Head(s) vs. Head(s)

[DEMOREP1 at aol.com]

A short summary of elections seen as being Head(s) vs. Head(s) using Number
Voting (rank order) (preference) (1, 2, etc.) voting --

All of the combinations of N candidate(s) [A] are paired against 1 candidate
[B] with the second or later choices of the other remaining candidates [C]
being counted for (transferred to) the [A] or [B] candidate(s) whomever came
first.

A candidate when in the [A] group will (1) win, tie or lose (2) in all or
some of the comparisons against each [B] candidate (3 x 2 = 6 possibilities).

Legislative bodies using proportional representation, N members, 1 vote per
voter (i.e. The one highest choice on a ballot is effective).

Example elect 5 members, 12 candidates

Each combination of 5 candidates would be paired against 1 candidate with the
second choice on the ballots of the other candidates going to one of the 5 or
the 1 whomever came first.
Example-
B,D,G,H,J   versus C    Others A,E,F,I,L
The second choice on a E first choice ballot is for H.
The 5 winners would have a voting power in the legislative body equal to the
number of votes each receives in being elected.  [i.e. no fixed quotas]

Executive/Judicial offices

As an initial matter, candidates for such offices either have or do not have
majority yes/no approval (since Number Voting only shows relative approval).
 This is not the case with p.r. legislative bodies since no one p.r candidate
need have a majority of the votes. 

Elect one, 1 vote per voter (i.e. The one highest choice on a ballot is
effective). This is the standard Condorcet case.

Example elect 1 mayor, 4 candidates
Assume that each of the 4 has majority yes/no approval.
M versus P    Others S, T
On a first choice ballot for T, the second choice is for S, the third choice
is for M.

Elect two or more (M), M votes per voter (i.e. The M choices highest on each
ballot are effective).

Example elect 3 clerks, 8 candidates
Assume that each of the 8 has majority yes/no approval.
B,C,F   versus A      Others are D,E,G,H
On a first choice ballot for G, choices 2, 4 and 5 are for C,F and A. Choice
3 was for D so goes to the next choice.

Computers make doing the comparisons much easier. 
Due to the likely probability of ties when there are many candidates, tie
breaker rules are necessary.   One obvious tiebreaker is having the candidate
in ties who has the lowest number of first choice votes losing and redoing
the comparison math.  Note- a candidate who wins/loses in all of his/her
comparisons should win/lose.