Reverse Bucklin tiebreaker

[DEMOREP1 at aol.com]

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.