Black's method: Difference between revisions
Add example from ResearchGate
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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.
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:
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.
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