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: 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. [[User:BetterVotingAdvocacy|BetterVotingAdvocacy]] ([[User talk: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:
[deleted] How do I delete this whole thing? [[User:RalphInOttawa|RalphInOttawa]] ([[User talk:RalphInOttawa|talk]]) 0022: 4854, 1931 December 2023 (UTC) ▼
5 A>D>C
: Just do an edit and delete it, then save changes. [[User:Kristomun|Kristomun]] ([[User talk:Kristomun|talk]]) 17:03, 4 January 2024 (UTC)
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.
▲[[User:RalphInOttawa|RalphInOttawa]] ([[User talk: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 [https://munsterhjelm.no/km/rbvote/calc.html 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. [[User:Kristomun|Kristomun]] ([[User talk: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? [[User:RalphInOttawa|RalphInOttawa]] ([[User talk:RalphInOttawa|talk]]) 22:54, 31 December 2023 (UTC)
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