# Re: What is Data Model FOR?

From: Douglas W. Jones <jones_at_cs_dot_uiowa_dot_edu>
Date: Thu Apr 29 2004 - 14:28:38 CDT

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

>
> 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
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
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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