<html><body>Thank you all for your helpful responses. It will take me some time study them :-).<div><br></div><div>Luděk Belán<br><div><br></div></div><div><br></div><p>---------- Původní zpráva ----------<br>Od: Juho Laatu <juho.laatu@gmail.com><br>Komu: EM <election-methods@lists.electorama.com><br>Datum: 16. 9. 2016 9:26:03<br>Předmět: Re: [EM] Example with tie in Schulze method but not in IRV (small
correction)</p><br><blockquote><div style="word-wrap: break-word; -webkit-nbsp-mode: space; -webkit-line-break: after-white-space;"><div><blockquote><div>On 16 Sep 2016, at 03:28, Forest Simmons <<a href="mailto:fsimmons@pcc.edu">fsimmons@pcc.edu</a>> wrote:</div><br><div><div><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></div></div></div></div></div></div></div></div></blockquote><div><br></div><div>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.</div><div><br></div><div>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).</div><div><br></div><div>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.</div><div><br></div><div>BR, Juho</div><div><br></div><br><blockquote><div><div><div><div><div><div><div><div><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><br><div>On Thu, Sep 15, 2016 at 12:52 AM, C.Benham <span><<a href="mailto:cbenham@adam.com.au">cbenham@adam.com.au</a>></span> wrote:<br><blockquote style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
<div>
<div><br>
In defining Forest's algorithm I wrote:<br>
<br>
<blockquote> "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>
<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">http://wiki.electorama.com/ 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>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"><cbenham@adam.com.au></a><br>
Komu: <a href="mailto:election-methods@lists.electorama.com">election-methods@lists. 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>
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