Bart's "Median Rating" method?

[DEMOREP1 at aol.com]

Mr. Eppley wrote in part --

Here's an example which may make this even clearer:

   voter 1:  A=99, B=51
   voter 2:  A=99, B=51
   voter 3:  A=50, B=51
   voter 4:  A=49, B= 0
   voter 5:  A=49, B= 0

   One voter (#3) has a tiny preference for B.
   All the other voters "strongly" prefer A more than B.  

Median Rating suggests that B is best, which suggests that 
Median Rating should not be used to evaluate ratings.
-----
D - More deja vu somewhat.    

A choice is or is not acceptable to a majority on a simple YES/NO vote.

Currently, each vote is deemed a 100 vote.

However, a vote has an intensity factor (on a 0 to 100 scale or on a +100 to 
-100 scale). 

Thus the above example can also be viewed as -
2 A (99) > B (51)
2 A (49) > B (  0)
1 B (51) > A (50)

Net 3 A > B

As I have suggested, medians might be used as a tiebreaker if there was no 
head to head winner.  In the above, A is obviously the head to head winner 
(i.e. no need to look at the intensity factors or ratings).

However- a simple circular tie
Assume that A, B and C each are acceptable to a majority of the voters (at 
least 7).
5 ABC
4 BCA
3 CAB
12

8A>B4, 9B>C3, 7C>A5 or A>B>C>A

Rather than using simple YES votes as a tiebreaker, the intensity factors 
(i.e. medians) might be used.  

Thus, 3 tiebreaker extreme examples -

Tiebreaker 1
5 A (100)   B ( 51)   C (0)
4 B (100)   C ( 51)   A (0)
3 C (100)   A ( 99)   B (0)

5 A 100        4 B 100      3 C 100
3 A   99        5 B   51      4 C   51
4 A    0         3 B    0       5 C    0
        99                 51             51    medians -- A wins
--
Tiebreaker 2
5 A (100)   B ( 99)   C (0)
4 B (100)   C ( 51)   A (0)
3 C (100)   A ( 51)   B (0)

5 A 100        4 B 100      3 C 100
3 A   51        5 B   99      4 C   51
4 A    0         3 B    0       5 C    0
        51                 99             51    medians -- B wins
--
Tiebreaker 3
5 A (100)   B ( 51)   C (0)
4 B (100)   C ( 99)   A (0)
3 C (100)   A ( 51)   B (0)

5 A 100        4 B 100      3 C 100
3 A   51        5 B   51      4 C   99
4 A    0         3 B    0       5 C    0
        51                51             99    medians -- C wins

However, note that in each example
A 8 > 50
B 9 > 50
C 7 > 50

Choice B might be very unhappy with the use of a medians tiebreaker (in 
Tiebreaker 1 and 3).

Thus, the question becomes is a 100 or 99 intensity YES vote more important 
than a 51 intensity YES vote in a tiebreaker situation ??? 

Note in Tiebreaker 3 that if the votes 
4 B 100      3 C 100
5 B   51      4 C   99
3 B    0       5 C    0
are paired from top to bottom using the intensities (ratings) there would be

  B        C
100   100     tie
100   100     tie
100   100     tie
100      99     B
  51      99     C
  51      99     C  upper half
  51      99     C  lower half
  51      0       B
  51      0       B
    0      0      tie
    0      0      tie
    0      0      tie

Overall tie --- however 2 C > 1 B in the upper half and 2 B > 1 C in the 
lower half (showing the median effect- the higher median of C).   
Especially note however, the 4 B (100) first choice voters by their C (99) 
second choice votes helped defeat their first choice.

Thus, pair the votes using the ratings in circular tie cases ???

Other folks may want to produce more examples (noting that in real elections 
there might be all sorts of ratings other than 100, 99, 51 or 0) ???