Hi Kristofer and Richard, Richard
Fobes says that his VoteFair popularity ranking method is just another name for
the Condorcet-Kemeny or Kemeny-Young method.
Accordingly, Richard’s relevant pages in his book (Ending the Hidden
Unfairness…) are basically the same as the relevant part of the ‘Kemeny-Young’
Wikipedia article.

Still,
I have a question about the way these articles explain how the ranked alternatives
are to be scored in order to find the winner.
The ‘Kemeny-Young’ Wikipedia article offers the follow:

All
possible pairs

of choice names

Number
of votes with indicated preference

Prefer X over Y

Equal preference

Prefer Y over X

X = Selden

Y = Meredith

50

10

40

X = Selden

Y = Elliot

40

0

60

X = Selden

Y = Roland

40

0

60

X = Meredith

Y = Elliot

40

0

60

X = Meredith

Y = Roland

30

0

70

X = Elliot

Y = Roland

30

0

70

The sum of the
counts in each row must equal the total number of votes.

After the tally
table has been completed, each possible ranking of choices is examined in turn,
and its ranking score is calculated by adding the appropriate number from each
row of the tally table. For example, the possible ranking:

Elliot
Roland
Meredith
Selden

satisfies the
preferences Elliot > Roland, Elliot > Meredith, Elliot > Selden,
Roland > Meredith, Roland > Selden, and Meredith > Selden. The
respective scores, taken from the table, are

Elliot

Elliot

Elliot

Roland

Roland

Meredith

giving a total
ranking score of 30 + 60 + 60 + 70 + 60 + 40 = 320.

Calculating the overall ranking[edit]

After the scores for every possible ranking have
been calculated, the ranking that has the largest score can be identified,
and becomes the overall ranking. In this case, the overall ranking is:

Roland
Elliot
Selden
Meredith

with a ranking
score of 370.

S:  I had to confirm this as follows:

Roland>Elliot
70
2.      Roland>Selden  60
3.      Roland>Meredith70
4.      Elliot>Selden      60
5.      Elliot>Meredith  60
6.      Selden>
Meredith50
7.      _________________
Total Score      370
for this sequence
S:
My question is, why could we not always find the winning sequence
without have to score every possible
sequence as illustrated above?  Why
not simply add how many times the 100 voters have preferred each
alternative?  At least in this example,
the same winning sequence would have been discovered in this way:

Roland  200
Elliot      150
Selden    130
Meredith
110

My
second question responds to Kristofer’s answer in our 4th [EM] dialogue.
He suggested that ‘Ranked Pairs/MAM’ {i.e. maximize affirmed majorities} is  ‘a
simple yet good Condorcet method, how about [1]? It works like this:

1. Sort all the pairwise contests in
   order of strongest to weakest.

Discard those that are weaker than a
majority.

2. Go down the list and lock in a
   pairwise contest unless it contradicts

a contest you locked in earlier[2].

3. Once you're done, reassemble the
   pairwise contests into a ranking.

The candidate that is ranked first on it
wins.

Some additional details are required for
breaking ties, but I've left

those out here.’

S:  Is MAM
significantly different from Kemeny-Young/VoteFair?  Might VoteFair popularity ranking also Maximize
Affirmed Majorities as does MAM? If so, perhaps James
Green-Armytage’s following findings with
regard to MAM might equal characterize VoteFair:

The following part of JGA’s analysis of the different preferred results
when using the following different pairwise calculations to discover the winner,
her reports that:

“Judging from who beats whom, max. length, mean length, or sum
of

defeats, we get MAM

Judging from number of defeats (= Copeland score), we get

either MAM >
River+ > Beatpath

or MAM >
Beatpath > River+.”

[This is quotation from page 8 of JGA’s post 11
years ago:

From: James Green-Armytage

Subject: Heitzig method

Newsgroups: gmane.politics.election-methods

Date: 2004-09-20 06:29:21]

What do
you think?

On 10/01/2015 08:16 AM, steve bosworth wrote:

[snip]

Because it doesn't hold in every case. Kemeny is close to sum of
victories because it seeks to find an ordering

X1 > X2 > ... > Xn

so that the sum of
(X1 > X2) + (X1 > X3) + ... + (X1 > Xn) +
(X2 > X3) + (X2 > X4) + ... + (X2 > Xn) +
... +
(Xn-1 > Xn)

is maximized.

When there's no cycle, this is the same as sum of victories. However, if
there is a cycle, Kemeny breaks the cycle in the way that maximizes the
sum above. Sum of victories takes the sum over both sides of the cycle,
and that may alter the outcome.

Here's an example:

6: A > C > B
4: B > A > C
1: C > A > B
4: C > B > A

The magnitudes of the pairwise matchups are:

A>B:  7
A>C: 10
B>A:  8
B>C:  4
C>A:  5
C>B: 11

The majorities are:

A>C: 10
B>A:  8
C>B: 11

If we use wv (i.e. consider only majorities), then sum of victories gives:
A's score: 10
B's score: 8
C's score: 11

so its outcome is C>A>B.

However, Kemeny isn't permitted to consider both sides of a cycle, so if
we check the score for C>A>B along with another possible ordering...

Kemeny:
C>A>B: (C>A) + (C>B) + (A>B) = 0 + 11 + 0 = 11
A>C>B: (A>C) + (A>B) + (C>B) = 10 + 0 + 11 = 21

so A>C>B is a better Kemeny ordering. (Unless there's a bug in my voting
simulator, it is in fact the best Kemeny ordering.) Pretty much every
Condorcet method gives the win to A for this example - see
http://www.cs.wustl.edu/~legrand/rbvote/calc.html.

It's possible to construct similar examples if we consider every
pairwise result (not just majorities), but I suspect sum of victories
(sum of pairwise matchups) would in that case fail the Condorcet
criterion outright. But if you'd like to check anyway, try

5: A > B > C
2: B > A > C
5: C > A > B
3: C > B > A

The best Kemeny ordering is C>A>B but sum of pairwise matchups gives A>C>B.

Yes. RP/MAM passes independence of clones whereas Kemeny does not. It
does this because RP/MAM uses maximum instead of addition.

Consider a situation where Kemeny gives the same result as sum of
victories. Then there's a vote splitting problem: if A beats B, A might
go from losing to winning if B is cloned because A's victory against B
counts multiple times. The Kemeny vote-splitting example on the
Wikipedia page about independence of clones works in a similar way.

On the other hand, the Ranked Pairs mechanism doesn't have this problem.
Suppose C beats A (i.e. C>A). If A>B is locked before C>A before the
cloning, then no matter how many clones you add, A>Bk will be locked
before C>A is (where Bk is some clone) and this prevents Bk>A from being
locked later on.

Kemeny has no problem summing up the margins of victory of contests, and
that makes it possible to make some contests weigh more heavily than
others by adding clones. However, the Ranked Pairs mechanism takes the
strongest victory and in essence breaks the tie (among all orderings
consistent with the strongest victory) with the next strongest victory
and so on down (skipping contradictions). And since at least one clone's
relative position is locked when the original would have been, the
clones can only break ties consistent with the social ordering of the
first clone.

On the other hand, Kemeny passes reinforcement while neither RP nor MAM
does so. Reinforcement states that if you have the same social ordering
(ranking by the method) in every district and you combine the ballots to
get a nationwide result, then the nationwide social ordering will be the
same as that of the districts. Kemeny is the only Condorcet method that
passes that criterion.

By the logic above, I would imagine that judging by minimal sum of
defeats (same as maximal sum of victories when ballots are complete),
you'd get

Kemeny > MAM > River+ > Beatpath,

i.e. that Kemeny is closest to what sum-of-defeats provides. But from a
clone perspective, it's too close to sum-of-defeats.

In the Wikipedia clone example, MAM locks

B1 > B2, B1 > B3, B2 > B3
then B1 > C, B2 > C, B3 > C
and C>A, producing the ordering
B1 > B2 > B3 > C > A,
hence avoiding the clone problem.

The locking nature of RP/MAM also makes it easier to reason about RP/MAM
than about Kemeny. If you pick, say, A>B in Kemeny and there's a cycle
A>B>C>A, that might block a later path through C>B where, even though
A>B is stronger than C>B, the best path passing through C>B might give
you a higher score than the best path that passes through A>B. It
becomes rather like a puzzle where earlier decisions block off later
ones. On the other hand, when you do RP/MAM, it's easy: if A>B is
stronger than C>B, you affirm A>B before C>B, end of story, no lookahead
required. The only potential for needing to look ahead is if there's a
tie involving a cycle, but RP/MAM breaks that by random ballot (well,
random voter hierarchy) instead of doing any of that lookahead.