[EM] Challenge - give an example where FBC is violated for Condorcet methods

Peter Zbornik pzbornik at gmail.com
Mon Jun 6 12:02:57 PDT 2011


Kevin,

I am pretty sure you did not give an example, where generalized
symmetrical completion as defined (in the challenge) make the problem
worse.
You gave me an example where symmetrical completiion makes things
worse when it comes to FBC, but symetric completion and generalized
symmetric completion give different results.

With generalized symmetrical completion plurality holds trivially, as
margins and winning votes with give exactly the same results with
equalities resolved as 0.5 vs 0.5.

If you mean the example:
35:A>B
25:B
40:C
then standard Schulze Winning Votes gives B as the winner.
Generalized symmetrical completion gives also B as the winner.
Symmetrical completion gives the winner C.

Generalized symmetrical completion completes the election to
35:A>B>X>C
25:B>X>A=C
40:C>X>A=B

The stanard treatment of truncated ballots is the following:
(i) equal ranking of canditates explicitly not ranked,
(ii) lower ranking of candidates explicitly not ranked than candidates
explicitly ranked.
(iii) ties resolved 0:0, i.e. winning votes

Symmetric completion:
(i) equalities between candidates on the ballot (A=B) are resolved as
0.5 votes for each candidate in the pairwise comparison - i.e. margins
(ii) candidates not explicitly ranked on the ballot are ranked equally
below the null candidate.

Generalized symmetrical completion is defined as follows:
(i) equalities between candidates on the ballot (A=B) are resolved as
0.5 votes for each candidate in the pairwise comparison.
(ii) Ballots are completed with the null candidate ("none of the
above" X), who is ranked strictly lower than all candidates explicitly
ranked on the ballot,
(iii) candidates not explicitly ranked on the ballot (i.e. truncated
ballots, excluding the null candidate) are ranked equally below the
null candidate.
Example: election with candidates A, B C, the bullet-vote ballot A is
completed as A>X>B=C, A>B>C is completed as A>B>C>X).

Best regards
Peter Zborník

On Mon, Jun 6, 2011 at 8:18 PM, Kevin Venzke <stepjak at yahoo.fr> wrote:
> Hi Peter,
>
> --- En date de : Lun 6.6.11, Peter Zbornik <pzbornik at gmail.com> a écrit :
>> Personally I think that Condorcet methods (take Schulze as
>> reference)
>> with symmetrical ballot completion might satisfy FBC, but I
>> am not
>> sure, so that's why I ask for some guidance here.
>
> Actually I'm pretty sure I explained to you recently why symmetrical
> ballot completion makes the problem worse.
>
> Kevin Venzke
>
> ----
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>



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