Black's method: Difference between revisions
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Borda scores are A 185, B 205, C 210. A pairwise beats B beats C beats A, so there is no Condorcet winner (because everyone has at least one defeat). So the Borda winner, C, wins. |
Borda scores are A 185, B 205, C 210. A pairwise beats B beats C beats A, so there is no Condorcet winner (because everyone has at least one defeat). So the Borda winner, C, wins. |
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Note that the winner of the Borda count is not necessarily in the Smith set.<ref>{{Cite web| title = Is there formal proof that Duncan Black's Electoral System is a Condorcet System?| work = ResearchGate| accessdate = 2020-03-31| url = https://www.researchgate.net/post/Is_there_formal_proof_that_Duncan_Blacks_Electoral_System_is_a_Condorcet_System|date=March 12, 2014|last=di Fenizio|first=Pietro Speroni|archive-url=|archive-date=|url-status=live}}</ref> |
Note that the winner of the Borda count is not necessarily in the Smith set.<ref>{{Cite web| title = Is there formal proof that Duncan Black's Electoral System is a Condorcet System?| work = ResearchGate| accessdate = 2020-03-31| url = https://www.researchgate.net/post/Is_there_formal_proof_that_Duncan_Blacks_Electoral_System_is_a_Condorcet_System|date=March 12, 2014|last=di Fenizio|first=Pietro Speroni|archive-url=|archive-date=|url-status=live}}</ref> For example, if there are 8 candidates and 3 voters, with ballots: |
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A > B > X > D1 > D2 > D3 > D4 > C |
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B > C > X > D1 > D2 > D3 > D4 > A |
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C > A > X > D1 > D2 > D3 > D4 > B |
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Then A, B, and C form the Smith Set / Condorcet cycle (with 16 points each), while X is the Borda winner (with 18 points). |
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A [[Smith-efficient]] variant of Black's method could be used instead, to ensure someone in the [[Smith set]] will win. |
A [[Smith-efficient]] variant of Black's method could be used instead, to ensure someone in the [[Smith set]] will win. |