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) 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

On Sun, Mar 26, 2023, 10:59 PM Forest Simmons <forest.simmons21@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 >