# 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.