Bob Richard wrote:
Kristofer Munsterhjelm wrote:
> If we consider the votes as bullet votes, then we can expand to:
> 45: Able > Baker = Charlie
> 40: Baker > Able = Charlie
> 15: Charlie > Able = Baker
> which produces the matrix you gave above.
Able Baker Charlie
------- ------- -------
Able -- 45 45
Baker 40 -- 40
Charlie 15 15 --
OK, I was wrong when I said the cross-diagonal cells have to add up to
100. This way of accounting for tied rankings dictates otherwise.
Suppose, instead, we treat tied rankings as a half a vote for each
candidate:
22.5: Able > Bake > Charlie
22.5: Able > Charlie > Baker
20.0: Baker > Able > Charlie
20.0: Baker > Charlie > Able
7.5: Charlie > Able > Baker
7.5: Charlie > Baker > Able
Able Baker Charlie
------- ------- -------
Able -- 52.5 65.0
Baker 47.5 -- 62.5
Charlie 35.0 37.5 --
In another post in this thread, Raph Frank describes a third way of
representing tied rankings using proportions. Using the example above
instead of his example:
32.73: Able > Baker > Charlie
12.27: Able > Charlie > Baker
30.00: Baker > Able > Charlie
10.00: Baker > Charlie > Able
7.94: Charlie > Able > Baker
7.06: Charlie > Baker > Able
Able Baker Charlie
------- ------- -------
Able -- 52.94 75.00
Baker 47.06 -- 72.73
Charlie 25.00 27.27 --
In this example, Able is the Condorcet winner in all three matrices.
Several questions:
(1) Is this true in general, for all possible profiles? If there's a
Condorcet winner, is it always the same candidate no matter how you
treat tied rankings?
I think so, for the two first at least. What you've touched upon is the
wv versus margins argument.
Instead of altering the Condorcet matrices, you may count the victories
differently. That way you can use the same Condorcet matrix no matter
how your system counts votes.
Let's call cm[a][b] the strength of A vs B; that is, the number of
voters who preferred A to B. Then each Condorcet method acts upon a
different matrix, which we may call v (for victory), so that v[a][b] is
the victory of A over B.
WV means "winning votes". If you use wv, v[a][b] is cm[a][b] if cm[a][b]
> cm[b][a], otherwise 0.
Margins is defined similarly. v[a][b] is cm[a][b] - cm[b][a] unless this
value would be negative, in which case it's zero.
Finally, there's pairwise opposition, where v[a][b] is simply cm[a][b].
MMPO is Minmax(pairwise opposition), and doesn't actually pass Condorcet.
One may also define two other variants which are more uncommon: Ratio
margins, which is cm[a][b]/cm[b][a], and votes against, which is
total_votes - cm[b][a] if cm[a][b] > cm[b][a], otherwise 0. Something
like ratio margins is mentioned here:
http://listas.apesol.org/pipermail/election-methods-electorama.com/2001-March/005622.html
. Presumably there would have to be a way of handling infinities when
cm[b][a] = 0. A variant of ratio margins is given at
http://listas.apesol.org/pipermail/election-methods-electorama.com/2005-May/016023.html
and solves the infinity problem by having v[a][b] be equal to (cm[a][b]
- cm[b][a]) / (cm[a][b] + cm[b][a]).
Your second modification, where you count ties as 0.5 against both,
turns wv into margins.
(2) Are there profiles containing cycles for which different
Condorcet-completion methods would give different winners depending on
how the tied rankings are represented?
Yes, otherwise there would be no wv vs margins argument. I can't find a
concrete example, though, since my counting program doesn't handle tied
votes. (My simulator does, but it's a bit of a hack to get it to read
output from a file)
Perhaps somebody else can find an example, say for Ranked pairs as
compared to MAM? (Both use the same fundamental method, but ranked pairs
is margins, and MAM is wv, although some times "ranked pairs" is used on
this list to describe MAM)
(3) Going back to Dave Ketchum's original proposal that different voting
methods can be used in different subjurisdictions (e.g. states in the
case of NPV) and the matrices added together, could the method of
representing tied rankings ever affect the outcome in the jurisdiction
as a whole? I haven't tried to work this out, but intuitively it seems
to me that the answer is yes.
Yes, and mixing margins and wv explicitly would cause a mess. Therefore,
it's better to have one format for the Condorcet matrix itself, and just
translate it into the appropriate victory matrix according to what
method you're using.
(4) I gather that Kristofer's procedure is the one most frequently used
in discussions of Condorcet. Is that true, and what is the history or
reasoning behind this?
Once you've decoupled the condorcet matrix from the wv/margins choice,
it makes sense that "my" way of counting the Condorcet matrix would be
used. As for whether wv or margins is most common, I think wv is, and
that the reason is that it's less vulnerable to strategy (order reversal
and favorite betrayal). Also, Schulze(wv) meet some criteria that
Schulze(margins) do not, so the Schulze method's defined to use wv (as
far as I know).
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