Condorcet cyclic drop rule
DEMOREP1 at aol.com
DEMOREP1 at aol.com
Wed Mar 28 09:33:05 PST 2001
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.
---
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*.
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