User:BetterVotingAdvocacy/Negative vote-counting approach for pairwise counting: Difference between revisions
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* But negative counting only requires 3 marks: 1 each for A and B to indicate they are preferred in every matchup, and 1 to indicate that this isn't the case for B>A.
It is possible to compare the number of marks that must be made in either approach for certain elections, because their full ballot set has been published. Any election using ranked or rated ballots can be used for this purpose. The calculations and numbers used in this section may be slightly off for some examples, though general conclusions (should) still hold; because of this, it is suggested that the reader apply a ~10% margin of error or more when considering how superior one pairwise counting approach is to another.
Note that when equal-ranking isn't allowed, only the number of voters who ranked a certain number of candidates needs to be known.
(Vote totals taken from <ref>https://rangevoting.org/JLburl09.txt</ref>. The numbers and calculations are slightly off for this analysis.) This was an [[RCV]] election.
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{| class="wikitable"
!
!1 cand.
!2 cand.
!3 cand.
!4 cand.
!5 cand.
|-
|
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Based on the above table:
*The negative counting approach would require at least
**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
**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
** If counting 1st choices separately from all other ranks, then it would only require
***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.)
Also of interest may be the comparison to non-pairwise counting methods. The following are rough estimates for the number of marks:
*[[FPTP]]: ~
*[[RCV]]:
*Hypothetical:
**Approval voting: ~
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: ~
(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 and '''11''' candidates.
''Number of Ballots Ranking This Many Candidates:''
{| class="wikitable"
!
!1 cand.
!2 cand.
!3 cand.
!4 cand.
!5 cand.
!6 cand.
!7 cand.
!8 cand.
!9 cand.
!10 cand.
!11 cand.
|-
|
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|}
*Negative counting approach requires at least
**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
**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
**If 1st choices are counted separately, then this only requires
***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)
*[[RCV]]:
*Hypothetical:
**Approval voting:
**Score voting:
===2017 Green Party of Utah co-chair election===
{{Main|2017 Green Party of Utah officer election}}(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):
{| class="wikitable"
!
!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).
**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).
**Calculation: ([14+15+2] + [[2*12 + 1*7] + [2*1]] + [1*5])
*Regular approach requires at least 146 marks ('''4.29''' marks/ballot).
**If 1st choices are counted separately, then this only requires 88 marks ('''2.58''' marks/ballot).
***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])
====Non-pairwise methods====
* FPTP: 34 marks (''1'' mark/ballot)
* Score voting: 132 marks (''3.88'' marks/ballot)
** Score voting with averages: 134 marks (''3.94'' marks/ballot). 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)
** 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==
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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===
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