DEMOREP's Condorcet Tiebreaker

Mon Oct 19 17:53:55 PDT 1998

Mr. Davidson wrote-

 [1] Your solution appears to have two options. What determines which
option is to be used? The answer is important because in another example
the same candidate could be the winner in the first option and the loser in
the second option.

     Option one appears to be Approval Voting. If Condorcet fails, then
Approval Voting is tapped to provide a winner - in this example that would
be candidate C.
     But, 63 is not a majority of 300.

    [2]  In option two, I understand that candidate E is eliminated, but what
happens next? Do you transfer the 18 votes to the next choices and then go
back and try to work Condorcet a second time, hoping that it will not fail
     [3] Also, why is that you are willing to eliminate a candidate in
Condorcet, but you are not willing to eliminate a candidate in Choice
I mention again that either N/2 option is NOT my choice of a tiebreaker.  The
N/2 options lead me to my tiebreaker choice-- to repeatedly drop the last
numbered YES votes (Nth, N-1th, N-2th, etc.) from all ballots and to recheck
the head to head math.  

With that in mind--
[1] The idea is that there should be an incentive to reduce insincere votes in
the lower numbered places.
Plurality (i.e. voting in first place only) is defective because it obviously
causes many insincere votes for the winners.
The same may be said to a lesser extent for limited voting in more places
(such as voting only for a first place and a second place choice in a 10
choice situation).
Simple Approval Voting goes too far by permitting a *1* vote for all choices
but not number ranking the choices.
63 is a majority of 100.

[2] Due to truncated votes, there might be cases when option 1 would fail to
get any majority of all the votes winner.  
Choices not voted for on truncated ballots might be deemed to be in last place
for purposes of option 2. 
A choice thus will get a left or right majority of all the voters.
If a choice is eliminated using option 2, then the head to head math would be

[3] Since head to head math is not done in IRO, then the tiebreaker
elimination methods of Condorect and IRO cannot be compared.   IRO fans
somehow fail to see that 1 choice can beat each other choice head to head-
i.e. a Condorcet winner-- whether there are 2, 3 or N choices.

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