Condorcet cyclic drop rule

[DEMOREP1 at aol.com]

However, I'm not sure I agree with this.  How should we interpret tied

preferences (eg A=B>C>D=E)?  There is one view (with which I'm inclined to

agree) that gives each candidate in a pairwise tie 0.5 votes.  Truncating

your vote is the same as tieing all of the unranked candidates - the

previous example should be equivalent to A=B>C.  If you count the ballots in

this fashion, all of the ways to interpret pairwise defeats are equivalent -

ratios, winning votes, margins.
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D- Another reason to have 0.5 votes in pairings arises in part from a place 
votes table.

Example

18 ABC
17 BCA
16 CAB
48 Z
99

If the place votes matrix is to be *complete*, then

      1       2     3     4    

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

    99     99    99    99

That is,  A,B,C each get 48/3 votes in 2nd, 3rd and 4th 

The head to head table should also likewise be *complete*.