Ranking Array Math, Part 2

[DEMOREP1 at aol.com]

The recent Ossipoff/Anderson postings cause me to repeat some elementary
aspects of voting.
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A. Two choices or Yes/No Voting

The simple Yes or No (i.e. 2 choice) case may be made into a simple matrix.
X = diagonal to display matrix, Y =yes, N= no, YN= number of ballots ranking
yes over no, NY= number of ballots ranking no over yes.
X  Y  N
Y  X  NY
N  YN  X

Obviously YN or NY will be a majority of all the ballots (assuming each voter
votes and there is no tie).

B. Approval/Disapproval Voting

If there are multiple choices and the yes/no method is applied to each
choice, then there is Approval Voting (A) for yes and Disapproval Voting (D)
for no. That is, for each choice the matrix becomes
X  A  D
A  X  DA
D  AD  X

Each choice thus gets a majority or a minority of approval or disapproval
votes (assuming each voter votes and there is no tie).

An approval vote in effect combines all yes type rankings that might
otherwise occur. 

However, by combining such rankings Approval Voting can fail a majority of
first choice votes test (as if rank voting was actually being used) and a
head to head test.

Example- Votes for each choice, T=Total
    Choices
   1  2  3  T
A 20 31  2  53
B 20 32  2  54
C 51  1  0  52
D  8 35 53  96
With approval voting D wins with 96 of 99 ballots. With a majority of first
choice votes test C wins with 51 of 99 first choice votes. C would also win
each head to head pairing. Note that 3rd choices are truncated and there are
no 4th choices in the example.

A majority disapproval vote removes unacceptable choices so that the
remaining choices would be at least tolerable to the majority.

C. Three Choices- Single Winner

With 3 choices the matrix becomes as follows (excerpt from Ranking Array
Math,(Part 1).

Using relative rank order voting (1, 2, 3, etc.) produces an array (or
matrix). In the below the winner in a pairing is at the top and the loser is
at the left. Thus AB means the number of ballots that rank A over B. CT=
Column Totals, RT= Row Totals
(View table in 9 point Monaco)
X  A   B   C   RT
A  X   BA  CA  ART
B  AB  X   CB  BRT
C  AC  BC  X   CRT 

Assuming that AB>BA, BC>CB and CA>AC there is the well known tie of A beats
B, B beats C and C beats A or A>B>C>A.
The question is what tie breaker to use -- one or more of the numbers in the
matrix, one or more of the row or column totals, one or more differances in
the totals (such as ACT-ART) or some other combination.

Since the tie situation possibility begins with the three choice case any tie
breaker at the three choice case should apply with four or more choices.

D. Truncated votes and Unranked Choices

The matrix suggests that a ranking for a choice is a vote over each unranked
choice.

The simple case-- 3 candidates A, B and C-- 2 voters vote for B, 1 voter
votes A,C.  B obviously wins. In the matrix each vote for a choice must be
deemed a vote over each unranked candidate to produce the common sense
result.

X  A  B  C  RT
A  X  2  0  2
B  1  X  1  2
C  1  2  X  3 

BA 2>AB 1, B beats A
BC 2>CB 1, B beats C

A question is what value in the matrix, if any, that unranked candidates
should have regarding each other (e.g. the value, if any, for A and C on each
ballot where B is ranked first). The value has a direct effect on any tie
breaker using the matrix.

If the value is zero (0), then the result is the above. If the value is 0.5,
then the result is

X  A  B  C  RT
A  X  2  1  3
B  1  X  1  2
C  2  2  X  4 

Note that the 9 grand total is the number of ballots (3) times the number of
rankings on each ballot (3)--(1st over 2nd, 1st over 3rd and 2nd over 3rd). 

Again, where C is the number of choices on a ballot, then there are (C x
(C-1))/2 relative rankings on such ballot (the sum of 1 + 2 + ... + C) --
2-1, 3-3, 4-6, 5-10, etc.

E. Multiple Choices

The matrix also applies to multiple choice elections (such as electing 2 or
more sheriffs). Example- Choosing 2 of 5 choices (A,B,C,D,E) (i.e. 2 choices
on a ballot are to be effective).
The cells for each combination of two would be matched against each other
candidate. Example
   B  C
A  BA CA

and the transposed amounts
   A 
B  AB 
C  AC

If 2 candidates beat each other candidate, then they would be the Condorcet
winners.
However, in most cases there will be multiple ties and so the need for a tie
breaker.
Note that the BC and CB amounts do not matter when B and C are a team. 

F. Proportional Representation
The matrix also applies to a proxy proportional representation system in
which each voter has 1 vote. Example- Choosing 3 of 5 choices
The cells for each combination of three would be matched against each of the
other 2, one at a time. Example
Example- 3 candidates of 5 (M, R, T, S, W) to be elected.
  M   R   T
S MS  RS  TS

and the transposed amounts
    S
M  SM
R  SR
T  ST

Three candidates are the Condorcet winners if each beats each of the other
candidates. Each vote for a final loser would go to the winning candidate
ranked highest on a ballot. Once again the probability of ties requires a tie
breaker.

G. Tie Breakers
If the values in a matrix are not symetrical,then any value based matrix tie
breaker would seem to work. Which tie breaker is *best* would seem to be a
matter of opinion.

If time is available I will try to put the various ranking methods into a
matrix form to see if they produce majority rule results.