[EM] help w. planning/working on a monte-carlo simulation?
David L Wetzell
wetzelld at gmail.com
Fri May 17 08:28:17 PDT 2013
The Droop quota is often presumed for proportional representation
over the Hare quota that is more proportional, due to how the Hare quota
can result in a minority being in power. (I guess the majority get in
power only a
majority of the time with a Hare Quota. )
And since the amount of proportionality with a droop quota
gets watered down as the number of contested seats is reduced, this has
led
some activists/experts, like Douglas Amy, to insist that PR use at least 5
seats.
This is often coupled with an insistence on rank choice voting due to the
problems with party lists.
So I'd like to simulate the effects of using 3-seat LR Hare for a 13 seat
city council election, like in MInneapolis, MN.
We'd consider 7 cases:
1. 13 FPTP elections.
2. 13 IRV elections, as are used now.
3. Four 3-seat LR Hare elections with 1 at-large seat with IRV.
4. A 6 and a 7 seat with Droop quota election.
5. A 6 and a 7 seat with Hare quota election.
6. A 13 seat with Droop quota election.
7. A 13 seat with Hare quota election.
I'd like to measure relative proportionality and the probability of a
majority getting a ruling majority, the portion of close/competitive
elections, and maybe some other stuff that cd be of interest.
Anybody interested?
My intuition is that smaller-order PRs retain the constituent-legislator
relationship and would be preferred by many who like having their
council-person. I also think that the Hare quota is more important for
increasing the likelihood of having a competitive election and giving
minority groups a higher chance of being swing voters. If this is paired
with the use of an at-large seat or some other way of establishing a
hierarchy who can get things done, it might be a winning combination.
dlw
On Wed, May 15, 2013 at 2:02 PM, <
election-methods-request at lists.electorama.com> wrote:
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> Today's Topics:
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> 1. My cycle definition of the Schwartz set was incorrect
> (Michael Ossipoff)
>
>
> ----------------------------------------------------------------------
>
> Message: 1
> Date: Tue, 14 May 2013 16:18:25 -0400
> From: Michael Ossipoff <email9648742 at gmail.com>
> To: election-methods at electorama.com
> Subject: [EM] My cycle definition of the Schwartz set was incorrect
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>
> I wanted to express the beatpath definition of the Schwartz set in a
> simpler and more compelling or appealing way, and the cycle definition
> (that I've posted here) seemed such a simplification.
>
> But the cycle definition doesn't define the Schwartz set. A candidate
> that doesn't have a defeat that isn't in a cycle isn't necessarily in
> the Scwhartz set (as defined by the unbeaten set definition and the
> beatpath definition].
>
> Of the two definitions (unbeaten set and beatpath), the beatpath
> definition desn't have much compellingness. For compellingness, I much
> prefer the unbeaten set definition.
>
> Let me state both definitions here:
>
> Unbeaten set definition of the Schwartz set::
>
> 1. An unbeaten sets is a set of alternatives none of which are beaten
> by anything outside the set.
>
> 2. An innermost unbeaten set is an unbeaten set that doesn't contain a
> smaller unbeaten set.
>
> 3.The Schwartz set is the set of alternatives that are in innermost
> unbeaten sets.
>
> [end of unbeaten set definition of Schwartz set]
>
> ---------------------------------------
>
> Beatpath definition of Schwartz set:
>
> There is a beatpath from X to Y if X beats Y, or if X beats A and
> there is a beatpath from A to Y.
>
> If there is a beatpath from Y to X, but not from X to Y, then X is not
> in the Schwartz set.
>
> Otherwise X is in the Schwartz set.
>
> [end of beatpath definition of the Schwartz set]
>
> Michael Ossipoff
>
>
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> End of Election-Methods Digest, Vol 107, Issue 9
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