Halfway Bucklin Tiebreaker

DEMOREP1 at aol.com DEMOREP1 at aol.com
Mon Jun 29 16:23:12 PDT 1998

```I have mentioned earlier that, regarding a Condorcet circular tiebreaker, (a)
plain Bucklin is somewhat defective since it might cause some early insincere
voting (especially for second choice) in order to get an early bare majority
and (b) ordinary Approval counts too high of a majority (by counting all of
the choices of all of the voters).

Thus, I suggest a possible halfway tiebreaker that would count the choices at
the halfway level.

Theory-
Assume all first choices have roughly an equal number of votes (100 percent of
voters/ N (number of choices)).

At the N/2 (if N is even) or (N+1)/2 (if N is odd) choice level, there should
be various majorities.  The winner would be the choice with the highest
majority.

Example 7 candidates in a circular tie (difficult to imagine in a real
election).  All 7 candidates get around 14 first choice votes.
At the Bucklin 4th choice level, 1 or more of the choices would have
majorities of the total number of voters.  That is, several of the candidates
might have accumulated 1st to 4th choice majorities in the 51 to 70 range.
The candidate with the highest majority would win.

The practical effect would enable minorities to sincerely vote for their early
choices but then have to vote for the lesser of the other evils (in their
minds) or to truncate their votes.

```