Luděk,

Yes, you are right.

Chris Benham

On 9/12/2016 7:47 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@adam.com.au
Komu: election-methods@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:

opinion is result tie.

Election-Methods mailing list - see http://electorama.com/em for list info"

On 09/14/2016 09:57 AM, Luděk Belán wrote:

The Condorcet matrix is (row defeating column)

• A B C
  A - 3 2
  B 2 - 3
  C 3 2 -

So the candidates are completely indistinguishable from the pairwise
matrix alone. No method that uses only the pairwise data can tell the
candidates apart.

I think you have three options:

a. Use a random tiebreak (MAM at http://mam.hostei.com/default.php does
that for your ballot set).
b. Accept the tie.
c. Use a Condorcet method that use more than just the pairwise data
(e.g. Condorcet-IRV hybrids like Woodall), or break the tie with a
non-Condorcet method.

For your ballots, every candidate is part of the Smith set. If you break
the tie with IRV, A wins. If you break the tie with Bucklin, C wins.

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:

In defining Forest's algorithm I wrote:

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:

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@adam.com.au wrote:

This one made me think of also other alternatives. Picking just one of the voters to make the decision sounds a bit "random". How about breaking the tie by removing one vote at a time until there is a Condorcet winner? This is in a way a softer version of the Random Ballot method. Eventually there would be only one vote left (if you will not et a Condorcet winner before that), and this softened method reduces to a hard Random Ballot method.

Another (completely different) classical approach to breaking ties would be to add votes until there is a Condorcet winner. The candidate who needs the smallest number of additions will win. This approach is fair and neutral in the sense that we don't eliminate any voter opinions but just introduce additional imaginary ones. The problem with this approach is that it brings us close to the infamous(?) Minmax margins (MMM) method :-) . This one has also the problem that it is not a complete tie solver since ties are still possible, so we may need to complement this by adding one more tie breaking mechanism on top of it. The random ballot and random ballot elimination approaches could be used as final tie breakers. Or if you want to have a probability lottery, then maybe give additional votes to each candidate in random order until one of them can reaches the Condorcet winner level (without considering the votes that others have already collected). This would mean probability in proportion to the needed additional votes to become a Condorcet winner. In the example above this would be 33.3%A + 33.3%B + 33.3%C. Actually this (final tie breaker for MMM) would always be a lottery with equal weights (between those candidates that need the least number of additional votes).

But what if we would jump to the lottery approach without the intermediate (MMM) phase of declaring the candidate that needs the least additional votes as the winner. In this case we would just start casting random votes to the candidates whenever there is no Condorcet winner, until one of them would reach the required additional votes limit (without considering how many additional votes the others have so far received). This approach might have some beneficial impact on the strategies. It might not pay off to try to bury some candidate under an unwanted candidate since the probability of that candidate winning would rise. I can't tell yet how this variant would relate to the other lottery based approaches. Ffs to me.

BR, Juho

Thank you all for your helpful responses. It will take me some time study
them :-).

Luděk Belán

---------- Původní zpráva ----------
Od: Juho Laatu juho.laatu@gmail.com
Komu: EM election-methods@lists.electorama.com
Datum: 16. 9. 2016 9:26:03
Předmět: Re: [EM] Example with tie in Schulze method but not in IRV (small
correction)

"

"
On 16 Sep 2016, at 03:28, Forest Simmons <fsimmons@pcc.edu
(mailto:fsimmons@pcc.edu)> wrote:

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

"

This one made me think of also other alternatives. Picking just one of the
voters to make the decision sounds a bit "random". How about breaking the
tie by removing one vote at a time until there is a Condorcet winner? This
is in a way a softer version of the Random Ballot method. Eventually there
would be only one vote left (if you will not et a Condorcet winner before
that), and this softened method reduces to a hard Random Ballot method.

Another (completely different) classical approach to breaking ties would be
to add votes until there is a Condorcet winner. The candidate who needs the
smallest number of additions will win. This approach is fair and neutral in
the sense that we don't eliminate any voter opinions but just introduce
additional imaginary ones. The problem with this approach is that it brings
us close to the infamous(?) Minmax margins (MMM) method :-) . This one has
also the problem that it is not a complete tie solver since ties are still
possible, so we may need to complement this by adding one more tie breaking
mechanism on top of it. The random ballot and random ballot elimination
approaches could be used as final tie breakers. Or if you want to have a
probability lottery, then maybe give additional votes to each candidate in
random order until one of them can reaches the Condorcet winner level
(without considering the votes that others have already collected). This
would mean probability in proportion to the needed additional votes to
become a Condorcet winner. In the example above this would be 33.3%A + 33.3%
B + 33.3%C. Actually this (final tie breaker for MMM) would always be a
lottery with equal weights (between those candidates that need the least
number of additional votes).

But what if we would jump to the lottery approach without the intermediate
(MMM) phase of declaring the candidate that needs the least additional votes
as the winner. In this case we would just start casting random votes to the
candidates whenever there is no Condorcet winner, until one of them would
reach the required additional votes limit (without considering how many
additional votes the others have so far received). This approach might have
some beneficial impact on the strategies. It might not pay off to try to
bury some candidate under an unwanted candidate since the probability of
that candidate winning would rise. I can't tell yet how this variant would
relate to the other lottery based approaches. Ffs to me.

BR, Juho

"

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@adam.com.au
(mailto:cbenham@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
(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
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@adam.com.au(mailto:cbenham@adam.com.au)
Komu: election-methods@lists. electorama.com
(mailto:election-methods@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:

opinion is result tie.

(http://electorama.com/em) for list info

Election-Methods mailing list - see http://electorama.com/em
(http://electorama.com/em) for list info"

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