[EM] Condorcet question - why not bullet vote

[Peter Zbornik]

Hi,

a fourth method of measuring the strength of pairwise wins in Condorcet
would be a statistical test called Binomial test.
See: http://faculty.chass.ncsu.edu/garson/PA765/binomial.htm
Excel calculations at: http://udel.edu/~mcdonald/statexactbin.html

Short explanation:
Each vote for candidate A is coded as 1 and a vote for B is coded sa 0
n is the number of voters, who are not indifferent between A and B.
The hypothesed (H0) value of the mean is 0.5.
We compare the proportion of votes for A against the hypothesized mean 0.5
and calculate the probability that this prortion of votes for A is 0.5 (we
can use single or two tail tests, it doesn't matter, as the distribution is
symmetric).
The lower the probability that the proportion is 0.5, the bigger the
strength of the win for A.
Thus the rule to eliminate the weakest undropped link in point 3 at
http://en.wikipedia.org/wiki/Schulze_method#The_Schwartz_set_heuristic will
amount to dropping the win, with the highest p-value (i.e. the win with the
highest probability that, for an infinite number of voters from the same
distribution as the current voters, the proportion of voters for candidate A
is 0.5).

I would prefer this method before proportional completion (let's call it
Schulze Condorcet (binomial test), at least as an initial hunch.
This method would give a penalty to wins with many indifferent voters due to
few observations, which proportional completion does not.

I have no idea of what criterias
<http://en.wikipedia.org/wiki/Schulze_method#Comparison_with_other_preferential_single-winner_election_methods>will
be satisfied or failed.

Best regards
Peter Zborník

On Thu, Jun 17, 2010 at 8:51 AM, Peter Zbornik <pzbornik at gmail.com> wrote:

> Hi,
>
> Kevin, thanks for the comment.
> Well, it is true, that Schulze writes in
> http://m-schulze.webhop.net/schulze1.pdf, page 154, that "There has been
> some debate about how to define D [Schulze ranking relation] when it is
> presumed that on the one side each voter has a sincere linear ordering of
> the candidates, but on the other side some voters cast only a partial
> ordering because of strategic considerations. We got to the conclusion that
> the strength (N[e,f],N[f,e]) of the pairwise win ef ∈ A × A should be
> measured primarily by the absolute number of votes for the winner of this
> pairwise defeat N[e,f] and secondarily by the absolute number of votes for
> the loser of this pairwise defeat N[f,e]."
>
> However, for Schulze STV, proportional completion is used for incomplete
> orderings (see page 42 in http://m-schulze.webhop.net/schulze2.pdf).
> I thought, that Schulze STV reduces to Schulze Condorcet in the case where
> there is one seat.
> Now, this seems not to be the case when we have incomplete ballots (i.e. we
> allow for equal ranking of candidates), as Schulze Condorcet uses winning
> (and losing) votes and Schulze STV uses proportional completion before
> deciding upon winning votes.
>
> Maybe Markus Schulze could comment on this himself.
> I think proportional completion could be used in Schulze Condorcet, but
> there is obviously one big open question in this respect.
> Does Schulze Condorcet (proportional completion) meet the same criteria as
> Schulze (WV),
> http://en.wikipedia.org/wiki/Schulze_method#Satisfied_and_failed_criteria?
>
> Kevin, you seem to say that Shulze Condorcet (proportional completion) does
> meet the same criteria as Minimax(margins), quote "If you are using
> proportional completion (or "symmetric completion") then you're not using
> winning votes, you're using margins".
> Are you sure about this?
>
> Schulze Condorcet (proportional completion) gives different results than
> Schulze Condorcet (margins).
>
> For instance: Say we have two pairwise defeats and 100 voters - A vs B.
> First defeat A-B, 1-5. Margin gives 4 as the strength of the win.
> Proportional completion gives: 1+94*1/6 - 5+94*5/6=16,67-83,33, i.e. a
> margin of 66,67 (94 voters, each split into two with proportional weights).
> Second defeat A-B, 48-52. The margin is 4 both with proportional completion
> and without.
>
> Thus, it seems that proportional completion gives different results from
> both the winning (losing) votes approach and the margin approach for
> truncated Condorcet ballots.
> The natural question is:
> What are the differences in satisfied and failed criteria (
> http://en.wikipedia.org/wiki/Schulze_method#Satisfied_and_failed_criteria)
> between Schulze Condorcet (proportional completion) and, Schulze Condorcet
> (WV)?
>
> Best regards
> Peter Zborník
>
> On Wed, Jun 16, 2010 at 11:29 PM, Kevin Venzke <stepjak at yahoo.fr> wrote:
>
>> Hi Peter,
>>
>> --- En date de : Mer 16.6.10, Peter Zbornik <pzbornik at gmail.com> a
>> écrit :
>> >Thus: "If the three C voters will truncate then they will win instead of
>> B
>> >in winning votes based Condorcet methods."
>> >
>> >This is correct, if proportional completion is not used (see page 42
>> >in http://m-schulze.webhop.net/schulze2.pdf)
>> >If proportional completion is used (which I would recommend) then B wins.
>>
>> If you are using proportional completion (or "symmetric completion") then
>> you're not using winning votes, you're using margins.
>>
>> Juho advocates MinMax(margins) which is why he posted this example
>> (Schulze is usually assumed to use winning votes), and also why he didn't
>> like it when I pointed out that clone independence and ISDA were the
>> probable answers to your criteria question
>>
>> Kevin Venzke
>>
>>
>>
>>
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>> info
>>
>
>
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