<div dir="ltr"><div><div><div><div><div><div><div><div><div>Something else interesting about this example:<br><br></div><div>First write out the five individual ballots<br></div><div><br></div> BCA<br></div><div> BCA<br></div> CAB<br></div> ABC<br></div> ACB<br><br></div>In the spirit of Dodgson find which (if any) simple preference reversal(s) will produce a ballot set with a (ballot) Condorcet Winner:<br><br></div>Candidate A becomes a CW if any of the three CA pairs is reversed.<br></div>Candidate B becomes the CW if either of the two AB pairs is reversed.<br></div>Candidate C becomes the CW if any of the three BC pairs is reversed.<br><br></div><div>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.<br><br></div><div>So the appropriate lottery to break the Dodgson tie is<br></div><div> 37.5%A+ 25%B+37.5%C<br><br></div><div>And the appropriate lottery for breaking either the beat-path or Borda tie is<br></div><div> 30%A + 30%B + 40%C<br></div><div><br></div>[all under the assumption that a random tie breaker is desired]<br><div><div><div><div><div><div><br></div><div>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.<br><br></div><div>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)<br><br></div><div>Candidate C is preferred over the Dodgson lottery on the first three ballots, but not on the other two.<br><br></div><div>One might also consider the Random Ballot ("benchmark") lottery as a tie breaker:<br><br></div><div>40%A + 40%B + 20%C<br><br></div><div>If we use the "Toby" transform to convert the ballots to weighted sums of approval ballots, and then use random approval, we get<br><br></div><div>[31(A+B)+28C]/90, which is very close to "random candidate" (A+B+C)/3 .<br></div><div><br></div><div>Whom would the US Supreme Court choose? <br><br></div><div>Answer: the one most connected to wealth and power.<br><br></div><div>Best Wishes,<br><br></div><div>Forest<br></div><div><br><br></div><div><div><div><div><br></div></div></div></div></div></div></div></div></div></div><div class="gmail_extra"><br><div class="gmail_quote">On Thu, Sep 15, 2016 at 12:52 AM, C.Benham <span dir="ltr"><<a href="mailto:cbenham@adam.com.au" target="_blank">cbenham@adam.com.au</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
<div bgcolor="#FFFFFF" text="#000000">
<div><br>
In defining Forest's algorithm I wrote:<br>
<br>
<blockquote type="cite"> "For as long as it is possible, keep
adding to the chain the most approved candidate that pairwise
beats the last candidate added."</blockquote>
<br>
The "pairwise beats" is wrong. It should be ' <b>covers</b>'.<br>
<br>
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. <br>
<br>
These mistakes are fixed in text below.<br>
<br>
Chris Benham<br>
<br>
<br>
On 9/15/2016 3:25 AM, C.Benham wrote:<br>
</div>
<blockquote type="cite">
<div>Luděk,<br>
<br>
One of the methods I like for public elections is a
Condorcet-IRV hybrid (the simplest of several) that has been
dubbed "Benham".<br>
<br>
Voters strictly rank from the top however many candidates they
wish. If a Condorcet winner exists, he/she wins. Otherwise
continue<br>
with IRV, checking before each elimination for a CW among
remaining candidates and electing the first one to appear.<br>
<br>
A few others I like use "approval" information, interpreting
above-bottom ranking as approval (unless the ballot rules
allow/invite<br>
voters to explicitly specify an approval threshold in their
rankings). <br>
<br>
One of these methods that works here is Approval Margins Sort:
Line up the candidates in order of approval from highest to
lowest<br>
as the first tentative ordering of the candidates (with the
Approval winner highest in the order). Look for adjacent pairs
of candidates<br>
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<br>
candidate highest in the order wins.<br>
<br>
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.<br>
If among the out-of-pairwise-order adjacent pairs there is
gap-size tie, flip the order of lowest ordered pair.<br>
<br>
Repeat until each candidate not bottom in the order pairwise
beats the candidate just below it. Then elect the
highest-ordered candidate.<br>
<br>
<a href="http://wiki.electorama.com/wiki/Approval_Sorted_Margins" target="_blank">http://wiki.electorama.com/<wbr>wiki/Approval_Sorted_Margins</a><br>
<br>
<br>
1: A>B<br>
2: B>C<br>
1: C>A<br>
1: C>A<br>
<br>
Approval Scores: C4 > A3 = B3<br>
<br>
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<br>
pair of candidates (C>A or A>B) is out of pairwise order
so that order is final and C wins.<br>
<br>
Another one of the methods I like for public office is
Smith//Approval: Elect the most approved member of the Smith
set. The "Smith<br>
set" is the smallest set of candidates that pairwise-beat all
(if any) of the candidates outside the set.<br>
<br>
Here all the candidates are in the Smith set, and the most
approved of them is C, so again C wins.<br>
<br>
Something a bit more elegant that might give a different result
in a complicated example with more than 3 candidates in the
Smith set<br>
is a suggestion of Forrest Simmons: Construct a "chain" of
candidates thus: begin with the most approved candidate. If no
candidate<br>
"covers" this candidate A (i.e. pairwise beats A and also
pairwise beats every candidate that A pairwise beats and doesn't
pairwise lose<br>
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<br>
as it is possible, keep adding to the chain the most approved
candidate that covers the last candidate added. When no new
candidate<br>
can be added, elect the last added candidate.<br>
<br>
In this example all the voters have submitted a full ranking, so
maybe they (and/or you) think that interpreting above-bottom
ranking as<br>
approval is arbitrary and too much focused on the bottoms of the
ballots. <br>
<br>
In that case a possible alternative is Borda Margins-Sort
Elimination: Using Borda scores (on each ballot each candidate
gets a point-score<br>
equal to the number of candidates ranked below it. Candidates
ranked the same should perhaps be handled in a away that meets
"Symmetric<br>
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<br>
the candidates as in Approval-Margins Sort. <br>
<br>
Then eliminate the lowest ordered candidate, recalculate new
Borda scores without the eliminated candidate. Repeat until the
last remaining 3 candidates<br>
are finally ordered and then elect the one highest in that
order.<br>
<br>
In your example the candidates' Borda scores are the same (5),
so here this method is also a tie.<br>
<br>
Chris Benham<br>
<br>
<br>
<br>
On 9/14/2016 5:27 PM, Luděk Belán wrote:<br>
</div>
<blockquote type="cite">Chris,
<div>thank you for confirmation. Do you know any variant of
Schulze method (or of other Condorcet method) to determine the
winner in this example?</div>
<div>Thank you.</div>
<div><br>
</div>
<div>Luděk Belán</div>
<div><br>
</div>
<div><br>
<p>---------- Původní zpráva ----------<br>
Od: C.Benham <a href="mailto:cbenham@adam.com.au" target="_blank"><cbenham@adam.com.au></a><br>
Komu: <a href="mailto:election-methods@lists.electorama.com" target="_blank">election-methods@lists.<wbr>electorama.com</a><br>
Datum: 12. 9. 2016 18:36:36<br>
Předmět: Re: [EM] Example with tie in Schulze method but not
in IRV</p>
<br>
<blockquote>Luděk,<br>
<br>
Yes, you are right.<br>
<br>
<br>
Chris Benham<br>
<br>
<br>
<br>
On 9/12/2016 7:47 PM, Luděk Belán wrote:<br>
> Dear all,<br>
><br>
> excuse my bad English.<br>
><br>
> Example for discusion:<br>
> Candidates: A, B, C<br>
> Ballots (count: order):<br>
> 1: A>B>C<br>
> 2: B>C>A<br>
> 1: C>A>B<br>
> 1: A>C>B<br>
><br>
> In Instant Runoff Voting wins candidate A, but in
Schulze method in my opinion is result tie.<br>
> It's true, please?<br>
><br>
> Best regards<br>
><br>
> Luděk Belán<br>
> ----<br>
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