[EM] Example with tie in Schulze method but not in IRV
C.Benham
cbenham at adam.com.au
Wed Sep 14 10:55:37 PDT 2016
Luděk,
One of the methods I like for public elections is a Condorcet-IRV hybrid
(the simplest of several) that has been dubbed "Benham".
Voters strictly rank from the top however many candidates they wish. If
a Condorcet winner exists, he/she wins. Otherwise continue
with IRV, checking before each elimination for a CW among remaining
candidates and electing the first one to appear.
A few others I like use "approval" information, interpreting
above-bottom ranking as approval (unless the ballot rules allow/invite
voters to explicitly specify an approval threshold in their rankings).
One of these methods that works here is Approval Margins Sort: Line up
the candidates in order of approval from highest to lowest
as the first tentative ordering of the candidates (with the Approval
winner highest in the order). Look for adjacent pairs of candidates
where the candidate lower in the order pairwise beats the one higher in
the order. If there are none then the order is confirmed and
candidate highest in the order wins.
If there are adjacent pairs of candidate who are "out of pairwise
order", flip the order of the one (of those pairs) with the smallest
approval-score gap.
Repeat until each candidate not bottom in the order pairwise beats the
candidate just below it. Then elect the highest-ordered candidate.
http://wiki.electorama.com/wiki/Approval_Sorted_Margins
1: A>B
2: B>C
1: C>A
1: C>A
Approval Scores: C4 > A3 = B3
Here B and A have the same approval scores but are out of order
pairwise, so the next tentative order is C > A > B. Now no adjacent
pair of candidates (C>A or A>B) is out of pairwise order so that order
is final and C wins.
Another one of the methods I like for public office is Smith//Approval:
Elect the most approved member of the Smith set. The "Smith
set" is the smallest set of candidates that pairwise-beat all (if any)
of the candidates outside the set.
Here all the candidates are in the Smith set, and the most approved of
them is C, so again C wins.
Something a bit more elegant that might give a different result in a
complicated example with more than 3 candidates in the Smith set
is a suggestion of Forrest Simmons: Construct a "chain" of candidates
thus: begin with the most approved candidate. If no candidate
"covers" this candidate A (i.e. pairwise beats A and also pairwise beats
every candidate that A pairwise beats and doesn't pairwise lose
to any candidate A pairwise ties with) then A wins. Otherwise add to the
chain the most approved candidate that covers A. For as long
as it is possible, keep adding to the chain the most approved candidate
that pairwise beats the last candidate added. When no new candidate
can be added, elect the last added candidate.
In this example all the voters have submitted a full ranking, so maybe
they (and/or you) think that interpreting above-bottom ranking as
approval is arbitrary and too much focused on the bottoms of the ballots.
In that case a possible alternative is Borda Margins-Sort Elimination:
Using Borda scores (on each ballot each candidate gets a point-score
equal to the number of candidates ranked below it. Candidates ranked the
same should perhaps be handled in a away that meets "Symmetric
Completion", so that say a "A=B > C" ballot should give 1.5 points
each to A and B and zero points to C.) instead of Approval scores, order
the candidates as in Approval-Margins Sort.
Then eliminate the lowest ordered candidate, recalculate new Borda
scores without the eliminated candidate. Repeat until the last remaining
3 candidates
are finally ordered and then elect the one highest in that order.
In your example the candidates' Borda scores are the same (5), so here
this method is also a tie.
Chris Benham
On 9/14/2016 5:27 PM, Luděk Belán wrote:
> Chris,
> thank you for confirmation. Do you know any variant of Schulze method
> (or of other Condorcet method) to determine the winner in this example?
> Thank you.
>
> Luděk Belán
>
>
> ---------- Původní zpráva ----------
> Od: C.Benham <cbenham at adam.com.au>
> Komu: election-methods at lists.electorama.com
> Datum: 12. 9. 2016 18:36:36
> Předmět: Re: [EM] Example with tie in Schulze method but not in IRV
>
>
> Luděk,
>
> Yes, you are right.
>
>
> Chris Benham
>
>
>
> On 9/12/2016 7:47 PM, Luděk Belán wrote:
> > Dear all,
> >
> > excuse my bad English.
> >
> > Example for discusion:
> > Candidates: A, B, C
> > Ballots (count: order):
> > 1: A>B>C
> > 2: B>C>A
> > 1: C>A>B
> > 1: A>C>B
> >
> > In Instant Runoff Voting wins candidate A, but in Schulze method
> in my opinion is result tie.
> > It's true, please?
> >
> > Best regards
> >
> > Luděk Belán
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