User:BetterVotingAdvocacy/Negative vote-counting approach for pairwise counting
The negative counting approach is an alternative method of doing pairwise counting. It is faster (i.e. requires less marks and tallying), depending on implementation, when voters rank multiple candidates last. Rather than counting a voter's preference for a candidate they ranked (i.e. over lower-ranked candidates), it counts that voter's lack of preference for that candidate (i.e. over the candidate themselves and higher-ranked candidates). In other words, negative pairwise counting a) treats a voter as having "approved" all of the candidates they ranked above last-place, b) counts the voter's pairwise preferences only among the candidates they "approved", and c) aggregates those two pieces of information for all voters (such that for each candidate, you have the total number of voters who "approved" them along with the relevant pairwise preferences), and then calculates the final pairwise totals using the aggregated information.
An example of negative pairwise counting: if 5 voters ranked a candidate (A), and 3 of them didn't rank A above some other candidate (B), then 2 voters must have ranked A above or equal to B. When a voter only ranks candidates as their 1st choice(s) or last choices (i.e. uses equal-ranking with only two ranks), then negative pairwise counting becomes essentially equivalent to Approval voting's vote-counting procedure for that voter's ballot.
Semi-negative pairwise counting, which is theoretically even faster, is based on using both of the regular and negative pairwise counting techniques. There are also vote-counting techniques based on similar principles that can be used in non-pairwise contexts, such as for Score voting and various PR methods.
Description[edit | edit source]
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.
- Vote-counting: The precinct vote-counters count the following values for a given candidate:
- The number of voters who ranked/rated/marked a candidate on their ballot.
- 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.)
- (This can be considered as, for a given ballot that ranks two candidates A and B as B>A, either:
- 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:
* 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).A>=*B = A - A<B
- This can be succinctly summarized as, for finding the vote totals in a matchup between candidates A and B:
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.^{[1]}
Note: 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; see #Dealing with last-ranked candidates.
Example[edit | edit source]
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 | 0 votes |
B | -10 votes | -15 |
becoming:
A | B | |
---|---|---|
A | --- | 10 (votes) |
B | 5 | --- |
Dealing with equal-ranking[edit | edit source]
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).
- To get this value, one must consider a voter to "prefer A and B simultaneously" in the matchup between the two. It may be easier to think of this as the voter "supporting" or "liking" both, rather than preferring them; this is the type of support measured by Score voting and the rated pairwise preference ballot in 2-candidate elections (with no write-ins), real or simulated.
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[edit | edit source]
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[edit | edit source]
No extra work needs to be done.
Equal-ranked candidates don't get votes[edit | edit source]
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. (Note that this may require more values to be recorded, since not only are values recorded for the number of votes on both sides of a matchup, but also for each individual matchup)
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.
Semi-negative counting procedure[edit | edit source]
When a voter ranks more than half of the candidates, it is possible to further reduce the number of marks counted. This is because when a voter ranks one candidate below more than half of the candidates, instead of counting negative marks in that candidate's matchups against all of those candidates, it would be faster to count positive (regular) marks for that candidate against the fewer than the half of the candidates that they're ranked above. Example of this approach with a voter ranking all of 5 candidates:
Number of marks made
(smaller number = better) |
Voter's ranking | A | B | C | D | E | |
---|---|---|---|---|---|---|---|
1 vs 4 | 1st (choice) | A | 1 ballot | 1 | 1 | 1 | 1 |
2 vs 3 | 2nd | B | -1 | 1 ballot | 1 | 1 | 1 |
3 vs 2 | 3rd | C | -1 | -1 | 1 ballot | 1 | 1 |
4 vs 1 | 4th | D | -1 | -1 | -1 | 1 ballot | 1 |
5 vs 0 | 5th | E | -1 | -1 | -1 | -1 | 1 ballot |
Counting work for this example:
- With regular counting, this ballot would require at least 10 marks to count.
- The same applies to negative counting when last-ranked candidates aren't counted.
- With semi-negative counting, it only required 6 marks to count.
With semi-negative counting, it is possible to reduce the number of marks required to count a ballot by roughly half or greater, depending on which of the other two approaches it is compared to, and how many of the candidates are ranked by the voter. The positive and negative marks can either be stored as separate values, or they can be combined during the counting (i.e. 5 positive votes for A>B and 3 negative votes for A>B can be stored as 2 votes for A>B. This means incrementing and decrementing the same number during the count, so it could be confusing).
Dealing with last-place candidates[edit | edit source]
In negative pairwise counting, 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. One way of thinking about this is that if the voter hadn't ranked that candidate, then no marks would've been made to count them, and in effect the voter would still be treated as ranking them last. This is equivalent to doing semi-negative pairwise counting with regular pairwise counting only applied to the last rank.
This can save a great deal of work in some cases; if a voter ranks all of 10 candidates, and ranks 5 of them last (i.e. they are all ranked 6th), then 5*(5+1)=30 fewer marks need be made if not counting them (explanation for the math: there are 5 such candidates, and they are each ranked under 5 candidates, with 1 additional mark made to record that they were ranked by the voter). In fact, if not using this trick, negative counting can actually require more marks than regular counting in some cases.
Write-in candidates[edit | edit source]
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.^{[2]} This can be compared to the regular approach to dealing with write-ins at Pairwise counting#Dealing with write-in candidates.
Regular pairwise counting but done by counting first choices separately[edit | edit source]
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. 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. 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, a [(10)-2=] 8-mark difference.
- 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 other vote-counting procedures[edit | edit source]
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.
- 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.
Comparison to the regular pairwise counting approach[edit | edit source]
All pairwise counting approaches are based on counting the preferences of the voters for the candidates they ranked. 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.
It's actually possible to visualize, in a pairwise counting matrix, what values have to be recorded for each approach; this is because the cells of the pairwise matrix used in each approach are disjoint.
Example with a voter voting A>B=C>D with 6 candidates:
A | B | C | D | E | F | |
---|---|---|---|---|---|---|
A | 1 ballot | 1 | 1 | 1 | 1 | 1 |
B | -1 | 1 ballot | (1 equal-ranking) | 1 | 1 | 1 |
C | -1 | (1 equal-ranking) | 1 ballot | 1 | 1 | 1 |
D | -1 | -1 | -1 | 1 ballot | 1 | 1 |
E | ||||||
F |
The negative counting values can clearly be used to reproduce the values created by the regular approach (see the Math part of the #Description), whereas the regular approach leaves each of the cells used by the negative approach blank (i.e. with a value of 0). In either approach, the values in each cell can be added up per ballot; see the #Comprehensive example section.
The "(1 equal-ranking)" part indicates that one voter equally ranked B and C; this value is optional in the negative counting approach (and is never used in the regular approach), but can be included if it is desired to have traditional pairwise totals, by using it to give both B and C negative votes in the B vs C matchup (see the #Dealing with equal-ranking section).
Formula for counting the required number of marks to be made[edit | edit source]
(Note: See table below for definition of N and R) 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):
- 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.
- 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.
- This is because, if the negative counting approach is thought of as "count the number of times a ranked candidate is "not preferred over some candidate" (except possibly when equal-ranking is involved), then the regular approach doesn't require counting that 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 "not preferred over 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 "not preferred to themselves", and not preferred over 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.
- 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 "not preferred over 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.
- The formulas for regular counting w/ 1st choices counted separately, are derived by subtracting N-2 from the corresponding formula for the regular approach.
- 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.
- This is because, if the negative counting approach is thought of as "count the number of times a ranked candidate is "not preferred over some candidate" (except possibly when equal-ranking is involved), then the regular approach doesn't require counting that information, but does require counting the number of times a candidate is preferred over each of the other candidates
- 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 explains why semi-negative counting has the same worst-case as negative counting for N>=2R, since they become equivalent in that case.
- 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:
- When N also equals 2, then in both approaches, at least 1 mark needs to be made (to count the 1st choice; the 2nd choice can be skipped because they are ranked last).
- 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).
Here are some examples for the first numbers in each series (with the upper bound bolded):
Number of candidates ranked ("R") | Regular approach
(2nd column shows 1st choices counted separately) |
Negative counting | Semi-negative counting | |
---|---|---|---|---|
1 | N-1 [0, 1+] | 1 [0, 1] | 1 | 1 |
2 | 2N-3 [1, 3+] | N-1 [1, 2+] | [1, 3] | [1, 2, 3] |
3 | 3N-6 [3, 6+] | 2N-4 [2, 4+] | [3, 6] | [2, 4, 5, 6] |
4 | 4N-10 [6, 10+] | 3N-8 [4, 7+] | [6, 10] | [4, 6, 8, 9, 10] |
5 | 5N-15 [10, 15+] | 4N-13 [7, 11+] | [10, 15] | [6, 9, 11, 13, 14, 15] |
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.
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[edit | edit source]
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.
Burlington 2009 mayoral election[edit | edit source]
(Vote totals taken from ^{[3]}. 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 cand. | 2 cand. | 3 cand. | 4 cand. | 5 cand. | |
---|---|---|---|---|---|
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, for an average of 6.25 marks per ballot (=1.25 marks/ballot/number of non-write-in candidates).
- 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 or 0.87 marks/ballot/candidate).
- 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 or 1.64 marks/ballot/candidate).
- If counting 1st choices separately from all other ranks, then it would only require 46,642 marks (5.19 marks/ballot or 1.038 marks/ballot/candidate).
- 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.
- Calculation: (4*1481 + 7*1912 + 9*1799 + 10*873 + 10*2944). (This is with number of candidates, N, being 5.)
- If counting 1st choices separately from all other ranks, then it would only require 46,642 marks (5.19 marks/ballot or 1.038 marks/ballot/candidate).
Non-pairwise methods[edit | edit source]
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 or 0.2 marks/ballot/candidate)
- RCV: 12,309 marks (1.37 marks/ballot or 0.274 m/b/c. Derived by counting 8,980 1st choices + 3,329 vote transfers^{[4]})
- Hypothetical:
- Approval voting: ~20,000 marks (2.22 marks/ballot or 0.444 m/b/c^{[5]})
- Score voting: ~20,000 to ~30,000 marks (2.22 to 3.34 marks/ballot or 0.444 to 0.668 m/b/c)
Evanston, IL 2020 Democrat endorsement[edit | edit source]
(Vote totals from ^{[6]}) This was an RCV election with no equal-ranking.
There were 249 ballots and 11 candidates.
Number of Ballots Ranking This Many Candidates:
1 cand. | 2 cand. | 3 cand. | 4 cand. | 5 cand. | 6 cand. | 7 cand. | 8 cand. | 9 cand. | 10 cand. | 11 cand. | |
---|---|---|---|---|---|---|---|---|---|---|---|
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 4,482 marks (18 marks/ballot or 1.63 m/b/c).
- 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 or 1.17 m/b/c).
- 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 or 3 m/b/c).
- If 1st choices are counted separately, then this only requires 5,982 marks (24.02 marks/ballot or 2.18 m/b/c).
- 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)
- If 1st choices are counted separately, then this only requires 5,982 marks (24.02 marks/ballot or 2.18 m/b/c).
Non-pairwise methods[edit | edit source]
- RCV: 354 marks (average of 1.42 marks/ballot or 0.129 m/b/c. It is derived by counting 249 voters' 1st choices + 105 votes transferred throughout^{[7]})
- Hypothetical:
- Approval voting: 996 marks (4 marks/ballot or 0.36 m/b/c)
- Score voting: 1245 marks (5 marks/ballot or 0.45 m/b/c)
2017 Green Party of Utah co-chair election[edit | edit source]
(Numbers are slightly off for this analysis.) This was an election with scored ballots, so equal-ranking was allowed.
There were 34 ballots and 4 candidates.
Number of Voters scoring this many candidates above-last at each rank, with number of voters scoring a certain number of candidates above that rank provided in parentheses ("Irrelevant" numbers are left out; for example, any voter who ranked 2 candidates as their 2nd choices and ranked 2 candidates above those candidates have essentially ranked those candidates last, since there are only 4 candidates total, ignoring write-in candidates. Example for reading this table: 20 voters ranked only 1 candidate 2nd and above-last, with 12 of them ranking only 1 candidate above the 2nd rank, and 8 of them ranking 2 candidates above 2nd place):
1 cand. | 2 cand. | 3 cand. | |
---|---|---|---|
1st rank | 14 | 15 | 2 |
2nd rank | 19 (12, 7) | 1 | |
3rd rank | 5 |
- Negative counting approach requires at least 95 marks (2.85 marks/ballot or 0.69 m/b/c).
- Calculation: ([14+15+2] + [[2*12 + 3*7] + [2*2*1]] + [3*5])
- Semi-negative approach requires at least 69 marks (2.02 marks/ballot or 0.5 m/b/c).
- Calculation: ([14+15+2] + [[2*12 + 1*7] + [2*1]] + [1*5])
- Regular approach requires at least 146 marks (4.29 marks/ballot or 1.07 m/b/c).
- If 1st choices are counted separately, then this only requires 88 marks (2.58 marks/ballot or 0.64 m/b/c).
- Calculation: ([1*14 + 2*15 + 3*2] + [[2*12 + 1*7] + [2*1]] + [1*5])
- Calculation: ([3*14 + 4*15 + 3*2] + [[2*12 + 1*7] + [2*1]] + [1*5])
- If 1st choices are counted separately, then this only requires 88 marks (2.58 marks/ballot or 0.64 m/b/c).
Non-pairwise methods[edit | edit source]
- FPTP: 34 marks (1 mark/ballot or 0.25 m/b/c)
- Score voting: 132 marks (3.88 marks/ballot or 0.97 m/b/c)
- Score voting with averages: 134 marks (3.94 marks/ballot or 0.98 m/b/c). This is because there is one voter who explicitly scored 2 candidates at 0, so they would've required an additional 2 marks to count.
- RCV: At most 102 marks (3 marks/ballot or 0.75 m/b/c)
- There were 34 1st choices, and at most all ballots have to transfer their votes twice before only 2 candidates remain. Note: This is a non-tight upper bound.
Note: This election illustrates how, with pairwise counting, the assumption that each voter wants to have a full-strength vote in each pairwise matchup (i.e. be counted as 0 or 1 vote) can sometimes reduce the amount of counting work necessary. For example, with Ballot #4 (from the spreadsheet), the scores of (9, 9, 8, 8) become reduced to (1st, 1st, last, last), and so only the two first 1st choice preferences need to be counted, despite the voter explicitly expressing that they have a very minimal preference between their 1st choices and last choices. The fact that Score voting would require 4 marks to count this ballot, as opposed to only 2, is because it's keeping track of the voter's desire to weaken their vote.
Using negative pairwise counting while collecting additional non-pairwise information[edit | edit source]
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. ^{[8]}
Cardinal information[edit | edit source]
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:
- 3 voters scored Candidate A a 3/5
- 2 of those 3 voters scored A below B
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:
- 5 score A a 5/5
- 2 score A a 4/5
- 3 voters score A a 3/5
- 0 voters scored A any other score
then 10 voters total (explicitly) scored A. This allows you to figure out not only the number of voters scoring a candidate above-last, but also the score of the candidate (since you can multiply the number of voters giving the candidate a particular score by that score, and add this up for all possible scores).
- Note that the amount of cardinal information collected here allows you to also compute the results of cardinal methods like Majority Judgement.
- This requires just as many marks as negative pairwise counting done without collecting scores, because in either vote-counting scheme, one mark is made to count that a voter scored a candidate.
- Data-wise, this requires collecting [[number of candidates]*[number of possible scores]] data values rather than only [number of candidates] values to know how many voters scored a candidate (i.e. because rather than all of those voters being considered to be one group, they are separated into [number of possible scores] groups).
Bucklin (number of voters giving a particular ranking to a candidate)[edit | edit source]
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.
IRV and FPTP (1st choice-based methods)[edit | edit source]
A special case of counting the number of voters who give a particular ranking to a candidate are IRV and FPTP, and other Category:FPTP-based voting methods. With FPTP, only the 1st choice rankings need to be counted, while with IRV, an additional complication can exist depending on which of the Equal-ranking methods in IRV is chosen:
- If whole-votes equal ranking or no equal-ranking are used, then no additional complication exists.
- If using fractional equal-ranking, it's necessary to differentiate between two voters who rank the same candidate 1st based on how many candidates total they ranked 1st. For example, the first voter would only give their 1st choice 1/3rd of a vote if they ranked 3 candidates 1st, and the second voter would give their 1st choice 1/5th of a vote if they ranked 5 candidates 1st.
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).
Connection to cardinal methods[edit | edit source]
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.
Using with strength of preference[edit | edit source]
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.
Negative counting used for non-pairwise methods[edit | edit source]
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[edit | edit source]
- 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). This means using only 2 marks rather than 3.
- 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.
Negative vote-counting approach for sequential PR methods[edit | edit source]
- 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. ^{[9]}
Notes[edit | edit source]
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.
Negative pairwise counting requires slightly more values (= N^2) to be recorded than regular pairwise counting. Regular pairwise counting records N fewer values, because it doesn't record values for the "matchups" between each candidate and themselves.
Note that just because a certain vote-counting method requires fewer marks to count, that doesn't necessarily make it faster. Vote-counting can involve a vote-counter keeping track of several things which, while they don't have to mark them down, slow them down. Thus, a vote-counting procedure that is simpler or otherwise is easier for a vote-counter to follow may have an advantage.
Alternative ways to frame negative pairwise counting[edit | edit source]
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).
Inspiration[edit | edit source]
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.
Comprehensive example[edit | edit source]
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:
A | B | C | D | E | F | |
---|---|---|---|---|---|---|
A | (Ranked on) 1 ballot | |||||
B | -1 vote | 1 ballot | ||||
C | -1 vote | 1 ballot | ||||
D | -1 vote | -1 vote | -1 vote | 1 ballot | ||
E | ||||||
F |
A second voter who voted C>D=E>A would have their ballot processed into a matrix like this:
A | B | C | D | E | F | |
---|---|---|---|---|---|---|
A | (Ranked on) 1 ballot | -1 vote | -1 vote | -1 vote | ||
B | ||||||
C | 1 ballot | |||||
D | -1 vote | 1 ballot | ||||
E | -1 vote | 1 ballot | ||||
F |
The two ballots can be combined into this matrix:
A | B | C | D | E | F | |
---|---|---|---|---|---|---|
A | (Ranked on) 2 ballots | -1 vote | -1 vote | -1 vote | ||
B | -1 vote | 1 ballot | ||||
C | -1 vote | 2 ballots | ||||
D | -1 vote | -1 vote | -2 votes | 2 ballots | ||
E | -1 vote | 1 ballot | ||||
F |
Supposing that these two voters were the only voters in the precinct, this matrix would be the matrix containing all of the necessary information to get the final vote totals for this precinct, so the precinct could send it along. Each different precinct's matrix could be summed in the way shown above for the two voters. Once all precinct matrices have been summed, the following is the math done to get the final vote totals:
A | B | C | D | E | F | |
---|---|---|---|---|---|---|
A | 2 ballots | 2 (votes) | 1 | 1 | 1 | 2 |
B | 0 | 1 ballot | 1 | 1 | 1 | 1 |
C | 1 | 2 | 2 ballots | 2 | 2 | 2 |
D | 1 | 1 | 0 | 2 ballots | 2 | 2 |
E | 1 | 1 | 0 | 1 | 1 ballot | 1 |
F | 0 | 0 | 0 | 0 | 0 | 0 |
For example, look at the row for candidate A. Because A was ranked on 2 ballots, they received 2 votes in every head-to-head matchup against another candidate. But because they were ranked below C, D, and E on 1 ballot, their votes against C, D, and E were reduced by 1 vote, so that they only had 1 vote in favor of them against those 3 candidates in the respective head-to-head matchups.
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.
Dealing with truncation[edit | edit source]
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[edit | edit source]
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.
See also[edit | edit source]
References[edit | edit source]
- ↑ "Possible solution to the Condorcet write-in problem".
- ↑ "Negative vote-counting approach for pairwise counting". The Center for Election Science. 2020-04-27. Retrieved 2020-09-15.
- ↑ https://rangevoting.org/JLburl09.txt
- ↑ "2009 Burlington mayoral election", Wikipedia, 2020-05-05, retrieved 2020-05-20
- ↑ "RangeVoting.org - Burlington Vermont 2009 IRV mayoral election". rangevoting.org. Retrieved 2020-05-16.
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.
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at position 1031 (help) - ↑ "r/EndFPTP - Analysis of the Ballots for the Democratic Party of Evanston, IL Presidential Endorsement Vote". reddit. Retrieved 2020-05-19.
- ↑ https://i0.wp.com/evanstondems.com/wp-content/uploads/2020/02/RCVPrez-Results.png?fit=1024%2C341&ssl=1
- ↑ 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.
- ↑ "Possible trick for counting SPAV (and cardinal PR) faster". The Center for Election Science. 2020-05-26. Retrieved 2020-07-21.
- ↑ "r/EndFPTP - New pairwise counting approach based on Approval voting". reddit. Retrieved 2020-09-15.