[EM] Defeat strength, Winning Votes vs. Margins, what to do with equal-ranks on the ballot?
stepjak at yahoo.fr
Wed May 29 21:54:16 PDT 2019
I know that you feel it's adequate that there is a probabilistic relationship between (x-y) and (x+y). I think
the relationship to expect in reality is unclear. Consider that in real world experience major contests that
involve the strongest candidates and most of the voters are often close races, in which case (x-y) doesn't
predict (x+y) well at all.
Personally I would say that (x-y) just reveals the lowest possible value for (x+y).
Your arguments about balancing decisiveness and size/salience are unique I think. I think it's more
common to argue that salience is irrelevant or imaginary. One issue in particular is that margins is
supposed to incentivize voters to cast strict rankings. If this works, then every contest will have about the
same salience anyway.
>Le mercredi 29 mai 2019 à 01:36:41 UTC−5, robert bristow-johnson <rbj at audioimagination.com> a écrit :
>> You have to square p to have p in there at all, because r is defined as m/p. So r*p = (m/p)*p; it just cancels out the division.
>that ain't how i look at it.
>probabilistically, the bigger WV+LV is, we can expect a bigger WV-LV in magnitude.
>the expectation value (or mean) of |WV-LV| increases as WV+LV does.
>hmmm, i gotta think about this a little. the probability a voter votes for the winning candidate is p=[WV/(WV+LV)] and the probability the voter votes >for the losing candidate is q=[LV/(WV+LV)] consider a binomial distribution, with n, p and q=1-p,
>the variance of WV is (WV+LV)*[WV/(WV+LV)]*[LV/(WV+LV)] which is also the variance on LV. the standard deviation of both are sqrt( >WV*LV/(WV+LV) ) and the numerator increases as the square of the denominator. so I would guess that |WV-LV| goes up with the sqrt(WV+LV). >if the decisiveness (which Juho calls "r") remains the same, if you double the size of an election, the s.d. of the |WV-LV| margin is expected to >increase by 41%. but the means also go up.
>so the standard deviation goes up with sqrt(WV+LV). but i still expect the mean of |WV-LV| to go up proportionately with participation if the mind
>the electorate remains the same and the participation increases.
>the more i keep reading this thread (that i guess that i started), the more convinced that i am that, considering simplicity as an asset to sell to both >policy makers and the public, i am convinced that the "Best Single-Winner Method" must be Ranked-Choice ballots (not FPTP, not Score, not >Approval) and Condorcet-compliant (because the alternative to electing the CW is electing a candidate when explicitly more voters marked their >ballots preferring a different specific candidate) and i think that RP with simple Margins (WV-LV) is the most meaningful and simplest.
> WV-LV = [(WV-LV)/(WV+LV)] x (WV+LV)
>the factor [(WV-LV)/(WV+LV)] (or "r") is the measure of the decisiveness of an election. this is what we mean by "Brexit wins by 3.8% in 2016."
>the factor WV+LV (or "p") is the measure of how big the election is. how many people are affected by it enough to weigh in on it.
>the Defeat Strength is the product of the two. Bigger elections count more. And more decisive elections count more. i dunno.
>perhaps, to sell a Condorcet method, rather than RP, we could sell IRV-BTR to make the IRV crowd happy. we sorta lose the precinct summability >(actually precincts can report pairwise defeat totals which can still be used to check up on the official central counting in the likely case there is no >cycle). then this Defeat Strength discussion becomes moot again.
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