[EM] Net Wins Tiebreaker, Supp. 4

[DEMOREP1 at aol.com]

The Net Wins are thus in a even- odd math pyramid related to Pascal's 
triangle---
C = Number of Choices, W/L = Wins or Losses on each ballot (if each choice is 
voted for in a different place)

C    W/L        Net Wins by Place

2     1                       1  -1
3     3                    2   0  -2
4     6                  3   1  -1  -3
5    10               4   2   0   -2  -4 
6    15             5   3   1  -1  -3  -5
etc. etc.                     etc.

The pyramid also suggests that a tiebreaker should be tried on cases 
involving at least 4 or 5 choices (to see if it can collapse into a 3 choice 
or 2 choice case).

The following--

22 ABCDE
21 BCDEA
20 CDEAB
19 DEABC
18 EABCD
100

produces a net wins table of--

      A          B        C          D         E       Total
A      X       58       18      -20      -56         0
B    -58      X         60       22       -14       10
C    -18     -60       X        62         26       10
D      20     -22     -62       X          64         0
E      56      14       -26     -64          X      -20
Tot  0       -10      -10        0          20        0  

If the combinations of two versus the other three are done (such as BC versus 
ADE), then the following results --

AB 10
AC  10
AD  0
AE -20
BC  20
BD  10
BE  -10
CD  10
CE  -10
DE  -20

Total 0

Each combination amount is the sum of the individual totals from the net wins 
table.   The cross amounts cancel each other out (e.g. -  A over B cancels 
out B over A).  

Thus, the highest combination of votes will prevail.

Example -- with 7 choices, the combination of three choices with the highest 
total of net wins will prevail.