# [EM] Example with tie in Schulze method but not in IRV (small correction)

Forest Simmons fsimmons at pcc.edu
Thu Sep 15 17:28:48 PDT 2016

```Something else interesting about this example:

First write out the five individual ballots

BCA
BCA
CAB
ABC
ACB

In the spirit of Dodgson find which (if any) simple preference reversal(s)
will produce a ballot set with a (ballot) Condorcet Winner:

Candidate A becomes a CW if any of the three CA pairs is reversed.
Candidate B becomes the CW  if either of the two AB pairs is reversed.
Candidate C becomes the CW if any of the three BC pairs is reversed.

No other single reversal will create a CW, i.e. neither simple reversal in
the last ballot ACB will create a CW.  However, in that ballot if the AC
pair is reversed, then candidate C becomes the beat-path (and Borda)
winner.  And if the CB pair is reversed, then candidate B becomes the
beat-path (and Borda) winner.

So the appropriate lottery to break the Dodgson tie is
37.5%A+ 25%B+37.5%C

And the appropriate lottery for breaking either the beat-path or Borda tie
is
30%A + 30%B + 40%C

[all under the assumption that a random tie breaker is desired]

Also, it is interesting that the Dodgson lottery expectation is preferred
over the sure A option by the first three ballots, assuming equally spaced
utilities.

Sure candidate B is preferred over the Dodgson lottery on the first two
ballots, liked less on the two ballots where it is ranked bottom, and
equally on the other ballot. (so a tie)

Candidate C is preferred over the Dodgson lottery on the first three
ballots, but not on the other two.

One might also consider the Random Ballot ("benchmark") lottery as a tie
breaker:

40%A + 40%B + 20%C

If we use the "Toby" transform to convert the ballots to weighted sums of
approval ballots, and then use random approval, we get

[31(A+B)+28C]/90, which is very close to "random candidate"  (A+B+C)/3 .

Whom would the US Supreme Court choose?

Answer: the one most connected to wealth and power.

Best Wishes,

Forest

On Thu, Sep 15, 2016 at 12:52 AM, C.Benham <cbenham at adam.com.au> wrote:

>
> In defining Forest's algorithm I wrote:
>
>   "For as long as it is possible, keep adding to the chain the most
> approved candidate that pairwise beats the last candidate added."
>
>
> The "pairwise beats" is wrong. It should be  ' *covers*'.
>
> Also I  omitted  the detail in the Margins-Sort algorithm of what to do
> when score gaps between pairwise out-of-order candidates are the same.
>
> These mistakes are fixed in text below.
>
> Chris Benham
>
>
> On 9/15/2016 3:25 AM, C.Benham wrote:
>
> 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.
> If among the out-of-pairwise-order adjacent pairs there is gap-size tie,
> flip the order of lowest ordered pair.
>
> 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 covers the last candidate added. When no new 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 ----------
> 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.
> >
> > Best regards
> >
> > Luděk Belán
> > ----
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