*To*: "Debian Vote Discussion List" <debian-vote@lists.debian.org>*Cc*: "Darren O. Benham" <gecko@debian.org>*Subject*: Re: An ammendment (Re: Formal CFV: General Resolution to Abolish Non-Free)*From*: "Norman Petry" <npetry@cableregina.com>*Date*: Fri, 16 Jun 2000 01:27:18 -0600*Message-id*: <006f01bfd764$4ea06860$bf04a8c0@mandos.nga.sk.ca>*Reply-to*: "Norman Petry" <npetry@accesscomm.ca>

Darren O. Benham wrote: >> >> Except that Chris Lawrence is mistaken as to how Condorcet's method is >> implemented within the Debian Project. Yesterday he wrote: >> >Chris is not mistaken. That's strange... in my post I gave an example from one of Debian's previous elections proving (I think) that all ranked options score a vote against each unranked option. I double-checked my initial result and cannot find any errors in my analysis, which suggests that one of the following must be true: 1) Debian's implementation has changed since the Vote 0002 ballots were counted, so the results no longer apply. 2) Debian's implementation of the Concorde method is not working as intended, or as you believe it to be working (although it does work as it should, imho) 3) I'm still overlooking something, and my analysis is wrong. Here's another example which shows why I think explicitly ranked options score a vote against each unranked candidate: Suppose we have 3 ballots: Option: ABC Voter #1 [1--] Voter #2 [-1-] Voter #3 [2-1] To determine which options are dominated, and by whom, we can construct a 'pairwise matrix' such that P(row,col) contains the total number of votes for the 'row' option against the 'col' option. If we assume that explicitly ranked options score one vote each against all unranked options, the pairwise matrix (P) looks like this: P A B C --- --- --- A | X | 2 | 1 | --- --- --- B | 1 | X | 1 | --- --- --- C | 1 | 1 | X | --- --- --- Note that there are a total of 7 votes in the pairwise matrix. If instead we assume that explicitly ranked candidates *do not* score votes against unranked candidates, then the pairwise matrix looks like this: P A B C --- --- --- A | X | 0 | 0 | --- --- --- B | 0 | X | 0 | --- --- --- C | 1 | 0 | X | --- --- --- C scores a single vote against A from ballot 3. The first 2 ballots yield no pairwise votes. In this case, there is a total of 1 pairwise vote in the matrix. If we now count the number of 1st, 2nd, and 3rd place rankings in the ballots, we get: A-B-C Total 1st preferences 1-1-1 3 2nd preferences 1-0-0 1 3rd preferences 0-0-0 0 no preference 1-2-2 5 If explicitly ranked candidates *do* score votes against unranked candidates, the corresponding number of pairwise votes (in a 3-way contest) should equal: 2*total_1st_preferences + total_2nd_preferences = 2*3+1 = 7 votes -- check. If they do not, total pairwise votes will be somewhat less than that. Carrying out this check with one of Debian's actual elections, I found that the totals match, which can only mean that ranked candidates are scoring votes against unranked candidates (unless the implementation has changed since that time). Please note that this is a *desireable* outcome, in my opinion -- because people will commonly truncate a ranked ballot after ordering only a few of the more important choices, it is important that unranked choices are treated as though they had been ranked lower, otherwise unpopular/unimportant alternatives will gain an unintended advantage. It is not a flaw in the method or the implementation, but merely an unstated assumption that is apparently causing confusion, which is why I raised the matter after Chris had commented on it. If there is something I've overlooked which makes this conclusion incorrect, please let me know. Norm Petry p.s. Congratulations on the new baby! I'm sure you've got more pressing matters to deal with right now than voting methods trivia, so please reply at your leisure (you may not have any for the next few months, if what they say about newborn infants is true...)

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