4,208
edits
Black's method: Difference between revisions
Fixing typo in Category:Condorcet methods
Psephomancy (talk | contribs) (Created page with "'''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 w:Duncan Black)...") |
(Fixing typo in Category:Condorcet methods) |
||
| (8 intermediate revisions by 3 users not shown) | |||
Line 1:
{{wikipedia|Black's method}}
'''Black's method''' chooses the [[Condorcet winner]] if it exists, but uses the [[Borda count]] instead if there is an ambiguity. This method is named for [[Duncan Black]].<ref>{{Cite book| isbn = 978-94-009-4225-7| last = Black| first = Duncan| title = The Theory of Committees and Elections| accessdate = 2020-03-31| date = 1987| url = http://link.springer.com/openurl?genre=book&isbn=978-94-010-8375-1|publisher=|year=|location=|pages=66}}</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>
== Examples ==
=== Example one ===
In this example, the candidates are ranked as follows:
25: A>B>C
40: B>C>A
35: C>A>B
==== Stage 1 of example one ====
Evaluating these ballots pairwise, we see the following
A (1 win, 1 loss): defeats B (60 to 40) but loses to C (75 to 25)
B (1 win, 1 loss): defeats C (65 to 35) but loses to A (40 to 60)
C (1 win, 1 loss): defeats A (75 to 25) but loses to B (65 to 35)
This demonstrates a Condorcet cycle (where A defeats B, B defeats C, and C defeats A), so we advance to the Borda stage.
==== Stage 2 of example 1 ====
The points allocated in the Borda stage can be expressed using [[ABIF]] in the following text:
25: A/2 > B/1 > C/0
40: B/2 > C/1 > A/0
35: C/2 > A/1 > B/0
Counting up the Borda scores using the points above, we see:
A: ((25 * 2) + (35 * 1)) = 85
B: ((40 * 2) + (25 * 1)) = 105
C: ((35 * 2) + (40 * 1)) = 110
C earns 110 points in the second round, so C is the winner.
=== Example two ===
What follows is an example of Borda winner that is not in the Smith set, with 8 candidates ("A", "B", "C", "De", "Df", "Dg", "Dh", and "X") and 3 voters. The rankings are expressed below:
1: A > B > X > De > Df > Dg > Dh > C
1: B > C > X > De > Df > Dg > Dh > A
1: C > A > X > De > Df > Dg > Dh > B
==== Stage 1 of example two ====
A, B, and C form a Condorcet cycle. which can be seen by removing the candidates other than A, B, and C:
1: A > B > C
1: B > C > A
1: C > A > B
So the next stage is a Borda count.
==== Stage 2 of example two ====
The points allocated in the Borda round are expressed using [[ABIF]] in the following text:
1: A/7 > B/6 > X/5 > De/4 > Df/3 > Dg/2 > Dh/1 > C/0
1: B/7 > C/6 > X/5 > De/4 > Df/3 > Dg/2 > Dh/1 > A/0
1: C/7 > A/6 > X/5 > De/4 > Df/3 > Dg/2 > Dh/1 > B/0
This results in the following point allocations
* A receives 7+0+6 points (13 points)
* B receives 6+7+0 points (13 points)
* C receives 0+7+6 points (13 points)
* X receives 5+5+5 points (15 points)
This X wins the election in the Borda-based tiebreaker, even though X wasn't part of the Condorcet cycle in the first round.
== Similar methods==
A [[Smith-efficient]] Condorcet method could be used in the first stage (such as [[Copeland's method]]), where only the candidates that are tied using the first method advance to the second method. This would ensure that someone in the [[Smith set]] advances from round one, only resorting to a Borda-based tiebreaker when a [[Condorcet winner criterion|Condorcet winner]] doesn't defeat all other candidates pairwise.
==References==
<references />
[[Category:Condorcet-Borda hybrid methods]]
[[Category:Condorcet methods]]
[[Category:
| |||