# [EM] An attempt at a method satisfying FBC and SDSC

Kevin Venzke stepjak at yahoo.fr
Sun May 22 18:34:51 PDT 2005

```Hello,

Consider this method:

The voter ranks the candidates. All preference orders are admissible.
Candidates ranked above-last are considered "approved."
Let v[a,b] signify the number of voters ranking A over B.
Let t[ab] signify the number of voters ranking A and B in equal-first
(possibly with other candidates).
Let maxfor[a] signify the greatest value of v[a,x]+t[ax] where X is
another candidate.
Let maxagainst[a] signify the greatest value of v[x,a] where X is
another candidate.
Define a set S containing every candidate A for whom maxfor[a]>=maxagainst[b].
Elect the member of S with the greatest approval.

First let me point out that this method isn't clone-proof. I do believe
it's monotonic, though.

This method satisfies SDSC: If more than half of the voters rank A>B
and rank B above no one, then maxfor[b]<50% while maxagainst[b]>50% so
that B is not a member of S.

My argument that this satisfies FBC is the same as my argument that the
three-slot Condorcet//Approval variant (with the special tie rule)
satisfies FBC. That argument could be stronger, though, so I'll rehash
it here:

Raising A to equal-first cannot bump any candidate B also in equal-first
out of the set S. At first glance it seems that it may decrease maxfor[b].
But every vote deducted from v[b,a] is added to t[ab], so that there is
no effect on maxfor[b].

Raising A to equal-first can't move any lower-ranked candidate C into
the set S, because it just increases v[a,c] and possibly maxagainst[c].

If raising A does bump C out of S, causing a less-liked candidate D to
be elected, then raising C to equal-first with A will result in C
having at least the same maxfor and at most the same maxagainst (as
before A was raised), so that C will be in S again.

And the tie-breaker, again, is Approval, which clearly satisfies FBC.

Thoughts?

Kevin Venzke

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