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. |
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Revision as of 00:53, 1 April 2020
Black's method chooses the Condorcet winner if it exists, but uses the Borda count instead if there is an ambiguity (the method is named for Duncan Black).[1]
Example:
25 A>B>C 40 B>C>A 35 C>A>B
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.
Note that the winner of the Borda count is not necessarily in the Smith set.[2] For example, if there are 8 candidates and 3 voters, with ballots:
A > B > X > D1 > D2 > D3 > D4 > C B > C > X > D1 > D2 > D3 > D4 > A C > A > X > D1 > D2 > D3 > D4 > B
Then A, B, and C form the Smith Set / Condorcet cycle (with 16 points each), while X is the Borda winner (with 18 points).
A Smith-efficient variant of Black's method could be used instead, to ensure someone in the Smith set will win.
References
- ↑ Black, Duncan (1987). The Theory of Committees and Elections. p. 66. ISBN 978-94-009-4225-7. Retrieved 2020-03-31.
- ↑ di Fenizio, Pietro Speroni (March 12, 2014). "Is there formal proof that Duncan Black's Electoral System is a Condorcet System?". ResearchGate. Retrieved 2020-03-31.