Talk:Dominant mutual third set

Possible multi-winner generalizations

User:Kristomun, I think this may have some kind of STV-PR generalization. In a 2-seat election, we know that anyone who has over 1/4th of the active votes at any point in STV is guaranteed to be one of the final 3 remaining candidates, since it's impossible for 3 other candidates to each have more votes than this candidate (since they'd each have to have over 1/4th of the active votes, resulting in more than 100% of votes total being allocated to different candidates), which is what would enable them to survive elimination longer. So, we can say that when over 1/3rds of the voters prefer someone from the "dominant mutual quarter" set (DMT but for 1/4th of the electorate) over anyone else who survives until the final round, then the dominant mutual quarter candidate must win. In general, someone who is preferred by a solid coalition of 1/(k+2)th of the voters (k being the number of seats) and preferred by 1/(k+1)th of all voters over any other given rival must win. I'm not sure if there's a way to extract more from this insight, though. BetterVotingAdvocacy (talk) 03:48, 27 March 2020 (UTC)

Smith-efficiency

User:BetterVotingAdvocacy, I don't think what you said is true: that electing from the DMT set implies Smith when the Smith set is a subset of the DMT set is. Consider e.g. an election where more than a third of the voters vote ABCD in some order above everybody else, and that there are say, 26 candidates. Suppose furthermore that each voter in a majority votes (some random permutation of a random subset of candidates E..Z) > (some permutation of A, B, and C) > D > everybody else. Now {A,B,C,D} is the smallest DMT set, but D is beaten pairwise by A, B, and C, and thus D is not in the Smith set. So the Smith set is {A,B,C} which is a subset of the smallest DMT set {A,B,C,D}. Then our contrived DMT-passing method could elect D, which would be in the DMT set but not the Smith set.

E.g.

12: D>B>C>A>E>F>G>H>I
11: A>B>C>D>E>F>G>H>I
11: C>A>B>D>E>F>G>H>I
20: E>A>B>C>D
20: F>B>C>A>D
20: G>C>A>B>D

Or the obligatory anti-IRV example:

12:D>A>B>C>E>F>G
11:D>B>A>C>E>F>G
10:C>D>A>B>E>F>G
 1:C>B>A>D>E>F>G
21:E>A>B>C>D>F>G
22:F>B>C>A>D>E>G
23:G>C>A>B>D>E>F

Here the smallest DMT set is {ABCD}. IRV elects D. The Smith set is {ABC}. Kristomun (talk) 22:10, 8 May 2020 (UTC)

Speedup is not a speedup

I've removed the part where IRV is sped up by aborting early after finding the DMT winner. This because you need the pairwise matrix to determine if the candidate is a DMT winner, and compiling that pairwise matrix takes a longer time than just running IRV to completion. I've replaced it with a more general statement about potential speedups. Kristomun (talk) 19:28, 22 May 2020 (UTC)

I did explicitly write "This never requires more rounds of counting than the regular IRV approach (ignoring the discovery of the pairwise comparison matrix)," Even if you don't want to call it a speedup, why not preserve the example in some form? Part of the reason I prefer to mention that DMT can be used to reduce rounds of counting is because it helps provide a regularity or predictability to these methods, which is important because their fickle order of elimination often makes it hard to understand them. In other words, if there's an IRV election that requires 10 rounds of counting under the regular approach but 5 with DMT, then it's less cognitively burdensome to look at it using the DMT approach. BetterVotingAdvocacy (talk) 21:48, 22 May 2020 (UTC)

Electing a candidate outside the DMT set must be allowed for an honest vote.

I have a small but cleverly composed example:

5 A>D>C

3 B>E>C

2 B>A>C

5 C>E

2 D>E>A

1 D>C>E>B

2 E>B>C

IRV elects B. Condorcet methods, like Copeland, find a Condorcet Winner, electing C.

In addressing the shortcomings of IRV, I have a runoff voting process that will elect E despite Candidate C being the Condorcet winner. If being a Condorcet Winner was an overriding condition of victory, the 3 B > E > C supporters will be extremely reluctant to vote for C. When these three votes change to B>E, there is no Condorcet winner. Voters should not be given any reason to do that. This example demonstrates that a fair voting process must allow itself to fairly count honest opinions as cast on the ballots and accept the possibility that a candidate from outside the Dominant Mutual Third Set can be elected.

RalphInOttawa (talk) 00:48, 19 December 2023 (UTC)

While C is no longer a Condorcet winner after this modification, he is still a member of the Smith set, and the Smith set is a subset of the DMT set. Thus a Condorcet method that passes Smith (and thus DMT) could elect C both before and after, thus giving no incentive to use that strategy. (Also note that when there is a Condorcet winner, he is always part of the DMT set.) For instance, Schulze, which passes Smith, elects C after truncation: see Rob LeGrand's ranked-ballot voting calculator.

More broadly speaking, Condorcet is incompatible with later-no-help and later-no-harm. So you can't pass Condorcet and never have situations where truncation pays off. But that's not related to the DMT set as such - IRV itself passes both later-no-help and later-no-harm and elects from the DMT set. Kristomun (talk) 13:16, 20 December 2023 (UTC)

Okay, I take it back. Yes there is a Condorcet winner, but they don't have more than a third of the 1st preferences as needed to be a DMT of one. In my example, I believe that the DMT includes all 5 of my candidates, since everybody beats somebody. Therefore E is within the DMT, and I now know better. How do I delete this whole thing? RalphInOttawa (talk) 22:54, 31 December 2023 (UTC)