On Jan 21, 2010, at 6:30 PM, James Gilmour wrote:
robert bristow-johnson > Sent: Thursday, January 21, 2010 6:49 AM
but breaking it down to piles regarding every conceivable permutation
of candidate preference is *still* breaking it down to a finite
number of piles. for 3 candidates, that number is 9. if you or
Kathy say it's 15, then you're wrong (and it's your slip that's
showing). for 4 candidates the number of necessary piles is
40.
for N candidates, the number of piles necessary, P(N) is
N-1
P(N) = SUM{ N!/n! }
n=1
not
N-1
P(N) = SUM{ N!/n! }
n=0
I do not intend to comment on your formula, but I calculate the
numbers of possible unique preference profiles for increasing
numbers of candidates (N) as follows:
N Unique Preference Profiles
2 4
3 15
...
then your calculation is mistaken. the fact that you ostensibly need
4 piles when there are only two candidates should serve as a clue.
--
r b-j [email protected]
"Imagination is more important than knowledge."
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