juho.laatu at gmail.com
Mon Mar 26 14:18:35 PDT 2018
> On 26 Mar 2018, at 01:36, Curt <accounts at museworld.com> wrote:
> Hi Juho, thank you for this example ballot set. I have added it to the codebase as a test case with some documentation. (Both Smith and Schwartz should identify A, B, and C.)
> A, B, and C all defeat each other 66:33. And they each defeat d 50:49. I understand the urge to award d the win, given those numbers. But I believe that that urge ascribes “intensity of preference” to A, B, and C - when for Condorcet, which is purely ordinal, we have no idea. If I imagine those voters as voting stoically, dispassionately, poker-face, then I have no idea whether they passionately prefer A to B, or if, for instance, they are completely torn but have some consistent but trivial reason to pick one over the other. So in the absence of intensity-of-preference data, we really don’t have enough data to conclude that d should be the winner - for all we know, there actually is more passion in those 50:49 splits.
We don't know the strengths of preferences (strength), but we know the number of voters (count) on each side in every pairwise preference. A, B and C are preferred over d weakly (count). A is preferred over B strongly (count). In the following examples preferences vary a lot if we study the strengths of preferences (strength).
In the following example d is weak (strength). (only ">>>" added to the original example)
17: A > B >>> d > C
16: A >>> d > B > C
17: B > C >>> d > A
16: B >>> d > C > A
17: C > A >>> d > B
16: C >>> d > A > B
In the following example d is strong (strength). (only ">>>" added)
17: A > B > d >>> C
16: A > d >>> B > C
17: B > C > d >>> A
16: B > d >>> C > A
17: C > A > d >>> B
16: C > d >>> A > B
Votes in the original example are almost symmetric in the sense that A, B and C are about as often below and above d, and the votes are about symmetric whether you read them from left to right or from right to left. This makes the ABC group and d about equal when thinking in terms of having them listed "on the right" or "on the left". The Smith Set candidates are in the ballots only marginally more "on the left" than candidate d. Therefore also their preference strengths (strength) are with good probability (under some randomness assumptions) about the same.
In summary, I don't see how to make any conclusions on the relative preferences (strength) of the ABC group and d. Just like Condorcet Winners could be popular or not (strength), also Smith Set members and non Smith Set members could be popular or not (strength). The example shows a situation where preferences can be whatever or about the same (strength), or in favour of d (count). There are three candidates that beat d, but that should not carry much weight since number of pairwise victories is known to be a poor criterion (because of clone problems).
> So I am having trouble seeing it as a flaw with the Smith Set concept itself. I do agree that it points to some sort of flaw, but I think the proper identification of the flaw’s home requires zooming out and looking at the framework.
I think the interesting question is if electing from the Smith Set makes sense as often as electing a Condorcet Winner. Many people on this list think that in many elections it would make sense to always elect the CW (knowing that the strengths of preferences (strength) are not well known). People who think that way should ask themselves if they require the winner to come from the SS in the given example, of if d would be a better choice. If d is ok (again assuming no knowledge of the strengths of preferences (strength)), then SS should not be seen as a requirement (although 99.9% of the elections would still elect from the SS, since the situation given in the example is very rare).
To me that example tells that sometimes candidates outside the Smith Set can indeed be almost ideal (2 votes short of being a Condorcet Winner), and candidates in the Smith Set can sometimes be quite poor (creating a strong unified opposition with interest to change winner A to C). This kind of ballot sets are however very unusual. Practical Condorcet methods should therefore elect almost always from the Smith Set. But as far as I'm concerned, not necessarily as a requirement, because of special cases like this.
> Each time we step close to determining an election winner, we make a choice to lose fidelity in some fashion.
> 1. We have a need to collectively decide something, and so we voice that need by identifying a question to answer. But in identifying or voicing that question, we risk losing some essential part of the real question. In other words, we risk a failure of specification. But, we need to move forward, so we accept that risk, and move forward with the question as asked.
> 2. In asking the question, we identify options for a solution - the candidates in an election. But in doing so, we risk omitting some of the proper solution spectrum. The collected candidates may still be insufficient in some way. But, we have to draw the line somewhere, so voters are restricted to choosing between a potentially imperfect slate of candidates.
> 3. We set a time to choose. But in doing so, we risk the deadline being too soon for some voters, in that they might not be finished with their decision process. So we lose some fidelity there in measuring voter preference exactly.
> 4. We want to protect against bullying, intimidation, and some people being convinced to make their vote count “less” than someone else’s, and so we decide to protect the ideal of “one person, one vote” where every person’s vote counts the same. But there we lose some fidelity in measuring intensity of preference among voters.
> So I’m inclined to think that the example below is more about the costs associated with something like #4 above.
I don't see any strong connection to strength of preferences (strength). The example is intended to follow the "one person, one vote" and "majority" tradition of explaining how to design pure ranked (Condorcet) methods. To me the key question thus is, when measuring only pairwise preferences, are there situations where the best winner might come outside of the Smith Set. The given example is the most obvious and most extreme to me. And the key point there is that d is two votes short of being a Condorcet Winner. Let's assume that we all would like d to win if it was a CW. A, B and C on the other hand are far from being Condorcet winners. This can be seen as a question of having one major defeat (A, B, C) to some other candidate, or having three marginal defeats (d) to other candidates.
That's just my two cents. I'm just trying to encourage people to think if Smith Set should be seen as a requirement, or just as a the most common outcome when there is a top cycle.
P.S. I note once more that the Smith Set should not be visualised as a group _above_ d (although that is the way people always draw it). Cyclic preferences do not have any obvious geometric presentation on a 2D paper. An alternative drawing approach would be to draw all the candidates as far from the winner position as they have distance to being a Condorcet Winner. That's how I tend to see the given example. (There are however few alternative ways to count the distance of each candidate to becoming a Condorcet Winner.)
> In fact, the scenario seems very similar to me, to the one I sketched out earlier - if, in a two-candidate election, A defeats B 50:49, but A’s support is lukewarm and B’s is passionate, should B win? I think you can defend that if you value “social utility” and “intensity of preference” more than “one-person-one-vote” or "majority", but I wouldn’t count that as a flaw with the Condorcet method in particular, since it’s a majority-type of voting method.
>> On Mar 25, 2018, at 7:05 AM, Juho Laatu <juho.laatu at gmail.com> wrote:
>>> On 25 Mar 2018, at 06:30, Curt <accounts at museworld.com> wrote:
>>> What do you believe the Smith Set signifies? Is it meaningless to you other than something from which a winner should be algorithmically selected?
>> To me Smith Set is a criterion that in some sense and at first sight looks natural, but on second thought does not cover all possible scenarios well. I mean that if there is a group of candidates that are a unified group, and they beat all others, then yes, one of them should at least in most cases win. This is related to clones. If all the Smith Set candidates can be considered to be clones, then nominating only one of them would probably lead to electing that candidate as a Condorcet Winner.
>> What are the problems then? One problem is that those Condorcet methods that are based on rankings only, and possibly a pairwise matrix only, can carry only limited information on what the preferences of the voters are. There can be multiple explanations to what the voter preferences might have been. There are scenarios where electing from the Smith Set may not be natural.
>> Another problem of the Smith Set is that it may look more natural than it is, when one draws the end results (in paper or in one's mind) so that all the Smith Set candidates are at top, and all others below. This drawing technique to some extent hides the defeats within the Smith Set from the eye.
>> The best I can do to demonstrate these problems is to give you one particular (old) example scenario where selecting the winner outside of the Smith Set seems quite natural. In some extreme situations Smith Set may thus not be the right choice.
>> 17: A > B > d > C
>> 16: A > d > B > C
>> 17: B > C > d > A
>> 16: B > d > C > A
>> 17: C > A > d > B
>> 16: C > d > A > B
>> This example is a classic strong cycle of A, B and C, with one more candidate (d) added. Candidates A, B and C are not clones since they are not next to each others in the ballots. Candidate d is not in the Smith Set, but is very close to being a Condorcet Winner (2 votes short). Candidates A, B and C are very far from being Condorcet Winners.
>> BR, Juho
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