[EM] Re: Hand counting election methods

Ernest Prabhakar ernest at drernie.com
Wed Nov 19 21:28:24 PST 2003

Um, wouldn't that require 10! people = 3,628,800 for ten candidates?

-- Ernie P.

On Nov 17, 2003, at 5:39 PM, Gervase Lam wrote:

>> Date: Mon, 17 Nov 2003 14:11:01 -0500
>> From: Dave Ketchum <davek at clarityconnect.com>
>> Subject: Re: [EM] Electronic Voting Bill of Rights?
>> CONDORCET:  Choices here also:
>>       Count as in Plurality?  Tempting, since a majority winner would
>> end the task.
> That's it!  You've reminded me of a way to do a Condorcet hand count.
> An explanation follows, the logistics of which could be improved.  For
> brevity, I have not taken into account tied votes nor do I know how to
> atm.  I want to get to bed as soon as possible.
> Suppose there are 10 candidates.  Create 10 large teams.  Each team
> represents a candidate who was given the top rank.  Each of the teams
> obtain the relevant ballots.  From this, you are now able to fill in a
> large proportion of the pairwise matrix as you know that a top ranked
> candidate is a beats all candidate.
> Each large team has 9 sub-teams, the 9 being for each 2nd ranked
> candidate.  Each sub-team receives the ballots from its parent large 
> team.
> Again, from this, you are now able to fill in a large proportion of the
> pairwise matrix, a part of which has already been filled in by the 
> parent.
> The 9 sub-teams each have 8 sub-teams etc, etc...
> Therefore, you are able to complete a Condorcet hand count in N - 1
> iterations, where N is the number of candidates.
> Thanks,
> Gervase.
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