[EM] Bullet voting/truncation in Condorcet elections (was Re: NPV vs Condorcet)

Dave Ketchum davek at clarityconnect.com
Wed Oct 22 18:02:17 PDT 2008


On ed, 22 Oct 2008 23:43:34 +0200 Kristofer Munsterhjelm wrote:
> 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.

Following examples are different but the above, only, is what I proposed.
>>
>> 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?
> 
DOES NOT MATTER:
      This proposal is going down the NPV path.
      But do Condorcet instead of FPTP.
      Provide for scaling N*N matrices, assuming that may help in killing 
the Electoral College.
      Provide for merging in counts from states which are not ready to do 
Condorcet, and therefore stay, temporarily, with FPTP.
      But this is ONE election that will analyze the national N*N matrix. 
Its definition will involve topics such as margins and wv, but all that 
state matrices are used for is adding together in the national matrix. 
State matrices can, of course, be published for whatever value that may have.
      There is expectable desire by some states to use other methods such 
as Range.  My claim (not yet proven) is that doing such well would be more 
pain than it is worth.
> 
> 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?
> 
Again DOES NOT MATTER, for there is only one national matrix to be completed.
> 
> 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).
-- 
  davek at clarityconnect.com    people.clarityconnect.com/webpages3/davek
  Dave Ketchum   108 Halstead Ave, Owego, NY  13827-1708   607-687-5026
            Do to no one what you would not want done to you.
                  If you want peace, work for justice.






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