# 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.^{[1]} Note that the winner of the Borda count is not necessarily in the Smith set.^{[2]}

## Examples[edit | edit source]

### Example one[edit | edit source]

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[edit | edit source]

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[edit | edit source]

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[edit | edit source]

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[edit | edit source]

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[edit | edit source]

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[edit | edit source]

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[edit | edit source]

- ↑ 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.