[EM] Survey of voting methods obeying Mike Ossipoff's "Favorite Betrayal Criterion"

Kevin Venzke stepjak at yahoo.fr
Fri Nov 18 20:10:54 PST 2005


--- Warren Smith <wds at math.temple.edu> a écrit :
> now available at
> http://math.temple.edu/~wds/crv/FBCsurvey.html

The "MDD" process is inherently incompatible with clone
independence. No matter what you do after MDD, it's possible
that cloning the winner, such that there's a majority-strength
cycle among the clones, will cause the win to necessarily move
to some other candidate.

>Perhaps also other voting methods may be converted to be FBC-
>compliant by using Venzke's "tied at the top trick." (I am not 
>quite sure when this works.) 

I can think of two very different ways. In ICA, which seeks
to disqualify candidates with pairwise losses, it is never
beneficial to {a,b} for there to be a pairwise win for one over
the other. The "tied at the top" interpretation makes it so
that you need not strictly rank A over B (or vice versa) to
eliminate a win between them. It's always at least as effective
to rank them equally.

For another way, consider a (rather bad) method which elects
the candidate with the greatest number of rankings over some
other candidate. (That is, if the matrix has no value greater
than v[x,y], elect X.) By itself, this method doesn't satisfy
FBC because it could be that introducing a strict ranking
between X and Y could cause v[x,y] to be the greatest win and
move the win to X from some other candidate.

But we can use the "tied at the top" rule to fix this. Just say
that when X and Y are "tied at the top" on a given ballot, this
vote counts to both v[x,y] and v[y,x]. Then all that can be
"gained" by introducing a strict X>Y ranking is that v[y,x] is

These applications are kind of opposites. ICA uses the trick to
delete pairwise wins. This latter method uses the trick to beef
up the maximum scores of each tied candidate against the other,
which is harmless since a high score doesn't hurt the pairwise

Kevin Venzke


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