Majority/Quota Combinations

[DEMOREP1 at aol.com]

Many of my earlier postings are definitely incorrect due to the following
regarding majority or quota combinations.

The early combinations--
The dash (-) means *versus*
2    1-1
3    1-2
      2-1
4    1-3
      2-2
      3-1
5    1-4
      2-3
      3-2
      4-1
6    1-5
      2-4
      3-3
      4-2
      5-1
Etc.
There is a plurality/ majority number of votes on the left of each set. 
If there is a majority with 1 on the left, then such 1 is an obvious first
choice majority winner.
The problem obviously is when a majority is divided with 3 or more choices (in
executive and judicial elections).

Where a number is more than 1, it means all of the combinations.

Example- With 4 choices, A, B, C, D (the slash means a combination, e.g. B/D
means B > D or D > B), the 2-2 combinations are--
Left     Right
A/B      C/D
A/C      B/D
A/D      B/C
B/C      A/D
B/D      A/C
C/D      A/B

A majority on the left may happen before the N versus the Others combinations.
Example- 6 choices (all have YES majorities), 3 sheriffs to be chosen, 100
voters
Assume only the D/F combination gets a majority in all of the 2-4
combinations-

35  D > F  > [all combinations of ABCE]
23  F > D  > [all combinations of ABCE]
58
Both D and F should be chosen.  That is, the D/F voters are the first majority
(with D and F being a partial twin).

The remaining position then becomes a 4 choice case with 1 to be chosen.
The fewest YES votes could be used as a final tiebreaker to remove losers.

The above general case math plainly indicates that the tiebreaker in the
infamous 3 choice circular tie case should be the highest majority of the
first two-

N1  AB     C
N2  BA     C
N3  AC     B
N4  CA     B
N5  BC     A
N6  CB     A

One of the combinations (N1+N2, N3+N4, N5+N6) will be the highest majority
(assuming no truncated votes).
The choice on the right loses. The 2 remaining go head to head. Much confusion
has happened since the 2-1 combination is the first combination with a divided
majority.

The preceding also applies to STV p.r. elections.

P = Party or Group of parties, A, B = Party candidates or Party, T =
Independent candidate
                YES    NO       Number Vote
P1
   P1A
   P1B
P2
   P2A
   P2B
   P2C
T1
T2
T3

A YES/NO vote on each choice might produce a YES majority which should get a
majority of the voting power in the legislative body.
If a quota method is used, then after doing the head to head math the above
could be used to produce quota winners.

Example---1000 voters, 25 seats, 110 candidates, Hare quota (1000/25 = 40)
Parties X and Y each get YES Majorities (by mutual support of their voters).
Their initial number of quotas would be dependent on the number of first
choice votes for each.
X gets an initial 8 seats
Y gets an initial 7 seats.
In head to head math (all combinations of 25 candidates versus 1 candidate
with 84 test loser candidates), assume 4 X party and 2 Y party candidates get
quotas when in all of their 25 or 1 sets.

The remaining seats (the other 4 X and 5 Y initial party seats and the other
10 seats) would be determined in like manner among the remaining parties/
party candidates/ independent candidates with a quota replacing a majority.

The X and Y parties might get additional seats depending on the head to head
and/or tiebreaker and/or residual math results. 

I again suggest that no quotas be used but that the p.r. winners have a voting
power equal to the number of votes they each finally receive.

A computer would have to do the math in most elections.