[EM] Net Wins Tiebreaker, Supp. 3

[DEMOREP1 at aol.com]

If the Wins and Losses are added to a place votes table, then there is--
        
     4   3  2  1  0  Wins
     0  1   2  3  4  Losses
     4  2   0  -2 -4  Net Wins
22 A  B  C  D  E
21 B  C  D  E  A
20 C  D  E  A  B
19 D  E  A  B  C
18 E  A  B  C  D

The above shows the totally dangerous and evil nature of simple Plurality - 
it counts only the first (and only) place Wins.

Bucklin sums the Wins (while also overrating the 2nd, etc. place Wins).

Simple Approval Voting counts all of the Wins (while also overrating the 2nd, 
etc. place Wins (by ignoring true relative places)).

So-called Instant Run-off drops the choice with the fewest first choice wins 
(i.e. ignores later wins).

Various point methods use the wrong number of points per place (such as 5 
points in the above for a first place win).

All of the above ignore the Losses (especially Plurality).

Multiplying the Net Wins by the votes produces a Net Wins matrix--

            4         2     0        -2        -4
22     88A     44B     0     -44D     -88E
21     84B     42C     0     -42E     -84A
20     80C     40D     0     -40A     -80B
19     76D     38E     0     -38B     -76C
18     72E     36A     0     -36C     -72D

or
A 88 + 36  -40 -84 = 0
B 84 + 44  -38 - 80 = 10
C 80 + 42  -36 -76 = 10
D 76 + 40  -44 -72 = 0
E 72 + 38  -42  -88 = -20

See the head to head table in Supp. 1.  

Note - with an odd number of choices, one can divide the Net Wins by 2 for 
simpler math (e.g.  2   1   0   -1  -2).  

Not so with an even number of choices-
example 4 choices-- 
       3     2     1     0  Wins
       0     1     2      3  Losses
       3     1  - 1    -3  Net Wins

Applying the above to the standard simple 3 choice circular tie--
       2   1   0  Wins
       0   1   2   Losses
       2    0  -2  Net Wins
NI   A     B   C
N2   B    C   A
N3   C   A    B

or

A  2N1 - 2N2
B  2N2 - 2N3
C  2N3 - 2N1
A circular result- not surprising (draw a circle with A, B and C sectors).

With N1= 5, N2= 4, N3 = 3, there is
A 2
B 2
C -4

A versus B head to head-
8 A, 4 B  or A wins (who just happens to have a first choice simple 
plurality).  C just happens to lose with the fewest first choice votes.

We build on the works of our illustrious predecessors--- in this case, Pascal 
(for his 1, 3, 6, 10, 15, etc. triangle (see the pins in your local bowling 
alley or the balls in a pool table triangle) and the binomial formula)), 
Condorcet, Borda, Bucklin, and others.