# [EM] Verification of a voting outcome for VoteFair.

Richard Fobes ElectionMethods at VoteFair.org
Mon Apr 2 22:22:38 PDT 2012

```On 4/2/2012 11:41 AM, Kristofer Munsterhjelm wrote:
> Hello,
>
> could you verify that when VoteFair is given the ballot set:
>
> 2: A > B > C > D
> 1: A > C > D > B
> 1: B > A > C > D
> 1: B > C > D > A
> 2: C > B > D > A
> 2: C > D > B > A
> 1: D > A > B > C
> 1: D > A > C > B
> 2: D > B > A > C ,
>
> which implies a row-beats-column pairwise matrix of
>
> 0 5 8 4
> 8 0 7 6
> 5 6 0 9
> 9 7 4 0 ,
>
> it provides a social ordering of C>B>D>A?

This is a very interesting case.

Calculating all the sequence scores finds that there are two sequences
with the highest score of 44:

[below counts apply to sequence: B , C , D , A ] [seq score =     44]

[   ---       7       6       8  ]
[     6     ---       9       5  ]
[     7       4     ---       9  ]
[     5       8       4     ---  ]

[below counts apply to sequence: C , D , B , A ] [seq score =     44]

[   ---       9       6       5  ]
[     4     ---       7       9  ]
[     7       6     ---       8  ]
[     8       4       5     ---  ]

On this basis, the Condorcet-Kemeny method does not yield any particular
ranking.  As a result, it does not specify who should win the election.

This ambiguity arises because the Condorcet-Kemeny method does not
specify what to do if more than one sequence has the same highest
sequence score.

VoteFair ranking calculations find both of the sequences that have the
matching highest sequence score.

Among the (two) highest-score sequences, here are the highest and lowest
rankings for each choice:

[choice A has been at highest rank level 4 and lowest rank level 4]
[choice B has been at highest rank level 1 and lowest rank level 3]
[choice C has been at highest rank level 1 and lowest rank level 2]
[choice D has been at highest rank level 2 and lowest rank level 3]

VoteFair ranking resolves these ambiguities and calculates these results:

choice C is at ranking level 1
choice B is at ranking level 2
choice D is at ranking level 3
choice A is at ranking level 4

This matches what Kristofer Munsterhjelm presents as his result:

On 4/2/2012 11:41 AM, Kristofer Munsterhjelm wrote:
> it provides a social ordering of C>B>D>A?

However, this "declared-winning" sequence yields a sequence score of 43,
which is one less than the score of 44 that applies to the two
highest-score sequences.

Here is the arrangement of pairwise counts in the "winning" sequence:

[below counts apply to sequence: C , B , D , A ] [seq score =     43]

[   ---       6       9       5  ]
[     7     ---       6       8  ]
[     4       7     ---       9  ]
[     8       5       4     ---  ]

Here are my conclusions:

Conclusion 1:  This case is not fully resolvable according to the method
defined by John Kemeny because there is more than one sequence with the
same highest score.  The "Kemeny" method does not specify how to resolve
this kind of situation.  (This "dead end" is analogous to the (original)
Condorcet method not specifying what to do if there is no Condorcet winner.)

Conclusion 2:  The two sequences that have the highest sequence score
agree that choice A is least popular, but otherwise the results are
ambiguous.

Conclusion 3:  The two highest ranking sequences rank the other choices
as B , C , D and C , D , B.  Neither ranks choice D in first place, and
one ranks choice B as first and the other ranks choice C as first.  This
information suggests that either B or C is most popular.

Conclusion 4:  Choices B and C are not tied because one highest-ranking
sequence ranks choices B and C in the top two positions, yet the other
sequence ranks choices C and B at the first and third positions.  (If
instead the two highest-ranking sequences ranked choices B and C in
first and second place in both cases, then of course they would be tied
for first place.)

Conclusion 5:  Choice C makes sense as the winner because it is in
either first or second place in the two highest-ranking sequences.
Choice B makes sense as not being the winner because it is in first
place or third place in those two sequences.

Conclusion 6:  The pairwise counts for choices B and C indicate that 7
voters prefer choice B over choice C, whereas 6 voters prefer choice C
over choice B.  This information suggests that choice B is preferred
over (winning) choice C, but only by one vote.

Conclusion 7:  The plurality counts below suggest that choice B is not
popular because it is the first choice of only two voters.

plurality count for choice C is 4
plurality count for choice D is 4
plurality count for choice A is 3
plurality count for choice B is 2

Conclusion 8:  The voter preferences are unclear because they involve a
high level of circular ambiguity.  In particular, least-popular choice A
is preferred over the winning choice C by 8 to 5.

Conclusion 9:  The Smith set contains all four choices because there is
no smaller subset that pairwise beats all the remaining choices.

Conclusion 10:  If there was a runoff election between choices B and C,
the outcome would be difficult to predict.  Although the B-versus-C
pairwise counts suggest that B would win by one vote, the other
preferences are so unclear that we cannot see a clear pattern of
preferences.  And when preference patterns are unclear, there may have
been strategic voting going on, and the strategies will disappear in a
runoff election.  Also, the muddled preferences suggest that a runoff
election will cause some voters to realize that they have changed their
opinions about their relative preference for the two runoff candidates,
and a single voter switching their preference can alter the runoff results.

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.)

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 score.

Thank you Kristofer for this interesting case!

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.

Richard Fobes

--------------------------------------------------------

BTW, as confirmation that the data I entered matches the supplied data,
this is the starting arrangement of pairwise counts:

[   ---       5       8       4  ]
[     8     ---       7       6  ]
[     5       6     ---       9  ]
[     9       7       4     ---  ]

On 4/2/2012 11:41 AM, Kristofer Munsterhjelm wrote:
> ...
> which implies a row-beats-column pairwise matrix of
>
> 0 5 8 4
> 8 0 7 6
> 5 6 0 9
> 9 7 4 0 ,

```

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