Reverse Bucklin tiebreaker
DEMOREP1 at aol.com
DEMOREP1 at aol.com
Wed Jun 17 12:42:43 PDT 1998
Supplement 6
The math with 4 or more choices becomes much more complex.
With 3 or more choices (M) there are M factorial possible combinations of the
choices (assuming a voter votes for all choices-- i.e. no truncated votes).
Each choice would possibly appear in M factorial / M or (M -1) factorial times
in each vertical choice column.
Examples--
4 Choices, 4 factorial = 24, 3 factorial = 6
5 Choices, 5 factorial = 120, 4 factorial = 24
A rough example of 4 tied candidates (artificial limit of each choice only
once in each choice column) --
27 A>B>C>D
26 B>C>D>A
25 C>D>A>B
24 D>A>B>C
102, majority 52
The last 2 choices are
A 51
B 49
C 51
D 53 majority against, loses
Head to head math would be done for A, B, C.
A rough example of 5 tied candidates
N1 A>B>C>D>E
N2 B>C>D>E>A
N3 C>D>E>A>B
N4 D>E>A>B>C
N5 E>A>B>C>D
100, majority is 51
Each N amount is less than 51.
The last 2 choices--
A N2 + N3
B N3 + N4
C N4 + N5
D N5 + N1
E N1 + N2
200
Assuming a "normal" distribution of the N amounts, there might be 1 or more 51
majorities against. If the N amounts are nearly equal, then it would be
necessary to sum the amounts for the last 3 choices to get majorities against.
How often a 4 or more choice tie for a single office would occur in a real
public election is a matter of major conjecture.
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