[EM] Bush v Gore

Richard Lung voting at ukscientists.com
Tue Jun 7 10:16:50 PDT 2022

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).


Richard Lung.

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