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21159 |
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Date: March 09, 2020 at 13:26:33
From: Skywise, [DNS_Address]
Subject: Roger, question re Jones p for fractional windows |
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The Jones p is computed from how many consecutive
windows have hits divided by how many of those
windows there are. Easy when, say, it's a 5 day window
tested against 500 days of record - 100 windows.
But say it was a 6 day window? That's 83 1/3
consecutive windows. What's the math for that? I'm
trying different ideas but I'm getting non-sensical
results.
Brian
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21165 |
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Date: March 09, 2020 at 16:34:48
From: Roger Hunter, [DNS_Address]
Subject: Re: Roger, question re Jones p for fractional windows |
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Brian;
You just have to adjust the test range until it comes
out even.
If it's long enough it shouldn't matter anyway.
Roger
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21166 |
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Date: March 09, 2020 at 19:45:29
From: Skywise, [DNS_Address]
Subject: Re: Roger, question re Jones p for fractional windows |
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That's what I thought, as I couldn't come up with any sort of weighting factor for the partial window that would come up with the right answer.
I'm still messing around with small samples, and in this case, situations where I already know what the answer is.
BTW, in such a case, my overlapping windows method works and comes up with the right probability. But obviously much more work. I agree it's easier to simply adjust the search record to fit even multiples of the prediction window and use the consecutive window method.
Brian
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21167 |
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Date: March 09, 2020 at 20:14:15
From: Roger Hunter, [DNS_Address]
Subject: Re: Roger, question re Jones p for fractional windows |
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Skywise;
Remember too, the question is how many times in
history has this prediction been true. If you mess
with the window, shifting it a day at a time,
eventually it will miss the quake.
Roger
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21168 |
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Date: March 09, 2020 at 20:47:14
From: Skywise, [DNS_Address]
Subject: Re: Roger, question re Jones p for fractional windows |
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I've tested that empirically. (9 different scenarios so far)
My overlapping windows gets the same result as Jones'
consecutive windows *IF* the average of all distributions
is taken.
THAT is the real argument I have. Taking only a single
distribution does not result in an accurate probability.
I'm working on putting together the proof.
Brian
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21169 |
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Date: March 09, 2020 at 21:06:27
From: Roger Hunter, [DNS_Address]
Subject: Re: Roger, question re Jones p for fractional windows |
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Skywise;
How are you determining the correct probability?
Roger
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21170 |
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Date: March 09, 2020 at 22:33:25
From: Skywise, [DNS_Address]
Subject: Re: Roger, question re Jones p for fractional windows |
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In attempting to write a quick summary example for you,
I think I made have made an error somewhere.
This is good. It means I'm right to not be too sure of
myself just yet. Hence why I keep trying different
scenarios.
The whole point is, I want to make sure I truly understand
the math involved, rather than just using it blindly.
Brian
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Responses:
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21172 |
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Date: March 10, 2020 at 07:04:19
From: Roger Hunter, [DNS_Address]
Subject: Re: Roger, question re Jones p for fractional windows |
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Brian;
I asked Dr Jones if he had a paper describing it but
he does not.
Probability is the number of times a thing happens
divided by the number of times it could have
happened.
The "thing" here is a quake of a certain size and
location within a specific 10 day window. By
stepping thru windows a day at a time you're
involving different, unspecified windows.
Roger
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Responses:
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21173 |
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Date: March 10, 2020 at 07:10:54
From: Roger Hunter, [DNS_Address]
Subject: Re: Roger, question re Jones p for fractional windows |
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Brian;
What I don't understand is that the number of quakes
in a predicted window does not matter.
It seems to me, if window A has 100 quakes while
window B has only 1, the odds are different.
Dr Jones says no.
Roger
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21174 |
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Date: March 10, 2020 at 07:53:57
From: Skywise, [DNS_Address]
Subject: Re: Roger, question re Jones p for fractional windows |
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And I think you ARE right.
What got me started was I was just following a curiosity
hunch.
Using some samples from one of my tests...
A prediction has a 4 day window.
The 'historical record' is 12 days long.
This means there are 12/4 = 3 consecutive windows.
That means there are 3 test windows that can have a hit.
If for example we salt the 12 day 'historical record'
with 3 days of quakes, there are 220 possible 12 day
patterns that will have quakes on 3 days.
So let's look at the Jones p for some of those 12 day
patterns.
HHHm mmmm mmmm = 0.333
HmmH mmHm mmmm = 0.667
Hmmm mmHm mHmm = 1.000
As you can see, the Jones p changes wildly depending on
the exact distribution of the quakes in the record.
Given we salted the record to have 3 days out of 12
with quakes, the odds SHOULD be 3/12 or 0.250.
What I just realized when I was trying to give an example
before was that the average p method I was using was
NOT giving the expected odds of 0.250. So I realized I
was on the wrong track or missed something. I think I'm
still right that JUST using the p value of the given
historical record will not be correct. It may by chance
be close, but it may still be way off depending on how
the quakes cluster.
I'm now looking at a new idea.
Brian
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Responses:
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21175 |
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Date: March 10, 2020 at 08:09:19
From: Roger Hunter, [DNS_Address]
Subject: Re: Roger, question re Jones p for fractional windows |
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Skywise;
The historical record is all we have so it better be
right!
But the probability can change over time as new areas
become active or active areas die down.
An exact answer isn't possible.
Roger
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Responses:
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21176 |
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Date: March 10, 2020 at 08:34:55
From: Skywise, [DNS_Address]
Subject: Re: Roger, question re Jones p for fractional windows |
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Yes, the historical record is all we have for a sequence
of events with truly unknown probability.
We may not be able to get an exact answer, but we should
always strive to be as accurate as possible. We should
always question ourselves and our methods. Basic science
methodology.
Even if I'm going down a wrong path, at least I myself
will have a better understanding of the problem.
Brian
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Responses:
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21177 |
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Date: March 10, 2020 at 09:10:07
From: Skywise, [DNS_Address]
Subject: Re: Roger, question re Jones p for fractional windows |
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Occam's Razor. The simplest answer is most likely the
right one.
Using my same example from before of a 12 day 'historical
record' with 3 days having quakes - well, the simple
answer is 3/12 = .25, which is the expected answer.
The probability is simply the number of days with quakes
divided by the total number of days (or whatever the time
resolution is). No need for complicating things with
consecutive or overlapping windows. This also declusters
the catalog as it doesn't matter what the distribution is.
They could all happen in a clump or be spread evenly
across the record.
Which makes me ask, what is the logic behind the
consecutive window thing?
Brian
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Responses:
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21180 |
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Date: March 10, 2020 at 10:23:05
From: Roger Hunter, [DNS_Address]
Subject: Re: Roger, question re Jones p for fractional windows |
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Skywise;
The consecutive windows are all the possible windows.
If they overlapped you'd have too many possible
windows.
The search advances by years. For a given prediction
there is only one correct answer; the window contains
1 or more quakes or it does not. The "or more" cancels
clusters.
Roger
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Responses:
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21181 |
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Date: March 10, 2020 at 11:17:08
From: Skywise, [DNS_Address]
Subject: Re: Roger, question re Jones p for fractional windows |
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That's not the question.
Why use the consecutive windows thing at all? What is
the justification or logic behind it?
If a given quake history of 1000 days has 137 quake days,
why can't we just say the probability is .137? Simple math.
Number of days with quakes divided by total number of days.
In all the scenarios I've looked at so far, rarely does
the Jones p equal the actual real known p. And I know
the real p because I created the history to reflect
a chosen p. In fact, I looked at all possible variations
of the history that have that chosen p.
Again, my example of a 12 day 'history'. I chose a p of
.250, which means there are 3 of those 12 days with a
quake. Simplicity says 3/12, p = .250.
If the prediction window is 4 days, that means there
are 3 consecutive Jones windows. So that means the only
possible Jones probabilities are 0.000, 0.333, 0.666,
and 1.000. NEVER 0.250 which is the known actual value
of p.
Yes, in a large enough 'history' the Jones p _could_
approach the real p. But it will never equal it except
in certain circumstances, specifically...
If by chance the prediction window is some integer
multiple of the number of days with a quake in the
history, then yes it's possible to have a Jones p the
same as the actual p, but it's still not guaranteed as
it still depends on the distribution of the quake days,
again because the Jones p changes depending on the
actual distribution of those quake days in the history.
For example, if in the 12 day 'history' with a chosen
p of .250, thus 3 of 12 days with a quake, and we instead
chose a 3 day prediction window, this means there are
now 4 Jones windows, and the possible Jones p values
are 0.000, 0.250, 0.500, 0.750, and 1.000. So if by
chance the 3 quake days are distributed so that one and
only one Jones window has a hit, then yes the Jones p
will be the same as the real known p. But given there
are 220 possible history patterns of 3 quake days in
this 12 day 'history', and only 4 of them will have all
3 quake days in one Jones window, then the odds (pardon
the pun) are very much against the Jones p equaling the
real known value of p.
Brian
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Responses:
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21182 |
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Date: March 10, 2020 at 12:42:07
From: Roger Hunter, [DNS_Address]
Subject: Re: Roger, question re Jones p for fractional windows |
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Skywise;
While posting with Dr Jones the light finally came
on in my head!
We take the entire study period and divide it up
into consecutive windows of the predicted length.
Then we look at the quakes for the study period. If
a quake of the right location and mag happens we
drop it into the appropriate time window.
Repeat for all quakes.
Then we look at the windows and count how many have
quakes. That divided by total windows is the odds on
a selected window containing an appropriate quake.
Your sliding windows gives multiple hits.
Roger
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Responses:
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21185 |
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Date: March 10, 2020 at 13:52:00
From: Skywise, [DNS_Address]
Subject: Re: Roger, question re Jones p for fractional windows |
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Forget the sliding windows. I am NOT doing that any more.
I agree that it is not entirely correct, and even if it
were, it's unnecessary work.
I am doing the Jones consecutive windows.
And again, WHY do we do consecutive windows in the first
place? THAT is my question. I don't see the reason for
it. Using it I am getting unexpected answers.
Let me try to illustrate using the limits of a text only
message.
I am making a prediction with a 4 day window.
We are testing against a catalog with 12 days in it.
So this means there are 3 consecutive 4 day Jones
Windows.
We good so far?
Now let's say the 12 day catalog is of the following
pattern for quakes, where 'o' is no quake and 'X' is a
quake for that day.
oXoX oooo oXoo
I've already divided it up into those 3 consecutive
Jones windows. As can plainly be seen, there are two
Jones windows with a hit, and one without. That's 2
out of 3. So the Jones probability is .667.
Right?
OK.
However, shouldn't the probability actually be 3/12 = .250?
After all, there are 3 quake days across 12 days. 3
divided by 12 equals .250.
Let me illustrate why.
For 12 days with exactly 3 of them as quake days, there
are 220 possible combinations. I have written out all
220 combinations in a spread sheet so I can figure out
the Jones p for each one.
Here's a few examples:
XXXo oooo oooo 0.333
oXoo XoXo oooo 0.667
ooXo oooX Xooo 0.667
oooo ooXX Xooo 0.667
oooo XoXX oooo 0.333
Xooo ooXo oooX 1.000
oooX oooX Xooo 1.000
Since there are 3 Jones windows, that means there are
2^3-1 possible combinations of Jones windows with at
least one hit, where m = miss and H = hit, and
following each is the resulting Jones p.
mmH = 0.333
mHm = 0.333
mHH = 0.667
Hmm = 0.333
HmH = 0.667
HHm = 0.667
HHH = 1.000
(0.000, all three misses is not included because that
means there are no quakes, and we are looking only at
possibilites that have exactly 3 quakes)
Still following me?
So there are 3 possible Jones p results. Listed below
are those three results followed by how many of those
220 possible combinations mentioned above result in
that particular Jones p value.
0.333 = 12
0.667 = 144
1.000 = 64
As you can plainly see, depending on which of the 220
possible distributions of exactly 3 quake days in a 12
day catalog is evaluated, there could be any one of
three possible Jones p values.
Now remember, we are looking at 3 quake days out of 12
days. Does that not mean the probability is 3/12 or
0.250? It doesn't matter the distribution of those three
quake days. It could be any one of those 220 possible
variations. They all have 3 out of 12.
If so, in this particular case, the Jones p value is
NEVER right.
Yes, in reality, there is only one quake history. But
if a real 12 day catalog only contains 3 quake days,
what is the correct p value?
One of the three possible Jones p values which depend
on the particular distribution of the quakes?
Or simply 3/12 = 0.250?
Brian
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Responses:
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21187 |
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Date: March 10, 2020 at 14:14:39
From: Roger Hunter, [DNS_Address]
Subject: Re: Roger, question re Jones p for fractional windows |
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Skywise;
It all depends on how they happen.
The windows are consecutive so that all days are
covered.
Days don't count; windows count. There are 3 in your
example.
If all quakes fall in 1 window odds are 1/3
all in 2 days = 2/3
all in 3 windows, 3/3
Those are the only possibilities.
Roger
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Responses:
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21188 |
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Date: March 10, 2020 at 15:10:59
From: Skywise, [DNS_Address]
Subject: Re: Roger, question re Jones p for fractional windows |
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Yet, we know the real odds are 1/4.
Not 1/3. Not 2/3. Not 3/3.
Brian
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21191 |
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Date: March 10, 2020 at 18:24:57
From: Roger Hunter, [DNS_Address]
Subject: Re: Roger, question re Jones p for fractional windows |
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Wrong.
It's window probability, not day probability.
Roger
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21193 |
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Date: March 10, 2020 at 20:06:56
From: Skywise, [DNS_Address]
Subject: Re: Roger, question re Jones p for fractional windows |
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I've written code to simulate large scenarios. I can
adjust record size, prediction window size, and
predetermined probability at will.
When using record sizes of thousands, if not tens of
thousands, yeah, I sorta see it. I just don't quite
understand it.
Statistics is not intuitive nor easy.
Brian
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21194 |
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Date: March 10, 2020 at 20:21:06
From: Roger Hunter, [DNS_Address]
Subject: Re: Roger, question re Jones p for fractional windows |
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Brian;
Look at it this way; you want the probability of what
is predicted, not something else similar.
So it's THIS location +/- 3 degrees, not something
+2/-4 degrees.
Roger
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21195 |
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Date: March 10, 2020 at 20:53:31
From: Skywise, [DNS_Address]
Subject: Re: Roger, question re Jones p for fractional windows |
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"Look at it this way; you want the probability of
what is predicted, not something else similar."
Granted. But how do we know the probability?
I don't think you are quite understanding my question.
WHY is the method Jones uses the correct one? That is
what I'm struggling with. It doesn't make sense that
the distribution of the quakes can change the
probability.
If I have a 10,000 day catalog and test one of Shan's
20 day windows, that's 500 consecutive windows.
What if I change the start day of my analysis? Guess
what? The Jones probability changes because I've just
shifted the distribution of the quakes.
Do you not see this?
Hell, try it yourself. You say you use the USGS catalog
from 1974. Try shifting the start date by a day at a
time and watch what happens to the resulting
probability for a given prediction.
Brian
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Responses:
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21196 |
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Date: March 11, 2020 at 08:33:25
From: Roger Hunter, [DNS_Address]
Subject: Re: Roger, question re Jones p for fractional windows |
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Skywise;
Of course; the probability depends on the dataset.
So is doing it your way.
You expect absolute truth?
That's why we use a very large historical record, to
get the best approximation possible within reason.
Roger
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21197 |
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Date: March 11, 2020 at 08:37:34
From: Roger Hunter, [DNS_Address]
Subject: Re: Roger, question re Jones p for fractional windows |
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PS: Have you checked your answer using a large dataset
compared to Jones for the same dataset?
They should be very close.
Roger
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21200 |
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Date: March 11, 2020 at 13:49:08
From: Skywise, [DNS_Address]
Subject: Re: Roger, question re Jones p for fractional windows |
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I've been a silly boy.... my 'expected' probability was
all wrong. I was looking at it the wrong way. My mistake
is that although I may be salting the test record with
a 10% probability, I failed to note that that is the
DAILY probability. But what is the probability across
the length of the prediction window? ooooops.
This just goes to show that statistics is a tricky bugger.
And it's easy to make simple mistakes.
I've got the right formula now, and the Jones p does come
very close to this number.
I just did a VERY large test run now to see how well it
converges. I averaged 1,000 runs of 100,000 day records
with exactly 10% random probability analyzing a 20 day
window.
The expected probability was 0.878423.
min Jones p: 0.868000
avg Jones p: 0.878562
max Jones p: 0.889400
As you can see, it's only off by about 0.016% on
average, and at most about 1.25%.
Brian
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21203 |
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Date: March 11, 2020 at 15:41:34
From: Roger Hunter, [DNS_Address]
Subject: Re: Roger, question re Jones p for fractional windows |
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Skywise;
Good enough for government work as we used to say when
I was a government worker.
Roger
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Responses:
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