Hi Kristofer,
De : Kristofer Munsterhjelm <[email protected]>
>À : Michael Ossipoff <[email protected]>
>Cc : [email protected]
>Envoyé le : Mercredi 9 mai 2012 9h54
>Objet : Re: [EM] "FBC vs Condorcet's Criterion"
>
>
>On 05/08/2012 08:46 PM, Michael Ossipoff wrote:
>> Since Richard wants to make a "which one wins" comparison between
>> FBC and Condorcet's Criterion (CC), then I'll remind him that, when
>> FBC failure sufficiently makes its problem, CC compiance becomes
>> quite meaningless and valueless. And there is good reason to believe,
>> as described in my previous post, that Condorcet's FBC failure _will_
>> fully make its problem in our public elections.
>
>I'll get to your larger post later, but it seems what need isn't FBC as such,
>but rather u/a FBC.
>
>Here's a Condorcet method I think meets u/a FBC: Each voter submits a ranked
>ballot with an Approval cutoff. The most Approved candidate in the Smith set
>wins.
>
>If everybody ranks Approval style, then this becomes Approval. So let's see if
>there's any reason to favorite betray instead of ranking Approval style.
>
>If there is no cycle, then you can't make an acceptable have a greater chance
>of winning over an unacceptable by ranking Compromise over Favorite versus
>ranking Favorite over Compromise.
>
>If there's a cycle and Favorite is in the Smith set, but Compromise is not,
>then the only reason for getting Compromise into the Smith set would be to
>defend against an unacceptable candidate winning. However, you can do that by
>just voting Approval style. Since Smith set members are "only beaten by other
>Smith set members", Favorite vs Compromise doesn't enter into it as long as
>you put both above the cutoff and all the unacceptables below it.
>
>If there's a cycle and Compromise is in the Smith set, but Favorite is not,
>then because this is an u/a election, it doesn't matter. You'll still get an
>acceptable.
>
>If there's a cycle and neither Compromise nor Favorite is in the Smith set,
>then voting Approval style will make Compromise and Favorite both maximally
>work to push the unacceptables out of the Smith set.
>
>Hence it seems that the method above meets u/a FBC. By the time people get
>past u/a, they'll no longer be overcompromising and so "proper" FBC failure
>doesn't matter. So Condorcet can meet u/a FBC.
>
>I'm not saying Smith,Approval is necessarily a good method, but I only have to
>show a single method to disprove that u/a FBC and Condorcet is incompatible.
>
Did you cover the scenario where both Favorite and Compromise are in the Smith
set? If not, is there some reason
for u/a FBC (whose definition I don't actually know) why this scenario doesn't
matter?
In general the way C//A methods fail FBC is by creating incentive to make
Compromise beat Favorite because "Worst"
will win the approval tiebreaker.
Thanks.
Kevin
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