Borda Count to the median

[DEMOREP1 at aol.com]

Mr. Cretney wrote-

Consider the following example:


45 A B C

25 B C A

30 C B A
----------
D-

Place Votes
    1     2     3  

A   45    0    55
B   25   75     0
C   30   25    45

   100   100   100

Subtracting the 3rd place from the 1st place votes ---
Net Place Votes
   1-3     2

A  -10     0  
B   25    75
C  -15    25

Tot  0   100

Expand for 4 or more choices (with lots of truncated votes).

When there is a divided majority (as in Mr. Cretney's example), it is rather 
obviously defective to place unequal point values based on place vote 
positions.

That is, Borda on its face is very highly defective (due to having divided 
majorities).  Borda hangs on for the same defective reason that plurality 
hangs on-- it is simple but simply wrong.

The clone problem is also lurking with Borda.

N1   AB
N2   BA
N3   D


N1 plus N2 = majority, N3 = minority

C comes along who may or may not be a 100 percent clone of A or B OR that A 
or B becomes a 100 percent clone of C.

N1.1  CAB
N1.2  ACB
N1.3  ABC

N1 subtotal

N2.1  CBA
N2.2  BCA
N2.3  BAC

N2 subtotal

N3    D


Another example--

27 ABCD
26 BCDA
24 CDAB
23 DABC
100

Place Votes
        1     2     3     4

A      27    23    24    26
B      26    27    23    24
C      24    26    27    23
D      23    24    26    27
      100   100   100   100

Net Place Votes (1 minus 4, 2 minus 3)

      1-4    2-3  Sum

A       1    -1    0
B       2     4    6
C       1    -1    0
D      -2    -4   -6

        2    -2    0


Possible two member clone sets--

74 AB 26
76 BC 24
77 CD 23
73 DA 27

Should D lose (a possible highest clone) ???


Another example-

18 AMBC
17 BMCA
16 CMAB

51 subtotal

49 D

100 total

M is second on all of the divided majority ballots.

Expand for larger divided majorities (such as M is in all third place votes 
with a divided five majority).

As usual, I note that ALL relative place votes do NOT show *absolute* support.