[EM] Condorcet for public proposals - Tounament

[Ernest Prabhakar]

Hi Dave,

Thanks for the input, very helpful.  More comments below:

On Jan 27, 2004, at 7:29 PM, Dave Ketchum wrote:
> I like what Ernest writes, though I see a bit of room for improvement 
> and suggest "tournament" as a less foreign-sounding title (even though 
> its ancestry is also French).

Hmm, maybe.  It is better than Condorcet, but to me tournament evokes 
an image of knights jousting on horses.

Someone (sorry, I forgot who) suggested the word Pairwise is important. 
    I could live with the name Pairwise Matchup Voting (PMV). Pairwise 
by itself seems too vague, somehow

>> Yeah, my friends (on the radical centrist list) are unanimous that 
>> the term Condorcet has to go. :-)
>> I have been proposing the term 'Instant Matchup Voting', or IMV, by 
>> analogy with Instant Runoff Voting. I compare it to a round-robin 
>> tournament, which most people have direct experience with.  I think 
>> this leads to a simple, easy to visualize definition:
>
> Ahead of much that I have seen, but I suggest tournament as even 
> easier to visualize from.  My definition will follow yours.

Well, tournament does have the idea of a series of matches, but not 
necessarily individual pairwise matchups, I don't think.    We could 
use the term Instant Round-Robin, which is much more explicit, but IRR 
is too close to IRV. :-(

> IMV:
>
>> 1. Each rank-ordered ballot is interpreted as a series of "Instant 
>> Matchups"
>>     That is A > B > C, implies one point each for the three pairwise 
>> Matchups A > B, B > C, and A > C
>>     Note that "A>B" is counted separately from "B>A" (i.e., winning 
>> votes)
>> 2. Tally up the N * (N-1) Matchups, for each ordered pair of 
>> candidates
>> 3.  If one candidate beats everyone, that's the absolute winner
>> 4. If there is a 'rock-paper-scissors' tie (A >= B, B >= C, C >= A),
>>     the tiebreaking winner is the candidate from that group with the 
>> 'least greatest defeat'
> Tournament:
> 0. Voters simply rank as many of the candidates as they choose, 
> starting with their most-preferred.
> 1. Each rank-ordered ballot is interpreted as a series of matches 
> among all
> candidates in the election:
>     That is, ranking A > B > C, and D and E not ranked by this voter,
> implies each ranked candidate winning over each candidate ranked 
> later, and
> over each unranked candidate.
>
>      Thus unranked candidates do not get counted as ranked over each 
> other.

That's a good point. I don't think we usually spend enough time 
explaining how the ranking is supposed to work, so it would be good to 
be more explicit.

>     Note that "A>B" is counted separately from "B>A" (i.e., winning 
> votes).
> 2. Tally up the number of wins for each ordered pair of candidates in 
> an
> N*N array (with an empty diagonal, for candidates do not play against 
> themselves).

Good point, N*N does reduce explanation.
>
> 3. If one candidate wins when compared with each other candidate, 
> that's
> the absolute winner.
> 4. If no absolute winner, we have a 'rock-paper-scissors' near tie 
> such as
> (A >= B, B >= C, C >= A), and the tiebreaking winner is the candidate 
> from
> that group with the 'least greatest defeat'.
>
> NOTE:  I consider 'least greatest defeat' unacceptably opaque for this 
> purpose, and ask for help in providing simpler words.

Fair enough.    How about "whose worst loss is the smallest"?    Or 
simply "lost by the smallest margin" (a little ambiguous, but sounds 
simpler) - can always go into more detail elsewhere.


> BTW:  Debatable whether voters should be permitted to rank candidates 
> as equal.

Is there any good reason not to?  Implicit equal ranking certainly 
makes it clearer about how unlisted candidates are counted.   Any if at 
all possible, it seems good to give people the option of equality 
rather than forcing a random choice.    Has anyone presented a clear 
argument for or against equal ranking?

>   If so then, for each pair of equal candidates, count 1/2 win for 
> each (thus if two voters rank A=B=C then A>B, B>A, A>C, C>A, B>C, and 
> C>B each get credited one full win).

That doesn't make any sense to me.   If two candidates are ranked, I 
think that neither should get the win -- at least if we're doing 
winning votes (wv) For example, if all the candidates that most people 
don't rank at the bottom of the list get a win against each other, then 
one single vote in favor could make that person the 'wv' winner!  
Right?

Any more thoughts on the implications of Smith PC on strategy, assuming 
we can hammer out a decent, simple explanation?

-- Ernie P.