User:BetterVotingAdvocacy/Negative vote-counting approach for pairwise counting: Difference between revisions
User:BetterVotingAdvocacy/Negative vote-counting approach for pairwise counting (view source)
Revision as of 22:06, 15 September 2020
, 3 years ago→Regular pairwise counting but done by counting first choices separately
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This advice is less relevant when write-ins are allowed, however, because even if a voter ranks a candidate last among the candidates named on their ballot, they are still implicitly ranking that candidate above all of the write-in candidates they didn't rank on their ballot. So if last-ranked candidates aren't counted, then it may be necessary to modify how the calculation is done, or otherwise mention caveats in the final result, to avoid giving the impression that the vote totals are accurate for the matchups involving write-in candidates (however, the totals will only be off by the number of voters that rank a non-write-in candidate last and don't rank the write-in candidate for a given matchup between such candidates).
It is possible to make a special marking for a last-choice candidate that indicates they are not preferred over any of the on-ballot (regular) candidates, but that they are preferred over all write-in candidates. It would then only be necessary to record negative votes for matchups involving write-in candidates who are ranked above the last-choice candidate on some ballots. This would mean making at most two additional marks for every last-ranked candidate on a ballot, because in practice, in most elections, voters are only allowed to write in at most one candidate.<ref>{{Cite web|url=https://forum.electionscience.org/t/negative-vote-counting-approach-for-pairwise-counting/644|title=Negative vote-counting approach for pairwise counting|date=2020-04-27|website=The Center for Election Science|language=en-US|access-date=2020-09-15}}</ref> This can be compared to the regular approach to dealing with write-ins at [[Pairwise counting#Dealing with write-in candidates]].
== Semi-negative counting procedure ==
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== Comparison to other vote-counting procedures ==
It can also be useful to compare these results to the amount of vote-counting work that would be done in other voting methods. See also [[Summability_criterion#Summability_of_various_voting_methods]].
* For [[FPTP]], one mark is made per ballot.
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* However, when voters equally rank multiple candidates 1st, then whether or not the modification is still applied can make a significant difference in overall number of marks i.e. a voter ranking 2 candidates 1st can either be counted using 1 mark for each candidate, or using [number of candidates - 2] marks for each, as in the regular approach.
** If 1 mark is used per 1st choice candidate, then the matchups involving two 1st choice candidates will have each of them get 1 vote against the other. This is the same dilemma as mentioned in this article for [[#Dealing with equal-ranking]] in negative pairwise counting.
*This is essentially equivalent to doing semi-negative pairwise counting with negative counting only applied to 1st choices and regular pairwise counting applied to all other candidates.
===
Verbal comparison between the regular approach and negative counting:
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The formula is a series (though it is only accurate when equal-ranking isn't allowed; when it is allowed, then depending on implementation, this series may provide an upper bound on the number of marks that are to be made):
*
**
*
**This is because, if the negative counting approach is thought of as "count the number of times a candidate is "preferred to themselves" (i.e. the number of ballots they're ranked on) and the number of ballots that prefer them less than each other candidate", then the regular approach doesn't require counting either of those pieces of information, but does require counting the number of times a candidate is preferred over each of the other candidates
***For example, with 10 candidates, for a ballot with one 1st choice that doesn't rank any of the other candidates, the negative counting approach only requires counting the fact that the 1st choice is "preferred to themselves", whereas the regular approach requires counting that the 1st choice is preferred over each of the other 9 candidates; so there are either 1 or 9 (10-1) marks to be made.
**** If this ballot also ranked a 2nd choice, the negative approach requires indicating that the 2nd choice is also "preferred to themselves", and preferred less than the 1st choice, while the regular approach requires indicating that the 2nd choice is preferred over each of 8 candidates i.e. all of the 10 candidates that aren't the 1st choice or 2nd choice.
**The formulas for regular counting w/ 1st choices counted separately, are derived by subtracting N-2 from the corresponding formula for the regular approach.
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|4N-10 [6, '''10+''']
|3N-8 [4, '''7+''']
| [6, '''10''']
|[4, 6, 8, 9, '''10''']
|-
| 5
|5N-15 [10, '''15+''']
|4N-13 [7, '''11+''']
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Notes:
*
*
**
▲** For counting last-ranked candidates, see [[#Dealing with last-place candidates]]. If no marks are made for them, then a ballot that ranks all candidates requires the same number of marks as a ballot that ranks at least one less candidate i.e. all candidates except the last-ranked candidate(s). For example, a ballot that ranks 5 candidates when there are 5 candidates total can be thought of as ranking the top 4 candidates, and leaving the 5th candidate unranked.
*For both negative and semi-negative counting, the numbers provided are a series that include both lower and upper bounds on the number of marks that have to be made, depending on N (which starts at R, and sequentially increases by one, up to twice that i.e. 2R), and how last-ranked candidates are counted.
**Semi-negative counting's numbers can be derived as such: the voter's N/2 (rounded down to nearest integer) highest-ranked candidates are counted with negative counting's worst-case, and the remaining candidates are counted with regular pairwise counting separately. So if a voter ranks 5 out of 7 candidates, half of the candidates (3.5) rounded down (=3) are counted with negative counting (negative counting for 3 candidates requires at most 6 marks), and the remaining 2 ranked candidates are counted with regular pairwise counting (that means having to do regular pairwise counting for R=2 and N=4, because out of the 7 candidates, you ignore the 3 candidates already counted by negative counting; so plugging that into 2N-3 yields: 2(4)-3= 5 marks). So that is 6+5=11 marks for R=5 and N=7 with semi-negative counting.
***The reasoning for this can be derived as follows: separate the voter's ranked candidates into their N/2 most-preferred candidates (their "top-ranked half") and their N/2 least-preferred candidates (the "bottom-ranked half"); if there are an odd number of candidates, consider the median candidate to be in the bottom-ranked half. Any candidate in the top-ranked half will be ranked above more candidates than they are ranked below, so it will be easier or equally easy to use negative counting to count them. Any candidate in the bottom-ranked half will be in the opposite situation, so regular counting is easier for them.
***This
**Semi-negative counting's performance is always better than or equal to negative counting's performance for the same N and R. As an example for R=2:
***
*** If 3 candidates run, then in negative counting, the 2nd choice is counted with 2 additional marks, so at least ''3'' marks are made; but in semi-negative counting, only 1 additional mark needs to be made (because 2nd choice is ranked above only 1 candidate), so at least '''2''' marks are made.
*** If 4 candidates run, then in both approaches, '''3''' marks are made (2nd choice is ranked above 2 candidates and below 1 candidate).
▲** Note that regular pairwise counting's formula can be derived as follows: when only 1 candidate is ranked, the formula is N-1. When 2 candidates are ranked, it is the formula for when only 1 candidate is ranked plus another formula i.e. (N-1)+(N-2), which becomes 2N-3. For 3 candidates ranked, it is (2N-3)+(N-3), which is 3N-6, etc.
Note that negative counting is faster when voters rank only a few of all candidates, and potentially slower otherwise.
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For example, a voter who votes A>B when there are 10 candidates can be assumed to vote for A and B in every matchup, except they don't prefer B>A:
*
* But negative counting only requires 3 marks: 1 each for A and B to indicate they are preferred in every matchup, and 1 to indicate that this isn't the case for B>A.
===
It is possible to compare the number of marks that must be made in either approach for certain elections, because their full ballot set has been published. Any election using ranked or rated ballots can be used for this purpose. The calculations and numbers used in this section may be slightly off for some examples, though general conclusions (should) still hold; because of this, it is suggested that the reader apply a ~10% margin of error or more when considering how superior one pairwise counting approach is to another.
Note that when equal-ranking isn't allowed, only the number of voters who ranked a certain number of candidates needs to be known.
====
(Vote totals taken from <ref>https://rangevoting.org/JLburl09.txt</ref>. The numbers and calculations are slightly off for this analysis.) This was an [[RCV]] election.
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|-
|
| 1481
|1912
|1799
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Based on the above table:
*
**
**
*The semi-negative approach would require at least '''39,114''' marks (avg = 4.35 marks/ballot).
**Calculation: (1*1481+3*1912+5*1799+6*873+6*2944). This is found by observing that, for example, when a voter ranked 2 candidates, it was most efficient to count their 2nd choice with the negative approach rather than the regular approach, but when they ranked their 3rd choice, it was faster to use the regular approach (i.e. mark that they're ranked above 2 candidates) rather than the negative approach (3 values, because the candidate is ranked below 2 candidates, and a 3rd mark has to be made to show that they're ranked by the voter).
*
** If counting 1st choices separately from all other ranks, then it would require '''46,642''' marks (5.19 marks/ballot).
***Calculation: (1*1481 + 4*1912 + 6*1799 + 7*873 + 7*2944)
**
**
=====
Also of interest may be the comparison to non-pairwise counting methods. The following are rough estimates for the number of marks:
*[[FPTP]]: ~''8,980'' marks (1 mark/ballot)
*[[RCV]]: ''12,309'' marks (1.37 marks/ballot. Derived by counting 8,980 1st choices + 3,329 vote transfers<ref>{{Citation|last=|first=|title=2009 Burlington mayoral election|date=2020-05-05|url=https://en.wikipedia.org/wiki/2009_Burlington_mayoral_election#Results|work=Wikipedia|volume=|pages=|language=en|access-date=2020-05-20}}</ref>)
*Hypothetical:
**Approval voting: ~''20,000'' marks (2.22 marks/ballot. Derived because one possible upper bound on number of approvals is ~20,000 (i.e. each voter approves an average of ~2 candidates)<ref>{{Cite web|url=https://rangevoting.org/Burlington.html|title=RangeVoting.org - Burlington Vermont 2009 IRV mayoral election|last=|first=|date=|website=rangevoting.org|url-status=live|archive-url=|archive-date=|access-date=2020-05-16|quote=We do not know who Range & Approval voting would have elected because we only have rank-order ballot data – depending on how the voters chose their "approval thresholds" or numerical range-vote scores, they could have made any of the Big Three win (also Smith). However it seems likely they would have elected Montroll. Here's an analysis supporting that view: Suppose we assume that voters who ranked exactly one candidate among the big three would have approved him alone; voters who ranked exactly two would have approved both, and voters who ranked all three would have approved the top-two a fraction X of the time (otherwise approve top-one alone). The point of this analysis, suggested by Stephen Unger, is that voters were allowed to vote "A>B," which while mathematically equivalent to "A>B>C" among the three candidates A,B,C, was psychologically different; by "ranking" a candidate versus "leaving him unranked" those voters in some sense were providing an "approval threshhold." Then the total approval counts would be
Montroll=4261+1849X, Kiss=3774+1035X, and Wright=3694+741X.
Note that Montroll is the most-approved (and Wright the least-approved) regardless of the value of X for all X with 0≤X≤1.}}</ref>)
*
====
(Vote totals from <ref>{{Cite web|url=https://www.reddit.com/r/EndFPTP/comments/f3h0hh/analysis_of_the_ballots_for_the_democratic_party/|title=r/EndFPTP - Analysis of the Ballots for the Democratic Party of Evanston, IL Presidential Endorsement Vote|website=reddit|language=en-US|access-date=2020-05-19}}</ref>) This was an RCV election with no equal-ranking.
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|}
*
**
*
**Calculation: (48 + 3*35 + 6*22 + 10*39 + 15*25 + 20*16 + 24*10 + 27*6 + 29*4 + 30*4 + 30*40)
*Regular approach requires at least '''8,223''' marks (33.02 marks/ballot).
**
***Calculation: (1*48 + 10*35 + 18*22 + 25*39 + 31*25 + 36*16 + 40*10 + 43*6 + 45*4 + 46*4 + 46*40)
**
=====
*[[RCV]]: ''354'' marks (average of 1.42 marks/ballot. It is derived by counting 249 voters' 1st choices + 105 votes transferred throughout<ref>https://i0.wp.com/evanstondems.com/wp-content/uploads/2020/02/RCVPrez-Results.png?fit=1024%2C341&ssl=1</ref>)
*Hypothetical:
**Approval voting: ''996'' marks (4 marks/ballot)
**Score voting: ''1245'' marks (5 marks/ballot)
==
It is possible, while collecting information about the number of voters who rank a particular candidate, to simultaneously (without using more marks) count how many voters gave that candidate a particular score or rank. <ref>If you have any confusion on how this works, take a look at [[Summability criterion#Median methods]] and how Bucklin/Majority Judgement can be counted precinct-summably. </ref>
===
In some contexts (i.e. when precinct-summably counting [[STAR voting]]) it may be useful to collect both pairwise counts and scores for candidates. In such circumstances, one can take advantage of the fact that if:
*
**
then this can be recorded by, for Candidate A, tallying not just the number of voters who scored them, but specifically the number of voters who scored them at a specific score (except optionally the minimum score, which is usually 0). So if you know that:
*
*
* 3 voters score A a 3/5
* 0 voters scored A any other score
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* Note that the amount of cardinal information collected here allows you to also compute the results of cardinal methods like [[Majority Judgement]].
*
**
===
For [[Bucklin voting]] or any median-based ranked voting method, instead of counting the number of voters who ranked a candidate, you can count specifically the number of voters who ranked that candidate 1st, 2nd, etc. To tally this, count only the number of voters who ranked each candidate 1st, and if no candidate has a majority, add in the number of voters who ranked them 2nd, and repeat for 3rd, 4th, etc.
====
A special case of counting the number of voters who give a particular ranking to a candidate are IRV and FPTP, and other [[:
* If whole-votes equal ranking or no equal-ranking are used, then no additional complication exists.
*
With IRV, the 1st choice information collected with this approach only guarantees one knows who to eliminate in the first round; additional rounds of counting may be required, which would not involve re-doing any pairwise counts (since the pairwise preferences between two candidates can't change if a third candidate is eliminated).
==
[[File:Connection between negative pairwise counting and other vote-counting procedures.png|thumb|1654x1654px|Also see [[rated pairwise preference ballot]].]]
This approach can be considered an [[Approval voting]]-based or [[cardinal]] approach, because when using the Approval-style approach for equal-rankings (explicit equal-rankings are counted as a vote for both candidates in the matchup), then each voter that votes Approval-style (i.e. explicitly ranks some candidates 1st and leaves all other unranked, which is implicitly treated as ranking them last) will have their ballot counted like an [[Approval ballot]] (i.e. the preference for each approved candidate on the ballot will be counted with one mark per candidate, and no marks will be used to count disapproved candidates). Further, in some implementations, it allows a voter to give support to both candidates in a matchup, just like in cardinal methods.
===
Negative vote-counting can be used to count weak pairwise preferences (i.e. if a voter only wants to give 0.4 votes in a matchup, rather than 1 vote; see [[Rated pairwise preference ballot#Implementations]]) by counting only a "partial ballot" marking a candidate, and partial (i.e. weighted or fractional) negative votes in certain matchups. In other words, it is treated as if only a partial voter or ballot supported a candidate (see [[KP transform]]). Note that this is equivalent to using Score voting in a matchup, if using a scale of 0 to 1 rather than the more familiar 0 to 5 or 0 to 10 scales.
== Notes
[[File:Negative vote-counting approach to pairwise counting.png|thumb|1114x1114px|Negative vote-counting approach for pairwise counting (Note: Regular approach may be better in some use cases; see cited discussions in text to the left).Here is a discussion clarifying some confusion that may arise from seeing the above picture: <ref>{{Cite web|url=https://www.reddit.com/r/EndFPTP/comments/gh7fix/new_pairwise_counting_approach_based_on_approval/|title=r/EndFPTP - New pairwise counting approach based on Approval voting|website=reddit|language=en-US|access-date=2020-09-15}}</ref>]]
In practice, to make pairwise counting easier, voters could be provided with two or fewer ranks than the number of candidates, with equal-ranking being allowed so that voters could do [[preference compression]]. This way, a voter who would usually indicate a preference that would have to be counted between two candidates would have to indicate no preference between them instead.
===
An alternative way to do the negative approach, which is more similar to the regular approach, is to, when candidate B is explicitly ranked below A on a ballot, instead of counting -1 votes for B>A, count 1 vote for A>B, and later on, when the math is done, the number of votes for B>A is the number of ballots ranking B minus the number of votes for A>B. In other words, a part of the regular pairwise counting approach is used, but only in matchups where both candidates are explicitly ranked by the voter (i.e. a voter who voted A>B and left C unranked would have their vote for A>B counted, but not their vote for A>C, because later on it will be inferred that they must have preferred A to C by virtue of having ranked A but not C).
===
[[Approval voting]] can be thought of as a [[Smith-efficient]] [[Condorcet method]] where, when a voter approves a candidate, they are assumed to vote for them in every head-to-head matchup (see [[Self-referential Smith-efficient Condorcet method]]). Further, approving a candidate can be thought of as ranking them 1st, while disapproving a candidate can be thought of as ranking them last. Given that connection, and that in Approval voting, the vote-counting is done by counting the number of ballots that approve/mark a candidate, rather than by harder [[pairwise counting]], it is clear that a similar counting procedure could be applied to pairwise counting itself. The only complexity is that when voters rank candidates, they are allowed to express that they don't prefer certain candidates in certain matchups; thus, negative votes are necessary to communicate that lack of preference for specific matchups.
This has the advantage of, when every voter does [[bullet voting]], being counted exactly like an [[FPTP]] election (one mark per ballot for the candidate it marked), which also shows that FPTP can be thought of as a constrained form of Approval.
===
Here is a more comprehensive picture of how the negative approach works with an example: for a voter who voted A>B=C>D with 2 unranked candidates E and F, their ballot would be processed into this matrix:
{| class="wikitable"
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|-
|B
| -1 vote
|''1 ballot''
|
Line 612 ⟶ 619:
|
|-
| D
|
|
Line 654 ⟶ 661:
|
|-
| B
| -1 vote
|''1 ballot''
Line 670 ⟶ 677:
|
|-
| D
| -1 vote
| -1 vote
| -2 votes
|''2 ballots''
|
|
|-
| E
|
|
| -1 vote
|
|''1 ballot''
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!D
!E
! F
|-
| A
|''2 ballots''
|2 (votes)
Line 710 ⟶ 717:
|1
|1
| 2
|-
|B
Line 720 ⟶ 727:
|1
|-
| C
|1
|2
Line 726 ⟶ 733:
|2
|2
| 2
|-
|D
Line 737 ⟶ 744:
|-
|E
| 1
|1
|0
Line 744 ⟶ 751:
|1
|-
| F
|0▼
|0
|0
▲| 0
|0
|0
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Note: In the B vs C matchup, B has 1 vote and C has 2 votes; yet in reality, 1 of the 2 voters equally ranked B and C (giving 0 votes to either of them), and the other ranked C over B (giving 1 vote to C and 0 to B); this discrepancy can be explained as being because the equal-ranking voter was treated as voting for both B and C, rather than neither of them.
=== Negative counting used for non-pairwise methods
Negative numbers can be used to count Approval voting, and can also be used in the context of most PR methods.
==== Negative vote-counting approach for Score voting ====
* For Approval, when a voter approves more than half of the candidates, they can be considered to disapprove fewer than half the candidates. That is, a ballot approving (A, B, and C), but not D, can be counted as a ballot that approves everyone except D (i.e. gives everyone a positive vote and additionally gives D a negative vote). ▼
** Note that the same trick works for Score, but doesn't always speed the counting up there. For example, a voter who scores a candidate a 3 out of 5 can be thought of as giving them 5 positive points and -2 negative points as well.▼
* For sequential PR methods, some selection algorithm generally picks the a winner in a given round, then reweighting is applied to the votes, and this repeats. Because reweighting generally only alters some votes, it's often possible to treat the votes in one round as being the votes from a previous round, except some candidates have lost some support. [[SPAV]] is one such example. <ref>{{Cite web|url=https://forum.electionscience.org/t/possible-trick-for-counting-spav-and-cardinal-pr-faster/657|title=Possible trick for counting SPAV (and cardinal PR) faster|date=2020-05-26|website=The Center for Election Science|language=en-US|access-date=2020-07-21}}</ref>▼
▲*
=== Dealing with truncation ===▼
▲**
==== Negative vote-counting approach for sequential PR methods ====
▲*
When voters aren't allowed to do [[truncation]] (and [[Write-in candidate|Write-in candidat]]<nowiki/>es aren't allowed), then it can be useful to skip the part of the negative counting procedure where the vote-counters mark how many ballots a candidate is ranked on, and instead assume a candidate is preferred in every matchup on every voter's ballot. The negative votes would then be applied as usual.
*
*
=== Independence of unranked candidates
The negative approach doesn't require additional marks to be made for a given ballot when candidates are added to the election that that ballot doesn't vote for. For example, at most 3 marks need be made for a voter whose ballot is A>B, regardless of whether there are 2 candidates in the election or 100.
==
<references />
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