smith/schwartz/landau
Here's yet one more example that explains my line of thought.
This is an example of a weak (strength) Condorcet Winner. This is the same as the earlier example of a weak (strength) d, but numbers "16" and "17" have been swapped. The point is to demonstrate that Condorcet Winners can be weak or strong (strength) in the same way as candidates outside the Smith Set can be, and the difference is sometimes not big (swap of "16" and "17"). You can also change the location of the ">>>" marks if you want to make d strong (strength) as a Condorcet Winner or as a non Smith Set member.
16: A > B >>> d > C
17: A >>> d > B > C
16: B > C >>> d > A
17: B >>> d > C > A
16: C > A >>> d > B
17: C >>> d > A > B
Also this example should be evaluated assuming that the pairwise preferences are exactly known (count), the individual ballots are possibly known (count), and the strengths of preferences (strength) are totally unknown (i.e. forget the ">>>" marks). They (strengths) could be anything, but some distributions are more common than some others (for statistical reasons, although we know nothing bout the voters).
BR, Juho
On 27 Mar 2018, at 00:18, Juho Laatu juho.laatu@gmail.com wrote:
On 26 Mar 2018, at 01:36, Curt accounts@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.
-
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.
-
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.
-
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.
-
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.
BB, Juho
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.
Regards,
Curt
On Mar 25, 2018, at 7:05 AM, Juho Laatu juho.laatu@gmail.com wrote:
On 25 Mar 2018, at 06:30, Curt accounts@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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On 03/26/2018 11:18 PM, Juho Laatu wrote:
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).
Isn't this analogous to majority vs utility? Suppose you have a ranked
election like this:
51: A>B>C
48: B>C>A
2: C>B>A
A is the majority winner, but it only takes two voters changing from
A>B>C to B>A>C to make B the majority winner, analogous to how {ABC} is
the Smith set but {d} is close to being the CW. A voting method that
passes the majority criterion will always elect A, but a more
consensus-focused method like Borda elects B.
If we knew the utilities, then we could determine whether this is a
51: A: 100, B: 1, C: 0
48: B: 100, C: 1, A: 0
2: C: 100, B: 1, A: 0
election (A should win if we're counting utilities), or a
51: A: 100, B: 99, C: 0
48: B: 100, C: 99, A: 0
2: C: 100, B: 99, A: 0
election (B should win). But we don't, so we can't. Almost any situation
where A is the superior candidate can be matched by a parallel situation
where B is the superior candidate. Lacking utility information, we can't
establish which is correct, and so there seem to be only two ways to get
out of the problem.
The first way is to say that we have a model of how the electorate
behaves, which lets us determine which rated election is most likely
given the ranked data. (E.g. in a very polarized electorate, it's more
likely to be the former than the latter.)
The second way is to say "we have no idea of which may be true, but we
want to pass the majority criterion, so that settles the matter for us".
In that case, the hypothetical "we" would choose a method that elects A.
If the Smith situation is analogous to majority, then it seems that the
most consistent choice with respect to majority (as opposed to utility)
is to elect from the Smith set. A method that doesn't elect from the
Smith set may be superior in a utilitarian sense, but it may also not
be. There's no way to know (short of taking the first approach, or by
proving that e.g. for a very wide range of utility models, some
non-Smith method has better utilitarian performance than some Smith method).
I suppose what I'm saying is that if we choose to have a method that
passes majority on the grounds on the grounds that the majority is
right, then the argument can be adapted pretty easily to Condorcet and
then to Smith. Any counter of the sort that electing from Smith can get
the wrong winner can be similarly "ported back" to the majority
situation by showing that the majority criterion can also get it wrong.
Of course, if there's a more indirect argument behind wanting majority,
then that can break the symmetry. For instance, one could argue that
passing majority is a sort of DSV; in strategic Range, everybody votes
Approval style and so the majority winner wins anyway, so why not ease
the burden of the voter? Or "a large majority has the backing to get
what it wants, better let it have what it wants on the ballot than risk
a riot". It's not obvious how such arguments would support Smith,
whereas it is pretty clear that they would support the majority criterion.
But then that argument has to be made, first; in the absence of such an
argument, Smith seems a reasonable extrapolation from majority (by way
of Condorcet).
On 27 Mar 2018, at 18:11, Kristofer Munsterhjelm km_elmet@t-online.de wrote:
On 03/26/2018 11:18 PM, Juho Laatu wrote:
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).
Isn't this analogous to majority vs utility? Suppose you have a ranked election like this:
Yes, my count vs strength corresponds to majority vs utility. I should have already introduced new standard terms for "better if measured in utility" and "better if measured in majority/plurality".
51: A>B>C
48: B>C>A
2: C>B>A
A is the majority winner, but it only takes two voters changing from A>B>C to B>A>C to make B the majority winner, analogous to how {ABC} is the Smith set but {d} is close to being the CW. A voting method that passes the majority criterion will always elect A, but a more consensus-focused method like Borda elects B.
Condorcet fans tend to emphasize the majority approach. One key reason is the strategy concerns.
If we knew the utilities, then we could determine whether this is a
51: A: 100, B: 1, C: 0
48: B: 100, C: 1, A: 0
2: C: 100, B: 1, A: 0
election (A should win if we're counting utilities), or a
51: A: 100, B: 99, C: 0
48: B: 100, C: 99, A: 0
2: C: 100, B: 99, A: 0
election (B should win). But we don't, so we can't. Almost any situation where A is the superior candidate can be matched by a parallel situation where B is the superior candidate. Lacking utility information, we can't establish which is correct, and so there seem to be only two ways to get out of the problem.
The first way is to say that we have a model of how the electorate behaves, which lets us determine which rated election is most likely given the ranked data. (E.g. in a very polarized electorate, it's more likely to be the former than the latter.)
This approach is good for studying the results (based on various assumptions on voter preferences). Not that easy if one would have to decide the winner based on an agreed model on what the preferences of the voters are.
The second way is to say "we have no idea of which may be true, but we want to pass the majority criterion, so that settles the matter for us". In that case, the hypothetical "we" would choose a method that elects A.
I believe many Condorcet fans think that getting honest rankings is about as much honest information as we can get. That means that one is to some extent forced to follow the majority logic.
On the other hand people believe in the principle of one person one vote, and in the right on majorities to decide. That makes the majoritarian approach of Condorcet again very natural.
If the Smith situation is analogous to majority, then it seems that the most consistent choice with respect to majority (as opposed to utility) is to elect from the Smith set.
If the Smith Set forms some sort of clear majority, then yes, but that need not be the case. If A, B and C would beat each others by max n votes, and they all would beat d by more than n votes, then one could say that someone from the Smith set should win to respect majority.
One can also study the examples assuming that some of the candidates are clones. In my example A, B and C certainly are not clones. But there could be also ballots where A, B and C would be next to each others in all ballots (technical clones). This does not yet guarantee that those candidates are also real life clones (candidates of one party, in friendly terms, under the assumption the utilities of the voters are about the same for all three).
In the example that I gave, we should probably assume that there are four parties that compete with each others. If we would elect A, there would be a 66:33 opposition saying that we should have elected C. If we elect d, there would be only a weak (majority) opposition saying that we should have elected one of the others. Although A, B and C form a Smith Set, they probably do not form any kind of ideological set that should be given the right to claim victory. I.e. defeats within the Smith Set are just as damaging as defeats of d. Being part of the (technical) Smith Set does not give any additional privileges.
A method that doesn't elect from the Smith set may be superior in a utilitarian sense, but it may also not be. There's no way to know (short of taking the first approach, or by proving that e.g. for a very wide range of utility models, some non-Smith method has better utilitarian performance than some Smith method).
I think the utility uncertainties and arguments have no specific role in the given example. Condorcet Winners can have low or hight utility. Same with members of Smith Set, and candidates outside of the Smith Set. My arguments to consider d to be a reasonable winner are based on the majoritarian pairwise comparisons only. Shortly, d is two votes short of being a majoritarian Condorcet Winner (and the others are far behind in this kind of measurements).
I think human intuition is a problem here. The Smith Set with its three candidates seems to be between d and the winning position (that is drawn above the Smith Set). That makes d look bad. But I claim that this image hides all the defeats within the Smith Set, and it ignores the fact that d is actually very close to being a Condorcet Winner, not far from it.
I suppose what I'm saying is that if we choose to have a method that passes majority on the grounds on the grounds that the majority is right, then the argument can be adapted pretty easily to Condorcet and then to Smith. Any counter of the sort that electing from Smith can get the wrong winner can be similarly "ported back" to the majority situation by showing that the majority criterion can also get it wrong.
Majority criterion is not a solution to all problems, but we can take it as granted in a traditional majoritarian political system. I would not say that the fact that Smith Set members beat all the other candidates is a valid majority based argument supporting election from the Smith Set. The reason is that this statement would ignore all the defeats within the Smith Set. We must see the Condorcet election as a competition between individual candidates, and in that competition all pairwise victories and defeats do count.
I note one more problem with human intuition. Humans have a tendency to force group opinions to form a linear preference order. It looks natural that d is behind the Smith Set, and that also all members of the Smith Set should be forced to form a linear preference order. I think this would be a big mistake since we know that group preferences are not linear. The winner should be chosen based on the ballot or matrix preferences only, not based on any imagined linear preference order.
In my example electing someone from the Smith Set is a violation of one of the 66:33 majorities.
Of course, if there's a more indirect argument behind wanting majority, then that can break the symmetry. For instance, one could argue that passing majority is a sort of DSV; in strategic Range, everybody votes Approval style and so the majority winner wins anyway, so why not ease the burden of the voter? Or "a large majority has the backing to get what it wants, better let it have what it wants on the ballot than risk a riot". It's not obvious how such arguments would support Smith, whereas it is pretty clear that they would support the majority criterion.
But then that argument has to be made, first; in the absence of such an argument, Smith seems a reasonable extrapolation from majority (by way of Condorcet).
I tried to avoid all more complex argumentation and stick to the pairwise majority comparisons (and potentially information from the ballots too).
I don't see Smith Set as an extension of majority. It would be if we would decide to see the Smith Set members as one unified block of candidates of one party, and their defeats of each others would be either minor, or we would decide that their defeats do not matter even if they are big (assuming a small difference in utility although there are many voters preferring one over the other).
BR, Juho
Hi Juho, if you extrapolate Kristofer’s 51-48-2 example (quoted below) to your 66:33 example, the point is that if the objective is to only use ordinal rankings and not incorporate utility, there really isn’t any reason to consider a 66:33 cycle a “strong” cycle, or 50:49 a “weak” majority. This is because the concepts of “strong” and “weak” suggest intensity of preference. Without any supporting utility data, we can’t infer anything about the public’s willingness to accept “d” over any of A, B, and C. Considering “d” to be a reasonable winner appears to be based on more than pairwise comparisons. It appears to also be based on ascribing probable utility to the voters given the vote margins.
It’s true that these might be entirely sane assumptions given the numbers. 66:33 probably does correlate in some fashion to intensity of preference or utility. You’re probably right there. But the point is that concluding that d should probably be awarded the winner is the same thing as incorporating utility into the calculation of the vote.
For those votes based off of the majority criterion or “one person one vote”, that’s improper. It’s not a flaw with Condorcet or Smith Set; it’s just applying the argument that utility should matter - which is a different subject. In other words, advocating Condorcet is basically accepting the axiom that 50:49 should win even if 49 has more passion, and, similarly, that someone from the ABC 66:33 Smith Set should win over d since they all defeat d 50:49. You can always disagree with an axiom, but by doing so you are entering into the conversation of whether utility should matter in Condorcet voting.
(That said, even for Condorcet voting, I am curious about capturing score preferences on the ballots, and then using them only to select from a multi-candidate Smith Set.)
Curt
On Mar 27, 2018, at 10:54 AM, Juho Laatu juho.laatu@gmail.com wrote:
If we knew the utilities, then we could determine whether this is a
51: A: 100, B: 1, C: 0
48: B: 100, C: 1, A: 0
2: C: 100, B: 1, A: 0
election (A should win if we're counting utilities), or a
51: A: 100, B: 99, C: 0
48: B: 100, C: 99, A: 0
2: C: 100, B: 99, A: 0
election (B should win). But we don't, so we can't. Almost any situation where A is the superior candidate can be matched by a parallel situation where B is the superior candidate. Lacking utility information, we can't establish which is correct, […]
On 28 Mar 2018, at 01:44, Curt accounts@museworld.com wrote:
Hi Juho, if you extrapolate Kristofer’s 51-48-2 example (quoted below) to your 66:33 example, the point is that if the objective is to only use ordinal rankings and not incorporate utility, there really isn’t any reason to consider a 66:33 cycle a “strong” cycle, or 50:49 a “weak” majority. This is because the concepts of “strong” and “weak” suggest intensity of preference.
I think the difference is between measuring the number of voters vs measuring the strength of opinion of those voters. The former we can do (quite reliably). The latter the pure ranked methods do not measure (one reason being that one could not do that in a reliable way anyway). Numbers 66, 33, 50 and 49 can be picked from the pairwise matrix. Numbers 51, 48 and 2 can be derived from the ballots (but probably we are more interested in the matrix that those ballots will produce).
Without any supporting utility data, we can’t infer anything about the public’s willingness to accept “d” over any of A, B, and C. Considering “d” to be a reasonable winner appears to be based on more than pairwise comparisons. It appears to also be based on ascribing probable utility to the voters given the vote margins.
The rankings / pairwise preferences are available. Utilities are not known and we can ignore them. My intention was to present d as a potential winner based on the rankings only.
It’s true that these might be entirely sane assumptions given the numbers. 66:33 probably does correlate in some fashion to intensity of preference or utility. You’re probably right there. But the point is that concluding that d should probably be awarded the winner is the same thing as incorporating utility into the calculation of the vote.
I tried to avoid linking d to the utilities.
For those votes based off of the majority criterion or “one person one vote”, that’s improper. It’s not a flaw with Condorcet or Smith Set;
I don't think Smith Set is flawed, but I question the idea that it would be a natural extension of the majority rule to require that the winner should always be picked from the Smith Set (in majority oriented elections).
it’s just applying the argument that utility should matter - which is a different subject. In other words, advocating Condorcet is basically accepting the axiom that 50:49 should win even if 49 has more passion, and, similarly, that someone from the ABC 66:33 Smith Set should win over d since they all defeat d 50:49. You can always disagree with an axiom, but by doing so you are entering into the conversation of whether utility should matter in Condorcet voting.
There are some sensible Condorcet methods that can use also utility based information, but in this discussion I try to live under the basic assumption that Condorcet methods are pure ranked methods. That includes reasoning about d.
(That said, even for Condorcet voting, I am curious about capturing score preferences on the ballots, and then using them only to select from a multi-candidate Smith Set.)
Maybe James Green-Armytage's cardinal-weighted pairwise method would be of interest to you.
BR, Juho
Curt
On Mar 27, 2018, at 10:54 AM, Juho Laatu <juho.laatu@gmail.com mailto:juho.laatu@gmail.com> wrote:
If we knew the utilities, then we could determine whether this is a
51: A: 100, B: 1, C: 0
48: B: 100, C: 1, A: 0
2: C: 100, B: 1, A: 0
election (A should win if we're counting utilities), or a
51: A: 100, B: 99, C: 0
48: B: 100, C: 99, A: 0
2: C: 100, B: 99, A: 0
election (B should win). But we don't, so we can't. Almost any situation where A is the superior candidate can be matched by a parallel situation where B is the superior candidate. Lacking utility information, we can't establish which is correct, […]
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Hi Juho, you question whether it is a “natural extension of the majority rule to require that the winner should always be picked from the Smith Set in majority oriented elections”, and it appears your argument depends on the concept of a “strong” cycle.
My view is that we cannot assume a 66:33 Smith Set is a “strong” cycle. And that the only way to see 66:33 as a “strong” cycle is to imbue it with utility or strength-of-preference data.
Because rankings are ordinal, whether the margin is 66:33, 98:1, or 50:49, we cannot infer anything about “strength of victory” in those margins. All we know is that the one with more votes is majority preferred over the other. We have zero conception of “how much” they are preferred by. Ordinal rankings cannot be used to simulate utility or strength of preference. 98:1 could be a whisker, and 50:49 could be a gulf.
Since ABC is preferred to d by 50:49, ABC being the winner makes sense to me because it is the winner by definition. It is axiomatic (when we are talking about Condorcet-style elections).
I respect the view that utility would be helpful data in an example such as the 66:33 example, but I also see it as intrinsically related to the view that utility data would be helpful in a two-candidate 50:49 election.
To illustrate how we cannot rely on 66:33 to indicate any information supporting d as a winner, here are two examples of the 66:33 scenario with utility data added in:
Here, any of A, B, or C are clearly preferred to d, and ABC are clearly tied in utility despite the 66:33 margins:
17: A:100, B:99, d:2, C:1
16: A:100, d:3, B:2, C:1
17: B:100, C:99, d:2, A:1
16: B:100, d:3, C:2, A:1
17: C:100, A:99, d:2, B:1
16: C:100, d:3, A:2, B:1
A: 1700 + 1600 + 17 + 16 + 1683 + 32 = 5048
B: 1684 + 32 + 1700 + 1600 + 17 + 16 = 5048
C: 17 + 16 + 1683 + 32 + 1700 + 1600 = 5048
d: 34 + 48 + 34 + 48 + 34 + 48 = 246
Here, d is clearly preferred to A, B, and C, but ABC are still tied:
17: A:4, B:3, d:2, C:1
16: A:100, d:99 B:2, C:1
17: B:4, C:3, d:2, A:1
16: B :100, d:99, C:2, A:1
17: C:4, A:3, d:2, B:1
16: C:100, d:99, A:2, B:1
A: 1600 + 16 + 32 + 68 + 17 + 51 = 1784
B: 1600 + 16 + 32 + 51 + 68 + 17 = 1784
C: 1600 + 16 + 32 + 17 + 51 + 68 = 1784
d: 4752 + 34 + 34 + 34 = 4854
Curt
On Mar 27, 2018, at 5:15 PM, Juho Laatu juho.laatu@gmail.com wrote:
On 28 Mar 2018, at 01:44, Curt <accounts@museworld.com mailto:accounts@museworld.com> wrote:
Hi Juho, if you extrapolate Kristofer’s 51-48-2 example (quoted below) to your 66:33 example, the point is that if the objective is to only use ordinal rankings and not incorporate utility, there really isn’t any reason to consider a 66:33 cycle a “strong” cycle, or 50:49 a “weak” majority. This is because the concepts of “strong” and “weak” suggest intensity of preference.
I think the difference is between measuring the number of voters vs measuring the strength of opinion of those voters. The former we can do (quite reliably). The latter the pure ranked methods do not measure (one reason being that one could not do that in a reliable way anyway). Numbers 66, 33, 50 and 49 can be picked from the pairwise matrix. Numbers 51, 48 and 2 can be derived from the ballots (but probably we are more interested in the matrix that those ballots will produce).
Without any supporting utility data, we can’t infer anything about the public’s willingness to accept “d” over any of A, B, and C. Considering “d” to be a reasonable winner appears to be based on more than pairwise comparisons. It appears to also be based on ascribing probable utility to the voters given the vote margins.
The rankings / pairwise preferences are available. Utilities are not known and we can ignore them. My intention was to present d as a potential winner based on the rankings only.
It’s true that these might be entirely sane assumptions given the numbers. 66:33 probably does correlate in some fashion to intensity of preference or utility. You’re probably right there. But the point is that concluding that d should probably be awarded the winner is the same thing as incorporating utility into the calculation of the vote.
I tried to avoid linking d to the utilities.
For those votes based off of the majority criterion or “one person one vote”, that’s improper. It’s not a flaw with Condorcet or Smith Set;
I don't think Smith Set is flawed, but I question the idea that it would be a natural extension of the majority rule to require that the winner should always be picked from the Smith Set (in majority oriented elections).
it’s just applying the argument that utility should matter - which is a different subject. In other words, advocating Condorcet is basically accepting the axiom that 50:49 should win even if 49 has more passion, and, similarly, that someone from the ABC 66:33 Smith Set should win over d since they all defeat d 50:49. You can always disagree with an axiom, but by doing so you are entering into the conversation of whether utility should matter in Condorcet voting.
There are some sensible Condorcet methods that can use also utility based information, but in this discussion I try to live under the basic assumption that Condorcet methods are pure ranked methods. That includes reasoning about d.
(That said, even for Condorcet voting, I am curious about capturing score preferences on the ballots, and then using them only to select from a multi-candidate Smith Set.)
Maybe James Green-Armytage's cardinal-weighted pairwise method would be of interest to you.
BR, Juho
Curt
On Mar 27, 2018, at 10:54 AM, Juho Laatu <juho.laatu@gmail.com mailto:juho.laatu@gmail.com> wrote:
If we knew the utilities, then we could determine whether this is a
51: A: 100, B: 1, C: 0
48: B: 100, C: 1, A: 0
2: C: 100, B: 1, A: 0
election (A should win if we're counting utilities), or a
51: A: 100, B: 99, C: 0
48: B: 100, C: 99, A: 0
2: C: 100, B: 99, A: 0
election (B should win). But we don't, so we can't. Almost any situation where A is the superior candidate can be matched by a parallel situation where B is the superior candidate. Lacking utility information, we can't establish which is correct, […]
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On 28 Mar 2018, at 08:43, Curt accounts@museworld.com wrote:
Hi Juho, you question whether it is a “natural extension of the majority rule to require that the winner should always be picked from the Smith Set in majority oriented elections”, and it appears your argument depends on the concept of a “strong” cycle.
Yes. A strong cycle among all the Smith Set members makes the non Smith Set candidates more appealing.
My view is that we cannot assume a 66:33 Smith Set is a “strong” cycle. And that the only way to see 66:33 as a “strong” cycle is to imbue it with utility or strength-of-preference data.
In terms of plurality (number of voters) It is hard to make strong cycles any stronger than that. And the 50:49 preferences are as weak as you can get with 99 voters.
If you observe the strength of comparisons from utility point of view, that could make those 66:33 preferences either stronger or weaker. But my argumentation (in favour of considering the non Smith Set members as potential winners) is based on plurality only, not on utilities.
Because rankings are ordinal, whether the margin is 66:33, 98:1, or 50:49, we cannot infer anything about “strength of victory” in those margins.
Talking still about utilities here. There can be many kind of utility data, but if we assume that the voters have a finite range of integer values to give to the candidates, then the difference of sums of voter utilities in 66 vs 33 could be higher than in the sums of 50 vs 49.
(In some sense pure ordinal methods make the one-man-one-vote assumption and behave as if utilities correspond directly to the plurality difference (of some other function of the pluralities).)
All we know is that the one with more votes is majority preferred over the other. We have zero conception of “how much” they are preferred by. Ordinal rankings cannot be used to simulate utility or strength of preference. 98:1 could be a whisker, and 50:49 could be a gulf.
Since ABC is preferred to d by 50:49, ABC being the winner makes sense to me because it is the winner by definition. It is axiomatic (when we are talking about Condorcet-style elections).
If ABC was one candidate, then yes. ABC is however a group of candidates that also win each others (strong preferences) in the strong cycle.
I respect the view that utility would be helpful data in an example such as the 66:33 example, but I also see it as intrinsically related to the view that utility data would be helpful in a two-candidate 50:49 election.
Utility data would be useful. Sincere utility data could be used in different methods e.g. to make a Condorcet Winner lose, or to make a majority winner lose. But pure ranked Condorcet methods are popular since it is difficult to collect sincere utility opinions.
To illustrate how we cannot rely on 66:33 to indicate any information supporting d as a winner, here are two examples of the 66:33 scenario with utility data added in:
Here, any of A, B, or C are clearly preferred to d, and ABC are clearly tied in utility despite the 66:33 margins:
17: A:100, B:99, d:2, C:1
16: A:100, d:3, B:2, C:1
17: B:100, C:99, d:2, A:1
16: B:100, d:3, C:2, A:1
17: C:100, A:99, d:2, B:1
16: C:100, d:3, A:2, B:1
A: 1700 + 1600 + 17 + 16 + 1683 + 32 = 5048
B: 1684 + 32 + 1700 + 1600 + 17 + 16 = 5048
C: 17 + 16 + 1683 + 32 + 1700 + 1600 = 5048
d: 34 + 48 + 34 + 48 + 34 + 48 = 246
This one is about the same as one of the examples that I gave.
Here, d is clearly preferred to A, B, and C, but ABC are still tied:
17: A:4, B:3, d:2, C:1
16: A:100, d:99 B:2, C:1
17: B:4, C:3, d:2, A:1
16: B :100, d:99, C:2, A:1
17: C:4, A:3, d:2, B:1
16: C:100, d:99, A:2, B:1
A: 1600 + 16 + 32 + 68 + 17 + 51 = 1784
B: 1600 + 16 + 32 + 51 + 68 + 17 = 1784
C: 1600 + 16 + 32 + 17 + 51 + 68 = 1784
d: 4752 + 34 + 34 + 34 = 4854
Yes, works too. Includes "weak" votes that are not normalised to the 100...1 range.
My question is still "whether it is a natural extension of the majority rule to require that the winner should always be picked from the Smith Set, when observing the number of voters only”.
BR, Juho
Curt
On Mar 27, 2018, at 5:15 PM, Juho Laatu <juho.laatu@gmail.com mailto:juho.laatu@gmail.com> wrote:
On 28 Mar 2018, at 01:44, Curt <accounts@museworld.com mailto:accounts@museworld.com> wrote:
Hi Juho, if you extrapolate Kristofer’s 51-48-2 example (quoted below) to your 66:33 example, the point is that if the objective is to only use ordinal rankings and not incorporate utility, there really isn’t any reason to consider a 66:33 cycle a “strong” cycle, or 50:49 a “weak” majority. This is because the concepts of “strong” and “weak” suggest intensity of preference.
I think the difference is between measuring the number of voters vs measuring the strength of opinion of those voters. The former we can do (quite reliably). The latter the pure ranked methods do not measure (one reason being that one could not do that in a reliable way anyway). Numbers 66, 33, 50 and 49 can be picked from the pairwise matrix. Numbers 51, 48 and 2 can be derived from the ballots (but probably we are more interested in the matrix that those ballots will produce).
Without any supporting utility data, we can’t infer anything about the public’s willingness to accept “d” over any of A, B, and C. Considering “d” to be a reasonable winner appears to be based on more than pairwise comparisons. It appears to also be based on ascribing probable utility to the voters given the vote margins.
The rankings / pairwise preferences are available. Utilities are not known and we can ignore them. My intention was to present d as a potential winner based on the rankings only.
It’s true that these might be entirely sane assumptions given the numbers. 66:33 probably does correlate in some fashion to intensity of preference or utility. You’re probably right there. But the point is that concluding that d should probably be awarded the winner is the same thing as incorporating utility into the calculation of the vote.
I tried to avoid linking d to the utilities.
For those votes based off of the majority criterion or “one person one vote”, that’s improper. It’s not a flaw with Condorcet or Smith Set;
I don't think Smith Set is flawed, but I question the idea that it would be a natural extension of the majority rule to require that the winner should always be picked from the Smith Set (in majority oriented elections).
it’s just applying the argument that utility should matter - which is a different subject. In other words, advocating Condorcet is basically accepting the axiom that 50:49 should win even if 49 has more passion, and, similarly, that someone from the ABC 66:33 Smith Set should win over d since they all defeat d 50:49. You can always disagree with an axiom, but by doing so you are entering into the conversation of whether utility should matter in Condorcet voting.
There are some sensible Condorcet methods that can use also utility based information, but in this discussion I try to live under the basic assumption that Condorcet methods are pure ranked methods. That includes reasoning about d.
(That said, even for Condorcet voting, I am curious about capturing score preferences on the ballots, and then using them only to select from a multi-candidate Smith Set.)
Maybe James Green-Armytage's cardinal-weighted pairwise method would be of interest to you.
BR, Juho
Curt
On Mar 27, 2018, at 10:54 AM, Juho Laatu <juho.laatu@gmail.com mailto:juho.laatu@gmail.com> wrote:
If we knew the utilities, then we could determine whether this is a
51: A: 100, B: 1, C: 0
48: B: 100, C: 1, A: 0
2: C: 100, B: 1, A: 0
election (A should win if we're counting utilities), or a
51: A: 100, B: 99, C: 0
48: B: 100, C: 99, A: 0
2: C: 100, B: 99, A: 0
election (B should win). But we don't, so we can't. Almost any situation where A is the superior candidate can be matched by a parallel situation where B is the superior candidate. Lacking utility information, we can't establish which is correct, […]
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On 03/28/2018 10:26 AM, Juho Laatu wrote:
Yes, works too. Includes "weak" votes that are not normalised to the
100...1 range.
My question is still "whether it is a natural extension of the majority
rule to require that the winner should always be picked from the Smith
Set, when observing the number of voters only”.
I just thought of a way to see if the Smith criterion is a reasonable
extension of majority; or rather, if it's a reasonable extension of
Condorcet.
You say that if ABC were one candidate, he should win. What if we make
ABC originate as one candidate? One way of doing so would be by cloning
an initial A into ABC. Then we know (if we value clone independence)
that in any situation where A wins against d in a majority election
(that's where majority comes in), then if we clone A into ABC, the
winner should be one of ABC.
And if we can turn any ABCd election (Smith set of size three, plus a
Condorcet loser) into one where ABC are clones, then we get that any
clone independent Condorcet method must be Smith in the four candidate
case, just by extending majority, through Condorcet, into Smith.
Are there any clone-independent methods that meet Condorcet but not
Smith? The only one I can think of is the "obvious" method that
collapses clones and then runs Minmax. Perhaps also BTR-IRV? I'm not
sure. But both of these require access to the ballots themselves, which
makes me suspect that you can't tell a clone scenario from a non-clone
scenario if all you have is the pairwise matrix.
If I'm right, then for any given (3-cycle + 1) Condorcet matrix, I could
use my old tool, linear programming, to determine a clone ballot set
that would produce that matrix. It would be something like:
-
Because ABC is a clone set, the only permitted ballots are xD or Dx
where x is one of ABC ACB BAC BCA CAB CBA, i.e. ABCD ACBD BACD BCAD CABD
CBAD DABC DACB DBAC DBCA DCAB DCBA. -
We want the pairwise strengths to be as determined beforehand.
-
Usual constraints: no negative number of ballots, no infinities.
In your ABCd case, we have:
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
and the pairwise matrix is (for row, against column)
A B C d
A -- 66 33 50
B 33 -- 66 50
C 66 33 -- 50
d 49 49 49 --
Programming the LP gives the following clone solution:
33: A>B>C>d
17: B>C>A>d
16: d>B>C>A
33: d>C>A>B
that produces the same pairwise matrix as above.
If we collapse the clone set, we get
50: A>d
49: d>A
so any clone-independent Condorcet method that only knows the pairwise
matrix must elect from the ABC set here, and since it can't distinguish
the LP solution (where ABC is a clone set) from the original ballot set,
there's no way for it to elect d in your example.
I haven't actually proven that clone independence + Condorcet + only
look at the pairwise matrix implies 4-candidate Smith; that would
require more linear algebra than I feel like using today. But it's not
that implausible; and if it's true, that means that whatever makes a
Condorcet loser deserve to win, if anything, must come from information
not provided by the pairwise matrix.
On 28 Mar 2018, at 14:05, Kristofer Munsterhjelm km_elmet@t-online.de wrote:
On 03/28/2018 10:26 AM, Juho Laatu wrote:
Yes, works too. Includes "weak" votes that are not normalised to the 100...1 range.
My question is still "whether it is a natural extension of the majority rule to require that the winner should always be picked from the Smith Set, when observing the number of voters only”.
I just thought of a way to see if the Smith criterion is a reasonable extension of majority; or rather, if it's a reasonable extension of Condorcet.
You say that if ABC were one candidate, he should win. What if we make ABC originate as one candidate? One way of doing so would be by cloning an initial A into ABC. Then we know (if we value clone independence) that in any situation where A wins against d in a majority election (that's where majority comes in), then if we clone A into ABC, the winner should be one of ABC.
m: A=B=C > d
n: d > A=B=C
With these ballots, if n < m, we can say that majority extensions should not support electing d.
And if we can turn any ABCd election (Smith set of size three, plus a Condorcet loser) into one where ABC are clones, then we get that any clone independent Condorcet method must be Smith in the four candidate case, just by extending majority, through Condorcet, into Smith.
Let's say we have matrix M that can be derived either from ballot set B1 or B2, where B1 contains no technical clones and in B2 A, B and C are technical clones. Being a technical clone means that those candidates are ranked together in all the ballots (the usual clone definition). It doesn't say that those candidates would be similar of that the voters would consider them to be closely related political alternatives (political clones) (they could be as well from three fiercely competing parties).
I note that clone criterion is a problematic criterion in the sense that it does not cover only political clones (does of course often miss also them), but sometimes also competing groups. In methods that are based on the matrix only it forces also ballots B1 to be treated as if A, B and C would be clones, although they can be very far from being political or technical clones. This means that especially in matrix based methods clone criterion can be an overkill.
m: mix(A, B, C) > d
n: d > mix(A, B, C)
Lets use margins as the default measure of preference strength (without losing generality). Expression mix(A, B, C) refers to a mixture of any preference orders between A, B and C. The strongest unavoidable defeat (sud) of that mix ((or any set of candidates)) is defined as the strongest defeat of the candidate whose strongest defeat is smallest (minmax style). The interesting ABCd cases (from the perspective on my question) could be the ones where sud(A, B, C) > m:n. I note also that if sud(A, B, C) > m:n, one can ask if A, B and C are really political clones, even in the case that they are technical clones in the ballots (B2).
Are there any clone-independent methods that meet Condorcet but not Smith? The only one I can think of is the "obvious" method that collapses clones and then runs Minmax. Perhaps also BTR-IRV? I'm not sure. But both of these require access to the ballots themselves, which
makes me suspect that you can't tell a clone scenario from a non-clone scenario if all you have is the pairwise matrix.
Yes, the pairwise matrix hides lots of ballot information, including having candidates next to each others in the ballots. (M vs. B1 and B2)
If I'm right, then for any given (3-cycle + 1) Condorcet matrix, I could use my old tool, linear programming, to determine a clone ballot set that would produce that matrix. It would be something like:
-
Because ABC is a clone set, the only permitted ballots are xD or Dx where x is one of ABC ACB BAC BCA CAB CBA, i.e. ABCD ACBD BACD BCAD CABD CBAD DABC DACB DBAC DBCA DCAB DCBA.
-
We want the pairwise strengths to be as determined beforehand.
-
Usual constraints: no negative number of ballots, no infinities.
In your ABCd case, we have:
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
and the pairwise matrix is (for row, against column)
A B C d
A -- 66 33 50
B 33 -- 66 50
C 66 33 -- 50
d 49 49 49 --
Programming the LP gives the following clone solution:
33: A>B>C>d
17: B>C>A>d
16: d>B>C>A
33: d>C>A>B
that produces the same pairwise matrix as above.
If we collapse the clone set, we get
50: A>d
49: d>A
so any clone-independent Condorcet method that only knows the pairwise matrix must elect from the ABC set here, and since it can't distinguish the LP solution (where ABC is a clone set) from the original ballot set, there's no way for it to elect d in your example.
I haven't actually proven that clone independence + Condorcet + only look at the pairwise matrix implies 4-candidate Smith; that would require more linear algebra than I feel like using today.
Fine so far.
But it's not that implausible; and if it's true, that means that whatever makes a Condorcet loser deserve to win, if anything, must come from information not provided by the pairwise matrix.
Note that the problems between matrix and ballot information mainly emerge from the clone criterion. One could say that pure clone independence / existence of clone candidates can not be measured from the matrix only (without doing the "overkill"). The non Smith Set arguments are more neutral (e.g. minmax style arguments) with respect to using the matrix only vs also the ballots (matrix is enough). The "overkill" is the problem that forces d not to be elected also when there are no technical clones. (Smith Set criterion is close to the clone independence criterion.)
The arguments in favour of d can be valid also when we have technical clones (B2). The sud(A, B, C) > m:n argument/criterion may work also with B2 (since the cycle among the technical clones / possible political clones would still cause lots of dissatisfaction if one of the technical clones would be elected).
BR; Juho
On Mar 28, 2018, at 1:26 AM, Juho Laatu juho.laatu@gmail.com wrote:
If you observe the strength of comparisons from utility point of view, that could make those 66:33 preferences either stronger or weaker. But my argumentation (in favour of considering the non Smith Set members as potential winners) is based on plurality only, not on utilities.
I view a 66:33 Smith Cycle as having no additional useful information over a 60:40 Smith Cycle or a 50:49 Smith Cycle. The only way to see a 66:33 Smith Cycle as being “stronger” than the other cycles is to imagine that it corresponds to a strength of preference or depth of division in the electorate, when there is no data to support that.
(In some sense pure ordinal methods make the one-man-one-vote assumption and behave as if utilities correspond directly to the plurality difference (of some other function of the pluralities).)
I would argue that they actually don’t behave that way, and should not be looked at this way. It assumes probability, from an imagined distribution that is not that predictable in reality. Electorate preference is not evenly distributed. It clusters all over the place in ways that are not predictable. There is nothing about a 66:33 cycle that makes “d” a more appropriate winner than a 60:40 or 50:49 cycle. To argue otherwise is to imagine a correspondence with utility.
Number of voters in a margin is meaningless to Condorcet methods. All that matters is whether a majority is reached. Beyond that, it is impossible to characterize that majority.
Incidentally, this does point to what I believe is a major downside of a Condorcet method - it’s by definition entirely unsuitable for figuring proportional representation, because that again would imbuing vote margins with utility concepts.