[EM] Round Robin Tournament Showings

Susan Simmons suzerainsimmons at outlook.com
Thu Aug 5 19:01:52 PDT 2021


You're quoting the pre-corrected, hence superseded, version of RSM. If only I could delete the superseded version from the EM list ...:-)

Even so, if both the max and min entries were zero that would mean that X was voted bottom on every ballot, which would mean that X abstained from voting himself above any other candidate oh, an exceptional case we won't worry about right now!

In terms of the corrected version ...

... if both the max opposition and min support for X are zero, then nobody was ranked above or below X on any ballot ... so all candidates were truncated on all ballots!

If only the min support (the numerator) is 0, no problem ... the ratio of zero to any other number is zero, and the output of log(zero) (in any context where negative inputs make no sense) is (by convention among computer scientists, information theorists, etc) negative infinity, which means that this candidate is on the extreme unfavorable end of the agenda ... as he should be for voting somebody else above himself .. or not voting at all!

What about ties, etc?  It is customary to worry about these kinds of details later ... so as not to obscure the forest with too many trees :-)

Sorry about the confusion.


Sent from my MetroPCS 4G LTE Android Device


-------- Mensaje original --------
De: Ted Stern <dodecatheon at gmail.com>
Fecha: 5/8/21 4:19 p. m. (GMT-08:00)
A: Susan Simmons <suzerainsimmons at outlook.com>
CC: election-methods at lists.electorama.com
Asunto: Re: [EM] Round Robin Tournament Showings

What if worst score is zero?

What if best and worst score are both zero?

What if best and worst scores are both small, but Log(R) is relatively large?

WLOG, you could first restrict to the Smith Set, which would avoid such troubles, but it seems a little fragile to me.

On Wed, Aug 4, 2021 at 11:30 AM Susan Simmons <suzerainsimmons at outlook.com<mailto:suzerainsimmons at outlook.com>> wrote:
Note if we replace A(X) approval of X with
Log R(X), where R(X) is the ratio of X's best pairwise score to X's worst pairwise score, then ASM and RSM are the same



Sent from my MetroPCS 4G LTE Android Device


-------- Mensaje original --------
De: Susan Simmons <suzerainsimmons at outlook.com<mailto:suzerainsimmons at outlook.com>>
Fecha: 4/8/21 10:31 a. m. (GMT-08:00)
A: election-methods at lists.electorama.com<mailto:election-methods at lists.electorama.com>
Asunto: Round Robin Tournament Showings


After a Round RobinTournament concludes its pairwise contests how should we decide the finishing order (1st place, 2nd place, 3rd place, etc.) of the participating teams?

Here's a solution that's reminiscent of Approval Sorted Margins:

Since there is no precise analogue for a team's approval in this context we use the ratio R of its best score to its worst score to determine a tentative list order.

Then as long as some adjacent pair of teams is out of order pairwise, among such pairs transpose the one whose members' R ratios are closest,  i.e. with the smallest absolute value of log(R1/R2).

The CW and CL (when they exist) will appear at opposite ends of the sorted list.

And there is the same kind of reverse symmetry that ASM provides in the context of elections.

In fact, we could call this method RSM or Ratio Sorted Margins, where the margins are the absolute differences of form
 |log R1 - log R2|
in analogy to the approval margins of form |A1 - A2|.

People that are uncomfortable with approval cutoffs can use RSM instead of ASM ... no approval necessary ... ranked preference style ballots are perfectly adequate ...in fact, since it is a tournament method, the pairwise vote matrix is adequate by itself.

No more excuses for clone dependent, intractable Kemeny-Young: just use RSM!


Sent from my MetroPCS 4G LTE Android Device
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