dnow1 | 27 Apr 21:13 2016
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Number Votes, 10 Apr 2016

Number Votes - 10 April 2016


1. Use Number Votes (1, 2, 3, etc.) for each choice.

A Voter MUST put a number vote for ALL choices.


2. Vote YES or NO for each choice.


3. Ballot Form -

Numbered Boxes 1 2 3 etc. (i.e. total choices)

YES/NO Boxes  Name


4. Do Head to Head combinations math (would require computer voting in any large election).

Test Winner(s) (TW) -- Test Loser (TL) -- Test Others (TO)

Each TO 2nd or later choice vote goes to a TW or TL.


5. If a TW wins in all combinations, then it wins.


6. If a TW loses in all combinations, then it loses (and repeat step 5).


7. Tiebreaker - If no TW, then add 1st plus 2nd plus etc. choices to get a Droop Quota (majority for 1 seat winner).  If 2 or more get a Droop Quota, then any other non-Droop Quota choices lose (and repeat step 5).

8. OR -- have the lowest YES choice lose (and repeat step 5).

-----

Legislative bodies - each final winner would have a voting power equal to the final votes received.


For multiple executive/judicial offices (e.g. elect 3 judges) the N highest Number Votes count - for TO transfer purposes.

----

Theory - Condorcet in France in the 1780s (repeat 1780s) noted that a third choice could beat two existing choices head to head --

A beats B

C comes along.

C beats A and C beats B.

---

Divided majority example -

26 AB

25 BA

49 Z


The Z voters (i.e. all voters) would be required to make number votes for ALL choices.


i.e. example might result in --

26 ABZ

25 BAZ

26 ZBA

23 ZAB

Z is beat by both A and B and thus loses.

51 B beats 49 A.

-----

Possible circular tie -

34 ABC

33 BCA

32 CAB

99


Adding 1st plus 2nd votes -

34 + 32 = 66 A

33 + 34 = 67 B

32 + 33 = 65 C  Loses.

66 A beats 33 B.


OR use Approval Votes - Lowest would lose.

One of the two remaining would win.


Note - 

Number Votes are *relative* only.

Approval Votes are *absolute* only.

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Markus Schulze | 23 Apr 22:12 2016
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Re: Name of this Criterion

Hallo,

I have now added a proof that the Schulze method
satisfies this criterion. See section 4.12 of
my paper:

http://m-schulze.9mail.de/schulze1.pdf

Markus Schulze

 > Hallo,
 >
 > I remember that we discussed the following
 > criterion at this mailing list. Unfortunately,
 > I forgot the name of this criterion. Could
 > someone please tell me the name of this criterion?
 >
 >    Suppose M is the number of candidates.
 >
 >    Suppose there is a k with 2 <= k <= (M-1) such
 >    that candidate A wins every sub-election between
 >    candidate A and (k-1) other candidates. Then
 >    candidate A should also be the overall winner.
 >
 > Markus Schulze

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⸘Ŭalabio‽ | 19 Apr 22:04 2016

Re: 1/r as a voting system.

	Date: 	 Tue, 19 Apr 2016 09:45:39 -0400
	From: 	 Fred Gohlke <fredgohlke <at> verizon.net>
	Subject: 	 Re: 	 [EM] 1/r as a voting system.

> 	The problem with petition-type systems is that they are exploited by rabble-rousers; people who are
experts are inflaming the public passion on issues like women’s rights or terrorism.

	We have no way of preventing rich people from ending up on the ballot; but then again however, we do not want a
million names on the ballot, most of whom are people running for a lark.  The idea of having to reach the
number of signatures greater than the square-root of the population is that, in addition to the rich, poor
devoted people wanting to make a positive change in the world can achieve this goal:

	Let us suppose that one has average wealth and income and lives in a country with a population of
100,000,000 people.  One runs as a joke.  One probably would not bother to gather the myriad (10,000)
signatures, so probably would not get on the ballot.  If one is seriously committed to making the country a
better place and has average wealth and income, if one is willing to use all of one’s spare time for
gathering signatures, getting a myriad signatures in an years, is difficult, but doable.  Certainly, the
rich could easily pay a myriad people for their signatures, but once they are on the ballot, they can use
only public financing, which is a great equalizer because all candidates receive the same amount and can
only campaign using public financing.

	Once on the ballot, candidates can only use public money for campaigning, all candidates receive the same
about of money, and candidates cannot lie and stay on the ballot.  The limiting all candidates to the same
amount of public money is a great equalizer, allowing good candidates to bubble to the top.

	Allowing write-ins is a way of letting the voters choose someone not on the ballot because of chicanery. 
The vast majority of writ-ins probably will be rich people trying to buy elections, but I would not
write-in such a person:

	In my country, we have a man running for president who lies like a rug.  If candidates could not lie, he would
be struck from the ballot.  If he would be struck from the ballot, I am pretty sure that he would start an
expensive write-in campaign with his wealth.  No matter how-many of his advertisements I would see, I
would never write his name onto my ballot.

	The reason for the primaries with top-2 runoffs and general election with top-2 ruff is information-overload:

	Let us suppose that we have a score (20) of parties, each with a score of candidates and a score of
independents.  That is 420 candidates.  None could research that many candidates.  Each party has a primary
and the independents have a primary.  Each voter need only research 20 candies.  After the primary, each
voter need research only 2 candidates.

	For the general election, we have a score of parties, each running 1 candidate, and 1 independent.  That is
only 21 candidates for researching.  After the general election, we have only 2 candidates for
researching for the runoff.
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⸘Ŭalabio‽ | 17 Apr 23:20 2016

Re: 1/r as a voting system.

	Date: 	 Sun, 17 Apr 2016 14:43:35 -0400
	From: 	 Fred Gohlke <fredgohlke <at> verizon.net>
	Subject: 	 Re: 	 [EM] 1/r as a voting system.

> 	Can you suggest a practical way for a large electorate to sift among themselves to find the best advocates
of the public interest?

> 	Fred Gohlke

	This is technically beyond the scope of the post, but to get on the ballot, candidates would need to get more
signatures than the square-root of the population who will vote by the filing deadline:

	Let us suppose that a candidate runs for president of a country with a population of 100,000,000.  The
candidate would have to get over a Myriad (10,000) signatures by the filing deadline.

¿What if shenanigans keep a candidate off of the ballot?
	Always have a space on the ballot for write-ins.

	This is how I would imagine the election would work:

	0.	Gather signatures before the deadline.
	1.	Once on the ballot, candidates can use only governmental money for campaigning (we waste at least 10%,
probably 20%, and maybe as much as 30% on tax breaks, subsidies, and pork the rich buy in
campaign-contributions, so using 1% of the budget for campaigning, but requiring candidates to only use
that money and forbidding lying would be a great investment) and will be struck from the ballot for lying. 
Write-Ins are always allowed.
	2.	Hold a primary with the independents treated like a party.
	3.	Have a Top-2 RunOff.
	4.	Have the general election.
	5	Have a Top-2 RunOff.

	I hope that I answered your questions satisfactorily.
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⸘Ŭalabio‽ | 16 Apr 10:19 2016

1/r as a voting system.

	⸘Howdy‽

	This is not a new idea.  It has much prior art.  Still, its mathematical properties are such that one should
investigate it.  Before I continue, I stipulate that I have another idea, but this is important for
expressing that idea, so list this 1stly:

	1/r where r is rank.  Umm the votes.  Those for whom one supports receive positive rs; while those whom one
rejects, receives negative votes.  It could work thus:

Alpha:
	-1/6

Bravo:
	+½

Charlie:
	+1/1

Delta:
	-⅓

Echo:
	-1/7

Foxtrot:
	-1/1

Golf:
	+1/6

Hotel:
	+⅓

India:
	-1/9

Juliet:
	-⅛

Kilo:
	+1/7

Mike:
	+1/1

November:
	-¼

Oscar:
	+¼

Papa:
	-⅛

Quebec:

	+⅛

Romeo:
	+1/9

	For making the mathematics easier, multiply the numbers by 2,520, which is the least common multiple of
all natural numbers up to 9.  The voters need only give positive and negative ranks up to an absolute value of |9|:

	-5 means -1/5 which is -504.  We can limit the range of ranks to single-digit numbers because the reason for
an extended range is burying bad candidates like Hitler & Stalin, but we allow negative ratings, so this is unnecessary.

	We allow equal rankings, but require consecutive rankings and starting these consecutive ratings at
|1/1|.  This is a sanity-rule (ranking everyone either +9 (+1/9 (+280)) or -9 (-1/9 (-280)) would weaken
one’s vote to the point that one might as well not have voted at all).

	¡Peace!

--

-- 

	“⸘Ŭalabio‽” <Walabio <at> MacOSX.Com>

Skype:
	Walabio

An IntactWiki:
	http://intactwiki.org

	“You are entitled to your own opinion, but you are not entitled to your own facts.”
	——
	Senator Daniel Patrick Moynihan
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dnow1 | 15 Apr 08:49 2016
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Number Votes - 10 April 2016

Number Votes - 10 April 2016


1. Use Number Votes (1, 2, 3, etc.) for each choice.

A Voter MUST put a number vote for ALL choices.


2. Vote YES or NO for each choice.

    3. Ballot Form -

Numbered Boxes 1 2 3 etc. (i.e. total choices)

YES/NO Boxes  Name

    4. Do Head to Head combinations math (would require computer voting in any large election).

Test Winner(s) (TW) -- Test Loser (TL) -- Test Others (TO)

Each TO 2nd or later choice vote goes to a TW or TL.


5. If a TW wins in all combinations, then it wins.


    6. If a TW loses in all combinations, then it loses (and repeat step 5).

    7. Tiebreaker - If no TW, then add 1st plus 2nd plus etc. choices to get a Droop Quota (majority for 1 seat winner).  If 2 or more get a Droop Quota, then any other non-Droop Quota choices lose (and repeat step 5).
    8. OR -- have the lowest YES choice lose (and repeat step 5).

-----

Legislative bodies - each final winner would have a voting power equal to the final votes received.


For multiple executive/judicial offices (e.g. elect 3 judges) the N highest Number Votes count - for TO transfer purposes.

----

Theory - Condorcet in France in the 1780s (repeat 1780s) noted that a third choice could beat two existing choices head to head --

A beats B

C comes along.

C beats A and C beats B.

---

Divided majority example -

26 AB

25 BA

49 Z


The Z voters (i.e. all voters) would be required to make number votes for ALL choices.


i.e. example might result in --

26 ABZ

25 BAZ

26 ZBA

23 ZAB

Z is beat by both A and B and thus loses.

51 B beats 49 A.

-----

Possible circular tie -

34 ABC

33 BCA

32 CAB

99


Adding 1st plus 2nd votes -

34 + 32 = 66 A

33 + 34 = 67 B

32 + 33 = 65 C  Loses.

66 A beats 33 B.


OR use Approval Votes - Lowest would lose.

One of the two remaining would win.


Note - 

Number Votes are *relative* only.

Approval Votes are *absolute* only.

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⸘Ŭalabio‽ | 14 Apr 20:03 2016

[offtopic] The mathematics of a fair rational progressive IncomeTaxSystem with Mincome [/offtopic]

	We are all into mathematics, so I figured I would share something I wrote about the mathematics of a fair
rational progressive IncomeTaxSystem with Mincome I wrote for US-TaxDay (YYYYY-04-15):

	It is almost TaxDay in the USA (YYYYY-04-15).  I present a rational TaxSystem for IncomeTax:

	TaxOwed = ( Income / TaxRate ) - ( Constant + Deductions )

	It is that simple. Let us throw in some dumb numbers:

	*	TaxRate = 2
	*	Constant = 10,000.00 U$D

	Ignoring deductions, the TaxOwed can be seen here:

	http://wolframalpha.com/input/?i=y%3D(x%2F2)-10000

	As of can see, if one has an income of 0.00 U$D, one receives + 10,000.00 U$D, not enough to live, but will keep
one going until one gets back on one’s feet.  This is called income.

	As income increases, the TaxOwed approaches 50%. For high incomes, the tax is pretty flat but still a bit progressive.

	Let us look at how the deductions help the typical working person:

	Let us suppose that we have 2 people earning 30,000.00 U$D annually.  One lives rent free with parents;
while the other, pays 1,000.00 U$D / month in rent:

	The rent-free person has no deductions:

	( 30,000.00 U$D / 2 ) - 10,000.00 U$D = 5,000.00 U$D

	This person owes 5,000.00 U$D; so now, has 25,000.00 U$D after taxes.

	The rent-paying pony pays thus after deductions rent:

	( 30,000.00 U$D / 2 ) - ( 10,000.00 U$D + 12, 0000.00 U$D ) = - 7,000.00 U$D

	The rent-paying person 12,000.00 U$D on rent, so the 30,0000.00 U$D became 18,000.00 U$D after rent and
after receiving the - 7,000.00 U$D in tax, this person has 25,000.00 U$D.

	Both people have the same amount of money after taxes and rent.

	Rich people are always looking for loopholes to get out of paying their fair share.  I have hacks for fixing that:

	*	Reduce the total deductions on the books to 99.
	*	Allow only a score (20) deductions on each return.
	*	Limit the value of all deductions on a TaxReturn combined to 1,000,000.00 U$D.

	We have too many loopholes. Limiting the deductions on the books to double-digit numbers makes it hard to
slip in loopholes.  The average working person qualifies for only 5-10 deductions.  I figure that an honest
rich person should only qualify for some teen deductions (a dozen (12) to a score (20) deductions).  Rich
people overstate their deductions.  We should limit them to a total value of 1 million U$Ds so that rich
people will have to pay their fair share.
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Markus Schulze | 8 Apr 13:39 2016
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Name of this Criterion

Hallo,

I remember that we discussed the following
criterion at this mailing list. Unfortunately,
I forgot the name of this criterion. Could
someone please tell me the name of this criterion?

   Suppose M is the number of candidates.

   Suppose there is a k with 2 <= k <= (M-1) such
   that candidate A wins every sub-election between
   candidate A and (k-1) other candidates. Then
   candidate A should also be the overall winner.

Markus Schulze

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Richard Lung | 5 Apr 20:07 2016

Binomial STV counters Irish strategic voting.

To STV voting and to Election methods groups.


The Electoral Reform Society survey of the 2016 Irish general election describes a strategic voting practice, by which allied candidates seek to bolster the first preferences of their least prefered colleagues, to prevent their early exclusion.

My invention of (abstentions-inclusive keep-value averaged) Binomial STV would do away with the need for this insincere voting. It does this by a minimum of two complementary counts: an election count and an exclusion count. The latter is a rational count, in its own right, conducted on the voters preferences in reverse order, instead of an arbitrary exclusion, when the transferable surpluses run out, in the election count.

 

To ensure that the exclusion count is not given undue importance, compared to the election count, all preferences are counted, including abstentions, which are generally at the end of the ballot papers, when voters cease to express a preference.

Hence it is possible for the abstentions to reach a quota, in which case, a seat remains unfilled.

 

It is possible for consistently rational counts, both for election and for exclusion, by extending the Meek method use of the re-adjustable keep value, to candidates in deficit of a quota, as well as in surplus of a quota.

Each candidates election keep value and exclusion keep value, inverted to provide a back-up election keep value, are averaged to arrive at deciding keep values.

 

This describes only the simplest first order Binomial STV, corresponding to the first order of the binomial theorem, consisting of just two terms: one election count of preferences and one exclusion count of unpreferences.

But it is also possible to have a second-order Binomial STV, based on the four combinations of the second-order binomial theorem. And so on.

 

An example of how first and second order Binomial STV work is given here:

http://www.voting.ukscientists.com/Binomial_STV.html

 

The example is drastically over-simplified, for a system requiring computer programming, as does Meek method. Unlike Meek method, all the abstentions are counted, so there is no requirement to reduce the Droop quota, as the preferences run out. Otherwise, the way to code Binomial STV is to start from the Meek method program and adapt to the modified rules, extending the use of the keep value, and so forth.

I am looking for some organisation that might take up this work of implementing Binomial STV and running preliminary trials.

 

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robert bristow-johnson | 15 Mar 02:39 2016

how come i hadn't heard about "Meek STV" before on this list?

 

maybe it was because i wasn't paying attention, but i hadn't even heard of this until looking into elections for moderators at Stack Exchange.

 

also, are Jeff O'Neill and/or Chuck Sipos of OpaVote.com on this EM mailing list?  would you be willing to discuss here the ideas your website promotes regarding the "Best way to elect..."?

http://blog.opavote.com/2015/11/electing-single-person.html

http://blog.opavote.com/2016/02/best-methods-for-electing-group-of.html

 

also, what to others on this list think of Meek STV?  i would love to hear pros and cons?

and can we discuss Condorcet methods for multi-winner elections?


--


r b-j                  rbj <at> audioimagination.com

 


"Imagination is more important than knowledge."


 

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Sennet Williams | 29 Feb 05:25 2016
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The influence of each voter is what matters

The fact that some states now allow overseas military (only)  to use IRV will cause rational politicians to change their campaigns.   Candidates will make more effort to recruit military voters than non-military the GI's ballot will be more likely to be counted in the end.  That is why it is illegal to give some people better voting rights than others.  

   This is also why it would be irrational for individual states to change to PR to allocate electoral college "votes."   Candidates would practically ignore states that used PR so that they could focus on states where all the electors are at stake.   Regardless of how much effort/money they spent on each state, probably only 10-20%(?) of voters would switch their vote.  That is enough to win all the electors from the other states, but only 10-20% more of the electors in the PR states. 
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Gmane