User:BetterVotingAdvocacy/Negative vote-counting approach for pairwise counting: Difference between revisions
User:BetterVotingAdvocacy/Negative vote-counting approach for pairwise counting (view source)
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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.
This "modified regular approach" can outperform negative pairwise counting in some scenarios (the ballot counted in this example is A>B>C in a 3-candidate election):
* 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]]).▼
{| class="wikitable"
|+"Modified regular approach" values '''bolded''', negative approach values ''italicized''
!
!A
!B
!C
|-
|A
|'''''1 ballot'''''
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|B
|''-1''
|''1 ballot''
|'''''1'''''
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|C
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|}
* In this example, 2 tally marks are made when using the modified regular approach, while in negative pairwise counting, 3 marks must be made.
In other scenarios, the negative pairwise counting approach works better (ballot is A>B in a 5-candidate election):
{| class="wikitable"
|+Modified regular approach values '''bolded''', negative approach values ''italicized''
!
!A
!B
!C
!D
!E
|-
|A
|'''''1 ballot'''''
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|-
|B
|''-1''
|''1 ballot''
|'''1'''
|'''1'''
|'''1'''
|-
|C
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|-
|D
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|-
|E
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* 4 marks are made to count the modified regular approach, but only 3 marks are made in the negative pairwise counting approach.
Additional info:
▲*
* 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.
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Notes:
*The regular pairwise counting approach always performs worse than or as badly as the other pairwise counting methods, while semi-negative counting always performs better than or as well as the other pairwise counting methods.
*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.
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