Black's method: Difference between revisions
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{{wikipedia|Black's method}} |
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| ⚫ | '''Black's method''' chooses the [[Condorcet winner]] if it exists, but uses the [[Borda count]] instead if there is an ambiguity |
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| ⚫ | '''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> |
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Example: |
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== Examples == |
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=== Example one === |
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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. |
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In this example, the candidates are ranked as follows: |
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==== Stage 1 of example one ==== |
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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> |
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Evaluating these ballots pairwise, we see the following |
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A (1 win, 1 loss): defeats B (60 to 40) but loses to C (75 to 25) |
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B (1 win, 1 loss): defeats C (65 to 35) but loses to A (40 to 60) |
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C (1 win, 1 loss): defeats A (75 to 25) but loses to B (65 to 35) |
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This demonstrates a Condorcet cycle (where A defeats B, B defeats C, and C defeats A), so we advance to the Borda stage. |
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==== Stage 2 of example 1 ==== |
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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. |
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The points allocated in the Borda stage can be expressed using [[ABIF]] in the following text: |
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25: A/2 > B/1 > C/0 |
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40: B/2 > C/1 > A/0 |
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35: C/2 > A/1 > B/0 |
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Counting up the Borda scores using the points above, we see: |
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A: ((25 * 2) + (35 * 1)) = 85 |
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B: ((40 * 2) + (25 * 1)) = 105 |
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C: ((35 * 2) + (40 * 1)) = 110 |
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C earns 110 points in the second round, so C is the winner. |
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=== Example two === |
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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: |
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1: A > B > X > De > Df > Dg > Dh > C |
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1: B > C > X > De > Df > Dg > Dh > A |
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1: C > A > X > De > Df > Dg > Dh > B |
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==== Stage 1 of example two ==== |
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A, B, and C form a Condorcet cycle. which can be seen by removing the candidates other than A, B, and C: |
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1: A > B > C |
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1: B > C > A |
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1: C > A > B |
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So the next stage is a Borda count. |
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==== Stage 2 of example two ==== |
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The points allocated in the Borda round are expressed using [[ABIF]] in the following text: |
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1: A/7 > B/6 > X/5 > De/4 > Df/3 > Dg/2 > Dh/1 > C/0 |
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1: B/7 > C/6 > X/5 > De/4 > Df/3 > Dg/2 > Dh/1 > A/0 |
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1: C/7 > A/6 > X/5 > De/4 > Df/3 > Dg/2 > Dh/1 > B/0 |
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This results in the following point allocations |
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* A receives 7+0+6 points (13 points) |
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* B receives 6+7+0 points (13 points) |
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* C receives 0+7+6 points (13 points) |
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* X receives 5+5+5 points (15 points) |
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This X wins the election in the Borda-based tiebreaker, even though X wasn't part of the Condorcet cycle in the first round. |
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== Similar methods== |
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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. |
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<references /> |
<references /> |
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[[Category:Condorcet-Borda hybrid methods]] |
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[[Category:Condorcet methods]] |
[[Category:Condorcet methods]] |
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[[Category: |
[[Category:Monotonic electoral systems]] |
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Latest revision as of 06:09, 25 May 2023
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.[1] Note that the winner of the Borda count is not necessarily in the Smith set.[2]
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 doesn't defeat all other candidates pairwise.
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.