[EM] SODA

[Jameson Quinn]

2011/7/5, fsimmons at pcc.edu <fsimmons at pcc.edu>:
>
> I thought that A was required to make her approvals consistent with her
> ordering, i.e. to approve
> everybody ranked above her cutoff.  Doesn't that mean she is required to
> approve herself?
>
> Maybe I'm thinking of an older version of SODA.
>
> I hope you are right that there is nothing to fix.

Let's do this slowly. Here's the scenario:

 34 A>B>C
 35 B>C>A
 31 C>A>B,

B delegates first. B delegates to B,C. Totals are now C 66, B35, A34.
A's turn. If A does not delegate, C will be winning when it comes to
C's turn, and so C will not delegate. So A delegates to A,B. Totals
are now B69, C66, A34. C's turn. C is unhappy with B and so delegates
to C,A - but it's not enough. Final totals are B69, C66, A65.

I believe that the correct strategy for any combination of delegable
and undelegable votes (including minor, non-Smith candidates) in a
3-candidate Smith set is always for everyone to approve two members of
the Smith set if they care between the bottom two. This gives the same
result as minimax and most Condorcet methods. I haven't proven this,
and I don't have a general understanding of strategy for larger Smith
sets.

It is possible, when there are 3 or more near-clones A1, A2, A3...
running against a different candidate B with almost 50% - that is, B
can beat any combination of fewer than all the A's, and B has no
preference among the A's - that the true Condorcet winner among the
A's is subject to center squeeze, and the A's are forced to throw
their support to whichever of them has the most delegable votes, in
order to prevent B from winning. The upshot is that SODA, even
assuming candidates are honest in their pre-vote rankings and
strategic in their delegation, does not pass the Condorcet criterion,
but does pass the majority Condorcet criterion (that is, a pairwise
winner always wins if each of the pairwise wins constitutes a
majority). But I can't find any nonmonotonic scenario pairs, so this
"Plurality within the faction" is the worst result I can find. I think
that it's both unlikely and, really, not so bad.

JQ

>
>
> ----- Original Message -----
> From: Jameson Quinn
> Date: Tuesday, July 5, 2011 1:07 pm
> Subject: Re: [EM] SODA
> To: fsimmons at pcc.edu
> Cc: election-methods at lists.electorama.com
>
>> 2011/7/5
>>
>> > Jameson suggested that the SODA candidates make their approval
>> decisions> sequentially instead of
>> > simultaneously.
>> >
>> > The problem with this is that if a winning candidate moves to
>> first place
>> > in the sequence by an increase
>> > in support, she may become a losing candidate:
>> >
>> > Assume sincere preferences are
>> >
>> > 35 A>B>C
>> > 34 B>C>A
>> > 31 C>A>B
>> >
>> > If approval decisions are made in descending order of faction
>> size A, B, C,
>> > then B wins.
>> >
>> > If B gains more support so that the totals become
>> >
>> > 34 A>B>C
>> > 35 B>C>A
>> > 31 C>A>B,
>> >
>> > the sequential order becomes B, A, C, and the winner will be C.
>> >
>>
>> No. B still wins. If A feels that C is winning, then A can
>> delegate to B,
>> and then B cannot lose. So C cannot be the winner. And therefore
>> B will
>> delegate to C, to force A's hand. Whether or not C delegates
>> then is
>> irrelevant.
>>
>> Of course, if A actually prefers C to B, and has managed to keep
>> B ignorant
>> of this fact, then C will win. But then, in such a case, A could
>> have gotten
>> the same result by being honest from the start.
>>
>> >
>> >
>> > How can we fix this?
>> >
>> >
>> I don't think there's anything that needs fixing, though you may find
>> another example to show I'm wrong.
>>
>>
>> > How about allowing the largest faction (in this example 49 C)
>> to go second,
>> > and making the second
>> > largest faction (in this example 27 A>B) go first?
>> >
>> > That would also work in the example above. How bad would it
>> be in a worst
>> > case example?
>> > ----
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>> >
>>
>