User:BetterVotingAdvocacy/Negative vote-counting approach for pairwise counting: Difference between revisions

* A>B is (10-0)=10 votes.

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

 

=== Dealing with equal-ranking ===

* '''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.

 

 

* 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.

 

{| 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)

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.

 

It can also be useful to compare these results to the amount of vote-counting work that would be done in other voting methods.

 

''Number of Ballots Ranking This Many Candidates:''

{| class="wikitable"

!1

!2

!5

|-

|1481

|1912

|873

|2944

|}

Based on the above table:

 

===== Non-pairwise 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)<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

 

''Number of Ballots Ranking This Many Candidates:''

{| class="wikitable"

!1

!2

!11

|-

|48

|35

|4

|40

|}

 

===== Non-pairwise methods =====

 

 

== Connection to cardinal methods ==