From: Douglas W. Jones <jones_at_cs_dot_uiowa_dot_edu>

Date: Thu Apr 29 2004 - 14:28:38 CDT

Date: Thu Apr 29 2004 - 14:28:38 CDT

On Apr 29, 2004, at 11:25 AM, Arthur Keller wrote:

*> Mark, I'm confused by your answer.
*

*>
*

*> Since ranked preference voting says: A, B, C is different than B, A,
*

*> C, there are more than 6 possible vote combinations for 3 candidates.
*

*> (ABC, ACB, BAC, BCA, CAB, CBA are the full ones; also A, B, C, AB, BA,
*

*> AC, CA, BC, CB, and no choices selected. Wow, that's 16 choices. Does
*

*> someone have a formula in closed form for the number of possible
*

*> rankings for n candidates? For the full ones, its the number of
*

*> permutations of n candidates, or n! (n factorial).)
*

This works, if you're willing to allocate n! bins for your ranked

preference

ballot. Imagine doing that with the California Recall election with 140

or so candidates.

But, with weighted preference, there are simple reconciliation schemes.

If you have 3 candidates, A, B and C, you give your first candidate 3

votes, your second choice candidate 2 votes, and your third choice

candidate

1 vote. In sum, you have 6 votes to distribute over 3 candidates, so

your

reconciliation scheme can be based on making sure that the number of

votes

adds up to 6 times the number of ballots counted (and if I only vote

for two

candidates, I have 3 undervotes, while if I only vote for one, I have

one

undervote). This reconciliation rule is computationally more tractable

than

the n! rule.

Similarly, for STV/IRV systems, you treat the ABC race as 3 races, one

for

first place, one for second place, and one for third place, and

reconcile

votes that way as you're carrying votes forward from the precinct to the

center. These numbers aren't the overall winners, just a convenient

reconciliation rule. Again, cheaper than n!

Doug Jones

jones@cs.uiowa.edu

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Received on Fri Apr 30 23:17:24 2004

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