Multiple Same Choices

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Multiple Same Choices

Craig Carey-2
In an article, [hidden email] ("Steve Eppley") writes:

>Lucien Saumur wrote:
>>Steve Eppley writes:
>>>Lucien wrote:
>>          Or NOTR for None of the Rest.
>Or NOTB for None of these Bums...  
>I don't care which term is used.
>>>>It may also indicate that no candidate is ranked.
>>>I'm unclear on this.  Do you mean some are equally approved and the
>>>rest are equally disapproved?
>>          I mean that ballots would be produced indicating:
>>                    NO CANDIDATE IS RANKED
>>                    NO CANDIDATE IS ACCEPTABLE
> That's what is meant by simply voting:  1=NOTB
>>          Such ballots would be tallied by adding .5 vote
>>to every count of the matrix and by adding zero to the
>>acceptability count of every candidate.
>Like I said, you can produce the same result by treating NOTB as just
>another candidate in the matrix, simplifying the algorithm.  You
>don't need a separate data structure for approval counts.  This is
>just an implementation detail, though; what matters is what the
>voters see.

          Yes, what matters is what the voters see. The
voting process must appear clear, simple and attractive to
the man in the street who will have to use it. The man in
the street will vote but he will not do the tally although
he will see the tally results.

          My computer system attempts to simplify the
voting aspect by asking the voters to use a computer mouse
and to rank the candidates by clicking on a list of
candidates on a computer monitor screen. As the voter
clicks on the name of a candidate, the name is transferred
to another list of "ranked" candidates on the same screen.
When the voter is satisfied with his selection, he may then
click on a button to produce a printed ballot.

          It is important to keep the number of buttons to
its minimum and the complexity of the system to its least.
My system has a button to restart the ranking process to
correct a mistake. It has a button to end the ranking
process prematurely and another button to mark when the
rest of the candidates are unacceptable.

          I am now considering the addition of a button to
allow equal ranking and I am concerned that the system
would be made too complicated for most voters. However, I
think that it is possible to train people to use a more
complicated system, as was done with banking machine.
However, for demonstration purpose and to sell the idea, we
should not include all the bells and whistles that may
eventually have to be incorporated to the system.

          The voters do no need to know the precise
tallying procedure. The tally must be done by electoral
officials who may be expected to understand more of the
process than the ordinary voters. In my system, the ballots
are tallied in two stages. In the first stage, the
electoral official reenters the data on the ballot by
repeating the voting process. The data, as it is entered,
is retained on a computer file. The second stage will
process this file to produce the matrix of pairwise counts
which may then be explained to the voters. (My
demonstration version of the system does create a computer
file but produce the matrices immediately. There are two
matrices: one for the current ballot and another for all of
the ballots processed to-date.)

          It think that the voters will better understand a
separate acceptability count than they would understand
that NOTB has become one of the candidates.

>>>>My system is also designed to tally the ballots. When candidates
>>>>are not ranked (indicating equal preference) the related counts for
>>>>both candidates are updated with half (1/2) a vote.
>>>I haven't thought about this, but my guess is that whether you should
>>>score equally ranked choices as .5 and .5, or 1 and 1, or 0 and 0,
>>>depends on the choice of tie-breaking algorithm you plan on using.
>>>It may not matter--is it possible to specify a sensible tie-breaking
>>>method that doesn't care how the equal pairs are counted?
>>          A tie means a tie and it may be broken most
>>fairly by flipping a coin.
>I meant a "Condorcet circular tie", in which no candidate beat all
>the rest in the pairwise matchups.  We're not yet in a coin-flipping
>The tie-breaking should probably look at how badly each candidate is
>beaten overall.  If a voter ranks two choices equal, does that mean
>each is unbeat or half-beat?  Using .5 means you think each is
>half-beat.  (There was a message several days ago, from Rob or Mike
>I think, on this question.  I'll have to go back and reread it.)

          A Condorcet circular tie does not necessarily
involves all of the candidates. Inasmuch as it does not
involve the winners, then it is irrelevant. Inasmuch as it
does, then all the candidates who are not involved may be
considered eliminated leaving only the candidates involved
in the circular tie to be considered as potential winners.

          I do not think that it makes any sense to speak
of a candidate as being "beaten overall." The point is
that, in a circular tie, each candidate has beaten another
candidate and may therefore consider himself cheated if
that candidate is declared elected. I suppose that circular
tie mean that the voters have no great preference between
the candidates involved and that the winner should be
decided most fairly PURELY by chance (by flipping a coin).

          I am ready to look at an actual situation and to
test if my hunch is correct. To prove that it is correct,
we should have a runoff election which would ask the voters
to indicate if it is, or is not, true that they have no
great preference between the candidates involved in the
circular tie.

          [hidden email]

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Multiple Same Choices

Craig Carey-2
Replying to Saumur:

Of course it makes sense to speak of how beaten a candidate is overall.

If you're part of that majority that prefers Clinton to Buchanan, even
though you don't really like Clinton, and voted Nader 1st, then you're
helping Buchanan be beaten with a majority against him. Do you want
that majority to be counted or not? Do you want Buchanan to be denied
victory because you, & the rest of that majority, have said you all
like the same other candidate more?

As I've said, our initial goal, in single-winner reform, is the
cast a reliably-counted, full-strength vote for Clinton against
Buchanan, to prevent a Buchanan victory, without having to vote
Clinton 1st.

So, if we didn't count how beaten Buchanan is, we wouldn't be counting
you or the rest of that majority that he has against him.


You're assuming that every circular tie is the result of there not
being a Condorcet winner--that every circular tie happens because
every candidate actually has another candidate who is _preferred to_
him by more voters than vice-versa. We mustn't confuse _preferred to_
with _ranked over_.

It's often assumed that every voter sincerely ranks all of the candidates.
Or at least that every voter ranks all of the candidates. But that
never happens in real elections. In every rank-balloting election,
many people will only rank some of the candidates, sometimes only

But even if it's a natural circular tie, where every pairwise "beat"
coincides with overall public sentiment, there's still such a thing
as a candidate with a majority against him, a candidate instead of
whom a full majority have said they'd rather have someone else.
Choosing randomly from the circular tie ignores the wishes of that
majority. Maybe there's a candidate in the circular tie who, while
beaten, isn't "majority-rejected". Picking someone who is majority-rejected
in that election would violate majority rule.


Aside from that, randomly choosing from the circular tie isn't good

As I said, truncation (voting of a short ballot) or bullet-voting
(only voting for 1 candidate) are things that occur on a large
scale in any rank-balloting election. Truncation or bullet-voting
(which is extreme truncation) will often cause a circular tie, when
it lets someone beat the Condorcet winner. When that happens in, say,
a 3-candidate race, and you solve the circular tie randomly, there's
only a 1/3 chance that the result will be worse for the truncators
than the Condorcet winner, who would have won had they voted a
complete ranking.

So there's a strategic incentive to truncate. But whether the truncation
is strategically motivated or not, it will still steal victory from
the Condorcet winner sometimes. Condorcet's method systematically
gives the election to the Condorcet winner in that situation.

With a random solution, whether or not the truncation is strategically
intended, it creates a risk for the Condorcet winner, and if the
Republicans are likely to truncate, you might need to vote Clinton
in 1st place to protect him from Republicans who are willing to
take a 1/3 risk of electing Nader.

When the Random-Solution method picks Buchanan, as it, of course,
will do 1/3 of the time, when the Republicans truncate, then the
Random-Solution method is ignoring the fact that you're part of a
full majority who've ranked Clinton over Buchanan because all of
you don't want Buchanan to win, and are saying so by all ranking
the same person over him. Majority rule is violated, and the lesser-
of-2-evils problem is stil there, since your vote for Clinton over
Buchanan didn't count for diddly.


Your suggestion to solve circular ties by a 2nd balloting is ok,
and I wouldn't object to it. But no one likes a method that requires
a 2nd balloting. The 2nd balloting could be by Approval. It matters
little, if at all, what method is used in the 2nd balloting, since
the published results of the 1st balloting give voters very good
information about how far, if at all, they need to compromise in the
2nd balloting.

As I said, though, I never propose this 2-balloting method, which
I've called "Runoff-Pairwise", or "BeatsAll-Approval" anymore, because
people seem unanimous in not wanting a 2-balloting system. I propose
only Condorcet's method.

One use of Runoff-Pairwise would be in a committee where the
members were unable to agree on a circular tie solution. In tht
situation the natural thing would be to simply use a 2nd balloting
to solve circular ties.