[EM] Verification of a voting outcome for VoteFair.
ElectionMethods at VoteFair.org
Wed Apr 4 11:06:12 PDT 2012
My comments are interspersed as answers to specific questions/statements.
On 4/3/2012 12:53 AM, Kristofer Munsterhjelm wrote:
> On 04/03/2012 07:22 AM, Richard Fobes wrote:
>> Conclusion 11: VoteFair ranking is consistent with the Condorcet-Kemeny
>> method because the Condorcet-Kemeny method does not specify an overall
>> ranking, nor does it specify who should win this election. (It only
>> specifies that choice A is the lowest-ranked choice, and this is
>> consistent with the VoteFair ranking result.)
> Is that true? Speaking of Kemeny as an optimization method, the problem
> is specified as:
> Determine the transitive ordering of all candidates,
> so that the sum of, for each pairwise preference X consistent with the
> ordering, the number of voters agreeing with preference X, is maximized.
> (At least in absence of ties)
> That seems to make it pretty likely that Kemeny specifies a full
> ordering. It has to, in order to calculate the score that is to be
The Condorcet-Kemeny method does produce an unambiguous full ordering IF
there is only one "maximized" score.
When two or more sequences produce the same score and there is no
obvious way to resolve the match -- such as with a simple tie -- then
the "Kemeny" method does not specify a full ranking. (This situation
applies to the instance/case we are discussing.)
VoteFair ranking does resolve such ambiguous cases, and it produces a
full ranking regardless of how many sequences have the same highest
score, and regardless of how unsymmetrical those sequences are. (Of
course the ranking may involve ties.)
> But anyway, I'll try to find an example where:
> - VoteFair elects A,
> - VoteFair has no ties in its social ordering,
> - Kemeny finds another candidate X as the winner,
> - There is no Kemeny-optimal ordering that puts A first.
> Would that suffice to show that VoteFair isn't Kemeny?
No. As Jameson Quinn points out (in a message I haven't had time to
reply to yet), real-world elections typically involve no more than four
candidates in the Smith set. VoteFair ranking easily ranks Smith-set
candidates at the top (for reasons I plan to explain later). So, you
might be able to find a set of ballots for 50 candidates in which ALL
(or most) of the candidates are in the Smith set (and there are no
ties), for which VoteFair ranking identifies a "non-Kemeny" winner. But
as I've repeatedly said, if the voter preferences are that ambiguous,
then the difference is not significant.
>> Conclusion 12: VoteFair ranking calculates a fair result within the
>> limitations of the preference information available, and does so within
>> the context of the goal of maximizing the Condorcet-Kemeny sequence
> It doesn't actually maximize that sequence score, however; it falls one
> short. It does provide the same winner as one of the sequences that do,
> I see that.
Actually VoteFair ranking does find both the sequences that have the
same highest ("maximized") sequence score, but the "Kemeny" method --
even as you've defined it above -- does not specify how to resolve this
"tie" in the sequence scores.
Notice that in such cases the "Kemeny" method does not specify choosing
one of the sequences. (They have the same score, so they are equally
>> This is an excellent example of the unclear (muddled) preferences that I
>> have referred to in other messages. It clarifies what I've said before,
>> which is that if there were a 50-candidate election that had this kind
>> of circular ambiguity throughout the candidates (which is what can make
>> it harder to quickly find the highest sequence score), and if VoteFair
>> ranking failed to find the sequence with the highest score (assuming
>> only one such sequence), then a runoff election between the fully
>> calculated Condorcet-Kemeny winner and the VoteFair ranking winner would
>> be difficult to predict.
> You could say the same of other Condorcet methods. E.g. you could say
> "If Ranked Pairs fails to find the winner according to Kemeny, the
> outcome of a runoff election between the fully calculated Kemeny winner
> and Ranked Pairs would be difficult to predict". Still doesn't save
> Ranked Pairs from not being Kemeny, though! :-)
Yes, I should probably be more rigorous than just saying that a
runoff-election outcome would be difficult to predict.
I would rather say that if there were a difference between "Kemeny" and
"VoteFair ranking" (keeping in mind that it would have an unusually
large Smith set), then the _fairness_ of the VoteFair-ranking winner
matches the _fairness_ of the Kemeny winner. But of course the word
"fairness" is difficult (if not impossible) to define.
More specifically (as I tried to convey in another message), if a set of
ballots can produce such a difference, the difference will be small
compared to the difference between various possible voting methods, even
if those various possible voting methods are limited to the ones
supported in the Declaration.
I continue to appreciate that you are taking an interest in fully
understanding the Condorcet-Kemeny method.
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