[EM] Example with tie in Schulze method but not in IRV (small correction)
Juho Laatu
juho.laatu at gmail.com
Fri Sep 16 00:25:47 PDT 2016
> On 16 Sep 2016, at 03:28, Forest Simmons <fsimmons at 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 at adam.com.au <mailto: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 <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 at adam.com.au> <mailto:cbenham at adam.com.au>
>>> Komu: election-methods at lists.electorama.com <mailto: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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>>
>>
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