Lowest Number of First Choice votes tie breaker

[DEMOREP1 at aol.com]

Assume 3 Condorcet candidates are in a circular tie.

Example (left side is greater than right side on each line,1=first choice
votes, 2 = second choice votes)----

(A1 + C1A2) > (B1 + C1B2) (A beats B)
(B1 + A1B2) > (C1 + A1C2) (B beats C)
(C1 + B1C2) > (A1 + B1A2) (C beats A)

As a tie breaker I suggest that the candidate (in the tie group of
candidates) with the lowest number of first choice votes should be the
initial loser.

I note that if the fewest-votes-against-in-worst-defeat (FVA) method is used
as a tie breaker then some of the voters by voting sincerely may defeat their
first choice.

Example- 
40 + 15 > 35 + 10  (A beats B)
35 + 19 > 25 + 21  (B beats C)
25 + 31 > 40 + 4    (C beats A)

55 > 45 (A beats B)
54 > 46 (B beats C)
56 > 44 (C beats A)
With FVA, C wins.

The 21 A1C2 voters on the right side by voting sincerely for C have caused C
to win.

However, if 2 or more of such 21 A1C2 voters vote insincerely for B, then A
will win using FVA (B will beat C by at least 56 to 44).

If FVA is the tie breaker, then all candidates will play games and try to get
their first choice voters to vote insincerely.

To avoid such game playing, I suggest the candidate with the lowest number of
first choice (LNFC) votes should lose. Thus in the example, C loses and the
C1 voters can vote sincerely for A or B (A wins).

With 4 or more candidates in a tie and after each loser is eliminated the
head to head pairings among the remaining tied candidates would be looked at
again to see if there is a Condorcet winner. 

Eliminating just one candidate may produce a Condorcet winner among the
remaining candidates.

Note- LFC is not direct instant run off but only a tie breaker after doing
the standard Condorcet pairings. LFC is also easy to explain to the public if
there are ties.