[EM] Equal defeats in RP(m)

[MIKE OSSIPOFF]

Blake said:

If you want a result that doesn't depend on a random tie-breaker

I reply:

I don't think it makes sense to use a random tiebreaker when
RP encounters equally droppable defeats during its count. But if you
want to do it that way, it's your method. But you or the person who
designated RP(m) really must tell us how you want to deal with the
situations that I asked about, randomly or otherwise, if those situations
occur. Of course if they don't occur and then it's fine if you'd
rather not say how you'd deal with those situations. But if it happens
and you don't say within 48 hours of the announcement of the situation
and the final request for instructions, then we'll solve the situation
by declaring a tie among all the alternatives that could win by the
various possible solutions.

We don't want a random tiebreaker when a designated method returns more
than one winner. I'd suggested just having that ballot give a final
Approval vote to each of its designated method's winners. Richard
suggested something else, and I agreed to that. We only use RB for
if 2 alternatives get the same vote total in the final Approval count.
Of course, for most purposes, we don't need it there either, since we
can just say "The poll chose these methods:..." The application of
RB will just be so that there will be a single winner in case there's
some later need for one.

Blake continued:
,
then, I would do the following.  Find all the possible winners (based
on Zavist and Tideman's method

I reply:

Is Zavist & Tideman's method simply Ranked Pairs(margins), or do they
specify answers to the questions that I asked in my posting. If they
do, then I repeat my request that you tell us how you want it done,
or at least tell us where in the archives your posting about that is.
If they don't answer those questions, then, if one of those situations
occurs in the poll's RP(m) count, then either you answer those questions,
or the person who designated RP(m) answers them, or we've got an RP(m)
tie.

But if you don't want to answer the questions, that's probably perfectly
ok--it only matters if one of those situations actually occurs. If it
does, then we'll repeat the question, just one more time.

Blake continues:

, and with arbitrary TBRC

I reply:

That's great. What's a "TBRC"? A tiebrakeing random...[?]?

Blake continues:

).  That is,
don't assume that the TBRC has to be drawn from the ballots, a
possible winner is a winner for any TBRC.

I reply:

If you really want to use randomness in your RP(m) procedure, that's
for you to decide. But we'll need a little more detail than you've
given here.

Blake continues:

I can help if you have any
trouble calculating this.

I reply:

Actually, how you could be of help is, if any of the situations I
asked about occur, or if some similarly ambiguous Ranked Pairs situation
occurs, then you could be really helpful by telling us how you'd
like that situation to be dealt with. But we'll repeat that question
if such a situation occurs. If you don't answer the question, maybe
the person who designated RP(m) will answer it.

Blake continues:

For the purposes of Voter's Choice, score this as approval for all the
possible winners, and, of course, for anyone the voter likes more.

I reply:

Are you saying that if any of the situations that I asked about occur,
or if some similarly ambiguous RP(m) situation occurs, we should
declare as winners every alternative that would win by some possible
reasonable solution to that situation? Sure, that's the default
solution, if such a situation occurs, and if neither you nor the
person who designated RP(m) gives another solution within 48 hours
after our final request.

And yes, as I said, I'd suggested that if a designated method returns
more than one winner, we give, from the designating ballot, a final
Approval vote to each of the designated method's winners. Richard
preferred another solution for that, and I agreed to it, and so
if a designated method returns more than one winner, we'll deal with
that by the method that Richard & I agreed to. Once the designated
voting system has returned more than one winner, the choice of what
to do about that isn't part of the method's rules. It's part of this
poll's rules.

Blake continues:

Since I assume you will find this easier to carry out, this is the
tiebreaker I now give for the purposes of this Voter's Choice
election.

I reply:

I certainly don't know that it's easier, since it requires repeating
parts of the count with different defeats dropped, but if you're sure
that you want a tie between all the alternatives that could win under
various different interpretations of what to do in the situations that
I asked about, then that's fine, unless the person who designated RP has a 
different
suggestion. Let's say that if the person who designated RP(m) has
another request for those situations we should go with his request.
But lacking any  other request or suggestion from either of you, then
of course what you've just suggested is what we should go by, if
Rob LG doesn't consider it unreasonably awkward to carry out.

Mike Ossipoff


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