[EM] Best Single-Winner Method

[VoteFair]

On 5/22/2019 1:28 AM, C.Benham wrote:
 > ... I suppose you could have a bottom cycle and a
 > CW who is ranked bottom on the highest number of ballots.
 >
 > That combination would be very very unlikely, but why allow it?
 > Why not use my version?

If you are referring to Majority Judgement (MJ) then my answer is that 
MJ is difficult to understand -- from the perspective of just using 
spoken words while talking to someone who does not already understand 
terms such as "majority" -- not to mention "Smith set" and "Condorcet".

(If you are referring to something other than MJ, please specify what it 
is.)

Being easy to understand is extremely important -- as mentioned in item 
number 1 in an earlier message within this thread.

When talking to non-math people, I use the term "pairwise counting" and 
when someone does not immediately nod their head in recognition then I 
refer to the analogy of sports teams and the idea that if there is a 
team that loses every game against each other team then they clearly 
deserve to be eliminated.  I haven't encountered anyone who hasn't 
understood that.

Also I find that non-math folks seem to emotionally like the idea of 
working from the bottom up, eliminating one candidate at a time.  This 
might be why instant-runoff voting (IRV) is so popular.

If a "rock-paper-scissors cycle" (which most people immediately 
understand) occurs, I agree that there are better methods for resolving 
those, but they are more difficult to understand.

I figure if Instant Pairwise Elimination (IPE) gets used anywhere then 
it will be easy to later transition to something better, such as the 
Condorcet-Kemeny method, which is mathematically equivalent to VoteFair 
popularity ranking, which is part of the VoteFair system that I advocate 
(and which includes proportional representation).

In other words, it would be rare for IPE and Condorcet-Kemeny to yield 
different winners, so I figure this difference is not worth teaching to 
non-math people.  IPE gives them something they can quickly understand. 
And it helps them quickly realize why our current single-winner method 
is so primitive.

Richard Fobes


On 5/22/2019 1:28 AM, C.Benham wrote:
> Richard,
>
> To answer my own question, I suppose you could have a bottom cycle and a
> CW who is ranked
> bottom on the highest number of ballots.
>
> That combination would be very very unlikely, but why allow it?  Why not
> use my version?
>
> Chris Benham
>
> On 22/05/2019 3:15 pm, C.Benham wrote:
>>
>> Richard,
>>
>> Can you please explain, perhaps with an example, how your "Instant
>> Pairwise Elimination"
>> method can possibly "elect a non-Condorcet winner"?
>>
>> As I understand it, the method eliminates all the non-members of the
>> Smith set, and then if
>> more than one candidate remains, next eliminates the candidate who
>> (among remaining candidates)
>> is lowest ranked on the highest number of ballots, and then repeats
>> both steps until one candidate
>> remains.
>>
>> Is that right?
>>
>> From the page you linked to:
>>
>>> To appreciate how easy it is to understand, here’s a full description
>>> of the IPE method:
>>>
>>> “*Voters rank the candidates using as many ranking levels as there
>>> are candidates, or at least 5 ranking levels if ovals are marked on a
>>> paper ballot and space is limited and lots of the candidates are
>>> unlikely to win. During counting, each elimination round eliminates
>>> the candidate who loses every pairwise contest against every other
>>> remaining candidate. If an elimination round has no pairwise-losing
>>> candidate then, for that round, each ballot gives one count to the
>>> lowest-ranked remaining candidate on that ballot, and the candidate
>>> with the highest such count is eliminated. The last remaining
>>> candidate wins.*”
>>>
>>
>> Chris Benham