[EM] Am I still subscribed?

[Kristofer Munsterhjelm]

On 02/09/2017 10:56 AM, robert bristow-johnson wrote:
> 
> 
> ---------------------------- Original Message ----------------------------
> Subject: Re: [EM] Am I still subscribed?
> From: "Kristofer Munsterhjelm" <km_elmet at t-online.de>
> Date: Thu, February 9, 2017 4:41 am
> To: rbj at audioimagination.com
> "election-methods at lists.electorama.com"
> <election-methods at lists.electorama.com>
> --------------------------------------------------------------------------
> 
>> On 02/09/2017 08:26 AM, robert bristow-johnson wrote:
>>>
>>>
>>> ---------------------------- Original Message
> ----------------------------
>>> Subject: Re: [EM] Am I still subscribed?
>>> From: "VoteFair" <electionmethods at votefair.org>
>>> Date: Thu, February 9, 2017 1:39 am
>>> To: "election-methods at lists.electorama.com"
>>> <election-methods at lists.electorama.com>
>>>
> --------------------------------------------------------------------------
>>>
>>>> On 2/8/2017 1:59 PM, Kristofer Munsterhjelm wrote:
>>>>
>>>>> It seems simple enough to just do a New Zealand ballot:
>>>>> 1. Do you want to change the voting method? (yes/no)
>>>>> 2. If yes wins, what do you want to replace it with? (use either
>>>>> Plurality or Approval for the meta-election method here)
>>>>
>>>> Nope. In order to make the decision fair, the final vote needs to be
>>>> yes/no. That means the method needs to be picked first.
>>>>
>>>
>>> well, no. this is just like saying "no" is one of the run-off finalists
>>> and the other finalist must be picked first.
>>>
>>> but if "no" is just one option among the other voting methods to choose
>>> from, then this compound decision resolves exactly the same as ranking
>>> all of the options (including the "no" option) and, *if* there is a
>>> Condorcet winner, choosing the Condorcet winner. (if there isn't an
>>> Condorcet winner, that means the voting body prefer *something* over
>>> "no". but if there is a Condorcet winner among all options, that
>>> resolves the same as everyone choosing the "yes" option first (using
>>> Condorcet) and then stacking that up against "no" and see which option
>>> the voters decide.
>>>
>>> this is why i like Condorcet so much (and why i worry less about the
>>> cycle). Condorcet makes all of the options, including the status quo,
>>> all positioned on a flat playing field.
>>
>> A Condorcet-in-spirit referendum system could go like this:
>>
>> - Order the options randomly.
>> - Ask a yes/no about whether the voters prefer the first option to the
>> second.
>> - Ask a yes/no about whether the voters prefer the winner of the
>> previous round to the third option.
>> - ... and so on up to the last option, asking a yes/no about the winner
>> of the last round compared to the kth option.
> 
> in presentation, this is not flat.  the kth option is always in the
> "final round".
> 
> now, up to here, we're describing what happens on an individual ballot...

Well, what option is in the final round depends on who survived. The
challenger in the final round has the benefit that it hasn't had to go
through the entire gauntlet, but the drawback that it's facing the
winner that did go through it.

To take a concrete example, suppose there are three options: A, B, and
C. Suppose the random reordering goes like this: "C, A, B". And suppose
that, were you to hold a Condorcet election, the pairwise results would be:

A beats B and C (hence is the CW)
C beats B
B beats nobody,

and that the voters don't change their minds between the rounds.

Then the procedure works like this:

First round:

- The ballot asks the voters which they prefer of C and A (the first two
in the random ordering). Since the voters' pairwise preference for this
contest is that A beats C, A proceeds to the second round.

Second round:

- The ballot asks the voters which they prefer of A and B (the winner
from the previous round and the challenger according to the random
ordering). Since the voters' pairwise preference for this contest is
that A beats B, A wins.

>> Whichever option wins is the overall winner, and it must be the CW if it
>> exists (and voters never change their minds between the rounds). The
>> cycle tiebreaker is in effect to pick a random option in the Smith set.
> 
> ... but now we have tabulated voting results.  how did we get here?

I'm simply saying this under the caveat that the voters don't change
their minds between rounds; i.e. that if in hypothetical universe A you
held a Condorcet election, then in hypothetical universe B, a voter X
would say "I prefer A to B" on a yes/no question about A vs B iff he
ranked A ahead of B in universe A.

In a practical setting, the voters could change their minds between
rounds, but then you can't really speak of a CW or a Condorcet election
to compare to.

For similar reasons, it's hard to say that top two runoff fails
monotonicity, because the voters can change their minds between rounds.
But if you make the assumption that they don't, then top two runoff is
the same thing as the contingent vote, and the contingent vote does fail
monotonicity (for the same reason IRV does).
> 
>>
>> But it's really cumbersome.
> 
> i don't see it as equivalently flat as a simple ranked ballot decided by
> a Condocet-compliant method (and a CW existing).
> 
> in a referendum, why does the status-quo option have to be a "special
> option"?  why can't it be just one of the several options being chosen from?

It doesn't need to be. The status quo option could easily have been C
above, or B for that matter.