DEMOREP's Condorcet Tiebreaker
New Democracy
donald at mich.com
Mon Oct 19 00:14:14 PDT 1998
Greetings,
I asked DEMOREP to supply us with an example of his Condorcet
tiebreaker solution and he was kind enough to do so.
DEMOREP1 wrote:
Of some interest in one tiebreaker is the number of votes that each choice
gets at the N/2 (N even) or (N+1)/2 (N odd) horizontal choices level (both
left to right and right to left).
Donald wrote-
I am interested in this tiebreaker, but it is not clear how it works.
Could you give us a simple example?
----
[DEMOREP]-adding some clarifying words--
1 2 3 4 5 Votes in 1-3 places Votes in 5- 3 places
22 A B C D E A 59 A 60
21 B C D E A B 61 B 57
20 C D E A B C 63 max C 59
19 D E A B C D 60 D 61
18 E A B C D E 57 E 63 max
--- ---- ----
100 300 300
[option one] The theory- If there is no Condorcet winner, then the
compromise winner should get the highest majority in the first N/2 places.
[option two] If there is no Condorcet winner and if elimination of
candidates is to be
done, then the highest majority in the last N/2 places should lose.
Dear DEMOREP.
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.
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
again?
Also, why is that you are willing to eliminate a candidate in
Condorcet, but you are not willing to eliminate a candidate in Choice
Run-Off(IRV)?
Regards,
Donald Davison
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