STV for party candidate lists?

Fri Jul 31 23:30:33 PDT 1998

N = 14,  100 candidates, 3000 voters, STV Hare quota = 214.28  ---> 215
Average first choice votes per candidate = 3000/100 = 30

Assuming that no candidate gets a quota of first choice votes there would be
(for each set of 14 choices being matched each one other choice)
N = number of first choice votes, C = Choices

N1   C1
N2   C2
N14  C14
N15  C15
N16   C16
N 17   C17
N100  C100

The earliest (second, third, etc.) choices of the C16-C100 choices get
transferred to the C1-C15 choices.  To avoid possible bias, fractions should
be used.

20 C10
C10 is the second choice of 300 C90 to C100 voters.
215- 20 = 195
Each of the 300 votes should be multipled by 195/300.

If high accuracy is needed (especially with a low number of voters), then the
decimal places should be kept.
214.28- 20 = 194.28
Each of the 300 votes should be multiplied by 194.28/300.

Child's play if the voting is computerized (with a possible 85 factorial down
by 100 across matrix for each set of 85).  A major headache if done by hand.

How many choices can or cannot get a quota in all sets of N vs. 1 (such as 14
vs. 1) in a real public election is a matter of total speculation.  

Thus my suggestion for Bucklin (with a fractional modification) to break ties
for choices that do not always get quotas when in each set of N (except when N

Anybody for electing the 435 members of the U.S. House of Representatives
using STV Hare quotas- around 96 million voters in 1996 with possibly
thousands of candidates (Hare quota of around 221,000) ?

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