Re: current A.6 draft
- To: debian-vote@lists.debian.org
- Subject: Re: current A.6 draft
- From: Andrew Pimlott <andrew@pimlott.net>
- Date: Wed, 4 Dec 2002 18:32:26 -0500
- Message-id: <20021204233226.GA5936@pimlott.net>
- In-reply-to: <1038588533.fe8551c4@debian.org>
- References: <1038167671.7a204f7a@debian.org> <20021127190440.GB28363@pimlott.net> <1038534787.de38225b@debian.org> <20021129052232.GA8224@pimlott.net> <1038588533.fe8551c4@debian.org>
(I only have time for a quick reply, and I haven't read any of the
other recent discussion carefully.)
On Fri, Nov 29, 2002 at 11:48:53AM -0500, Raul Miller wrote:
> A has a 2:1 supermajority requirement, B has no special majority
> requirement, D is the default option, votes are
> 3 ABD
> 1 BDA
> 1 DBA
>
> A defeats B by 4:1
> B defeats D by 4:1
> D defeats A by 4:3
>
> Because D is the default option 4:3 cannot be an instance of the weakest
> defeat, so the weakest defeat is 4:1.
Your draft says:
c. A weakest defeat is a defeat that has no other defeat weaker
than it. There may be more than one such defeat.
In this example, none of the defeats has a defeat weaker than it.
Therefore, they are all weakest defeats, are are eliminated. That
is the problem I was pointing out.
So is it: A defeat by the default is always stronger than a defeat
by a real option?
> > It sounds like you're getting at something close to aj's
> > proposal, in which any option defeated by the default option has no
> > chance. If that's not what you mean to do, can you clarify the
> > difference?
>
> That's exactly what I mean. The difference between this draft and aj's
> earlier draft is that this characteristic of the default option doesn't
> cause us to lose information where an otherwise significant option is
> defeated by the default option.
I see. However, this system (like aj's) still rewards the strategy
of ranking the default option second, because a pairwise defeat by
the default is effectively fatal. So it seems too prone to abuse
for me.
YAExample: sincere preferences are
3 ABD
2 BAD
but voters vote strategically
3 ADB
2 BDA
We are deadlocked.
To remind, I suggest that the defeat D>A be scored for weakness
purposes as 2:3. In this case, if the A voters truncate[1]
3 A
2 BDA
then
D>A 2:3
A>B 3:2
B>D 2:0
and A wins. Although A has escaped its supermajority requirement, I
find this a lesser evil.
Andrew
[1] From http://www.electionmethods.org/evaluation.htm, in
Condorcet/CSSD,
"a majority never needs any more than truncation strategy to
defeat a particular candidate, even when countering offensive
order reversal by that candidate's voters."
I haven't verified that this is true of my proposed system, but I
believe it is.
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