From: Danny Swarzman <danny_at_stowlake_dot_com>

Date: Mon Nov 12 2007 - 09:12:50 CST

Date: Mon Nov 12 2007 - 09:12:50 CST

Let's consider 12 candidates. That's what we had for mayor in SF last

time.

How many combinations is that? I get 1453.

Here's how I figure it.

The number of completely filled ranks is the permutations of 12

things taken 3 at a time, 12! / (12-3)! or 1320.

Since all ballots with 2 ranks filled are equivalent, there are 12! /

(12-2)! such possibilities or 132.

And there are 12 possibilities with one choice marked or 12

And one empty or 1

1320+132+1 = 1453 combinations.

How big is the tally sheet now.

If there were 20 candidates, there would be 6840 possibilities for

completely filled ballots, plus some for partially filled ballots.

The number 12 comes from San Francisco's last mayoral contest. I

think that last year there were 22 candidates for one of the

supervisor positions.

A limited number, but an unwieldy number. If you follow the language

of SF's charter, the elections board could choose to offer more than

3 ranks. If 3 ranks is more democratic than 1, maybe 10 is more

democratic than 3. Want to try those numbers?

-Danny

On Nov 12, 2007, at 6:07 AM, Kaj Telenar wrote:

*>
*

*> Arthur Keller wrote:
*

*>
*

*>>> Could you imagine an IRV vote for Governor of California, where 8.5
*

*>>> million votes are cast? Would you tabulate that IRV race by
*

*>>> hand-count? How?
*

*>
*

*>
*

*>
*

*> There are only a limited number of possible sequences of votes. If the
*

*> buckets for that race are not just who ever is first, but the whole
*

*> series, then you would end up with a bunch of series with the
*

*> number of
*

*> votes cast for each series. Those series are much easier to manipulate
*

*> than stacks of ballots.
*

*>
*

*> First the ballots are put into the appropriate buckets and the
*

*> number of
*

*> ballots in each bucket is totaled. Each series could get its own 3x5
*

*> card. Then each of the series is allocated to the candidates based on
*

*> first choice. When the candidate with the lowest totals is eliminated,
*

*> then the cards are moved to the appropriate candidates pile and the
*

*> totals are adjusted.
*

*>
*

*> This system only gets difficult if you have multiple winners, such as
*

*> for a city council race. I assume that there is a limit to the
*

*> number of
*

*> choices someone can cast. For example, one can vote for 1st, 2nd, 3rd,
*

*> and 4th choice; but not 5th or beyond.
*

*>
*

*> example series:
*

*>
*

*> Let's assume 4 candidates for governor and at most 3 choices (1st,
*

*> 2nd,
*

*> and 3rd choice).
*

*> You would end up with the following series:
*

*>
*

*> A, B, C
*

*> B, A, C
*

*> B, C, A
*

*> C, A, B
*

*> C, B, A
*

*>
*

*> A, B, D
*

*> B, A, D
*

*> B, D, A
*

*> D, A, B
*

*> D, B, A
*

*>
*

*> A, C, D
*

*> C, A, D
*

*> C, D, A
*

*> D, A, C
*

*> D, C, A
*

*>
*

*> B, C, D
*

*> C, B, D
*

*> C, D, B
*

*> D, B, C
*

*> D, C, B
*

*>
*

*> A, B
*

*> B, A
*

*>
*

*> A, C
*

*> C, A
*

*>
*

*> A, D
*

*> D, A
*

*>
*

*> B, C
*

*> C, B
*

*>
*

*> B, D
*

*> D, B
*

*>
*

*> C, D
*

*> D, C
*

*>
*

*> A
*

*> B
*

*> C
*

*> D
*

*>
*

*> _______________________________________________
*

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*

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*

*>
*

*>
*

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Received on Fri Nov 30 23:17:17 2007

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