[EM] Winning-votes intuitive?

[Blake Cretney]

Adam Tarr wrote:

 > Despite the fact that this debate has been on the list since long
 > before I showed up, I really think we're making progress.

I agree.

 >
 > I wrote and Blake responded
 >
 >>> A beats B, 70% winning votes (25% losing)
 >>> B beats C, 52% winning votes (45% losing)
 >>> C beats A, 50% winning votes (40% losing)
 >>>
 >>> By virtue of a slight perturbation (the sort that would fall within
 >>> polling error margins in the real-world, nonzero-information case)
 >>> candidate A now wins the election.  In this case, if I (any many
 >>> others like me) randomly vote B over C, we change nothing, while if
 >>> I (any many others like me) randomly vote C over B, we may turn B's
 >>> victory over C into a defeat, and we turn C into a Condorcet
 >>> winner.  This causes the defeat of A, our favorite.
 >>
 >> When I talk about randomly filling out a ballot, it is something each
 >> person would do individually.  The whole group wouldn't decide to
 >> vote B over C or C over B en masse.  Now I understand what you meant
 >> by saying this would require co-ordination.
 >
 >
 > That's not what I meant when I said that (in an earlier message).
 > What I meant was that I doubt most voters will take to the idea of
 > random ballot completion on their own, so a faction's leaders would
 > have to encourage their voters to do so.  Even if folks decide
 > independently to randomly complete, some voters may have some
 > inclination to cast the vote for the candidate they prefer
 > ever-so-slightly between those bottom two.  So co-ordination from the
 > top (random voting schemes based on the last digit of your phone
 > number, for example) would probably be called for to guarantee a
 > faction doesn't lean their "random" ballot completion one way, and
 > hurt its chances.

That's an interesting question.  Does a lack of randomness hurt its chances?

Let's consider the following situation.  You have a certain number of
votes against your candidate.  There's nothing you can do to lower this
number.  Your only hope is that each of the other two candidates in the
Smith set will have a higher number of votes scored against it.

For this to be possible, you must have a pairwise victory over one of
them, otherwise, you wouldn't be in the Smith set, and you couldn't win.
  I'm only interested in the possibility that you can affect the outcome
by your vote.

One of the opposing candidates already has votes scored against him, if
this score isn't higher than the score against you, then there is no way
you can win.  So, I'll discount that possibility.

So, what it comes down to is that last victory between the candidates.
  It must be both high enough (higher than the score against you) and in
the right direction (against the otherwise undefeated candidate) for you
to win.

Now, your only way of influencing this outcome is by ranking or not
ranking these two candidates on your ballot.  My contention is that you
should rank them, and there is no need to do so randomly.  If you have
reason to use an order-reversal strategy, then go to it, but otherwise,
you can rank them arbitrarily, so long as you rank them.

The point is this.  Except for the tie case, which is rare enough that
we can discount it, your vote can only increase the size of the victory
between the two candidates.  The danger is that it may flip the victory
to go the wrong direction.  But since you don't know which direction
that is, it is just as likely to flip it in the right direction.

If the vote could have no effect on the absolute size of the victory, it
would be obvious that the effects cancel each other out.  For example,
if you don't know whether door number 1 or door number 2 has the big
prize, changing your pick doesn't alter your chances.   But the effect
of increasing the size of the victory is not canceled out.  So, you
should rank the candidates.  The same argument holds BTW even if you
think a number of people are going to do the same as you.

To give you another example, let's say I give you a choice of what's
behind door number 1 or door number 2.  Now, let's say I tell you that
if you switch your choice, I'll put an extra $1000 dollars behind the
correct door.  It's obvious that unless you think I know which is the
correct door (and am working against you), you should switch your
choice.  Of course, this change can hurt you, but that's not the right
way of looking at it.

 > One final thought.  on 3/20 I posted a message about Approval
 > Completed Condorcet.  The idea was to use a graded ballot (ABCDEF, for
 > example).  If there was not a Condorcet winner, then the candidate
 > with the most approval votes (A's, B's, and C's in the case of ABCDEF
 > ballots) wins the election.  In my initial analysis, this method
 > seemed at least as good as the other Condorcet methods we like to
 > discuss.  Furthermore, it seems like it could be an easier method to
 > pitch, since the cycle-breaker is VERY intuitive.  So I guess what I'm
 > asking is... does ACC render this whole debate meaningless?  Just a
 > thought.
 >
Someone could cause their candidate to defeat the CW by bullet voting,
but since Mike accepts this in approval, I suspect he would accept this
here as well.  I don't know if this method would pass his endless parade
of criteria.

As for me, I like the ranked ballot because it allows a limitless number
of candidates to be sorted.  It seems to me that with 6 candidates,
there would be a tendency to use all grades, and therefore approve of
exactly 3, which isn't necessarily what you want.

---
Blake Cretney (http://condorcet.org)