[EM] Something Different.[correction]

[Forest Simmons]

On Sun, Mar 26, 2023, 10:59 PM Forest Simmons <forest.simmons21 at gmail.com>
wrote:

> Our Friendly Method experiments have led in this direction:
>
> Remember Bottom  Count Dsapproval?
>
> The Bottom Count Disapproval of X ... BotCountD(X) is the number of
> ballots on which X outranks no candidate.
>
> The Bottom Count Anti-Favorite Lottery probabilities f'(X) are obtained
> from the Bottom Count Disapproval scores D(X) the same way the Martin
> Harper lottery probabilities f(X) are ontained from the Equal Top Approval
> scores A(X):
>
> f'(X) is the percentage of ballots B on which the disapproval D(X) is the
> largest disapproval of any candidate disapproved on ballot X ... meaning
> not ranked above any candidate on ballot B ... i.e. contributing to the
> Bottom Count Disapproval of X.
>
> For example ...
> 48 C
> 28 A>B
> 24 C
>
> D(A)=72
> D(B)=48
> D(C)=52
>
> So f'(A)=72, f'(B)=0, and f'(C)=28.
>
> Our first anti-favorite lottery method is to elect argmax Q(X), where Q(X)
> is the quotient Sum{f'(Y|X defeats Y}/D(X)
>
> In this case the sums reduce to one term each because no candidate X
> defeats more than one candidate Y.
>
> Q(A)=f'(B)/D(A)=0/72
> Q(B)=f'(C)/D(B)=28/48
> Q(C)=f'(A)/D(A)=72/72
>
> The winner is argmax Q(X) = C.
>
> For the less lottery minded, here's a different version:
>
> Let Q(X)=min{f'(Y)| X defeats Y}/D(X)
>

Should be
Let Q(X)=max{D(Y)| X defeats Y}/D(X)

Also note that if we leave out the denominator of the quotient, the method
becomes Landau efficient.

Elect argmaxQ(X).
>
> The result is the same ... because min S and sum S are the same when the
> set S has only one member.
>
> -Forest
>
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