>>>> contains a version of STAR3: >>>>> >>>>> https://www.chocolatepi.net/voteapp/ >>>>> >>>>> In this model, STAR3 picks the top three scoring candidates, then the >>>>> winner is chosen by pairing S1 against the winner of S2 vs S3. Which is the >>>>> same as the Smith//Score winner for three candidates. >>>>> >>>>> This method has most of the features I look for: >>>>> >>>>> * More resistance to clones. Obviously, a crowding attack could still >>>>> work, but it's much less likely. >>>>> >>>>> * Some resistance to chicken dilemma. Not perfect, but enough to make >>>>> the strategy unpredictable. >>>>> >>>>> * Bias toward score winner. >>>>> >>>>> * Simplicity. As easy to describe as Vote321. >>>>> >>>>> On Wed, Jun 15, 2022, 12:56 Ted Stern <dodecatheon@gmail.com> wrote: >>>>> >>>>>> For my final (?) cloneproof STAR proposal, I think the virtual three >>>>>> candidate primary needs modification. >>>>>> >>>>>> If the first two candidates are the top two score winners, which >>>>>> third candidate would maximize equity in the >>>>>> >>>>>> That candidate can be found by, for each ballot, subtracting >>>>>> (ballot_score(score winner) + ballot_score(score runner up)/maxscore from >>>>>> the ballot's voting strength [or reducing voting strength to zero if that >>>>>> quantity is greater than 1], then finding the score winner over all >>>>>> reweighted ballots. >>>>>> >>>>>> >>>>>> This approach also minimizes pushover incentive. >>>>>> >>>>>> The goal here is to have at least 2 distinct factions represented in >>>>>> the extreme case where the top two scoresum candidates are clones. >>>>>> >>>>>> To test this, does anyone have any non-Smith STAR examples? >>>>>> >>>>>> On Tue, May 31, 2022, 11:29 Ted Stern <dodecatheon@gmail.com> wrote: >>>>>> >>>>>>> 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@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@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@yahoo.fr> escribió: >>>>>>>>> >>>>>>>>>> Hi Forest, >>>>>>>>>> >>>>>>>>>> Le vendredi 11 mars 2022, 23:03:30 UTC−6, Forest Simmons < >>>>>>>>>> forest.simmons21@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 >>>>>>>>>> >>>>>>>>> ---- >>>>>> Election-Methods mailing list - see https://electorama.com/em for >>>>>> list info >>>>>> >>>>>