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, 3 June 2020
, 4 years ago→Comparison to the regular pairwise counting 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.
Note that regular pairwise counting can have its required number of marks reduced without using any negative numbers by counting 1st choices separately from all other ranks; see the section above [[Pairwise counting#Uses for first choice information]]. The regular approach requires [(number of candidates)-2] less marks if using this modification, though when voters equally rank multiple candidates 1st, then whether or not the modification is applied to count all of those 1st choices separately or not can make a significant difference in overall number of marks.
=== Election example comparisons ===
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**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 '''73,669''' marks.
** If counting 1st choices separately from all other ranks, then it would require '''46,642''' marks.
***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.)
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**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 '''8,223''' marks.
** If 1st choices are counted separately, then this requires '''5,982''' marks.
***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)
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