N factorial combinations (was The Davison Run-off)
DEMOREP1 at aol.com
DEMOREP1 at aol.com
Fri Dec 27 17:47:26 PST 1996
The simple general case with 3 candidates is--
N1 ABC
N2 ACB
N3 BCA
N4 BAC
N5 CAB
N6 CBA
where N1 .... N6 are numbers of voters.
Condorcet thus does
A vs. B or N1 + N2 + N5 vs. N3 + N4 + N6
B vs. C or N1 + N3 + N4 vs. N2 + N5 + N6
C vs. A or N3 + N5 + N6 vs. N1 + N2 + N4
---
Plurality does
A N1 + N2
B N3 + N4
C N5 + N6
and says that the highest sum wins.
---
Davidson does
A N3 + N6
B N2 + N5
C N1 + N4
and says that the highest sum loses the first round elimination.
I would suggest that it is equally improper to use only partial votes in
Plurality as in Davidson (2 of 6 *selections*).
Someone may wish to enlighten the list about other methods which do different
things with the N1 ... N6 and with 4 or more candidates where there are N x
(N-1) X (N-2) x .... x 2 or N factorial (N!) complete combinations.
Note- the N1 ... N6 example can be expanded for truncated votes of
N7 A
N8 B
N9 C
assuming that
AB is the same as ABC, AC is the same as ACB, etc. (thus making 9 numbers
involved for 3 candidates).
As Mr. Arrow informs us, choosing the *best* method involves various
tradeoffs in the strategies regarding the various N1.... Nmost values. I
will maintain that majority rule must be an absolute requirement in any
method.
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