[EM] Manual Construction of Smith Set

[Richard, the VoteFair guy]

My conclusion is that ALWAYS eliminating EVERY non-Smith-set candidate 
is too difficult to do as a visible process on an auditorium stage.  As 
Kristofer says:

On 1/16/2022 10:31 AM, Kristofer Munsterhjelm wrote:
 > But you still need to decide when to stop, which can be pretty
 > difficult. At first it'd seem to be obvious: if the bottom-most
 > candidate beats anyone else pairwise, he'd be in the Smith set, no?
 > But that doesn't quite work. Consider something like this: ...

It's much easier, and almost as "good," to eliminate pairwise losing 
candidates as they occur during elimination rounds.

As a reminder, a "pairwise losing candidate" is the candidate who would 
lose every one-on-one match against each and every remaining 
(not-yet-eliminated) candidate.

That would eliminate MOST, but not all, non-Smith-set candidates in MOST 
cases.

Thanks to those who helped get us close to a process that could be 
understood by an audience of typical voters.

Richard Fobes
The VoteFair guy


On 1/16/2022 10:31 AM, Kristofer Munsterhjelm wrote:
> On 15.01.2022 21:50, Richard, the VoteFair guy wrote:
>
>> Yet I believe that lots of audience members would give more trust to a
>> method that eliminates candidates one at a time. This is the simplicity
>> advantage that instant-runoff voting (IRV) has. In contrast, starting
>> with the Copeland winner might seem like magic to people who fear math.
>>
>> Would eliminating from the bottom always work?
>>
>> In the Wikipedia example below, it does work. It identifies B and E as
>> outside the Smith set, so the remaining candidates are in the Smith set.
>>
>> 5 wins for A
>>
>> 5 wins for D
>>
>> 4 wins for G
>>
>> 3.5 wins for C
>>
>> 2.5 wins for F
>>
>> 1 win for B
>>
>> 0 wins for E
>>
>> (Half indicates a tie.)
>>
>> So, my question is: Does elimination from the bottom always work?
>>
>> My guess is the answer is yes.
>
> I think so too, in theory. Suppose that A is in the Smith set and B is
> not. Then A beats everybody who B beats and then some. So B will be
> ranked below A by Copeland score. Hence the only way someone in the
> Smith set will at the bottom of the list is if everybody is in it.
>
> But you still need to decide when to stop, which can be pretty
> difficult. At first it'd seem to be obvious: if the bottom-most
> candidate beats anyone else pairwise, he'd be in the Smith set, no? But
> that doesn't quite work. Consider something like this:
>
> The candidates are ABCD. There's a Condorcet winner, A, who beats
> everybody. But there's also a loser three-cycle, B>C>D>B. Thus whoever
> you choose to eliminate first of B, C, and D, that candidate beats
> someone else despite being outside the Smith set. And batch-eliminating
> everybody with an identical least score fails in more complicated scenarios.
>
> To justify the Copeland winner being the initial candidate, I would just
> point to the reasoning above: any non-Smith set member beats or ties
> fewer people than every Smith set member (almost by definition). So the
> topmost by number of candidates beaten can't be outside the Smith set.
>
> If the voters are expected to know what a Smith set is, that leap
> shouldn't be too large. And if not, then every Smith//X method will look
> like magic to some degree. (Better use Benham or BTR-STV.)
>
> -km
>