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

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# Vote-counting: The precinct vote-counters count the following values for a given candidate:
# 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.
##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 also marked both candidates 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:
##*(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.
##**Counting the number of votes for A''<''B. This yields a ''positive'' value.
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# 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).
# 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:<blockquote>A>=B = A - A<B</blockquote>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).
#* This can be succinctly summarized as, for finding the vote totals in a matchup between candidates A and B:<blockquote>A>=B* = A - A<B</blockquote>* 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.<ref>{{Cite web|url=https://www.reddit.com/r/EndFPTP/comments/fsa4np/possible_solution_to_the_condorcet_writein_problem/|title=Possible solution to the Condorcet write-in problem|last=|first=|date=|website=|url-status=live|archive-url=|archive-date=|access-date=}}</ref>
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.<ref>{{Cite web|url=https://www.reddit.com/r/EndFPTP/comments/fsa4np/possible_solution_to_the_condorcet_writein_problem/|title=Possible solution to the Condorcet write-in problem|last=|first=|date=|website=|url-status=live|archive-url=|archive-date=|access-date=}}</ref>
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'''<big>Note</big>''': 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.
'''<big>Note</big>''': 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 ==
=== Example ===

If the votes are:<blockquote>10: A>B
If the votes are:<blockquote>10: A>B


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


== Dealing with equal-ranking ==
=== 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 votes being: <blockquote>2 A>B
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 votes being: <blockquote>2 A>B
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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.
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 ===
==== 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).
Suppose a voter had ranked 9 of 10 candidates as their 1st choices, and the 10th candidate was unranked (i.e. implicitly ranked last).


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In addition:
In addition:


==== Explicitly equal-ranked candidates both get a vote ====
===== Explicitly equal-ranked candidates both get a vote =====
No extra work needs to be done.
No extra work needs to be done.


==== Equal-ranked candidates don't get votes ====
===== Equal-ranked candidates don't get votes =====
For each matchup, the following number of markings can be made for two candidates A and B:
For each matchup, the following number of markings can be made for two candidates A and B:


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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.
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 ==
=== 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.
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 ===
==== 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.
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.


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=== Dealing with truncation ===
=== Dealing with truncation ===
When voters aren't allowed to do [[truncation]] (and [[Write-in candidate|Write-in candidat]]<nowiki/>es 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.
When voters aren't allowed to do [[truncation]] (and [[Write-in candidate|Write-in candidat]]<nowiki/>es 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 works because every candidate will actually be marked on every voter's ballot.

Revision as of 05:51, 16 May 2020

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.

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.[1]

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.

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 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:

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.

Comparison to the regular approach

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.

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.

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

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.

References

  1. "Possible solution to the Condorcet write-in problem".