# STV without Elimination

DEMOREP1 at aol.com DEMOREP1 at aol.com
Wed Sep 2 19:21:37 PDT 1998

```I repeat again--
Condorcet can be used for P.R. elections.

Each set of N candidates versus 1 (the other candidates are deemed losers)
would have to be done using number votes (1, 2, etc)- obviously using a
computer for large elections.   The problem of course is multiple ties-- i.e.
not having N Condorcet winners.  There could also be a YES/NO vote on each
candidate or party.   The lowest YES candidate/party would lose for tiebreaker
purposes and have the votes transferred to the voter's next remaining choice

Example-- Elect 5 in a district, 10 candidates.  2 candidates win in all of
their 5 versus 1 matches.   The lowest YES candidate/ party would lose.
Recheck head to head math. Repeat cycle.  Each winner would have a voting
power in the legislative body equal to the number of votes he/she receives.

A much more simple method uses simple IRO--

(a) An Elector may vote for one or more legislative candidates on the ballot
in a district (plus not more than [2] write-in votes) by voting "1", "2" and
so forth for his or her first, second and so forth choices. (b) If there are
more than [5] candidates (or remaining candidates) in the district, then the
candidate having the lowest number of votes shall be a losing candidate. (c)
Each vote for a losing candidate shall be transferred to the Elector's next
choice (if any) who is a remaining candidate in the district. (d) The two
prior steps shall be repeated until there are [5] remaining candidates in the
district who shall be elected. (e) A lottery shall be held if tie votes occur
in any step. (f) Each member of a legislative body (or his or her replacement)
shall have a voting power in the legislative body and its committees, in
person or by written proxy, equal to the votes that the member finally
receives in the election. (g) Example-
C = Candidates          Voting Power
C1  21          = 21 +   1   = 22
C2  20          = 20 +   5   = 25
C3  15          = 15 +   3   = 18
C4  12 + 5    = 17           = 17
C5  12 + 1    = 13 - 13   =   0
C6  11 + 3    = 14 +   2   = 16
C7   9  - 9     =   0          =   0
VNT 0            =   0 +   2  =   2
100            100           100
C7 Loses   C5 Loses