@keith-edmonds Yes, I anticipate that I should be back for the committee. I was going to post something in the loomio thread but I think that a lot of the goals that I was thinking of had already been said (and I don't think I need to try to come up with a whole lot of others since the goals other people stated seem like plenty to do already.)

@andy-dienes and rightfully so. Now how can we mathematically embody the intuition that throwing out pairs of votes, when compared to keeping all the votes and tallying them to arrive at scores (which may have pairwise indexing) that are sums over the votes, violates something?

[edit]

Suppose we have the system I described above, with the rule of matching ballots and throwing them out in pairs.

Let's say the candidates are Gore, Bush, and Nader and I vote Nader, 511. Let's account for the votes in terms of their effects on the pairs of candidates. In fact, let's ignore Nader-Bush and concentrate on Nader-Gore. What is the effect of my vote on the Nader-Gore accumulator? It has to be canceled by a vote of Nader, -511.

@rob If I finish this friggin' software, we may find out.

@jack-waugh said in Making Voters Equal in Power:

In principle, this is an infinite recursion.

And hence, "hall of mirrors."

In practice, my iterative vote simulator stops when the ordering of candidates doesn't change between two rounds. Since the ordering is all the "vote caster" algorithm looks at, there is no point continuing because it will always get the same result.

But keep in mind, that doesn't mean it has reached the one and only equilibrium. There can be multiple equilibria.

But again, I'd rather have a method that doesn't encourage this sort of thing. If I did a vote simulator for Condorcet, there would be very little, if any, iterating, since there isn't an obvious way to adjust your ballot even if you know how others will vote.

@rob I'm just going off of the definitions, no assumptions needed 🙂

Agreed that voters' ballots will likely change depending on the candidates in the race in Approval more with a higher probability than in most other voting rules.

@Toby-Pereira to answer your question I'm looking into it. Arrow's Theorem actually has quite a few (slightly different) formalizations, and it looks like what I said is technically not true for the version defined on Wikipedia since that one only allows (strict) linear orders, but I feel quite sure I saw a formalization where the domain was all (weak) linear orders. I will try to find it.

@cfrank said in The Condorcet Criterion:

@rob I revisited this and I think Delegated Condorcet is a really interesting idea, what are your latest thoughts about that concept?

I haven't given it a lot of thought in the years since. It seemed like a good idea, and probably would work pretty well if adopted, but I doubt people would trust it enough to adopt it in the first place.

It might be best for local elections where there isn't a huge amount of publicity and the public tends to only know one or two candidates out of a larger field.

@rob said in Are Equal-ranking Condorcet Systems susceptible to Duverger’s law?:

So what is that algorithm? I mean, it could be Condorcet, and I have no problem with that, but I can't see how the term "proportional representation" applies. It just sounds like multi-winner.... there isn't anything proportional about it.
The only way the term "proportional representation" would apply (by my understanding of the term) is if we assume that there are some number of parties, and each candidate and each voter is in one and only one party. If a 3rd of the voters are in the Bull Moose party, then a third of those elected should be in the Bull Moose party. The further you get away from that, the less "proportional representation" seems to be a meaningful descriptor.
All of my complaints regarding PR (and with so many people's insistence that it is so much better than single winner methods such as Condorcet methods) are based on the assumption that voter X below is considered to have "better representation" if d, f and e are elected (because candidate d is very close to voter X), than if a, b and c are elected.

For the algorithm, you will be aware of Single Transferable Vote, which gives PR without any mention of parties. If voters happen to vote along party lines, it gives party PR, of course. That's just one example, without having to mention obscure methods invented by people on this forum.

And as for a, b, c versus d, e, f, I discussed that in the other thread here. I'm not sure it's worth quoting though because it's quite long.

You are def not the first one to ask this: https://link.springer.com/content/pdf/10.1007/s003550100128.pdf Here they give a few examples that participation does not imply monotonicity.

@akazukin5151

floor( vote / droop_quota ) is the lower quota, ceil( votes / droop_quota ) is the upper quota.

I'd say that the Droop Quota isn't very good when the quota is so small.

I agree, I'm not allocating seats in a real life election so this is just purely theoretical.

A more realistic quota for most elections would be in the thousands, so rounding up to the next integer would be a very small percentage difference. I think this is largely where the problem stems from.

So that's the core reason, makes sense to me.

But if we're going with 4 as the quota anyway, I think we're just minimising the amount over quota they go. So we have the following seats "owed" to each party:

That's pretty smart. This is the algorithm as I understand it:

If there are still seats remaining to elect, award seats to the party with the smallest difference between seats won and "party quota" (seats_won - votes / quota). Repeat until all seats filled

(For simplicity I'm breaking ties in favour of the first party)

So my example would start from the automatic seats [5, 6, 12]. Award seats based on the largest remainders method as usual, stopping when all parties has been awarded a seat: [6, 6, 12], [6, 6, 13], [6, 7, 13].

The remaining 4 seats would be filled like this

[7, 7, 13], [7, 7, 14], [7, 8, 14], [8, 8, 14]

Thank you very much for the help!