User:BetterVotingAdvocacy/Negative vote-counting approach for pairwise counting

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).

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

There are two steps to the negative vote-counting approach: the vote-counters must capture certain pieces of information, and then some math is done on this information to find the final result.

1. Vote-counting: The precinct vote-counters count the following values for a given candidate:
   1. The number of voters who ranked/rated/marked a candidate on their ballot.
   2. In each head-to-head matchup, the number of voters who explicitly ranked that candidate below the other candidate ("explicitly" meaning they actually marked/ranked that candidate).
      • (This can be considered as, for a given ballot that ranks two candidates A and B as B>A, either:
        • Counting the number of votes for A<B. This yields a positive value.
        • Adding a negative vote (-1 votes) for A>B. This yields a negative value.)

1. 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:

     A>=*B = A - A<B

     * 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.

Note: When using this approach, there is an important caveat to consider when dealing with voters who explicitly rank two candidates equally; see the #Dealing with equal-ranking section below.

Example

If the votes are:

10: A>B 5: B

then each candidate is explicitly marked on:

• A: 10 ballots
• B: 15 ballots

with the number of ballots explicitly ranking one candidate below the other being:

• A<B: 0 ballots
• B<A: 10 ballots

The math that shows the final number of votes in favor of each candidate is then:

• A>B is (10-0)=10 votes.
• B>A is (15-10)=5 votes.

This can be demonstrated as:

	A	B
A	10 ballots	0 votes
B	-10 votes	15 ballots

becoming:

	A	B
A	---	10 (votes)
B	5	---

Dealing with equal-ranking

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:

2 A>B

1 B>A

5 A=B

then either it can be marked that A wins against B by:

• 2 votes to 1
• 7 votes to 6

This is because the voters who equally ranked A and B can be considered to, in the A vs B matchup, either be voting for:

• Neither of them (similar to choose-one FPTP voting; this is the traditional pairwise counting approach).
• Both of them (similar to Approval voting).

This is related to how, in Approval voting, if A has 30 approvals and B 20, and no other information is supplied, then it is impossible to know whether any of the 20 voters who approved B also approved A or not.

This can change who wins in certain Category:Pairwise counting-based voting methods; for example, in the winning-votes version of a Category:Defeat-dropping Condorcet method, not only does it matter who wins the matchup, but also exactly how many voters genuinely preferred the winner to the loser in each matchup.

Note: Voters who don't rank either of the candidates in a matchup are generally considered to equally rank them, but no implementation of negative counting would consider them to vote for both candidates in the matchup. Only voters who have marked both candidates can be counted that way.

Example of the two approaches to equal-ranking

Suppose a voter had ranked 9 of 10 candidates as their 1st choices, and the 10th candidate was unranked (i.e. implicitly ranked last).

At least 9 marks must be made in any approach to negative counting, to indicate that 1 voter has ranked each of the 9 candidates who are each one of the voter's 1st choice.

In addition:

Explicitly equal-ranked candidates both get a vote

No extra work needs to be done.

Equal-ranked candidates don't get votes

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 or equal 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.

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

It is not necessary to make any markings for a candidate a voter ranked as their last choice, because this means they wouldn't vote for that candidate in any matchups.

Write-in candidates

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.

It is perhaps possible to make a special marking for a last-choice candidate that indicates they are not preferred over any of the on-ballot 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.

Comparison to the regular approach

Verbal comparison between the regular approach and negative counting:

• The regular approach: The precinct vote-counters manually count all of the voter's preferences in each head-to-head matchup; in other words, a candidate is assumed to be preferred only in the matchups where the vote-counters mark them as being so.
  • This can be slow, and also can make it difficult to accommodate write-in candidates (see Pairwise counting#Dealing with write-in candidates), since the vote-counters won't know ahead of time who those candidates are, and thus won't be able to indicate preferences in those matchups.
• Negative counting approach: The vote-counters mark a candidate as being ranked on a ballot, assume the voter who ranked them prefers that candidate in every matchup, and then show which matchups this is not true for.

The formula for figuring out the number of marks that must be made in both approaches 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):

• In negative counting, the series starts at 0 when 0 candidates are ranked, and then the formula is that the number of marks that must be made for a given number of candidates that were ranked on a ballot is the value in the series for a ballot that had ranked one less candidate than that given number, plus the number of candidates ranked on the ballot.
  • This is because for each additional candidate added (ranked below all candidates already on a ballot), one mark is made to indicate that they were ranked, and [number of candidates already on ballot] marks are made to indicate the voter's preference for all of those already-ranked candidates over the newly ranked candidate.
• In the regular approach, take whatever number of marks would be produced in negative counting for a given number of ranked candidates on a ballot (see above bullet point), and then subtract it from the number of candidates that are ranked on a ballot multiplied by the number of non-write-in candidates in the election.

Here are some examples for the first numbers in each series:

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	Regular approach	Negative counting
1	N-1	1
2	2N-3	3
3	3N-6	6
4	4N-10	10
5	5N-15 (4N-10)*	15 (10)*

* 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).

Note that negative counting is faster when voters rank only a few of all candidates, and potentially slower otherwise.

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:

• 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.

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. 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 . 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").

Number of Ballots Ranking This Many Candidates:

	1	2	3	4	5
	1481	1912	1799	873	2944
Cumulative:	1481 (16.4%)	3393 (37.6%)	5192 (57.6%)	6065 (67.3%)	9009 (100%)

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.
  • This is if ignoring write-ins, which is the usual way to treat them.
  • Calculation: (4*1481 + 7*1912 + 9*1799 + 10*873 + 10*2944). (This is with number of candidates, N, being 5.)

Non-pairwise methods

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
• RCV: 12,309 marks (8,980 1st choices + 3,329 vote transfers)
• 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))
• Score voting: ~20,000 to ~30,000 marks

Evanston, IL 2020 Democrat endorsement

(Vote totals from ) This was an RCV election with no equal-ranking.

There were 249 ballots.

Number of Ballots Ranking This Many Candidates:

	1	2	3	4	5	6	7	8	9	10	11
	48	35	22	39	25	16	10	6	4	4	40
Cumulative:	48 (19.2%)	83 (33.3%)	105 (42.1%)	144 (57.8%)	169 (67.8%)	185 (74.2%)	195 (78.3%)	201 (80.7%)	205 (82.3%)	209 (83.9%)	249 (100%)

• 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)

Non-pairwise methods

• RCV: 354 marks (249 voters' 1st choices + 105 votes transferred throughout)

Connection to cardinal methods

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).

Using with strength of preference

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).

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

When voters aren't allowed to do truncation (and Write-in candidates 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.

• This works because every candidate will actually be marked on every voter's ballot.

• This would obviously be impractical when voters are allowed to truncate, however, because it could mean negative votes would have to be applied to each and every ballot for a candidate no voters ranked.

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.