Lowest YES tiebreaker

[DEMOREP1 at aol.com]

Supplement--

More elementary thoughts about 1 of 2, 2 of 3, 3 of 4, etc.

Going beyond the obvious 1 of 2 choices case, there is the N minus 1 choices
case when all N choices have majority acceptability (noting that in many
elections, 1 or more of the N choices may be unacceptable to many voters).

If each voter has N minus 1 YES votes for his/her choices in a tie (2 of 3, 3
of 4, 4 of 5, etc), then 1 of the N choices will have the lowest number of YES
votes (assuming no ties) and should lose if there is a circular tie AFTER
doing the head to head math.  

Once again for the umpteenth time, number votes do NOT indicate acceptability
but ONLY relative support.

A voter's vote might be--

A  NO 7
B  YES 3
C YES 4
D  NO  6
E  YES 1
F  NO 5
G  YES 2
For all voters 1 or more of the choices might get YES majorities.  If 3 or
more choices get YES majorities, then there is the ever present lurking
possibility that 3 or more choices will be in a circular tie using the number
votes.

With 4 or more choices in a circular tie, the YES votes might be very spread
out among such choices.

Thus, I suggest that in a tie case that the YES votes in the Nth place on each
ballot should be dropped (in ALL elections-  legislative, executive, judicial,
direct issue) and that the head to head math should be rechecked (repeatedly,
if necessary) among the remaining choices.

Once again I must complain that too much thinking has been given to the 3
choice tied case and not enough thinking to the 4 or more choice tied cases.

The above also brings up the idea of contingent YES votes-- that is, there
would be a YES vote for a choice only if certain things happen in the voting
(related to the withdrawal idea). 

The obvious problem is that if many voters have contingent YES votes that a
feedback crash/ lock loop might result.