Reverse Bucklin tiebreaker

[DEMOREP1 at aol.com]

Supplement 5

If Reverse Bucklin is applied to the 3 candidate tie general case (A>B>C>D),
then there is

N1   A>B>C
N2   A>C>B
N3   B>C>A
N4   B>A>C
N5   C>A>B
N6   C>B>A
T = Total = Sum of N1 to N6

The Reverse Bucklin majorities for the last 2 choices are--

A   T- the number of first choices for A
B   T- the number of first choices for B
C   T- the number of first choices for C

The choice with the highest Reverse Bucklin majority would lose.

In other words, in the 3 choice case the candidate with the lowest number of
first choice votes would lose which complies with common sense.

The 4 choice tie case will be in the next supplement but is  predictable as to
who the first loser would be.