[EM] : Chicken problem (was: SODA and the Condorcet

[Jameson Quinn]

Like IRV, tree approaches would not allow supporters of candidates from
other branches to help decide which of the "clones" on the winning branch
wins. They would also not allow a situation where A likes B but B doesn't
like A. In both cases, this leads to an IRV-like center-squeeze problem,
which, especially in one-dimensional scenarios, is quite costly in terms of
Bayesian Regret.

Perhaps you can think of ways to fix this, but if so, you'll have to be more
specific than "tree methods".

....

As to SODA; I included my proposed chicken-fix rule in the "optional rules"
section of the SODA page. And it's remarkably unsatisfying. Here is a fix
for what I think is the most significant practical problem scenario in all
of voting theory; and yet half the people would skip that section, half of
the people who read it wouldn't understand why it matters, and half the
people who did wouldn't understand why it works. So, although this is
something I'd love to be able to brag about more, I didn't even include
"fixes the chicken problem" anywhere among the top 15 advantages in the
advantages section.

Oh well.

Jameson Quinn

2011/8/7 Juho Laatu <juho4880 at yahoo.co.uk>

> On 7.8.2011, at 2.04, Jameson Quinn wrote:
>
>
>
> 2011/8/6 <fsimmons at pcc.edu>
>
>> Jan,
>>
>> IRV elects C like all of the other methods if the B faction doesn't
>> truncate.  But IRV elects A when the B
>> faction truncates.  Of course, with this knowledge, the B faction isn't
>> likely to truncate, and as you say C
>> will be elected.
>>
>> The trouble with IRV is that in the other scenario when the B faction
>> truncates sincerely because of
>> detesting both A and C, IRV still elects A instead of B.
>>
>
> Also, if the A faction votes A>B, then B clearly should win, but does not
> under IRV. So yes, IRV solves the chicken dilemma, but in so doing causes
> other problems. (This same argument, as it happens, works against tree-based
> methods.)
>
> I still claim that SODA is the only system I know of that can solve the
> chicken dilemma without over-solving it and making other problems.
>
>
> I wouldn't say that trees "over-solve" the problem. The tree approach to
> the chicken problem could be called "explicit clones". That's quite natural.
> Some candidates just announce that they are clones and that they will
> support each others. That sounds like a pretty exact solution, not an
> over-solution.
>
> Do trees "cause other problems" then? They do not allow the voter to
> support one of the clones without supporting the other. But this is exactly
> what the intention of the explicit clone approach is. Also the need to
> declare a branch in the tree could be considered to be a practical problem /
> increased complexity. And the need to identify the clones is an extra task /
> problem. But maybe not really. What other (more serious) problems would the
> trees cause?
>
> Juho
>
>
>
>
>
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