Reverse Bucklin tiebreaker

[DEMOREP1 at aol.com]

Supplement 7

Some clarification regarding the defeat of losers using Reverse Bucklin--

If all candidates lose at a certain Reverse Buckin level, then (a) the early
choices could be used to break the tie or (b) the candidate with the lowest
number of first choice votes would lose with the math being reexamined.

Example--

5 candidates in a circular tie
No candidates are defeated by majorities against using the 6th or the 6th and
5th choices using Reverse Bucklin.
The last 3 choices (6th plus 5th plus 4th) using Reverse Bucklin causes all 6
candidates to have majorities against them. 

The winner then might be (a) the candidate who got the highest majority using
the sum of the votes for each candidate in the first 4 choices (i.e. regular
Bucklin) or (b) as in (b) above. 

Note- There are odd-even groups in Reverse Bucklin (RB), namely, with the
indicated number of tied candidates, the last N choices would generally
produce a loser (assuming there are no earlier losers-- such as having a
Condorcet loser defeated by all other candidates).
 Cands.        Last RB choices
[ 2                1, head to head produces a winner]
3-4              2
5-6              3
7-8              4
etc.             etc.

With 5 or more candidates I would expect many more truncated votes with the
resulting quick defeats of many candidates (i.e. each voter might only rank
about 3 or 4 candidates and truncate the rest of the candidates-- that is, put
the truncated candidates equally as last choices).