[EM] Random Ballot Favorite Chain Climbing

[Forest Simmons]

One important (but easy) correction:

In order to make this method Monotone, we have to start the chain from the
bottom of the list ListF. That's what puts the "Climbing" in Random Ballot
Favorite Chain Climbing!

A comment on exposition for public consumption ... no mention of Condorcet
Smith, Landau, or Banks should be included in the method description, any
more than a brief introduction to IRV needs to explain what to do if the
last three remaining candidates have the same first place transferred vote
totals.  Every generic public ballot set will have a Banks member with at
least one first place vote, so no need to get people worried right off the
bat about what to do in the impossibly rare contrary case.  Stick with
generic conditions in the voters' pamphlets ... just make sure that the
rare exceptional possibilities are covered in the published official legal
definition, as well as the RAQ's (Rarely Asked Questions) if not the FAQ's!

I only mentioned it at all because of its earlier mention in related EM
list threads.

-Forest

On Thu, Oct 20, 2022, 11:48 AM Forest Simmons <forest.simmons21 at gmail.com>
wrote:

> First a preliminary procedure to make sure no single candidate defeats
> every member of the support of the random ballot favorite:
> As long as there is such a candidate, retain only candidates of this
> índole, recalibrating between elimination steps.
>
>
> Next: a non-deterministic lottery method ... Random Ballot Favorite Chain
> Climbing (RBFCC):
>
> Shuffle the ballots into some random order B1, B2, B3, ... and let ListF
> be a list of the candidates in the order induced by the first choices of
> the respective ballots in their order ... i.e. according to the order of
> their first appearance as a first choice on a ballot in the sequence B1,
> B2, B3, ...
>
> Now, Chain climb the list ListF by initializing the set variable CHAIN as
> the empty set, and then ....
> While some member of ListF defeats every member of CHAIN, add the first
> such candidate into CHAIN. EndWhile
>
> The head of the completed chain is the RBFCC (random trial) winner.
>
> Next, for each candidate X, let RBFCC(X) be the winning probability for X
> under this lottery.
>
> Finally, elect argmax RBFCC(X).
>
> Note that this method is Banks efficient, and obviously reduces to
> "fpA-SumfpC" in the eponymous three candidate case.
>
> On a practical note, should the computation of the RBFCC probabilities be
> intractable for some ballot set, then repeated trials in a MonteCarlo
> simulation of the lottery can be used to determine argmax RBFCC(X) with
> arbitrarily low error probability epsilon.
>
> Is this the simplest formulation of what we've been looking for?
>
> It doesn't seem like an easy method to "game".
>
> Other comments? Questions?
>
> Who can write this up in a way that Joe Q Public can easily relate to?
>
> -Forest
>
>
>
>
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