User:BetterVotingAdvocacy/Negative vote-counting approach for pairwise counting: Difference between revisions
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[[File:Adding ballot matrices in negative pairwise counting approach.png|thumb|1088x1088px|[[File:Pairwise counting negative counting with ranked ballot GIF.gif|thumb|454x454px|GIF for negative counting. Click on the image and then the thumbnail of the image to see the animation.]]]]
The negative counting approach is an alternative method of doing [[pairwise counting]]. It is faster, depending on implementation, when voters don't rank all of the candidates (or when they rank multiple candidates last), because it takes advantage of the fact that in most ranked voting elections, voters are assumed to prefer every candidate they ranked over every candidate they left unranked. Rather than considering a voter to only prefer their 2nd choice candidate over most of the candidates, it treats the 2nd choice as being only '''not''' preferred over the voter's 1st choice candidate(s).
A simple way of describing it is that, if you know that 5 voters ranked a candidate (A), and 3 of them ranked A below some other candidate (B), then 2 voters must have ranked A above or equal to B.
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The number of ballots marking each candidate can be placed in the blank cell comparing themselves to themselves in the pairwise matrix i.e. for candidate A, the cell A>A would contain the number of voters ranking A.<ref>{{Cite web|url=https://www.reddit.com/r/EndFPTP/comments/fsa4np/possible_solution_to_the_condorcet_writein_problem/|title=Possible solution to the Condorcet write-in problem|last=|first=|date=|website=|url-status=live|archive-url=|archive-date=|access-date=}}</ref>
'''<big>Note</big>''': When using this approach, there is an important caveat to consider when dealing with voters who explicitly rank two candidates equally which can alter the vote totals; see the [[#Dealing with equal-ranking]] section below.
Negative vote-counting can be used to yield accurate vote totals for pairwise matchups involving regular candidates versus write-in candidates, but this comes at the cost of greater complexity in terms of how last-ranked candidates are counted.
=== Example ===
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** It isn't possible to know for certain how voters would score the candidates when provided only ranked data. However, it is possible to guess; for example, if there were certain candidates known to be frontrunners, then strategic voters would likely do [[min-max voting]] among those frontrunners, and adjust their scores for candidates they preferred more or less than their preferred frontrunner(s) accordingly. In addition, it can be assumed some voters would do [[normalization]].
* With [[RCV]], there is some additional work involved in transporting the ballots to a centralized location, and in doing multiple rounds of counting, rather than one; ignoring that, however, at least one mark must be made per ballot (to indicate 1st choices), and then in each sequential round where votes are transferred, the number of transferred votes is the number of additional marks that must be made in that round.
Note that regular pairwise counting can have its required number of marks reduced, without using any negative numbers, by counting 1st choices separately from all other ranks; see the section above [[Pairwise counting#Uses for first choice information]]. The regular approach requires [(number of candidates)-2] less marks if using this modification i.e. a voter who ranks 2 candidates sequentially when there are 10 candidates only requires ''1''+8=9 marks rather than ''9''+8=17 marks to have their ballot counted,
* The 1st choice information allows one to determine the [[FPTP]] winner (so long as no voters equally ranked any candidates 1st), and the [[IRV]] winner in cases where some candidate is the [[Condorcet winner]] and has over 1/3rd of 1st choices (see [[Dominant mutual third]]).
* 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.
=== Comparison to the regular pairwise counting approach ===
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***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.
Here are some examples for the first numbers in each series (with the upper bound bolded):
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|+Number of marks required in each vote-counting approach ''when equal-ranking isn't allowed'' ("N" refers to total number of non-write-in candidates in election)
!Number of candidates ranked ("R")
! colspan="2" |Regular approach
''(2nd column shows 1st''
''choices counted separately)''
!Negative counting
!Semi-negative counting
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|1
|N-1 [0, '''1+''']
|1 [0, '''1''']
|1
|1
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|2
|2N-3 [1, '''3+''']
|N-1 [1, '''2+''']
|[1, '''3''']
|[1, 2, '''3''']
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|3
|3N-6 [3, '''6+''']
|2N-4 [2, '''4+''']
|[3, '''6''']
|[2, 4, 5, '''6]'''
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|4
|4N-10 [6, '''10+''']
|3N-8 [4, '''7+''']
|[6, '''10''']
|[4, 6, 8, 9, '''10''']
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|5
|5N-15 [10, '''15+''']
|4N-13 [7, '''11+''']
|[10, '''15''']
|[6, 9, 11, 13, 14, '''15''']
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Notes:
* For both regular counting approaches, a lower bound has been provided (when N=R), and since there is no upper bound, instead the number of marks required when N is one higher than R is provided.
* 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.
** 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.
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* Usually, this would require manually marking each of those positive preferences, resulting in 9 marks to show A being preferred to all other candidates, and 8 marks to show B preferred to all candidates except A, for a total of 17 marks.
* 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.
▲==== Regular pairwise counting but done by counting first choices separately ====
▲Note that regular pairwise counting can have its required number of marks reduced, without using any negative numbers, by counting 1st choices separately from all other ranks; see the section above [[Pairwise counting#Uses for first choice information]]. The regular approach requires [(number of candidates)-2] less marks if using this modification i.e. a voter who ranks 2 candidates sequentially when there are 10 candidates only requires 1+8=9 marks rather than 9+8=17 marks to have their ballot counted, an [(10)-2=] 8-mark difference.
▲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.
=== Election example comparisons ===
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Based on the above table:
* The negative counting approach would require at least '''56,181''' marks for any implementation, for an average of 6.25 marks per ballot.
** This assumes last-ranked candidates weren't counted with any marks (which means write-in candidates' pairwise matchups wouldn't have been counted accurately).
** The calculation is (1*1481 + 3*1912 + 6*1799 + 10*873 + 10*2944).
*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).
* The regular approach would require at least '''73,669''' marks (8.2 marks/ballot).
** 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)
** This is if miscounting write-ins' matchups, which is the usual way to treat them.
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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>)
*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>)
* Score voting: ~''20,000'' to ~''30,000'' marks (2.22 to 3.34 marks/ballot)
==== Evanston, IL 2020 Democrat endorsement ====
(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.
There were '''249''' ballots.
''Number of Ballots Ranking This Many Candidates:''
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|}
* Negative counting approach requires at least '''4,482''' marks (18 marks/ballot).
** Calculation: (48 + 3*35 + 6*22 + 10*39 + 15*25 + 21*16 + 28*10 + 36*6 + 45*4 + 55*4 + 55*40)
* Semi-negative approach requires at least '''3,208''' marks (12.88 marks/ballot).
**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).
** If 1st choices are counted separately, then this requires '''5,982''' marks (24.02 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)
** Calculation: (10*48 + 19*35 + 27*22 + 34*39 + 40*25 + 45*16 + 49*10 + 52*6 + 54*4 + 55*4 + 55*40)
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===== Non-pairwise methods =====
*[[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>)
== Using negative pairwise counting while collecting additional non-pairwise information ==
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