Consider a typical single member scenario, like the 2000
 USpresidential election. Say George W Bush (label B) gets 101. Al
 Gore (label A) gets 100. Ralph Nader (label C) gets 10. (After his
 vote got squeezed from strategic voting.)

 Bush is elected on single preference votes. If there were a second
 ballot, or instant run-off vote (IRV), Al Gore wins on second
 preferences of Nader, the “spoiler” who is eliminated.

 So, IRV passes on independence of irrelevant alternatives (IIA).
 But IRV fails on the Laplacelaw of preference gradation. (Orders
 of preference vote gradually fall off in count importance.) This
 also means that the IIA criterion is inconsistent with the Laplacelaw.

 Binomial STV avoids the dilemma of this inconsistency.

 Suppose IRV gives:

 100 A C _

 101 B _ _

 10 C A _

 With IRV, A picks up the second preferences of C, and wins with 110.

 Binomial STV counts abstentions, shown by the dash lines. The
 third preferences are all abstentions, and they do nothing to
 change the simple plurality count. It is conceivable that in a
 less contentious world than ours, this could be the case. But it
 is assumed that the voters have been informed that last
 preferences can be given to count against candidates. And the full
 slate of preferences is as follows:

 100 A > C > B

 101 B > A > C

 10 C > A > B

 The keep value quotient (kvq), of a candidate, is the election
 keep value, divided by the exclusion keep value:

 the keep value is the quota, 211/2 = 105.5, divided by candidate vote.

 kvq A = 0/100. Here, zero means close to 0, giving a very small
 fraction.

 B = 110/101

 C = 101/10

 Unity, or less, is the election (or exclusion) threshold of an
 election (or exclusion) keep value. Less than unity passes the
 threshold. So, A wins with binomial STV, without breaking the
 Laplacelaw, or IIA.

 Suppose, however, that Bush supporters decide to vote insincerely,
 by making Gore their last preference – even tho Nader is the last
 person they want to see win, but know he can’t, anyway.

 (However, this scenario might be sincere in UK, with A as Labour,
 B as Tory, and C as Liberal Democrat. In either case, a change in
 the vote, sincere or otherwise, will change the count in a toward
 manner, with good book-keeping.)

 The preference slate becomes:

 100 A > C > B

 101 B > C > A

 10 C > A > B

 Then, kvq becomes:

 A = 101/100

 B = 110/101

 C = 0/10

 There is a contradictory answer. B is closer to the quota but A is
 closer to the quotient.

 This is not a logical contradiction but a contingent
 contradiction. Decision, as to the winner, is not a democratic
 decision, but an administrative decision, at present FPTP, based
 on a convention, reached by previous agreement. (Single majority
 is the least democratic system, in the first place.)

 A first approximation of an administrative decision (It would be
 “spurious accuracy” to go further) is:

 B quota deficit: 105.5/101 ~ 1.0446.

 A quotient deficit: 101/100 = 1.01.

 Therefore, an administrative election is of Gore. (The Supreme
 Court also made an administrative decision.)

 The result would most likely be similar, if Binomial STV, used
 more accurate figures. To say nothing of the investigation, by
 Greg Palast, of the electoral roll. (The Best Democracy That Money
 Can Buy.)

 I repeat that for a democratic decision, as from those Keltic
 reports, prevously cited, I don’t recommend less than 4 or 5
 member Andrae/Hare system (at-large STV/PR).

 Regards,

 Richard Lung.



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 Election-Methods mailing list - see https://electorama.com/em for
 list info

Forest, My main concern is with the autocratic nature of single member monopolies, generally little or no better than minimally democratic, and unaccountable. They have not got beyond the ancient Greek tyranny, deserving of their reputation in the vulgar sense of tyranny. R.L. On 07/06/2022 19:06, Forest Simmons wrote: > I wish you could have made Nader win .. informal polls, including a > Time magazine poll that was continually updated online until it became > too embarrassing to the neoliberal establishment, showed that Nader > was the landslide Sincere Condorcet Winner. > > El mar., 7 de jun. de 2022 10:16 a. m., Richard Lung > <voting@ukscientists.com> escribió: > > > Consider a typical single member scenario, like the 2000 > USpresidential election. Say George W Bush (label B) gets 101. Al > Gore (label A) gets 100. Ralph Nader (label C) gets 10. (After his > vote got squeezed from strategic voting.) > > Bush is elected on single preference votes. If there were a second > ballot, or instant run-off vote (IRV), Al Gore wins on second > preferences of Nader, the “spoiler” who is eliminated. > > So, IRV passes on independence of irrelevant alternatives (IIA). > But IRV fails on the Laplacelaw of preference gradation. (Orders > of preference vote gradually fall off in count importance.) This > also means that the IIA criterion is inconsistent with the Laplacelaw. > > Binomial STV avoids the dilemma of this inconsistency. > > Suppose IRV gives: > > 100 A C _ > > 101 B _ _ > > 10 C A _ > > With IRV, A picks up the second preferences of C, and wins with 110. > > Binomial STV counts abstentions, shown by the dash lines. The > third preferences are all abstentions, and they do nothing to > change the simple plurality count. It is conceivable that in a > less contentious world than ours, this could be the case. But it > is assumed that the voters have been informed that last > preferences can be given to count against candidates. And the full > slate of preferences is as follows: > > 100 A > C > B > > 101 B > A > C > > 10 C > A > B > > The keep value quotient (kvq), of a candidate, is the election > keep value, divided by the exclusion keep value: > > the keep value is the quota, 211/2 = 105.5, divided by candidate vote. > > kvq A = 0/100. Here, zero means close to 0, giving a very small > fraction. > > B = 110/101 > > C = 101/10 > > Unity, or less, is the election (or exclusion) threshold of an > election (or exclusion) keep value. Less than unity passes the > threshold. So, A wins with binomial STV, without breaking the > Laplacelaw, or IIA. > > Suppose, however, that Bush supporters decide to vote insincerely, > by making Gore their last preference – even tho Nader is the last > person they want to see win, but know he can’t, anyway. > > (However, this scenario might be sincere in UK, with A as Labour, > B as Tory, and C as Liberal Democrat. In either case, a change in > the vote, sincere or otherwise, will change the count in a toward > manner, with good book-keeping.) > > The preference slate becomes: > > 100 A > C > B > > 101 B > C > A > > 10 C > A > B > > Then, kvq becomes: > > A = 101/100 > > B = 110/101 > > C = 0/10 > > There is a contradictory answer. B is closer to the quota but A is > closer to the quotient. > > This is not a logical contradiction but a contingent > contradiction. Decision, as to the winner, is not a democratic > decision, but an administrative decision, at present FPTP, based > on a convention, reached by previous agreement. (Single majority > is the least democratic system, in the first place.) > > A first approximation of an administrative decision (It would be > “spurious accuracy” to go further) is: > > B quota deficit: 105.5/101 ~ 1.0446. > > A quotient deficit: 101/100 = 1.01. > > Therefore, an administrative election is of Gore. (The Supreme > Court also made an administrative decision.) > > The result would most likely be similar, if Binomial STV, used > more accurate figures. To say nothing of the investigation, by > Greg Palast, of the electoral roll. (The Best Democracy That Money > Can Buy.) > > I repeat that for a democratic decision, as from those Keltic > reports, prevously cited, I don’t recommend less than 4 or 5 > member Andrae/Hare system (at-large STV/PR). > > Regards, > > Richard Lung. > > > > ---- > Election-Methods mailing list - see https://electorama.com/em for > list info >