robert bristow-johnson wrote:
On Feb 1, 2010, at 11:36 PM, robert bristow-johnson wrote:
at http://rangevoting.org/IrvNonAdd.html you say:
"In a C-candidate [IRV] race, each voting machine would have to have
C! pseudo-candidates and each precinct would have to pass C!
'subtotal' counts on to the central tabulator."
and you have a table that shows
3 4 5 6 ...
6 24 120 720 ...
is that correct?
is not a more correct answer
C-1
SUM{ C!/n! }
n=1
that'll teach me to do it from memory, without thinking about it.
He's assuming Australian style IRV, i.e. all candidates have to be
ranked and equalities are not allowed. Given those constraints, there
are C! ways of ordering C candidates.
Quoting from his other page, rangevoting.org/rangeVirv.html,
A lot of voters want to just vote for one candidate, plurality-style.
In range voting they can do that by voting (99,0,0,0,0,0). In IRV,
they can't do it. (Actually in some variants of IRV - with
"truncation" - they can. In many Australian elections, full orderings
of all candidates are required or your vote is invalid. But in
Ireland and Malta, just naming your first few choices and not the
others is allowed.
With truncation and more than two candidates, obviously the number of
piles are greater (because the set of fully-specified candidate
orderings, of which there are C!, is a subset of this). With truncation
and equalities, greater still.
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