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[[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).]]
[[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, 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.
 
== Description ==
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# Math: The final number of votes for the first candidate against the second candidate in each head-to-head matchup is then found by, if treating the second value as a positive number, subtracting the second value for the first candidate from the first value (addition is done instead of subtraction if the second value is treated as a negative number).
#* This can be succinctly summarized as, for finding the vote totals in a matchup between candidates A and B:<blockquote>A>=B*B = A - A<B</blockquote>* meaning that the number of ballots ranking A over or (explicitly) equal to B, is equal to the number of ballots ranking A, minus the number of ballots ranking A below B (i.e. B over A).
 
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>
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The negative counting approach, depending on implementation, can require even more markings when equal-ranking is allowed and it is desired to have traditional pairwise vote totals. Any implementation of negative counting will give accurate information about who won, tied, or lost each matchup (i.e. the pairwise [[margins]]), however. This is because if there are 2 candidates A and B, with the (explicit) votes being: <blockquote>2 A>B
 
1 B>A
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===== Equal-ranked candidates don't get votes =====
<blockquote>For each matchup, the following number of markings can be made for two candidates A and B:
 
* 2 markings can be made (1 negative vote for A>B and 1 for B>A).
**Note: In essence, this approach involves counting the number of voters who explicitly ranked a given candidate '''below''' '''<big>or equal</big>''' to the other candidate, rather than only '''below'''.
* 1 negative marking can be made for the A vs B matchup in general, which is later interpreted as a negative vote for both candidates.
</blockquote>In this example, there are 0.5*(9*8)=0.5*72='''36''' matchups to count between equally-ranked candidates. Accordingly, at least either 36*2='''72''' or '''36''' markings must be made.
 
In this example, there are 0.5*(9*8)=0.5*72='''36''' matchups to count between equally-ranked candidates. Accordingly, at least either 36*2='''72''' or '''36''' markings must be made.
 
=== Dealing with last-place candidates ===
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=== Election example comparisons ===
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 margin of error 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. When equal-ranking is allowed, depending on implementation, the following numbers may be an upper bound on number of markings to be made in either approach:
{| class="wikitable"
|+Number of marks required in each vote-counting approach ''when equal-ranking isn't allowed''
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|}
<nowiki>*</nowiki> The way in which last-ranked candidates are counted can change how many marks need to be made; 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 all candidates except the last-ranked candidate(s).
 
It can also be useful to compare these results to the amount of vote-counting work that would be done in other voting methods.
 
* For [[FPTP]], one mark is made per ballot.
 
* With [[Score voting]], the number of candidates that a ballot gives an above-minimum score to is the number of marks that are made for that ballot.
** Note that this will generally be strictly greater than with [[Approval voting]], because a voter is likely to score more candidates than they would approve; the general trend is that the more gradations/allowed scores there are, the more likely a voter is to score more candidates.
** 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.
 
==== Burlington 2009 mayoral election ====
(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.
 
Almost no voters used equal-ranking (i.e. because it wasn't allowed), so it will be ignored for this analysis. There were roughly '''8,980''' non-equal-ranking votes.
 
There were 5 on-ballot candidates in the election (6 if all write-ins are counted as one "supercandidate"). Here is the number of voters who ranked a given number of candidates:
 
* Ranked 1 candidate: 1481
* Ranked 2 candidates: 1912
* 3: 1799
* 4: 873
* 5: 2944
 
''Number of Ballots Ranking This Many Candidates:''
{| class="wikitable"
!1
!2
!3
!4
!5
|-
|1481
|1912
* 3: |1799
|873
* 5: |2944
|}
Based on the above table:
 
* The negative counting approach would require at least '''56,181''' marks for any implementation.
** 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 regular approach would require at least '''73,669''' marks.
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===== Non-pairwise methods =====
Also of interest may be the comparison to non-pairwise counting methods. The following are rough estimates.
 
*[[FPTP]]: ~8,980 marks
* Approval voting: ~''20,000'' marks (one possible upper bound on number of approvals is ~20,000 (i.e. each voter approves an average of ~2 candidates), with each approval requiring one mark<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 (there would have been more marks required than with Approval, because marks would have to be made for candidates that voters gave only partial approval (intermediate scores) to.)
 
==== 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:''
{| class="wikitable"
!1
!2
!3
!4
!5
!6
!7
!8
!9
!10
!11
|-
|48
|35
|22
|39
|25
|16
|10
|6
|4
|4
|40
|}
 
* Negative counting approach requires at least '''4482''' marks.
** Calculation: (48 + 3*35 + 6*22 + 10*39 + 15*25 + 21*16 + 28*10 + 36*6 + 45*4 + 55*4 + 55*40)
* Regular approach requires at least '''8223''' marks.
** Calculation: (10*48 + 19*35 + 27*22 + 34*39 + 40*25 + 45*16 + 49*10 + 52*6 + 54*4 + 55*4 + 55*40)
 
== Connection to cardinal methods ==
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== Notes ==
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 might indicate no preference between them instead.
 
=== Dealing with truncation ===
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