[EM] Markus's '98 Cloneproof SSD wording
Markus Schulze
schulze at sol.physik.tu-berlin.de
Sat Mar 17 09:51:49 PST 2001
Dear Mike,
you wrote (16 Mar 2001):
> I checked message #2291, and I noticed that its way of dealing with
> ties isn't what I'd believed. I must have misread it the 1st time.
>
> I thought that you'd said:
>
> If there are more than 1 winner, then eliminate the other candidates
> and repeat the count, from the start, with just those winners.
>
> Repeat till this doesn't reduce the set of winners. At that time
> pick the final winner by Random Ballot.
>
> That's what I thought it was, and that's what I've been telling people
> all this time.
>
> So what I've been advocating isn't the same as what you said. Would
> the procedure that I describe above cause a problem, such as
> nonmonotocity, or some other fault? If not, it's more decisive,
> since it doesn't resort to Random Ballot till re-application of the
> count rule among its ties no longer reduces the winner-set.
>
> My main question here is whether that procedure that I describe above
> has a fault.
Blake Cretney demonstrated in his 3 Nov 1998 mail that
monotonicity is violated when one simply re-applies this algorithm
(http://groups.yahoo.com/group/election-methods-list/message/1955).
Example 1:
3 voters vote A > B > C > D.
2 voters vote D > A > B > C.
2 voters vote D > B > C > A.
2 voters vote C > B > D > A.
The pairwise matrix looks as follows:
A:B=5:4
A:C=5:4
A:D=3:6
B:C=7:2
B:D=5:4
C:D=5:4
B and D are SSD winners. When SSD is re-applied among candidate B
and candidate D, then candidate B wins decisively.
Example 2:
One voter changes from D > A > B > C to A > D > B > C
3 voters vote A > B > C > D.
1 voter votes D > A > B > C.
1 voter votes A > D > B > C.
2 voters vote D > B > C > A.
2 voters vote C > B > D > A.
The pairwise matrix looks as follows:
A:B=5:4
A:C=5:4
A:D=4:5
B:C=7:2
B:D=5:4
C:D=5:4
A, B, and D are SSD winners. When SSD is re-applied among candidate
A, B, and D then no further reduction of the winner-set can be
achieved. Therefore, Random Ballot is used so that candidate D
wins with a probability of 1/3. This can be interpreted as
a violation of monotonicity.
************
You wrote (16 Mar 2001):
> I notice, in the message that I copy below, that you're saying to
> use Random Ballot immediately if there are more than 1 winner, and then
> , if Random Ballot doesn't reduce the winner set to 1, _then_ you'd
> repeat the count, from the start, among only those winners. But
> having just resorted to Random Ballot, then why not just stay with it
> and if it doesn't reduce the winner set to 1, draw another random ballot,
> till there's just 1 winner?
Usually, Random Candidate and Random Ballot are something very awkward.
To avoid manipulations the electoral laws usually say that the Secretary
of the Interior has to draw the lot in front of the parliament. And very
frequently there are very complicated regulations about the form of the
lots and the procedure of the drawing. On the one side, these regulations
minimize the possibility of manipulations. But on the other side, there
is the problem that it can happen that the drawing of the lot is declared
null and void because of formal reasons; and when then the drawing of the
lot is repeated and a different candidate is chosen then there will be a
loud cry.
Therefore -to my opinion- the number of random steps should be minimized.
Suppose that A, B, and C are potential winners. Suppose that a ballot
is chosen randomly and that this randomly chosen ballot is A = B > C.
Then -to my opinion- there is no need for a second randomly chosen ballot.
When candidate A pairwise beats candidate B, then candidate A should win.
When candidate B pairwise beats candidate A, then candidate B should win.
Only when there is a pairwise tie between candidate A and candidate B,
then a second random choice is necessary.
However, I don't consider the question (whether the algorithm should
be re-applied among the top ranked potential winners of this randomly
chosen ballot or whether Random Ballot should be re-applied among the
top ranked potential winners of this randomly chosen ballot) so
important.
************
You wrote (16 Mar 2001):
> As you said, your procedure sounds the same as Cloneproof SSD, but
> without mentioning cycles. But the price of not mentioning cycles
> seems to be a requirement for more paragraphs. And cycles are easy to
> check for while calculating the Schwartz set. But your wording might
> be more accepted by audiences who don't like to hear about cycles.
There are rather other reasons why my description is longer. For
example: I suggest that when there is more than one defeat with the same
number of votes for the winner then the defeat with the largest number
of votes for the loser should be dropped. I wrote in my 14 Nov 1998 mail
(http://groups.yahoo.com/group/election-methods-list/message/2291):
> The "weakest" pairwise inequality is that pairwise inequality
> with the smallest absolute number of votes for the winner of
> this pairwise inequality. If there is more than one pairwise
> inequality with the smallest absolute number of votes for the
> winner of this pairwise inequality, then the "weakest" pairwise
> inequality is that pairwise inequality (among those pairwise
> inequalities with the smallest absolute number of votes for the
> winner) with the largest absolute number of votes for the loser.
> If there is more than one pairwise inequality with the smallest
> absolute number of votes for the winner and the largest absolute
> number of votes for the loser, then all these inequalities are
> substituted with a pairwise equality simultaneously.
Markus Schulze
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