User:BetterVotingAdvocacy/Negative vote-counting approach for pairwise counting: Difference between revisions
User:BetterVotingAdvocacy/Negative vote-counting approach for pairwise counting (view source)
Revision as of 21:10, 16 May 2020
, 4 years ago→Write-in candidates
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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 ==
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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.
=== 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.
Note that when equal-ranking isn't allowed, only the number of voters who ranked a certain number of candidates needs to be known:
{| class="wikitable"
|+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)*
|}
<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).
==== 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.)
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
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.
** 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.
* 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.)
== Connection to cardinal methods ==
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