RL
Richard Lung
Thu, Feb 24, 2022 6:36 PM
“Monotonic” Binomial STV
I was told (hello Kristofer) that I could not say that binomial STV is
“monotonic”unlike traditional or conventional STV. But I gave my reasons
why I could say this, and they were not contradicted or even answered.
It is not tabu or forbidden to say, and say again, what there is good
reason to believe is true, whatever the prevailing view.
In conventional STV, the transfer of surpluses, over a quota, to next
preferences is monotonic. There is “later no harm” unlike the Borda
count. The intermediate Plant report quoted a non-monotonic test example
from Riker, to justify their rejection of STV. This was based solely on
the perverse outcome of a different candidate being last past the post,
for elimination.
Riker made the unsupported claim that STV is “chaotic.” From a century
of STV usage, he did not provide a single real case of this. The record
is that STV counts well approximate STV votes, all things considered.
A paper that tried to provide some doubt, of STV as a well-behaved
system, drew not on a conventional STV election of candidates, but on
NASA using STV for outer space engineers to vote on a set of best
trajectories (I forget where).
Traditional STV is not “chaotic”. It is not even wrong. It is just an
initial or first approximation of binomial STV, a zero order binomial STV.
Zero order STV is a uninomial count that does not clearly distinguish
between an election count or an exclusion count. In 1912, HG Wells said
of FPTP, we no longer have elections we only have Rejections. From first
order Binomial STV, the two counts, election and exclusion counts, are
clearly distinguished and both made operational.
Binomial STV does not exclude candidates during the count. It uses an
exclusion count, to help determine a final election. This exclusion
count is exactly the same or symmetrical to the (monotonic) transfer of
surplus votes in an election count.
In both election and exclusion counts, Gregory Method or the senatorial
rules are expressed in terms of keep values, which enable proper
book-keeping of all preferences. Keep values can keep track of all the
preference votes, including abstentions. So, no perverse results are
possible from the chance exclusion of preferences from this or that
candidate last past the post. This is also why binomial STV is one
complete dimension of choice.
Binomial STV has “Independence of Irrelevant Alternatives.” For
instance, it makes no difference what level the quota is set, to the
order of the candidates keep values, their order of election. It is just
that bigger quotas raise the threshold of election.
Regards,
Richard Lung.
“Monotonic” Binomial STV
I was told (hello Kristofer) that I could not say that binomial STV is
“monotonic”unlike traditional or conventional STV. But I gave my reasons
why I could say this, and they were not contradicted or even answered.
It is not tabu or forbidden to say, and say again, what there is good
reason to believe is true, whatever the prevailing view.
In conventional STV, the transfer of surpluses, over a quota, to next
preferences is monotonic. There is “later no harm” unlike the Borda
count. The intermediate Plant report quoted a non-monotonic test example
from Riker, to justify their rejection of STV. This was based solely on
the perverse outcome of a different candidate being last past the post,
for elimination.
Riker made the unsupported claim that STV is “chaotic.” From a century
of STV usage, he did not provide a single real case of this. The record
is that STV counts well approximate STV votes, all things considered.
A paper that tried to provide some doubt, of STV as a well-behaved
system, drew not on a conventional STV election of candidates, but on
NASA using STV for outer space engineers to vote on a set of best
trajectories (I forget where).
Traditional STV is not “chaotic”. It is not even wrong. It is just an
initial or first approximation of binomial STV, a zero order binomial STV.
Zero order STV is a uninomial count that does not clearly distinguish
between an election count or an exclusion count. In 1912, HG Wells said
of FPTP, we no longer have elections we only have Rejections. From first
order Binomial STV, the two counts, election and exclusion counts, are
clearly distinguished and both made operational.
Binomial STV does not exclude candidates during the count. It uses an
exclusion count, to help determine a final election. This exclusion
count is exactly the same or symmetrical to the (monotonic) transfer of
surplus votes in an election count.
In both election and exclusion counts, Gregory Method or the senatorial
rules are expressed in terms of keep values, which enable proper
book-keeping of all preferences. Keep values can keep track of all the
preference votes, including abstentions. So, no perverse results are
possible from the chance exclusion of preferences from this or that
candidate last past the post. This is also why binomial STV is one
complete dimension of choice.
Binomial STV has “Independence of Irrelevant Alternatives.” For
instance, it makes no difference what level the quota is set, to the
order of the candidates keep values, their order of election. It is just
that bigger quotas raise the threshold of election.
Regards,
Richard Lung.
FS
Forest Simmons
Sat, Feb 26, 2022 1:30 AM
Richard,
Here's an example of monotonicity failure in conventional single winner STV
as I understand it:
Original profile of ballots:
35 A>B>C
33 B>C>A
32 C>A>B
C eliminated and A wins.
New profile: two members of B faction defect to A faction:
37 A>B>C
31 B>C>A
32 C>A>B
Now B is eliminated and C wins.
How does Binomial STV avoid this monotonicity failure?
Thanks!
-Forest
El jue., 24 de feb. de 2022 10:36 a. m., Richard Lung <
voting@ukscientists.com> escribió:
“Monotonic” Binomial STV
I was told (hello Kristofer) that I could not say that binomial STV is
“monotonic” unlike traditional or conventional STV. But I gave my reasons
why I could say this, and they were not contradicted or even answered. It
is not tabu or forbidden to say, and say again, what there is good reason
to believe is true, whatever the prevailing view.
In conventional STV, the transfer of surpluses, over a quota, to next
preferences is monotonic. There is “later no harm” unlike the Borda count.
The intermediate Plant report quoted a non-monotonic test example from
Riker, to justify their rejection of STV. This was based solely on the
perverse outcome of a different candidate being last past the post, for
elimination.
Riker made the unsupported claim that STV is “chaotic.” From a century of
STV usage, he did not provide a single real case of this. The record is
that STV counts well approximate STV votes, all things considered.
A paper that tried to provide some doubt, of STV as a well-behaved system,
drew not on a conventional STV election of candidates, but on NASA using
STV for outer space engineers to vote on a set of best trajectories (I
forget where).
Traditional STV is not “chaotic”. It is not even wrong. It is just an
initial or first approximation of binomial STV, a zero order binomial STV.
Zero order STV is a uninomial count that does not clearly distinguish
between an election count or an exclusion count. In 1912, HG Wells said of
FPTP, we no longer have elections we only have Rejections. From first order
Binomial STV, the two counts, election and exclusion counts, are clearly
distinguished and both made operational.
Binomial STV does not exclude candidates during the count. It uses an
exclusion count, to help determine a final election. This exclusion count
is exactly the same or symmetrical to the (monotonic) transfer of surplus
votes in an election count.
In both election and exclusion counts, Gregory Method or the senatorial
rules are expressed in terms of keep values, which enable proper
book-keeping of all preferences. Keep values can keep track of all the
preference votes, including abstentions. So, no perverse results are
possible from the chance exclusion of preferences from this or that
candidate last past the post. This is also why binomial STV is one complete
dimension of choice.
Binomial STV has “Independence of Irrelevant Alternatives.” For instance,
it makes no difference what level the quota is set, to the order of the
candidates keep values, their order of election. It is just that bigger
quotas raise the threshold of election.
Regards,
Richard Lung.
Election-Methods mailing list - see https://electorama.com/em for list
info
Richard,
Here's an example of monotonicity failure in conventional single winner STV
as I understand it:
Original profile of ballots:
35 A>B>C
33 B>C>A
32 C>A>B
C eliminated and A wins.
New profile: two members of B faction defect to A faction:
37 A>B>C
31 B>C>A
32 C>A>B
Now B is eliminated and C wins.
How does Binomial STV avoid this monotonicity failure?
Thanks!
-Forest
El jue., 24 de feb. de 2022 10:36 a. m., Richard Lung <
voting@ukscientists.com> escribió:
>
> “Monotonic” Binomial STV
>
>
>
> I was told (hello Kristofer) that I could not say that binomial STV is
> “monotonic” unlike traditional or conventional STV. But I gave my reasons
> why I could say this, and they were not contradicted or even answered. It
> is not tabu or forbidden to say, and say again, what there is good reason
> to believe is true, whatever the prevailing view.
>
> In conventional STV, the transfer of surpluses, over a quota, to next
> preferences is monotonic. There is “later no harm” unlike the Borda count.
> The intermediate Plant report quoted a non-monotonic test example from
> Riker, to justify their rejection of STV. This was based solely on the
> perverse outcome of a different candidate being last past the post, for
> elimination.
>
> Riker made the unsupported claim that STV is “chaotic.” From a century of
> STV usage, he did not provide a single real case of this. The record is
> that STV counts well approximate STV votes, all things considered.
>
> A paper that tried to provide some doubt, of STV as a well-behaved system,
> drew not on a conventional STV election of candidates, but on NASA using
> STV for outer space engineers to vote on a set of best trajectories (I
> forget where).
>
> Traditional STV is not “chaotic”. It is not even wrong. It is just an
> initial or first approximation of binomial STV, a zero order binomial STV.
>
> Zero order STV is a uninomial count that does not clearly distinguish
> between an election count or an exclusion count. In 1912, HG Wells said of
> FPTP, we no longer have elections we only have Rejections. From first order
> Binomial STV, the two counts, election and exclusion counts, are clearly
> distinguished and both made operational.
>
> Binomial STV does not exclude candidates during the count. It uses an
> exclusion count, to help determine a final election. This exclusion count
> is exactly the same or symmetrical to the (monotonic) transfer of surplus
> votes in an election count.
>
> In both election and exclusion counts, Gregory Method or the senatorial
> rules are expressed in terms of keep values, which enable proper
> book-keeping of all preferences. Keep values can keep track of all the
> preference votes, including abstentions. So, no perverse results are
> possible from the chance exclusion of preferences from this or that
> candidate last past the post. This is also why binomial STV is one complete
> dimension of choice.
>
> Binomial STV has “Independence of Irrelevant Alternatives.” For instance,
> it makes no difference what level the quota is set, to the order of the
> candidates keep values, their order of election. It is just that bigger
> quotas raise the threshold of election.
>
> Regards,
>
> Richard Lung.
>
> ----
> Election-Methods mailing list - see https://electorama.com/em for list
> info
>
RL
Richard Lung
Sat, Feb 26, 2022 12:21 PM
Thank you, Forest,
Your example is the kind of example that Riker gave.
Here the quota equals 50 = 100/[1+1].
Original profile:
Election Keep value is quota/candidate vote:
for A 50/35
B: 50/33
C: 50/32
Exclusion keep value = quota/candidate reverse vote:
for A: 50/33
B: 50/32
C: 50/35
Final (geometric mean) keep values, divide election keep value by
exclusion keep value.
(This is equivalent to multiplying by the inverse exclusion keep value,
as a make-shift second opinion election keep value.)
for A: 50/35 x 33/50. And take their square root ~ ,971
for B: 50/33 x 32/50. As above, gives ~ .9847
for C: 50/32 x 35/50. ... gives ~ 1.0458
Keep values below unity are technically electable. A wins, with lowest
keep value.
New profile:
Election divided by exclusion keep values:
A: 50/37 x 31/50. Take square root of 31/37, for ~ .9153
B: 50/31 x 32/50. As above, ~ 1.016
C: 50/32 x 37/50. As above, ~ 1,075
Again, A is elected as before, and with a yet lower keep value, as the
extra preferences for A warrant.
Regards,
Richard Lung.
On 26/02/2022 01:30, Forest Simmons wrote:
Richard,
Here's an example of monotonicity failure in conventional single
winner STV as I understand it:
Original profile of ballots:
35 A>B>C
33 B>C>A
32 C>A>B
C eliminated and A wins.
New profile: two members of B faction defect to A faction:
37 A>B>C
31 B>C>A
32 C>A>B
Now B is eliminated and C wins.
How does Binomial STV avoid this monotonicity failure?
Thanks!
-Forest
El jue., 24 de feb. de 2022 10:36 a. m., Richard Lung
voting@ukscientists.com escribió:
“Monotonic” Binomial STV
I was told (hello Kristofer) that I could not say that binomial
STV is “monotonic”unlike traditional or conventional STV. But I
gave my reasons why I could say this, and they were not
contradicted or even answered. It is not tabu or forbidden to say,
and say again, what there is good reason to believe is true,
whatever the prevailing view.
In conventional STV, the transfer of surpluses, over a quota, to
next preferences is monotonic. There is “later no harm” unlike the
Borda count. The intermediate Plant report quoted a non-monotonic
test example from Riker, to justify their rejection of STV. This
was based solely on the perverse outcome of a different candidate
being last past the post, for elimination.
Riker made the unsupported claim that STV is “chaotic.” From a
century of STV usage, he did not provide a single real case of
this. The record is that STV counts well approximate STV votes,
all things considered.
A paper that tried to provide some doubt, of STV as a well-behaved
system, drew not on a conventional STV election of candidates, but
on NASA using STV for outer space engineers to vote on a set of
best trajectories (I forget where).
Traditional STV is not “chaotic”. It is not even wrong. It is just
an initial or first approximation of binomial STV, a zero order
binomial STV.
Zero order STV is a uninomial count that does not clearly
distinguish between an election count or an exclusion count. In
1912, HG Wells said of FPTP, we no longer have elections we only
have Rejections. From first order Binomial STV, the two counts,
election and exclusion counts, are clearly distinguished and both
made operational.
Binomial STV does not exclude candidates during the count. It uses
an exclusion count, to help determine a final election. This
exclusion count is exactly the same or symmetrical to the
(monotonic) transfer of surplus votes in an election count.
In both election and exclusion counts, Gregory Method or the
senatorial rules are expressed in terms of keep values, which
enable proper book-keeping of all preferences. Keep values can
keep track of all the preference votes, including abstentions. So,
no perverse results are possible from the chance exclusion of
preferences from this or that candidate last past the post. This
is also why binomial STV is one complete dimension of choice.
Binomial STV has “Independence of Irrelevant Alternatives.” For
instance, it makes no difference what level the quota is set, to
the order of the candidates keep values, their order of election.
It is just that bigger quotas raise the threshold of election.
Regards,
Richard Lung.
----
Election-Methods mailing list - see https://electorama.com/em for
list info
Thank you, Forest,
Your example is the kind of example that Riker gave.
Here the quota equals 50 = 100/[1+1].
Original profile:
Election Keep value is quota/candidate vote:
for A 50/35
B: 50/33
C: 50/32
Exclusion keep value = quota/candidate reverse vote:
for A: 50/33
B: 50/32
C: 50/35
Final (geometric mean) keep values, divide election keep value by
exclusion keep value.
(This is equivalent to multiplying by the inverse exclusion keep value,
as a make-shift second opinion election keep value.)
for A: 50/35 x 33/50. And take their square root ~ ,971
for B: 50/33 x 32/50. As above, gives ~ .9847
for C: 50/32 x 35/50. ... gives ~ 1.0458
Keep values below unity are technically electable. A wins, with lowest
keep value.
New profile:
Election divided by exclusion keep values:
A: 50/37 x 31/50. Take square root of 31/37, for ~ .9153
B: 50/31 x 32/50. As above, ~ 1.016
C: 50/32 x 37/50. As above, ~ 1,075
Again, A is elected as before, and with a yet lower keep value, as the
extra preferences for A warrant.
Regards,
Richard Lung.
On 26/02/2022 01:30, Forest Simmons wrote:
> Richard,
>
> Here's an example of monotonicity failure in conventional single
> winner STV as I understand it:
>
> Original profile of ballots:
>
> 35 A>B>C
> 33 B>C>A
> 32 C>A>B
>
> C eliminated and A wins.
>
> New profile: two members of B faction defect to A faction:
>
> 37 A>B>C
> 31 B>C>A
> 32 C>A>B
> Now B is eliminated and C wins.
>
> How does Binomial STV avoid this monotonicity failure?
>
> Thanks!
>
> -Forest
>
> El jue., 24 de feb. de 2022 10:36 a. m., Richard Lung
> <voting@ukscientists.com> escribió:
>
>
> “Monotonic” Binomial STV
>
> I was told (hello Kristofer) that I could not say that binomial
> STV is “monotonic”unlike traditional or conventional STV. But I
> gave my reasons why I could say this, and they were not
> contradicted or even answered. It is not tabu or forbidden to say,
> and say again, what there is good reason to believe is true,
> whatever the prevailing view.
>
> In conventional STV, the transfer of surpluses, over a quota, to
> next preferences is monotonic. There is “later no harm” unlike the
> Borda count. The intermediate Plant report quoted a non-monotonic
> test example from Riker, to justify their rejection of STV. This
> was based solely on the perverse outcome of a different candidate
> being last past the post, for elimination.
>
> Riker made the unsupported claim that STV is “chaotic.” From a
> century of STV usage, he did not provide a single real case of
> this. The record is that STV counts well approximate STV votes,
> all things considered.
>
> A paper that tried to provide some doubt, of STV as a well-behaved
> system, drew not on a conventional STV election of candidates, but
> on NASA using STV for outer space engineers to vote on a set of
> best trajectories (I forget where).
>
> Traditional STV is not “chaotic”. It is not even wrong. It is just
> an initial or first approximation of binomial STV, a zero order
> binomial STV.
>
> Zero order STV is a uninomial count that does not clearly
> distinguish between an election count or an exclusion count. In
> 1912, HG Wells said of FPTP, we no longer have elections we only
> have Rejections. From first order Binomial STV, the two counts,
> election and exclusion counts, are clearly distinguished and both
> made operational.
>
> Binomial STV does not exclude candidates during the count. It uses
> an exclusion count, to help determine a final election. This
> exclusion count is exactly the same or symmetrical to the
> (monotonic) transfer of surplus votes in an election count.
>
> In both election and exclusion counts, Gregory Method or the
> senatorial rules are expressed in terms of keep values, which
> enable proper book-keeping of all preferences. Keep values can
> keep track of all the preference votes, including abstentions. So,
> no perverse results are possible from the chance exclusion of
> preferences from this or that candidate last past the post. This
> is also why binomial STV is one complete dimension of choice.
>
> Binomial STV has “Independence of Irrelevant Alternatives.” For
> instance, it makes no difference what level the quota is set, to
> the order of the candidates keep values, their order of election.
> It is just that bigger quotas raise the threshold of election.
>
> Regards,
>
> Richard Lung.
>
> ----
> Election-Methods mailing list - see https://electorama.com/em for
> list info
>
KM
Kristofer Munsterhjelm
Sat, Feb 26, 2022 2:02 PM
On 26.02.2022 13:21, Richard Lung wrote:
Thank you, Forest,
Your example is the kind of example that Riker gave.
Thank you for providing information about how to calculate who wins
according to Binomial STV.
I have a few examples of my own. Could you tell me who wins, and how the
wins are calculated, for these single-winner elections?
First:
51: A>B>C
48: B>A>C
1: B>C>A
And second:
36: A>B>C
34: B>C>A
30: C>A>B
-km
On 26.02.2022 13:21, Richard Lung wrote:
>
> Thank you, Forest,
>
> Your example is the kind of example that Riker gave.
Thank you for providing information about how to calculate who wins
according to Binomial STV.
I have a few examples of my own. Could you tell me who wins, and how the
wins are calculated, for these single-winner elections?
First:
51: A>B>C
48: B>A>C
1: B>C>A
And second:
36: A>B>C
34: B>C>A
30: C>A>B
-km
KM
Kristofer Munsterhjelm
Sat, Feb 26, 2022 10:28 PM
On 26.02.2022 15:02, Kristofer Munsterhjelm wrote:
On 26.02.2022 13:21, Richard Lung wrote:
Thank you, Forest,
Your example is the kind of example that Riker gave.
Thank you for providing information about how to calculate who wins
according to Binomial STV.
I have a few examples of my own. Could you tell me who wins, and how the
wins are calculated, for these single-winner elections?
First:
51: A>B>C
48: B>A>C
1: B>C>A
And second:
36: A>B>C
34: B>C>A
30: C>A>B
Oops, I probably should've noticed that Forest's example is a variant of
my second election, so I don't think I need the outcome and calculations
for that one to verify whether the way I think the method works (for
three candidates) is right.
The first one would still be useful, though :-)
-km
On 26.02.2022 15:02, Kristofer Munsterhjelm wrote:
> On 26.02.2022 13:21, Richard Lung wrote:
>>
>> Thank you, Forest,
>>
>> Your example is the kind of example that Riker gave.
>
> Thank you for providing information about how to calculate who wins
> according to Binomial STV.
>
> I have a few examples of my own. Could you tell me who wins, and how the
> wins are calculated, for these single-winner elections?
>
> First:
>
> 51: A>B>C
> 48: B>A>C
> 1: B>C>A
>
> And second:
>
> 36: A>B>C
> 34: B>C>A
> 30: C>A>B
Oops, I probably should've noticed that Forest's example is a variant of
my second election, so I don't think I need the outcome and calculations
for that one to verify whether the way I think the method works (for
three candidates) is right.
The first one would still be useful, though :-)
-km
FS
Forest Simmons
Sun, Feb 27, 2022 3:55 AM
Thanks, Richard. That's very helpful and tantalizing!
El sáb., 26 de feb. de 2022 4:21 a. m., Richard Lung <
voting@ukscientists.com> escribió:
Thank you, Forest,
Your example is the kind of example that Riker gave.
Here the quota equals 50 = 100/[1+1].
Original profile:
Election Keep value is quota/candidate vote:
for A 50/35
B: 50/33
C: 50/32
Exclusion keep value = quota/candidate reverse vote:
for A: 50/33
B: 50/32
C: 50/35
Final (geometric mean) keep values, divide election keep value by
exclusion keep value.
(This is equivalent to multiplying by the inverse exclusion keep value, as
a make-shift second opinion election keep value.)
for A: 50/35 x 33/50. And take their square root ~ ,971
for B: 50/33 x 32/50. As above, gives ~ .9847
for C: 50/32 x 35/50. ... gives ~ 1.0458
Keep values below unity are technically electable. A wins, with lowest
keep value.
New profile:
Election divided by exclusion keep values:
A: 50/37 x 31/50. Take square root of 31/37, for ~ .9153
B: 50/31 x 32/50. As above, ~ 1.016
C: 50/32 x 37/50. As above, ~ 1,075
Again, A is elected as before, and with a yet lower keep value, as the
extra preferences for A warrant.
Regards,
Richard Lung.
On 26/02/2022 01:30, Forest Simmons wrote:
Richard,
Here's an example of monotonicity failure in conventional single winner
STV as I understand it:
Original profile of ballots:
35 A>B>C
33 B>C>A
32 C>A>B
C eliminated and A wins.
New profile: two members of B faction defect to A faction:
37 A>B>C
31 B>C>A
32 C>A>B
Now B is eliminated and C wins.
How does Binomial STV avoid this monotonicity failure?
Thanks!
-Forest
El jue., 24 de feb. de 2022 10:36 a. m., Richard Lung <
voting@ukscientists.com> escribió:
“Monotonic” Binomial STV
I was told (hello Kristofer) that I could not say that binomial STV is
“monotonic” unlike traditional or conventional STV. But I gave my
reasons why I could say this, and they were not contradicted or even
answered. It is not tabu or forbidden to say, and say again, what there is
good reason to believe is true, whatever the prevailing view.
In conventional STV, the transfer of surpluses, over a quota, to next
preferences is monotonic. There is “later no harm” unlike the Borda count.
The intermediate Plant report quoted a non-monotonic test example from
Riker, to justify their rejection of STV. This was based solely on the
perverse outcome of a different candidate being last past the post, for
elimination.
Riker made the unsupported claim that STV is “chaotic.” >From a century of
STV usage, he did not provide a single real case of this. The record is
that STV counts well approximate STV votes, all things considered.
A paper that tried to provide some doubt, of STV as a well-behaved
system, drew not on a conventional STV election of candidates, but on NASA
using STV for outer space engineers to vote on a set of best trajectories
(I forget where).
Traditional STV is not “chaotic”. It is not even wrong. It is just an
initial or first approximation of binomial STV, a zero order binomial STV.
Zero order STV is a uninomial count that does not clearly distinguish
between an election count or an exclusion count. In 1912, HG Wells said of
FPTP, we no longer have elections we only have Rejections. From first order
Binomial STV, the two counts, election and exclusion counts, are clearly
distinguished and both made operational.
Binomial STV does not exclude candidates during the count. It uses an
exclusion count, to help determine a final election. This exclusion count
is exactly the same or symmetrical to the (monotonic) transfer of surplus
votes in an election count.
In both election and exclusion counts, Gregory Method or the senatorial
rules are expressed in terms of keep values, which enable proper
book-keeping of all preferences. Keep values can keep track of all the
preference votes, including abstentions. So, no perverse results are
possible from the chance exclusion of preferences from this or that
candidate last past the post. This is also why binomial STV is one complete
dimension of choice.
Binomial STV has “Independence of Irrelevant Alternatives.” For instance,
it makes no difference what level the quota is set, to the order of the
candidates keep values, their order of election. It is just that bigger
quotas raise the threshold of election.
Regards,
Richard Lung.
Election-Methods mailing list - see https://electorama.com/em for list
info
Thanks, Richard. That's very helpful and tantalizing!
El sáb., 26 de feb. de 2022 4:21 a. m., Richard Lung <
voting@ukscientists.com> escribió:
>
> Thank you, Forest,
>
> Your example is the kind of example that Riker gave.
>
> Here the quota equals 50 = 100/[1+1].
>
> Original profile:
>
> Election Keep value is quota/candidate vote:
>
> for A 50/35
>
> B: 50/33
>
> C: 50/32
>
> Exclusion keep value = quota/candidate reverse vote:
>
> for A: 50/33
>
> B: 50/32
>
> C: 50/35
>
> Final (geometric mean) keep values, divide election keep value by
> exclusion keep value.
>
> (This is equivalent to multiplying by the inverse exclusion keep value, as
> a make-shift second opinion election keep value.)
>
> for A: 50/35 x 33/50. And take their square root ~ ,971
>
> for B: 50/33 x 32/50. As above, gives ~ .9847
>
> for C: 50/32 x 35/50. ... gives ~ 1.0458
>
> Keep values below unity are technically electable. A wins, with lowest
> keep value.
>
> New profile:
>
> Election divided by exclusion keep values:
>
> A: 50/37 x 31/50. Take square root of 31/37, for ~ .9153
>
> B: 50/31 x 32/50. As above, ~ 1.016
>
> C: 50/32 x 37/50. As above, ~ 1,075
>
> Again, A is elected as before, and with a yet lower keep value, as the
> extra preferences for A warrant.
>
> Regards,
>
> Richard Lung.
>
>
> On 26/02/2022 01:30, Forest Simmons wrote:
>
> Richard,
>
> Here's an example of monotonicity failure in conventional single winner
> STV as I understand it:
>
> Original profile of ballots:
>
> 35 A>B>C
> 33 B>C>A
> 32 C>A>B
>
> C eliminated and A wins.
>
> New profile: two members of B faction defect to A faction:
>
> 37 A>B>C
> 31 B>C>A
> 32 C>A>B
>
> Now B is eliminated and C wins.
>
> How does Binomial STV avoid this monotonicity failure?
>
> Thanks!
>
> -Forest
>
> El jue., 24 de feb. de 2022 10:36 a. m., Richard Lung <
> voting@ukscientists.com> escribió:
>
>>
>> “Monotonic” Binomial STV
>>
>>
>>
>> I was told (hello Kristofer) that I could not say that binomial STV is
>> “monotonic” unlike traditional or conventional STV. But I gave my
>> reasons why I could say this, and they were not contradicted or even
>> answered. It is not tabu or forbidden to say, and say again, what there is
>> good reason to believe is true, whatever the prevailing view.
>>
>> In conventional STV, the transfer of surpluses, over a quota, to next
>> preferences is monotonic. There is “later no harm” unlike the Borda count.
>> The intermediate Plant report quoted a non-monotonic test example from
>> Riker, to justify their rejection of STV. This was based solely on the
>> perverse outcome of a different candidate being last past the post, for
>> elimination.
>>
>> Riker made the unsupported claim that STV is “chaotic.” >From a century of
>> STV usage, he did not provide a single real case of this. The record is
>> that STV counts well approximate STV votes, all things considered.
>>
>> A paper that tried to provide some doubt, of STV as a well-behaved
>> system, drew not on a conventional STV election of candidates, but on NASA
>> using STV for outer space engineers to vote on a set of best trajectories
>> (I forget where).
>>
>> Traditional STV is not “chaotic”. It is not even wrong. It is just an
>> initial or first approximation of binomial STV, a zero order binomial STV.
>>
>> Zero order STV is a uninomial count that does not clearly distinguish
>> between an election count or an exclusion count. In 1912, HG Wells said of
>> FPTP, we no longer have elections we only have Rejections. From first order
>> Binomial STV, the two counts, election and exclusion counts, are clearly
>> distinguished and both made operational.
>>
>> Binomial STV does not exclude candidates during the count. It uses an
>> exclusion count, to help determine a final election. This exclusion count
>> is exactly the same or symmetrical to the (monotonic) transfer of surplus
>> votes in an election count.
>>
>> In both election and exclusion counts, Gregory Method or the senatorial
>> rules are expressed in terms of keep values, which enable proper
>> book-keeping of all preferences. Keep values can keep track of all the
>> preference votes, including abstentions. So, no perverse results are
>> possible from the chance exclusion of preferences from this or that
>> candidate last past the post. This is also why binomial STV is one complete
>> dimension of choice.
>>
>> Binomial STV has “Independence of Irrelevant Alternatives.” For instance,
>> it makes no difference what level the quota is set, to the order of the
>> candidates keep values, their order of election. It is just that bigger
>> quotas raise the threshold of election.
>>
>> Regards,
>>
>> Richard Lung.
>> ----
>> Election-Methods mailing list - see https://electorama.com/em for list
>> info
>>
>
RL
Richard Lung
Sun, Feb 27, 2022 1:04 PM
Thank you, Kristofer,
for first example.
The quota is 100/(1+1) = 50.
Election keep value is quota/(candidates preference votes)
for A: 50/51
B: 50/49
C: 50/0 Which, of course is infinite. It may be convenient, for tidy
book-keeping, that small elections require that each candidate votes for
themself. Then the keep value maximum simply equals the quota.
Generally, it is not necessary to make this stipulation, for large scale
elections, because no candidate, however miserable, ever gets no votes.
Exclusion keep value equals quota/(candidates reverse preference vote):
A: 50/1
B: 50/0
C: 50/99
Geometric mean keep value ( election keep value multiplied by inverse
exclusion keep value):
A: 50/51 x 1/50 ~ 0,0196
B: 50/49 x 0/50 = 0/49 is indeterminate. The closest determinate
approximation gives 1/49, not quite as low a keep value as 1/51 for A,
who is therefore the winner.
C: 50/0 x 99/50 = 99/0 is indeterminate. The closest determinate
approximation is 99/1. This keep value properly signifies huge
unpopularity, since an elective keep value is unity.
However, this example does illustrate the problem of the relative
importance of the election and exclusion counts. My guess is that the
problem would resolve itself in the entropy of preference voting, which
falls off exponentially with lower preferences.
Laws of physics are time-reversible in principle. But in practise, on
the classical or macroscopic scale, entropy intervenes, to give time a
one-way direction. (Binomial STV, in small scale elections, below the
level of much or any significance for parametric statistics, might be
likened to time-reversible quantum scale inter-actions, with regard to
the relative importance of election and exclusion, the direction of
choice, positive or negative.)
This falling off, of non-abstentions, would be countered, to some
extent, by the power of binomial stv to exclude, as well as elect,
candidates. And there is no reason in principle why a binomial stv
election might actually be more of an exclusion of generally disliked
candidates (just as often is FPTP). Still, it should be borne in mind,
that this very exclusive power of binomial stv should deter adversarial
candidate line-ups.
Regards,
Richard Lung.
On 26/02/2022 22:28, Kristofer Munsterhjelm wrote:
On 26.02.2022 15:02, Kristofer Munsterhjelm wrote:
On 26.02.2022 13:21, Richard Lung wrote:
Thank you, Forest,
Your example is the kind of example that Riker gave.
Thank you for providing information about how to calculate who wins
according to Binomial STV.
I have a few examples of my own. Could you tell me who wins, and how the
wins are calculated, for these single-winner elections?
First:
51: A>B>C
48: B>A>C
1: B>C>A
And second:
36: A>B>C
34: B>C>A
30: C>A>B
Oops, I probably should've noticed that Forest's example is a variant of
my second election, so I don't think I need the outcome and calculations
for that one to verify whether the way I think the method works (for
three candidates) is right.
The first one would still be useful, though :-)
-km
Thank you, Kristofer,
for first example.
The quota is 100/(1+1) = 50.
Election keep value is quota/(candidates preference votes)
for A: 50/51
B: 50/49
C: 50/0 Which, of course is infinite. It may be convenient, for tidy
book-keeping, that small elections require that each candidate votes for
themself. Then the keep value maximum simply equals the quota.
Generally, it is not necessary to make this stipulation, for large scale
elections, because no candidate, however miserable, ever gets no votes.
Exclusion keep value equals quota/(candidates reverse preference vote):
A: 50/1
B: 50/0
C: 50/99
Geometric mean keep value ( election keep value multiplied by inverse
exclusion keep value):
A: 50/51 x 1/50 ~ 0,0196
B: 50/49 x 0/50 = 0/49 is indeterminate. The closest determinate
approximation gives 1/49, not quite as low a keep value as 1/51 for A,
who is therefore the winner.
C: 50/0 x 99/50 = 99/0 is indeterminate. The closest determinate
approximation is 99/1. This keep value properly signifies huge
unpopularity, since an elective keep value is unity.
However, this example does illustrate the problem of the relative
importance of the election and exclusion counts. My guess is that the
problem would resolve itself in the entropy of preference voting, which
falls off exponentially with lower preferences.
Laws of physics are time-reversible in principle. But in practise, on
the classical or macroscopic scale, entropy intervenes, to give time a
one-way direction. (Binomial STV, in small scale elections, below the
level of much or any significance for parametric statistics, might be
likened to time-reversible quantum scale inter-actions, with regard to
the relative importance of election and exclusion, the direction of
choice, positive or negative.)
This falling off, of non-abstentions, would be countered, to some
extent, by the power of binomial stv to exclude, as well as elect,
candidates. And there is no reason in principle why a binomial stv
election might actually be more of an exclusion of generally disliked
candidates (just as often is FPTP). Still, it should be borne in mind,
that this very exclusive power of binomial stv should deter adversarial
candidate line-ups.
Regards,
Richard Lung.
On 26/02/2022 22:28, Kristofer Munsterhjelm wrote:
> On 26.02.2022 15:02, Kristofer Munsterhjelm wrote:
>> On 26.02.2022 13:21, Richard Lung wrote:
>>> Thank you, Forest,
>>>
>>> Your example is the kind of example that Riker gave.
>> Thank you for providing information about how to calculate who wins
>> according to Binomial STV.
>>
>> I have a few examples of my own. Could you tell me who wins, and how the
>> wins are calculated, for these single-winner elections?
>>
>> First:
>>
>> 51: A>B>C
>> 48: B>A>C
>> 1: B>C>A
>>
>> And second:
>>
>> 36: A>B>C
>> 34: B>C>A
>> 30: C>A>B
> Oops, I probably should've noticed that Forest's example is a variant of
> my second election, so I don't think I need the outcome and calculations
> for that one to verify whether the way I think the method works (for
> three candidates) is right.
>
> The first one would still be useful, though :-)
>
> -km
KM
Kristofer Munsterhjelm
Sun, Feb 27, 2022 1:41 PM
On 27.02.2022 14:04, Richard Lung wrote:
Thank you, Kristofer,
for first example.
The quota is 100/(1+1) = 50.
Election keep value is quota/(candidates preference votes)
for A: 50/51
B: 50/49
C: 50/0 Which, of course is infinite. It may be convenient, for tidy
book-keeping, that small elections require that each candidate votes for
themself. Then the keep value maximum simply equals the quota.
Generally, it is not necessary to make this stipulation, for large scale
elections, because no candidate, however miserable, ever gets no votes.
Another option is to just let infinities be worse than any alternative.
Since not every candidate can have a zero last preference count, at
least one candidate must have a finite value and so would be considered
better than every candidate with an infinite value.
Exclusion keep value equals quota/(candidates reverse preference vote):
A: 50/1
B: 50/0
C: 50/99
Geometric mean keep value ( election keep value multiplied by inverse
exclusion keep value):
A: 50/51 x 1/50 ~ 0,0196
B: 50/49 x 0/50 = 0/49 is indeterminate. The closest determinate
approximation gives 1/49, not quite as low a keep value as 1/51 for A,
who is therefore the winner.
Is 0/49 indeterminate? Shouldn't it just be zero? 0/x = 0 for x not
equal to zero, and the square root of zero is zero.
But let me in any case revise my example. Who wins in this one?
50: A>B>C
47: B>A>C
2: B>C>A
1: A>C>B
My calculations are as follows:
The quota is 50.
Election keep value is quota/candidate preferences:
A: 50/51
B: 50/49
C: infinity
Exclusion keep value equals quota/candidates reversed first preferences:
A: 50/2
B: 50/1
C: 50/97
Geometric mean:
A: square root of (50/51 x 2/50) ~ 0.198
B: square root of (50/49 x 1/50) ~ 0.143
C: ~= infinity (or very high)
So B wins, having the lowest keep value. Is this correct?
(You seem to have omitted the square root in your calculations, but it
shouldn't make a difference. Without the square root, A and B's values
are 0.0392 and 0.0204 respectively.)
-km
On 27.02.2022 14:04, Richard Lung wrote:
>
> Thank you, Kristofer,
>
>
> for first example.
>
> The quota is 100/(1+1) = 50.
>
> Election keep value is quota/(candidates preference votes)
>
> for A: 50/51
>
> B: 50/49
>
> C: 50/0 Which, of course is infinite. It may be convenient, for tidy
> book-keeping, that small elections require that each candidate votes for
> themself. Then the keep value maximum simply equals the quota.
> Generally, it is not necessary to make this stipulation, for large scale
> elections, because no candidate, however miserable, ever gets no votes.
Another option is to just let infinities be worse than any alternative.
Since not every candidate can have a zero last preference count, at
least one candidate must have a finite value and so would be considered
better than every candidate with an infinite value.
> Exclusion keep value equals quota/(candidates reverse preference vote):
>
> A: 50/1
>
> B: 50/0
>
> C: 50/99
>
> Geometric mean keep value ( election keep value multiplied by inverse
> exclusion keep value):
>
> A: 50/51 x 1/50 ~ 0,0196
>
> B: 50/49 x 0/50 = 0/49 is indeterminate. The closest determinate
> approximation gives 1/49, not quite as low a keep value as 1/51 for A,
> who is therefore the winner.
Is 0/49 indeterminate? Shouldn't it just be zero? 0/x = 0 for x not
equal to zero, and the square root of zero is zero.
But let me in any case revise my example. Who wins in this one?
50: A>B>C
47: B>A>C
2: B>C>A
1: A>C>B
My calculations are as follows:
The quota is 50.
Election keep value is quota/candidate preferences:
A: 50/51
B: 50/49
C: infinity
Exclusion keep value equals quota/candidates reversed first preferences:
A: 50/2
B: 50/1
C: 50/97
Geometric mean:
A: square root of (50/51 x 2/50) ~ 0.198
B: square root of (50/49 x 1/50) ~ 0.143
C: ~= infinity (or very high)
So B wins, having the lowest keep value. Is this correct?
(You seem to have omitted the square root in your calculations, but it
shouldn't make a difference. Without the square root, A and B's values
are 0.0392 and 0.0204 respectively.)
-km
RL
Richard Lung
Sun, Feb 27, 2022 7:08 PM
Kristofer,
Thank you for correcting me. I do tend to forget the square root, to
finish the average, the geometric mean.
In some formalisms, a zero numerator implies zero, but the geometric
mean, unlike the arithmetic mean, does not work wih zero, so that result
cannot be infered from it.
The trouble with just putting infinity is that there are different
infinities! One could require 1 vote for a candidate, from the
candidates themselves. Then we have a standard of comparison.
Glancing at your example, tho, I am reminded of an example with small
numbers, in which I had to reduce the 1 vote minimum to 0.1. Traditional
STV resorted to this expedient for small numbers elections, with the
Droop quota. They could not add plus one, because that made the quota
too hard for candidates to win. So, they resorted to a final plus 0,001,
I believe. But then ERS Ballot Services Major Frank Britton realised
that the final constant was never needed. And it is true, in this case,
of Binomial STV that a minimum candidate vote is not needed, as I think
you have suggested.
Moreover, if candidates have a minimum vote, they must also have the
same reverse preference minimum, for the sake of symmetrical treatment -
and perhaps as a neutralising factor.
I don't know yet what will turn out to be the most elegant count
instructions, in this and, no doubt, other instances. Your method
designating keep value infinity may be better, because such results are
nowhere in it, anyway. And it cuts out a troublesome added minimum
constant to candidates votes. I guess your result is correct. One has to
be a bit careful, in general, tho. An extremely bad election count may
be somewhat redeemed by a tolerable exclusion count, getting nowhere
near an exclusion quota. In that case, the not popular but also not
unpopular candidate is perhaps entitled to a quantitative tabulation.
This quandry reminds me of the caution, from an stv count expert, to use
floating point arithmetic in computer coding stv. Meek method, in New
Zealand, uses decimal point, but that might create future difficulties.
Anyway, I appreciate how important it is not to under-estimate the
possible ill consequences of casually considered operations.
The New Zealand government has not made its Meek method open source.
They denied access. Dr David Hill made his coding, of Meek method, open
source. He is a direct descendant of Thomas Wright Hill, who published
the first known instance of tranferable voting (barring the Gospel
Incident of the loaves and fishes). To celebrate the 200th anniversaey,
I re-published this and the public domain code by David Hill. Smashwords
does not allow publishing public domain works, so I had to put it on
Amazon, who charge a minimal fee. However, I could perhaps put it on
archive.org who don't charge, and where I put many of my e-books in pdf
format.
Dr David Hill wrote his code, for Meek method, in Pascal, an early
script. But I did not include his code text for eliminated candidates
last past the post, when the quota surpluses run out. Nor did I include
the code for reducing the quota, when preference voting gives way to
abstentions. Binomial STV does not use either of these expedients.
But Meek method does calculate the keep values of elected candidates:
the quota divided by their total transferable vote, which may increase
with further preferences after a quota is already achieved. The lower
the keep value below unity, the greater the popularity.
Binomial STV greatly extends the Meek method use of keep values, to all
candidates, and for their exclusion, as well as their election. However,
there is no difference in principle to these extended operations, of
transferble voting, by keep value.
The hand count version of binomial stv, tho, drops the distinctively
computerised Meek count of post-quota preference counting. And sticks to
the first order binomial count, making it simpler than traditional
counts, as well as Meek stv.
The New Zealand government hired two software coding firms, as back-up,
paying both, but only using one of them. This shows how arduous and
uncertain the results of their making Dr Hill coding texts executable.
Regards,
Richard Lung.
On 27/02/2022 13:41, Kristofer Munsterhjelm wrote:
On 27.02.2022 14:04, Richard Lung wrote:
Thank you, Kristofer,
for first example.
The quota is 100/(1+1) = 50.
Election keep value is quota/(candidates preference votes)
for A: 50/51
B: 50/49
C: 50/0 Which, of course is infinite. It may be convenient, for tidy
book-keeping, that small elections require that each candidate votes for
themself. Then the keep value maximum simply equals the quota.
Generally, it is not necessary to make this stipulation, for large scale
elections, because no candidate, however miserable, ever gets no votes.
Another option is to just let infinities be worse than any alternative.
Since not every candidate can have a zero last preference count, at
least one candidate must have a finite value and so would be considered
better than every candidate with an infinite value.
Exclusion keep value equals quota/(candidates reverse preference vote):
A: 50/1
B: 50/0
C: 50/99
Geometric mean keep value ( election keep value multiplied by inverse
exclusion keep value):
A: 50/51 x 1/50 ~ 0,0196
B: 50/49 x 0/50 = 0/49 is indeterminate. The closest determinate
approximation gives 1/49, not quite as low a keep value as 1/51 for A,
who is therefore the winner.
Is 0/49 indeterminate? Shouldn't it just be zero? 0/x = 0 for x not
equal to zero, and the square root of zero is zero.
But let me in any case revise my example. Who wins in this one?
50: A>B>C
47: B>A>C
2: B>C>A
1: A>C>B
My calculations are as follows:
The quota is 50.
Election keep value is quota/candidate preferences:
A: 50/51
B: 50/49
C: infinity
Exclusion keep value equals quota/candidates reversed first preferences:
A: 50/2
B: 50/1
C: 50/97
Geometric mean:
A: square root of (50/51 x 2/50) ~ 0.198
B: square root of (50/49 x 1/50) ~ 0.143
C: ~= infinity (or very high)
So B wins, having the lowest keep value. Is this correct?
(You seem to have omitted the square root in your calculations, but it
shouldn't make a difference. Without the square root, A and B's values
are 0.0392 and 0.0204 respectively.)
-km
Kristofer,
Thank you for correcting me. I do tend to forget the square root, to
finish the average, the geometric mean.
In some formalisms, a zero numerator implies zero, but the geometric
mean, unlike the arithmetic mean, does not work wih zero, so that result
cannot be infered from it.
The trouble with just putting infinity is that there are different
infinities! One could require 1 vote for a candidate, from the
candidates themselves. Then we have a standard of comparison.
Glancing at your example, tho, I am reminded of an example with small
numbers, in which I had to reduce the 1 vote minimum to 0.1. Traditional
STV resorted to this expedient for small numbers elections, with the
Droop quota. They could not add plus one, because that made the quota
too hard for candidates to win. So, they resorted to a final plus 0,001,
I believe. But then ERS Ballot Services Major Frank Britton realised
that the final constant was never needed. And it is true, in this case,
of Binomial STV that a minimum candidate vote is not needed, as I think
you have suggested.
Moreover, if candidates have a minimum vote, they must also have the
same reverse preference minimum, for the sake of symmetrical treatment -
and perhaps as a neutralising factor.
I don't know yet what will turn out to be the most elegant count
instructions, in this and, no doubt, other instances. Your method
designating keep value infinity may be better, because such results are
nowhere in it, anyway. And it cuts out a troublesome added minimum
constant to candidates votes. I guess your result is correct. One has to
be a bit careful, in general, tho. An extremely bad election count may
be somewhat redeemed by a tolerable exclusion count, getting nowhere
near an exclusion quota. In that case, the not popular but also not
unpopular candidate is perhaps entitled to a quantitative tabulation.
This quandry reminds me of the caution, from an stv count expert, to use
floating point arithmetic in computer coding stv. Meek method, in New
Zealand, uses decimal point, but that might create future difficulties.
Anyway, I appreciate how important it is not to under-estimate the
possible ill consequences of casually considered operations.
The New Zealand government has not made its Meek method open source.
They denied access. Dr David Hill made his coding, of Meek method, open
source. He is a direct descendant of Thomas Wright Hill, who published
the first known instance of tranferable voting (barring the Gospel
Incident of the loaves and fishes). To celebrate the 200th anniversaey,
I re-published this and the public domain code by David Hill. Smashwords
does not allow publishing public domain works, so I had to put it on
Amazon, who charge a minimal fee. However, I could perhaps put it on
archive.org who don't charge, and where I put many of my e-books in pdf
format.
Dr David Hill wrote his code, for Meek method, in Pascal, an early
script. But I did not include his code text for eliminated candidates
last past the post, when the quota surpluses run out. Nor did I include
the code for reducing the quota, when preference voting gives way to
abstentions. Binomial STV does not use either of these expedients.
But Meek method does calculate the keep values of elected candidates:
the quota divided by their total transferable vote, which may increase
with further preferences after a quota is already achieved. The lower
the keep value below unity, the greater the popularity.
Binomial STV greatly extends the Meek method use of keep values, to all
candidates, and for their exclusion, as well as their election. However,
there is no difference in principle to these extended operations, of
transferble voting, by keep value.
The hand count version of binomial stv, tho, drops the distinctively
computerised Meek count of post-quota preference counting. And sticks to
the first order binomial count, making it simpler than traditional
counts, as well as Meek stv.
The New Zealand government hired two software coding firms, as back-up,
paying both, but only using one of them. This shows how arduous and
uncertain the results of their making Dr Hill coding texts executable.
Regards,
Richard Lung.
On 27/02/2022 13:41, Kristofer Munsterhjelm wrote:
> On 27.02.2022 14:04, Richard Lung wrote:
>> Thank you, Kristofer,
>>
>>
>> for first example.
>>
>> The quota is 100/(1+1) = 50.
>>
>> Election keep value is quota/(candidates preference votes)
>>
>> for A: 50/51
>>
>> B: 50/49
>>
>> C: 50/0 Which, of course is infinite. It may be convenient, for tidy
>> book-keeping, that small elections require that each candidate votes for
>> themself. Then the keep value maximum simply equals the quota.
>> Generally, it is not necessary to make this stipulation, for large scale
>> elections, because no candidate, however miserable, ever gets no votes.
> Another option is to just let infinities be worse than any alternative.
> Since not every candidate can have a zero last preference count, at
> least one candidate must have a finite value and so would be considered
> better than every candidate with an infinite value.
>
>> Exclusion keep value equals quota/(candidates reverse preference vote):
>>
>> A: 50/1
>>
>> B: 50/0
>>
>> C: 50/99
>>
>> Geometric mean keep value ( election keep value multiplied by inverse
>> exclusion keep value):
>>
>> A: 50/51 x 1/50 ~ 0,0196
>>
>> B: 50/49 x 0/50 = 0/49 is indeterminate. The closest determinate
>> approximation gives 1/49, not quite as low a keep value as 1/51 for A,
>> who is therefore the winner.
> Is 0/49 indeterminate? Shouldn't it just be zero? 0/x = 0 for x not
> equal to zero, and the square root of zero is zero.
>
> But let me in any case revise my example. Who wins in this one?
>
> 50: A>B>C
> 47: B>A>C
> 2: B>C>A
> 1: A>C>B
>
> My calculations are as follows:
>
> The quota is 50.
>
> Election keep value is quota/candidate preferences:
>
> A: 50/51
> B: 50/49
> C: infinity
>
> Exclusion keep value equals quota/candidates reversed first preferences:
>
> A: 50/2
> B: 50/1
> C: 50/97
>
> Geometric mean:
>
> A: square root of (50/51 x 2/50) ~ 0.198
> B: square root of (50/49 x 1/50) ~ 0.143
> C: ~= infinity (or very high)
>
> So B wins, having the lowest keep value. Is this correct?
>
> (You seem to have omitted the square root in your calculations, but it
> shouldn't make a difference. Without the square root, A and B's values
> are 0.0392 and 0.0204 respectively.)
>
> -km
KV
Kevin Venzke
Sun, Feb 27, 2022 7:30 PM
Hi Kristofer/Richard,
I wonder not just about the square root, but also if the quota has some additional
role in the method, perhaps when there are 4+ candidates.
Because this expression:
( quota / keep ) * ( exclude / quota )
Appears to simplify to:
( exclude / keep )
This creates the appearance that the quota has no effect on the outcome.
Richard stated that final values below unity are electable. It looks like there
will always be an electable candidate, unless it's a complete tie, or perhaps if
there is some other rule not yet stated here.
It seems to me that the 3-candidate 1-winner case of this method is monotone.
It would help to see a four-candidate election resolved, too.
Kevin
Le dimanche 27 février 2022, 07:41:20 UTC−6, Kristofer Munsterhjelm km_elmet@t-online.de a écrit :
On 27.02.2022 14:04, Richard Lung wrote:
Thank you, Kristofer,
for first example.
The quota is 100/(1+1) = 50.
Election keep value is quota/(candidates preference votes)
Exclusion keep value equals quota/(candidates reverse preference vote):
Geometric mean keep value ( election keep value multiplied by inverse
exclusion keep value):
Geometric mean:
A: square root of (50/51 x 2/50) ~ 0.198
B: square root of (50/49 x 1/50) ~ 0.143
C: ~= infinity (or very high)
So B wins, having the lowest keep value. Is this correct?
(You seem to have omitted the square root in your calculations, but it
shouldn't make a difference. Without the square root, A and B's values
are 0.0392 and 0.0204 respectively.)
Hi Kristofer/Richard,
I wonder not just about the square root, but also if the quota has some additional
role in the method, perhaps when there are 4+ candidates.
Because this expression:
( quota / keep ) * ( exclude / quota )
Appears to simplify to:
( exclude / keep )
This creates the appearance that the quota has no effect on the outcome.
Richard says final values below unity are electable. It seems like there will
always be an electable candidate, unless it's a complete tie, or perhaps if there
is some other rule not yet stated here.
Kevin
Le dimanche 27 février 2022, 07:41:20 UTC−6, Kristofer Munsterhjelm km_elmet@t-online.de a écrit :
On 27.02.2022 14:04, Richard Lung wrote:
Thank you, Kristofer,
for first example.
The quota is 100/(1+1) = 50.
Election keep value is quota/(candidates preference votes)
Exclusion keep value equals quota/(candidates reverse preference vote):
Geometric mean keep value ( election keep value multiplied by inverse
exclusion keep value):
Geometric mean:
A: square root of (50/51 x 2/50) ~ 0.198
B: square root of (50/49 x 1/50) ~ 0.143
C: ~= infinity (or very high)
So B wins, having the lowest keep value. Is this correct?
(You seem to have omitted the square root in your calculations, but it
shouldn't make a difference. Without the square root, A and B's values
are 0.0392 and 0.0204 respectively.)
Hi Kristofer/Richard,
I wonder not just about the square root, but also if the quota has some additional
role in the method, perhaps when there are 4+ candidates.
Because this expression:
( quota / keep ) * ( exclude / quota )
Appears to simplify to:
( exclude / keep )
This creates the appearance that the quota has no effect on the outcome.
Richard stated that final values below unity are electable. It looks like there
will always be an electable candidate, unless it's a complete tie, or perhaps if
there is some other rule not yet stated here.
It seems to me that the 3-candidate 1-winner case of this method is monotone.
It would help to see a four-candidate election resolved, too.
Kevin
Le dimanche 27 février 2022, 07:41:20 UTC−6, Kristofer Munsterhjelm <km_elmet@t-online.de> a écrit :
> On 27.02.2022 14:04, Richard Lung wrote:
> > Thank you, Kristofer,
> >
> >
> > for first example.
> >
> > The quota is 100/(1+1) = 50.
> >
> > Election keep value is quota/(candidates preference votes)
> >
> > Exclusion keep value equals quota/(candidates reverse preference vote):
> >
> > Geometric mean keep value ( election keep value multiplied by inverse
> > exclusion keep value):
> Geometric mean:
>
> A: square root of (50/51 x 2/50) ~ 0.198
> B: square root of (50/49 x 1/50) ~ 0.143
> C: ~= infinity (or very high)
>
> So B wins, having the lowest keep value. Is this correct?
>
> (You seem to have omitted the square root in your calculations, but it
> shouldn't make a difference. Without the square root, A and B's values
> are 0.0392 and 0.0204 respectively.)
Hi Kristofer/Richard,
I wonder not just about the square root, but also if the quota has some additional
role in the method, perhaps when there are 4+ candidates.
Because this expression:
( quota / keep ) * ( exclude / quota )
Appears to simplify to:
( exclude / keep )
This creates the appearance that the quota has no effect on the outcome.
Richard says final values below unity are electable. It seems like there will
always be an electable candidate, unless it's a complete tie, or perhaps if there
is some other rule not yet stated here.
Kevin
Le dimanche 27 février 2022, 07:41:20 UTC−6, Kristofer Munsterhjelm <km_elmet@t-online.de> a écrit :
> On 27.02.2022 14:04, Richard Lung wrote:
> > Thank you, Kristofer,
> >
> >
> > for first example.
> >
> > The quota is 100/(1+1) = 50.
> >
> > Election keep value is quota/(candidates preference votes)
> >
> > Exclusion keep value equals quota/(candidates reverse preference vote):
> >
> > Geometric mean keep value ( election keep value multiplied by inverse
> > exclusion keep value):
> Geometric mean:
>
> A: square root of (50/51 x 2/50) ~ 0.198
> B: square root of (50/49 x 1/50) ~ 0.143
> C: ~= infinity (or very high)
>
> So B wins, having the lowest keep value. Is this correct?
>
> (You seem to have omitted the square root in your calculations, but it
> shouldn't make a difference. Without the square root, A and B's values
> are 0.0392 and 0.0204 respectively.)
