[EM] Friendly Voting: Some Criteria Compliance Proof Sketches

[Forest Simmons]

This "friendly approval" turns out to be pretty blah.

But here's a new method that has fpA-SumfpC as a first order approximation:

It is of the form elect argmax L(X), where L is a lottery on the candidates
defined by the following experiment:

Draw a random ballot A.

Let f(A) be the candidate most favored on ballot A.

Continue drawing ballots, and let C be the first ballot drawn such that
f(C) defeats f(A), provided there is such a ballot.

Continue drawing ballots, and let D1be the first ballot for which f(D1)
defeats both f(A) and f(C), if there is such a ballot.
...
Continue until you reach the last ballot DMax in this sequence ... the one
such that f(DMax) defeats all of the previous favorites in the sequence,
but is maximal in that regard ... no ballot has a favorite that defeats
both f(DMax) and all of the previous favorites in the sequence.

This last ballot favorite DMax is the lottery winner, which is the result
of a random experiment.

To get a deterministic method based on this experiment, we define L(X) as
the probability that X will be the lottery winner DMax. This probability is
not a random variable, but is completely determined by the voted ballots.

So our election method (elect argmaxL(X))  is deterministic.

If I am not mistaken, the first few terms in the calculation of L(A) are
fp(A) -fp(C1)-fp(C2) - ... -fp(Cn), where the Ck are the candidates that
defeat A.

The higher degree terms are beyond the scope of my tired brain.

But the thought experiment gives some probabilistic meaning to the original
fpA-SumfpC formula.

It might suggest how to alter the experiment to get a method with better
properties ... say when we get into the D's, start looking at lower
candidates, not just f(D) for defeaters of the previous D's.

-Forest

On Mon, Oct 17, 2022, 8:23 PM Forest Simmons <forest.simmons21 at gmail.com>
wrote:

> OK, here's what I would like to call Friendly Approval, a decisive
> approval score method based on friendly ideas:-)
>
> For each ballot B, let f(B) be B's favored first place choice.
>
> For each candidate X  let A(X) be the number of ballots B that rank X, for
> which X has a short beatpath to f(B).
> This A(X) is X's base level approval. If argmax A(X) has only one member,
> that member is to be elected.
>
> Otherwise, break the tie with E(X), defined as the number of ballots B
> that rank X for which X has a beatpath to f(B) of at most one step.
>
> If there are still tied candidates, break the tie with C(X) defined as the
> number of ballots B for which X is ranked on B and covers (or weakly
> covers) f(B).
>
> If there are still tied candidates, break the tie with F(X) defined as the
> number of ballots for which X=f(B).
>
> In other words elect argmax S(X), where S(X) is the sum given by
>
> A(X)/epsilon^3+E(X)/epsilon^2
> +C(X)/epsilon+F(X)
>
> For a weaker approval method, add to S(X) the term G(X)/epsilon^4, where
> G(X) is the number of ballots B on which X is ranked for which X has a
> finite beatpath to f(B).
>
> -Forest
>
>
> On Sun, Oct 16, 2022, 2:26 AM Kristofer Munsterhjelm <km_elmet at t-online.de>
> wrote:
>
>> On 10/16/22 03:08, Forest Simmons wrote:
>> > Suppose de granted each candidate one free bullet ballot .... that
>> would
>> > keep non-Smith candidates from being in the tied-for-win set ... but it
>> > creates a non-scaling problem.
>> >
>> > However, if we grant each candidate a gratuitous bullet ballot with
>> > positive weight epsilon, and shrink epsilon until further shrinkage
>> > stops changing the winner, then the method becomes scale invariant.
>>
>> Yeah, I thought about that. Basically you can add an epsilon to every
>> candidate's first preference count and then let epsilon go to zero. This
>> is equivalent to counting sum over friends A: fpA - sum over defeaters
>> C: fpC as a two-vector whose first element is just the sum and the
>> second is the number of friends minus the number of defeaters, and then
>> using leximax.
>>
>> The problem is that this fails clone independence. Suppose A and B are
>> tied even given the tiebreaker above, and let C be some friend of A
>> who's not a friend of B. Clone C, then A wins.
>>
>> -km
>>
>
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