# [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.

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

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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