Re: An ammendment (Re: Formal CFV: General Resolution to Abolish Non-Free)

```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,

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...)

```