[EM] Truncated Votes- Divided Votes

[DEMOREP1 at aol.com]

There has been some question about whether or not fractional votes should 
exist in head to head tables having truncated votes.

Based on a place votes table, such fractional votes would seem to be required 
-- such that any method which ignores such truncated votes is improper.

26 ABZ
25 BAZ
24 ZAB
24 ZBA

99

     1     2    3

A   26    49    24      99
B   25    50    24      99
Z   48     0    51      99

    99    99    99

  1-3     2

A   2    49    51
B   1    50    51
Z  -3     0    -3

    0    99    99 

There are now truncated votes---

26 AB [Z]
25 BA [Z]
48 Z [A=B]

99

In order to get the same results as the full votes example the truncated 
votes must be divided by the number of choices being truncated.

(i.e. the 48 Z votes are divided by 2)

That is, in a head to head table, it is NOT 26 AB 25 but is 50 (26+24) AB 49 
(25+24).

A larger example--

18 ABC [Z]
17 BCA [Z]
16 CAB [Z]
48 Z [A=B=C]

99

The 48 Z votes are divided by 3 (i.e. 16) for places 2, 3 and 4 for A, B and 
C.
 
      1     2      3     4    Total

A    18     32    33    16     99
B    17     34    32    16     99
C    16     33    34    16     99
Z    48      0     0    51     99

T    99     99    99    99

     1-4    2-3   Total

A     2     -1      1
B     1      2      3
C     0     -1     -1
Z    -3      0     -3

T     0      0      0

Also note--

    1+2    3+4   Total

A    50    49     99
B    51    48     99
C    49    50     99
Z    48    51     99

    198   198

That is, the left half of a place votes table will generally not be the same 
as the right half.

Thus, should the earliest place votes (first, first plus second, first plus 
second plus third, etc.) simply be summed to get a majority (if there is no 
Condorcet winner in a head to head table) ???

OR-- the reverse--- eliminate a choice that has the earliest majority vote 
against it in the right side of the place votes table (such as Z) and move up 
the votes for the remaining choices on the ballots involved ???

Note that a majority might be very highly divided (3, 4 or more choices) with 
divided minorities (2, 3 or more choices).

With a large number of candidates there would very likely be lots of 
truncated votes.  Thus, computers would generally be needed to quickly do the 
N1 truncated votes/ N2 truncated choices division, sums and differences math.

As usual, a place votes table shows ONLY relative support and NOT *absolute* 
support.  The use of YES/NO would eliminate lots of choices (leaving a place 
votes table with only YES majority choices).