Bart's "Median Rating" method?

[DEMOREP1 at aol.com]

Mr. Eppley wrote--

I think the definition of Bart's Median rating method needs 
clarification, since Bart's claim about highest median rating 
and majority appears dubious.  Here's an example to illustrate 
the problem:

   voter 1:  A=95, B=65
   voter 2:  A=85, B=60
   voter 3:  A=50, B=20
   voter 4:  A=40, B= 0
   voter 5:  A=45, B=55

   There is a majority (80%) who rank A ahead of B.

   Average rating for A = 63
   Average rating for B = 40

   Median rating for A  = 50?
   Median rating for B  = 55?

If I've interpreted correctly how Bart intends it to be 
tallied, B is the candidate with the highest "median rating."  
But I wouldn't agree that B is rated higher than A by a 
majority of voters.
----
D- This is deja vu all over again from late 1998.

I suggested that if medians are to be used, then (1) only the choices getting 
above 50 medians (on a 0 to 100 scale) should (2) go head to head (Condorcet) 
 (with (3) the highest median winning if there was no Condorcet winner).

Median = middle value of all values ranked high to low
A 65, 60, 55, 20, 0    median 55
B 95, 85, 50, 45, 40   median 50

Thus, B automatically wins (median 55) since A does not have an above 50 
median (i.e. stop at step (1)).

If voter 3 voted A=51, B=20, then the A median would be 51 so that A should 
beat B head to head (i.e. stop at step (2)).

The above 50 median test is a refinement of my suggested YES/ NO test (i.e. 
it gives some more info about the intensity of the support for a choice).