Patterns of Prediction

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Re: Patterns of Prediction

Post by RogerE »

On the "round numbers" theme, here are some very recent Stampboards examples:

Patterns of Prediction: 100 posts
Patterns of Prediction: 100 posts
.
Digital clock strikes 00:33:00
Digital clock strikes 00:33:00
.
Counting by palindromes starts a new K*
Counting by palindromes starts a new K*
.
Sequence of palindromes starts its last &quot;patch&quot;<br />before increasing in length
Sequence of palindromes starts its last "patch"
before increasing in length
.
So many "round" patterns to notice and enjoy!

/RogerE :D

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Re: Patterns of Prediction

Post by RogerE »

Body Mass Index = BMI

Screen Shot 2021-04-22 at 12.30.21 pm.png
.
Body mass index (BMI) is a value derived from the mass (weight) and height of a person. The BMI is defined as the body mass divided by the square of the body height, and is expressed in units of kg/sq.m, resulting from mass in kilograms and height in metres.

The BMI may be determined using a table or chart which displays BMI as a function of mass and height using contour lines or colours for different BMI categories, and which may use other units of measurement (converted to metric units for the calculation).

The BMI is a convenient rule of thumb used to broadly categorise a person as underweight, normal weight, overweight, or obese based on tissue mass (muscle, fat, and bone) and height. Commonly accepted BMI ranges are underweight (under 18.5 kg/sq.m), normal weight (18.5 to 25 kg/sq.m), overweight (25 to 30 kg/sq.m), and obese (over 30 kg/sq.m).

BMIs under 20 and over 25 have been associated with higher all-causes mortality, with the risk increasing with distance from the 20–25 range.
https://en.wikipedia.org/wiki/Body_mass_index
.
Historical background
.
Wikipedia wrote:Adolphe Quetelet, a Belgian astronomer, mathematician, statistician, and sociologist, devised the basis of the BMI between 1830 and 1850 as he developed what he called "social physics".
The modern term "body mass index" (BMI) for the ratio of human body weight to squared height was coined in a paper published in the July 1972 edition of the Journal of Chronic Diseases by Ancel Keys and others. In this paper, Keys argued that what he termed the BMI was "...if not fully satisfactory, at least as good as any other relative weight index as an indicator of relative obesity".
.
Limitations and "misuse"
.
In fact BMI is intended to be a statistical predictor of population health, not a measure of individual health.
.
Wikipedia wrote:The interest in an index that measures body fat came with observed increasing obesity in prosperous Western societies. Ancel Keys explicitly judged BMI as appropriate for population studies and inappropriate for individual evaluation. Nevertheless, due to its simplicity, it has come to be widely used for preliminary diagnoses. Additional metrics, such as waist circumference, can be more useful.
.
Consideration of mathematical features of the BMI helps to show why it is unsuitable for application to individuals:
Wikipedia wrote:BMI of an individual is proportional to mass and inversely proportional to the height squared. So, if all body dimensions double, and mass scales naturally with height cubed, then BMI doubles instead of remaining the same. This results in taller people having a reported BMI that is uncharacteristically high, compared to their actual body fat levels. In comparison, the Ponderal index is based on the natural scaling of mass with height cubed.

However, many taller people are not just "scaled up short people", but tend to have narrower frames in proportion to their height. Carl Lavie has written: "The B.M.I. tables are excellent for identifying obesity and body fat in large populations, but they are far less reliable for determining fatness in individuals."
Measures of predictive capacity and relevant criticisms based on scientific studies are being reported. For instance:
.
Healthline wrote: .
Is BMI a good indicator of health?

Despite concerns that BMI doesn’t accurately identify whether a person is healthy, most studies show that a person’s risk of chronic disease and premature death does increase with a BMI lower than 18.5 (“underweight”) or above 30.0 (“obese”).

For example, a 2017 retrospective study of 103,218 deaths found that those who had a BMI of 30.0 or greater (“obese”) had 1.5–2.7 times greater risk of death after a 30-year follow-up.

Another study showed that those in the “obese” BMI category had a 20% increased risk of death from all causes and heart disease, compared with those in the “normal” BMI category.

The researchers also found that those who were either in the “underweight” or “severely obese” and “extremely obese” categories died an average of 6.7 years and 3.7 years earlier, respectively, compared with those in the “normal” BMI category.

Other studies have shown that a BMI greater than 30.0 begins to significantly increase your risk of chronic health issues, such as type 2 diabetes, heart disease, breathing difficulties, kidney disease, non-alcoholic fatty liver disease, and mobility issues.

Furthermore, a 5–10% reduction in a person’s BMI has been associated with decreased rates of metabolic syndrome, heart disease, and type 2 diabetes.

Due to most research showing an increased chronic disease risk among people with obesity, many health professionals can use BMI as a general snapshot of a person’s risk. Still, it should not be the only diagnostic tool used.

SUMMARY
Though BMI has been criticised for its oversimplification of health, most research supports its ability to estimate a person’s risk of chronic disease, particularly one’s risk of early death and metabolic syndrome.
https://www.healthline.com/nutrition/is-bmi-accurate#effectiveness
.
Body Image Stigma

We are becoming more aware of stigma related to various perceptions of body image. In particular, an official Health Department commentary on possible stigma related to Body Mass Index includes the following (with references to expert papers):
health.gov.au wrote: .
Weight stigma
Determinants of heath, including weight status, involve complex interactions between multi-level factors including policy, community, socio-demographic, psychosocial, family and genetic influences. Awareness of weight stigma can help health professionals to provide care that recognises the complexity of these interacting factors and removes inadvertent blame from the individual for less desirable health outcomes. Among the general population, weight stigma is a recognised risk factor for adverse psychological and physical health issues, which can exacerbate unhealthy eating behaviours (such as binge eating) and weight gain (Yazdizadeh et al 2020).

There is evidence of perceived weight stigma felt by women receiving pregnancy care (regardless of their size) (Bombak et al 2016); (Incollingo Rodriguez et al 2019). This can be tied to the fact that high weight status has been linked to adverse pregnancy outcomes. High gestational weight gain has also been linked to increased vulnerability to weight stigmatisation (Incollingo Rodriguez et al 2019). Evidence from non-pregnant populations highlight the potential for stigma to reinforce an unhealthy weight gain cycle (Tomiyama 2014).
https://www.health.gov.au/resources/pregnancy-care-guideline ... mass-index
/RogerE :D

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Re: Patterns of Prediction

Post by RogerE »

In another thread I recently posed the following prediction problem:
https://www.stampboards.com/viewtopic.php?f=12&t=95017&start=84

It is related to aspects of Patterns of Prediction, so I hope to see some discussion of it here.

Rank predictions of race results
RogerE wrote:
03 May 2021 03:24
Consider a race of some sort in which there are 12 competitors. Before the race we don't know how it will turn out, but after the race is over we can assign the number 1 to the winner, 2 to the runner-up, 3 to third place-getter, and so on, down to 12 for the last competitor to finish. The correct sequence for the first three competitors is therefore (1, 2, 3).

Now suppose several people tried to predict the first three place-getters, in correct order:

Prediction A actually selected (2, 3, 4). That is, the competitor predicted to place first actually placed second, the predicted second place-getter actually finished third, and the competitor predicted to place third actually finished fourth.
Prediction B actually selected (3, 2, 1).
Prediction C actually selected (1, 3, 10).
How should those three predictions be ranked, best to worst? How do you decide?

I thought about this problem, and came up with several different ways of comparing predictions, but I wasn't convinced that any of my methods was very satisfactory...

Altogether there are 12 x 11 x 10 = 1320 possible predictions of the first three place-getters. The very best prediction would be (1, 2, 3), and the very worst prediction would be (12, 11, 10). I wonder if there is a satisfactory/satisfying way to rank all 1320 possible predictions.

Of course, we could simply prescribe some rule(s), and say that those wishing to make competitive predictions will simply have to accept that their predictions will be ranked as prescribed. That would allow the best predictor to be identified, but it might still not be a satisfactory/satisfying ranking when viewed objectively...

A lesser objective might be to find a way to rank predictions that would allow some predictions to be ranked as "of equal merit" rather than insisting on a ranking that takes any two predictions and always ranks one as better than the other [a "linear ordering"]. But it still seems difficult to come up with a "partial ordering" which is satisfactory/satisfying. For instance, a rule which says "(1, 2, 3) is the best prediction, and all other possible predictions are equally bad" certainly manages to rank the possible predictions. However, it wouldn't satisfy the participants, especially not someone who predicted (1, 2, 4) when comparing with A, B or C above.

How would you rank (1, 2, 4) in comparison with (1, 3, 2)?
.
Footnote

Notice the ranking problem I have posed is purely about trying to decide a basis for comparing predictions. It does not include any consideration about how to make a good prediction.

In "races" where totalisator betting is conducted there are published "odds" to suggest a likely ranking of the competitors. In the scenario described above there is no such information available. Indeed, all competitors can be considered equally likely to win. (The thread where I posed the ranking problem quoted here actually had a field of 24 competitors, and the outcome was to be determined by "random selection", with results robustly reported. That is equivalent to having all competitors equally likely to win. In other words, there was no element of skill/ knowledge that would assist in predicting the final outcome.)

/RogerE :D

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Re: Patterns of Prediction

Post by RogerE »

Rank predictions of Race Results: a Proposed Formula


Screen Shot 2021-05-05 at 2.36.53 am.png
Screen Shot 2021-05-05 at 2.37.25 am.png
Screen Shot 2021-05-05 at 2.37.56 am.png
Screen Shot 2021-05-05 at 3.37.06 am.png
Screen Shot 2021-05-05 at 3.37.39 am.png


Applying this to questions asked in the previous post

Prediction list (2, 3, 4) has payoff 43.75;
prediction list (3, 2, 1) has payoff 57.50;
prediction list (1, 3, 10) has payoff 62.51.
So these three prediction lists are increasingly good (according to my proposed payoff scheme).
Of course, none is as good as prediction list (1, 2, 3), with payoff 87.50.

Prediction list (1, 2, 4) has payoff 81.25;
prediction list (1, 3, 2) has payoff 81.25.
So these two prediction lists are equally good (according to my proposed payoff scheme).

Of course, different evaluation methods would likely produce different rankings of prediction lists, but I feel that the method I have produced here gives a "plausible" ranking. What do others think about its ranking of the sample prediction lists?

/RogerE :D

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Re: Patterns of Prediction

Post by RogerE »

Rank predictions of Race Results: a Proposed Formula cont.


Comparing some larger prediction lists

Here are a few sample prediction lists for the first six place-getters in a race:
How would you rank them (from most successful, down to least successful)?
.
Prediction list A = (3, 2, 5, 1, 6, 4)
Prediction list B = (3, 4, 1, 2, 5, 6)
Prediction list C = (2, 3, 1, 5, 6, 4)
Prediction list D = (3, 1, 6, 2, 4, 5)
Prediction list E = (6, 1, 2, 4, 5, 3)
Prediction list F = (5, 6, 1, 2, 3, 4)
Prediction list G = (5, 2, 1, 4, 3, 6)

Notation

Let X < Y mean that prediction list X is less successful than prediction list Y.
That notation will give us a compact way of summarising the results.

Payoffs for prediction lists A–G

Using the formula proposed in the previous post, we find that
.
A = (3, 2, 5, 1, 6, 4) has payoff P(A) = 53.93
B = (3, 4, 1, 2, 5, 6) has payoff P(B) = 53.44
C = (2, 3, 1, 5, 6, 4) has payoff P(C) = 67.50
D = (3, 1, 6, 2, 4, 5) has payoff P(D) = 66.67
E = (5, 6, 1, 2, 3, 4) has payoff P(E) = 44.02
F = (6, 1, 2, 4, 5, 3) has payoff P(F) = 69.04
G = (5, 2, 1, 4, 3, 6) has payoff P(G) = 62.16
.
It turns out that A and B are very close in their level of success. Lists C, D, F, G are fairly close, but all are clearly better than A and B. List E is evidently inferior to the other five lists. The payoffs reflect how well the contestants selected to finish #1, #2, #3 actually performed.

Resultant ranking of prediction lists A–G
.
E < B < A < G < D < C < F

Visual representation of ranking
Visual representation of ranking
.
How does this compare with your intuitive ranking of those prediction lists?
I hope some of our readers will post a thoughtful comment or two.

/RogerE :D

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Re: Patterns of Prediction

Post by RogerE »

Digit patterns in a power sequence

I found myself "playing" in the "counting randomly" thread by stringing together the terms of a familiar power sequence:
1
2
4
8
16
32
64
128
256
512
1024
2048
4096
8192
...
The resulting string of digits is
12481632641282565121024204840968192...


Notice that the even digits 2, 4, 8 appear frequently, as does the odd digit 1,
but 3 has only appeared once, 5 has briefly appeared twice, eventually 9 has
also appeared twice, and the even digits 6 and 0 have appeared, but were slow
to do so. What about 7?

Does every digit appear in the digit string?

After 8192 the sequence continues
16384
32768
65536
...
so the resulting digit string continues
[...8192]163843276865536...


At last 7 has made an appearance — it first appears as term number 43 in the digit string.
Let's write d(43) = 7 and f(7) = 43 to compactly report this, choosing "d" to remind us of "digit",
and "f" to remind us of "first".
For instance, d(10) = 4 because the tenth term in the digit string is 4.
Also f(3) = 7 because the first term equal to 3 is the seventh digit in the string.

The first appearance of each digit is:
.
f(0) = 21; f(1) = 1; f(2) = 2; f(3) = 7); f(4) = 3;
f(5) = 15; f(6) = 6; f(7) = 43; f(8) = 4; f(9) = 34.
.
Notice that the first pair of equal consecutive digits has just appeared: 55.
In terminology introduced elsewhere, this is a (1,2) repblock, a block of size 1repeated
a total of two times.
The digits comprising the first repblock 55 are d(47) = 5, d(48) = 5. By specifying where
this repblock starts in the digit string, let us extend the earlier notation thus: f(55) = 47.

Does every ordered pair of digits appear in the digit string?

By "ordered pair of digits" I not only mean pairs which are (1, 2) repblocks like 00, 11, 22, 33, ...
but all the other pairs as well, including 12 and 21 as distinct, 37 and 73 as distinct, and so on.
Because d(4) = 8 and d(5) = 1, we have f(81) = 4. That is, the ordered pair 81 first appears
in the digit string at the fourth term in the string. Also d(13) = 8 and d(14) = 2, so f(82) = 13.

After 65536 the sequence continues
131072
262144
524288
1048576
2097152
4194304
8388608
16777216
33554432
...
so the resulting digit string continues
[...65536]13107226214452428810485762097152419430483886081677721633554432...
.
Notice that 777 has now appeared, and soon after we find occurrences of 33, 55 and 44 consecutively.

First occurrences of (1, 2) repblocks

The digit string so far shows us f(55) = 47, f(22) = 56, f(44) = 61, f(88) = 67, f(77) = 99 = f(777), f(33) = 105.
Will 00, 11, 66 eventually also occur?

More generally, will every ordered pair of digits eventually occur? How about any finite ordered block of digits,
such as 3210 or 737373? At the other extreme of this type of question, is there any finite ordered block of digits that does not occur an unlimited number of times?

/RogerE :D

A footnote
When I was working in the "counting randomly" thread, I noticed that my total number of posts was nicely composed of the digits 2, 4, 8 — an apt use of powers of 2
.
Screen Shot 2021-05-07 at 3.48.02 am.png

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Re: Patterns of Prediction

Post by satsuma »

RogerE wrote:
07 May 2021 07:06

2097152
4194304
8388608
16777216
33554432
Note that the concluding digits are now repeating

2
4
8
16
32

What can be predicted from this fact?

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Re: Patterns of Prediction

Post by RogerE »

Let's look at the nice observation made by satsuma in the previous post.

I actually looked at two sequences of numbers, in decimal notation:
A = 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, ...
B = 124816326412825651210242048...
The terms of sequence A are the powers of 2, so first term is 1, second term is 2, ..., seventh term is 64, ...
The terms of sequence B are the digits when the powers of 2 are "abutted", that is, written without any break, so first term is 1, second term is 2, ..., seventh term is 3, ...
(Commas could also be inserted in B, but every term is a single digit, so no separator is required. On the other hand, commas are helpful separators in A because the "length" of the terms increases.)

My discussion focussed on B, and asked questions such as: Does every digit appear in B? Does every pair of digits appear as a block in B? Are there any finite blocks of digits which do not occur in B?

Sequence of final digits after doubling

To approach satsuma's question, let's look at a third sequence, the final digit of each power of 2:
C = 124862486248...
This sequence can be generated by starting with 1, then at each successive step multiplying the latest term by 2 and keeping only the last digit of the result. In turn, this gives us
1 —> 2 —> 4 —> 8 —> [1]6 —> [1]2 —> 4 —> ...
This sequence C can be seen as fitting into a wider context. If instead of starting with 1, we start with a different odd digit, the sequence settles down to one of two behaviours:
3 —> 6 —>[1]2 —> 4 —> 8 —> [1]6 —> [1]2 —> ...
5 —> [1]0 —> 0 —> 0 —> 0 —> ...
7 —> [1]4 —> 8 —> [1]6 —> [1]2 —> 4 —> ...
9 —> [1]8 —> [1]6 —>[1]2 —> 4 —> 8 —> [1]6 —> ...
This "action" on the ten decimal digits can be nicely represented as a directed graph:

&quot;Doubling modulo 10&quot;
"Doubling modulo 10"
.
Discarding all but the final digit (in decimal notation) is known as "modulo 10" representation.

Summary: Beginning with any decimal digit and successively doubling modulo 10 yields a periodic sequence in which only even digits appear in the periodic part, and the sequence either repeats the size four block 2486, or repeats the single digit block 0.

Applying this summary to C, notice that 2, 2², 2³, 2⁴ end in 2, 4, 8, 6 respectively, so this corresponds to the repeating block 2486 in the doubling modulo 10 sequence. Hence we can predict that the final decimal digit of 2 to the power n, for any positive integer n, will be
2, if n is 1 greater than a multiple of 4;
4, if n is 2 greater than a multiple of 4;
8, if n is 3 greater than a multiple of 4;
6, if n is a multiple of 4.
For instance, if n = 25 then 2 to the power n will have 2 as its final digit.
This is confirmed in a recent post, where I noted: 2^25 = 33554432.
Equally, 2^29, 2^33, 2^37, 2^41, ... will all have 2 as their final digit.

Sequence of final pair of digits after doubling

Now consider a sequence in which we pick a starting number, in decimal notation, and at each successive step continue by multiplying the latest term by 2 and keeping only the last two digits of the result. (Because we are keeping the last two digits, this is "doubling modulo 100".)
For instance, if we start with 16, the sequence produced is
16 —> 32 —> 64 —> [1]28 —> 56 —> [1]12 —> 24 —> 48 —> 96 —> [1]92
—> [1]84 —> [1]68 —> [1]36 —> 72 —> [1]44 —> 88 —> [1]76 —> [1]52 —> [1]04 —> 08 —> 16 —>...
Now that 16 has reappeared, the whole sequence must repeat, endlessly. The repeating block has 20 terms, each term a pair of digits. (I have written "04" and "08" rather than 4 and 8 to preserve the fact that we are dealing in the last two digits of a decimal number.)
That repeating block is: 16, 32, 64, 28, 56, 12, 24, 48, 96, 92, 84, 68, 36, 72, 44, 88, 76, 52, 04, 08
Notice that it must contain the repeating single (final) digit block 2486, so its length has to be a multiple of the length of that smaller block (and indeed, 20 is a multiple of 4, so the single digit period occurs five consecutive times in the two digit period).

Prediction: Since 2², 2³, 2⁴, ... have final two digits 04, 08, 16, ... , we can predict that the final two decimal digits of 2 to the power n, for any integer n ≥ 2, will be
04, if n is 2 greater than a multiple of 20;
08, if n is 3 greater than a multiple of 20;
16, if n is 4 greater than a multiple of 20;
32, if n is 5 greater than a multiple of 20;
etc.
76, if n is a multiple of 20;
52, if n is 1 greater than a multiple of 20.
.
For instance, if n = 25 then 2 to the power n will have 32 as its final two digits.
This is confirmed by 2^25 = 33554432. Equally, 2^45, 2^65, 2^85, ... will all have 32 as their final two digits.

Directed graph for doubling modulo 100

The directed graph for doubling modulo 10 has ten nodes, corresponding to the ten digits of decimal notation. The corresponding directed graph for doubling modulo 100 has a hundred nodes, corresponding to the ordered pairs of digits of decimal notation. We have just seen that it has a directed cycle containing 20 of these nodes.

What does the rest of the directed graph look like?

A node such as 01 or 11, which ends with an odd digit, cannot be the result of doubling, so is a "source" node with one arrow out, and no arrow in: therefore such a node cannot be part of a cycle. A node such as 02 or 22, which is a multiple of 2 but not of 4, has just one arrow in and that comes from a source, so these nodes cannot be part of a cycle. The node 00 has an arrow back to itself, so forms a cycle of size 1, and the component of the directed graph containing it has 25 and 75 as source nodes, each with an arrow to 50, which has an arrow to 00, and finally 00 has an arrow back to itself.

I will leave the rest of the graph to any reader interested to figure out the details as a private challenge.

The observation made by satsuma is really about the last two digits of the powers of 2. He notes the portion 52 —> 04 —> 08 —> 16 —> 32, which all lies in the period 20 cycle discussed here. In particular, there are infinitely many powers of 2 that end in 52, namely 2^n when n = 21, 41, 61, 81, ...
On the other hand, note that 2^1 is the only power of 2 that ends in 02, because 02 does not lie in a cycle in the directed graph for doubling modulo 100. The sources 01 and 51 have arrows to 02, which has an arrow to 04, part of the cycle of size 20 already discussed. Finally, no power of 2 ends in 22, 42, 62 or 82, while infinitely many powers of 2 end in 12, 32, 52, 72 or 92.

/RogerE :D

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Re: Patterns of Prediction

Post by castores »

RogerE wrote:
07 May 2021 03:06
Rank predictions of Race Results: a Proposed Formula cont.


Comparing some larger prediction lists

Here are a few sample prediction lists for the first six place-getters in a race:
How would you rank them (from most successful, down to least successful)?
.
Prediction list A = (3, 2, 5, 1, 6, 4)
Prediction list B = (3, 4, 1, 2, 5, 6)
Prediction list C = (2, 3, 1, 5, 6, 4)
Prediction list D = (3, 1, 6, 2, 4, 5)
Prediction list E = (6, 1, 2, 4, 5, 3)
Prediction list F = (5, 6, 1, 2, 3, 4)
Prediction list G = (5, 2, 1, 4, 3, 6)

Notation

Let X < Y mean that prediction list X is less successful than prediction list Y.
That notation will give us a compact way of summarising the results.

Payoffs for prediction lists A–G

Using the formula proposed in the previous post, we find that
.
A = (3, 2, 5, 1, 6, 4) has payoff P(A) = 53.93
B = (3, 4, 1, 2, 5, 6) has payoff P(B) = 53.44
C = (2, 3, 1, 5, 6, 4) has payoff P(C) = 67.50
D = (3, 1, 6, 2, 4, 5) has payoff P(D) = 66.67
E = (5, 6, 1, 2, 3, 4) has payoff P(E) = 44.02
F = (6, 1, 2, 4, 5, 3) has payoff P(F) = 69.04
G = (5, 2, 1, 4, 3, 6) has payoff P(G) = 62.16
.
It turns out that A and B are very close in their level of success. Lists C, D, F, G are fairly close, but all are clearly better than A and B. List E is evidently inferior to the other five lists. The payoffs reflect how well the contestants selected to finish #1, #2, #3 actually performed.

Resultant ranking of prediction lists A–G
.
E < B < A < G < D < C < F


Image
.
How does this compare with your intuitive ranking of those prediction lists?
I hope some of our readers will post a thoughtful comment or two.

/RogerE :D
When it comes to the various topics in math (algebra, logic, calculus, etc), probability would be my worst.

I follow all you wrote RogerE (though I do feel dizzy ) and I have to make two points.

1. I'm in awe over your posts and the time taken to educate us :oops: . I sometimes try to think what your room looks like where you do your pondering :?: :idea:

2. I read your post to the end and unfortunately you left your question till then. Once I had read your "prediction list" and then your allocated percentage points, there was no point in creating an "intuitive ranking" as my mind already had your input of percentage points invading. I now had information that affected my view on the prediction list. I hope that makes sense.

1.1 On a slightly different note but still about patterns - Sudoku, U Do? :lol:

There are a number of ways to attack a sudoku puzzle but whether you play easy, medium, hard or disgusting patterns do appear in each 9x9 box. You might start to notice, for example, that every digit 1,8,3 are adjacent (vertical, horizontal, diagonal). With this the probability of a 'guess' being correct is increased.

I'd say 6 out 7 times the 'guess' is correct :)

Have you noticed this RogerE?
Australia : Islands : various countries : Thematics : etc

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Re: Patterns of Prediction

Post by RogerE »

Thanks for your comments castores.

• Regarding ranking predictions about results of a race, there were several posts on the theme of "intuitive comparisons" and a proposed objective method of ranking. I assumed readers would think about their own intuitive comparisons of the sample predictions I suggested, and then read on to see how my proposed objective ranking stacked up. If you held off making an intuitive comparison until after I revealed the objective ranking my method produced, that means that the jury's verdict would have been affected by public commentary on the case, and would have to be dismissed as inadmissible, Your Honour. So, I accept what you said — sorry I spoiled your chance to compare objectively. (Wonders: Have I been watching too many episodes of Judge Deed lately?)

• Regarding sudoku: There was quite a bit of posting about sudoku on Stampboards in Jan 2020. Although you were a member back then, castores, perhaps that discussion did not attract your attention. You can find a post in the middle of that discussion at
https://www.stampboards.com/viewtopic.php?f=6&t=80434&p=6262627&hilit=sudoku#p6262627
.
• Further thoughts on sudoku: Is there a predictable pattern that 1, 3, 8 will "usually" appear adjacent to each other in a sudoku solution? It depends on how the sudoku is created in the first place. If a "randomised" construction method is used, there ought to be no such persistent patterns. On the other hand, if the construction method is done by a composer who has certain preferred patterns or combinations (deliberately chosen or subconsciously favoured) then your discernment of those patterns would give you an advantage in guessing correctly beyond the "logical forcing" that is supposed to be the sole operative principle in making each step forward in solving the puzzle. The fact that we have seen published sudoku with defects (multiple solutions, incorrect solutions, or even impossible solutions with duplicate entries) shows that "hand-made" puzzles have appeared, along with "computer-made" puzzles which have had built-in checks to the creation process to verify that they have a unique correct solution. Even a "computer-made" puzzle might have the original programmer's favourite biases or preferred patterns subtly present.

Summary: No, I hadn't noticed a "1, 3, 8" cluster pattern, but it might exist within a group of sudoku puzzles from a single source. If you go to a different source, I predict such a pattern will not be found.

/RogerE :D

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Re: Patterns of Prediction

Post by RogerE »

In another thread, the number 42 is about to appear. Inevitably this reminds fans of Douglas Adams about the answer to the Ultimate Question about Life, the Universe and Everything.

Even more so, it reminds us that occasionally people go looking for a pattern where there in none intended. In that case, inventive minds may "fill the much needed gap". For example:
There is the persistent tale that 42 is Adams' tribute to the indefatigable paperback book, and is the average number of lines on an average page of an average paperback. Another common guess is that 42 refers to the number of laws in cricket, a recurring theme of the books.
Another delightful line:
John Lloyd, Adams' collaborator on The Meaning of Liff and two Hitchhiker's fits, said that Adams has called 42 "the funniest of the two-digit numbers."
A fine discussion for enthusiasts and admirers (and the source of the quotes above) is
https://en.wikipedia.org/wiki/Phrases_from_The_Hitchhiker%27s_Guide_to_the_Galaxy

Hooray for Wikipedia.

/RogerE :D

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Re: Patterns of Prediction

Post by nigelc »

A former work colleague of mine believed he had inspired Douglas Adams' selection of the number 42 through its role in a lecture on brain function he had once delivered with Douglas Adams in the audience. :)
Nigel

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Re: Patterns of Prediction

Post by RogerE »

Apophenia

The Wikipedia discussion about Douglas Adams and the number 42 led me to this interesting link:
Apophenia /æpoʊˈfiːniə/ is the tendency to perceive meaningful connections between unrelated things.

The term (German: Apophänie) was coined by psychiatrist Klaus Conrad in his 1958 publication on the beginning stages of schizophrenia. He defined it as "unmotivated seeing of connections [accompanied by] a specific feeling of abnormal meaningfulness". He described the early stages of delusional thought as self-referential, over-interpretations of actual sensory perceptions, as opposed to hallucinations.

Apophenia has come to imply a human propensity to seek patterns in random information, such as gambling.
https://en.wikipedia.org/wiki/Apophenia
.
Footnote: Having a word/name for something gives us a peg on which to hang the idea, so focuses it and gives us a convenient place to go back to that idea if/when we wish to refer to it again.

/RogerE :D

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Re: Patterns of Prediction

Post by RogerE »

Digit patterns in another power sequence

Following the recent discussion on powers of 2 and occurrence of digits, or of final digits or of the final pair of digits (in decimal notation), it is interesting to compare what happens in other cases. Let us take just one: the powers of 7.

Powers of 7
1
7
49
343
2401
16807
117649
823543
5764801
40353607
282475249
1977326743
13841287201
96889010407
678223072849
4747561509943
33232930569601
232630513987207
1628413597910449
11398895185373143
79792266297612001
...
After abutting these numbers, the resulting string of digits is
17493432401168071176498235435764801403536072824752491977326743138412872019688901040767822307284947475615099433323293056960123263051398720716284135979104491139889518537314379792266297612001...
.
The first occurrence of each decimal digit in this sequence:
f(0) = 10, f(1) = 1, f(2) = 8, f(3) = 5, f(4) = 3,
f(5) = 26, f(6) = 13, f(7) = 2, f(8) = 14, f(9) = 4.
The last digit to appear is 5, occurring as term 26 in the sequence, contributed by 7^7 = 823543.

Does every ordered pair of digit appears in the digit string? "Probably" is the best we can say without full demonstration or proof (a much longer initial part of the sequence would be required to find the 100 first occurrences which probably occur).
But we can note first occurrences of "double digits" (except for 55, which we expect will occur, but hasn't appeared within the powers 7^n with n ≤ 20):
f(11) = 11, f(77) = 55, f(88) = 76, f(22) = 88, f(99) = 106,
f(33) = 109, f(44) = 152, f(66) = 178, f(00) = 186
f(55) = ??
Spoiler: 55 is "just around the corner".

Final digits of powers of 7

The sequence of final digits of 7^n for n = 0, 1, 2, ... is
1, 7, 9, 3, 1, 7, 9, 3, 1, ...
This is a recurring sequence with period 4; the recurring block is 1793. Hence we can predict
The final digit of 7^n, with n ≥ 0, is
1 whenever n is a multiple of 4;
7 whenever n is 1 greater than a multiple of 4;
9 whenever n is 2 greater than a multiple of 4;
3 whenever n is 3 greater than a multiple of 4.
For example, 7^10 will end in 9. This is confirmed by 7^10 = 282475249.
/RogerE :D

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Re: Patterns of Prediction

Post by RogerE »

Directed graph mod 10 for multiplying by 7

Consider the process of taking a positive integer a, multiplying by 7, and then discarding all the decimal digits except the last. For example, if a = 8, then 8 —> [5]6, so the result is 6. Equally, if a = 18, then 18 —> [12]6, so the result is again 6.

What happens when you begin with any single decimal digit as a, and repeat the calculating multiple times?
For example, beginning with a = 8, we have
8 —> [5]6 —> [4]2 —> [1]4 —> [2]8 —> [5]6 —>...
This is a recurring sequence. The directed graph representing the "whole story" is

&quot;Multiplying by 7 modulo 10&quot;
"Multiplying by 7 modulo 10"
The cycle 1793 corresponds to the repetitive sequence of final digits of powers of 7 (in decimal notation) discussed in the previous post. Note that it contains only odd digits. The cycle 2486 results, for example, if you double the consecutive powers of 7 and retain only the final digits. The single digit cycles 5 and 0 result, for example, if you multiply the consecutive powers of 7 by 5 or by 10, and retain only the final digits.

Last two digits after multiplying by 7

The modulo 100 graph for repeatedly multiplying by 7 and discarding all but the last two digits (in decimal notation) must contain the period 4 repetitive sequence
01 —> 07 —> 49 —> [3]43 —> [3]01 —> 07 —> ...
That sequence is exhibited by the final two digits of the powers of 7. Note that it contains exactly one period of the single digit sequence.

The modulo 100 directed graph has 100 nodes. The cycle just described contains only four of the nodes. What does the rest of the graph look like?
Multiplying by 2 or by 3 gives us
02 —> 14 —> 98 —> [6]86 —> [6]02 —> 14 —> ...
03 —> 21 —> [1]47 —> [3]29 —> [2]03 —> 21 —> ...
so these are more directed cycles of size 4.
Multiplying by 5 or by 10 or by 50 gives us
05 —> 35 —> [2]45 —> [3]15 —> [1]05 —> 35 —> ...
10 —> 70 —> [4]90 —> [6]30 —> [2]10 —> 70 —> ...
50 —> [3]50 —> [3]50 —> ...
The first two correspond to directed cycles of size 4, while the last is a directed cycle of size 1.

As mathematician John Conway (1937–2020) would remark in circumstances like this, the rest of the details are left to "our most diligent reader". ;)

/RogerE :D

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Re: Patterns of Prediction

Post by RogerE »

More about powers of 7

The length of each consecutive power of 7 either stays constant or grows by 1. Can we predict where the length stays constant, and where it grows by 1?

Here is a discussion of that question:

Screen Shot 2021-05-16 at 10.47.30 pm.png
Screen Shot 2021-05-16 at 10.47.52 pm.png
Screen Shot 2021-05-16 at 10.48.05 pm.png
Screen Shot 2021-05-16 at 10.48.20 pm.png
Screen Shot 2021-05-16 at 11.32.45 pm.png
.
/RogerE :D

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Re: Patterns of Prediction

Post by OldDuffer1 »

Thank you for this fascinating thread. Interesting to note that Pythagoras was, of course, actually meaning the shape "square" and not the mathematical function "squared" as we now know it. I presume they didn't have that concept at the time?

I still remember that we had to prove Pythagoras and other theorems at school. At the end we wrote "QED" which, being cool of course, we said stood for "quite easily done"! :D

We also used log tables (long before calculators!). Again it was considered very "uncool" to use the anti-logs- you had to use the log tables backwards!

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Re: Patterns of Prediction

Post by RogerE »

Thanks OldDuffer1.

Classical Greek mathematicians frequently treated number as a geometrical concept, so square was the area of a geometric figure (a square) and cube was the volume of a geometric figure (a cube), etc. They certainly appreciated that 1, 4, 9, 16, 25, ... and 1, 8, 27, 64, 125, ... were pure numbers, but also appreciated the geometric concreteness of treating them as areas and volumes.

They also manipulated numbers (integers) as lengths, or equally, the area of rectangles of unit width.
For example, the aliquot parts or aliquot divisors of a number N are the smaller rectangles of unit width with the property that multiple copies build a rectangle of length equal to N. Thus, the following diagram shows that 6 has aliquot divisors 1, 2 and 3, and finally it shows that 6 = 1 + 2 + 3, that is, 6 is the sum of its aliquot divisors. In Greek terminology, this shows the 6 is a perfect number.

6 is the sum of its aliquot parts, so is perfect
6 is the sum of its aliquot parts, so is perfect
.
Now be impressed: in classical Greek mathematics it was known that perfect numbers are rare, and the first four perfect numbers were known to be 6, 28, 496 and 8128.

Euclid's algorithm for finding the largest common divisor of two numbers M and N is essentially a process of lining up a unit width rectangle of length equal to the larger number, say M, against a line of abutting multiple copies of the smaller rectangle, N, and representing the "shortfall" as a smaller rectangle, say N'. Now repeat, with one copy of N and multiple abutting copies of N', representing the "shortfall" as an even smaller rectangle, say N''. This process continues until there is no shortfall. Then the latest small rectangle is an aliquot part of every preceding rectangle, so is a common divisor of M and N. The process also shows that no longer unit width rectangle could be a common divisor. So, the final small rectangle is the largest common divisor.

/RogerE :D

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Re: Patterns of Prediction

Post by RogerE »

Even more about powers of 7

The digits that occur in powers of 7 have still been giving me things to think about. Here I will show that any given finite block of decimal digits does occur in some power of 7. For example, the block 12345 will appear in some power of 7, and so will 54321. In fact, I will show that any given finite block of decimal digits occurs in infinitely many different powers of 7.


Screen Shot 2021-05-18 at 2.21.45 am.png
Screen Shot 2021-05-18 at 2.22.18 am.png
Screen Shot 2021-05-18 at 2.22.32 am.png
Screen Shot 2021-05-18 at 2.22.44 am.png
.
/RogerE :D

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Re: Patterns of Prediction

Post by RogerE »

Powers of 3

Recent posts about powers of 2 and powers of 7 show the "typical" features of power sequences in decimal notation. If we were to take powers of a different integer, or change the base of the representation (to binary or ternary, for example) the same general behaviours would become apparent.

That said, it might be of interest to briefly look at one more case. I choose the powers of 3, since I just used them to enter a string of posts in the random counting thread, so that made me pay attention to the way the features are displayed by this case. Here's how the sequence begins:

1
3
9
27
81
243
729
2187
6561
19683
59049
177147
531441
1594323
4782969
14348907
43046721
129140163
387420489
1162261467
3486784401
10460353203
31381059609
94143178827
.....
The corresponding concatenated sequence of decimal digits is

139278124372921876561196835904917714753144115943234782969143489074304672112914016338742048911622614673486784401104603532033138105960994143178827...
.
Every decimal digit has already made an appearance, and 0 is the slowest to enter: f(0) = 29.
Notice that the 2-block 11 occurs quite often. Can you "explain" that phenomenon?
.
Final digit sequence

The last digit of each consecutive power of 3 gives us yet again a 4-cycle of odd digits:

1 —> 3 —> 9 —> 7 —> 1 —> ....
.
Notice that it is the reverse of the 4-cycle of final digits of powers of 7.
That is because 3 + 7 = 10, so 3 behaves like "-7" when we discard all but the final digit in the powers in decimal notation. (More formally: 3 = -7 modulo 10.)


Sequence of final two digits

The last two digits of each consecutive power of 3 give us a 20-cycle, containing 5 iterations of the 4-cycle of the final digits:

01 —> 03 —> 09 —> 27 —> 81 —> 43 —> 29 —> 87
—> 61 —> 83 —> 49 —> 47 —> 41 —> 23 —> 69 —> 07
—> 21 —> 63 —> 89 —> 67 —> 01 —> ....
.
Number of digits in the powers of 3

Let L(n) denote the length of the decimal representation in 3 to the power n (= number of digits in the decimal representation of 3^n). The sequence of lengths begins
L(0) = 1, L(1) = 1, L(2) = 1, L(3) = 2, L(4) = 2, L(5) = 3, L(6) = 3, L(7) = 4, L(8) = 4, L(9) = 5, L(10) = 5, ....
Because 1 < 3 < 9 < 10, there will be at least two powers of 3 between any two consecutive powers of 10. Then
0 ≤ L(n+1) – L(n) ≤ 1
.
Fact: L(n+3) = L(n) + 1 holds for every n ≥ 0.
Proof: If 10^m < 3^n < 10^(m+1) then
10^(m+1) < 27x10^m < 3^(n+3) < 27x10^(m+1) < 10^(m+2). [End of proof]

On the other hand, if a = 3^n is a power of 3 just slightly larger than 10^m, we can have three consecutive powers of 3 fitting between two consecutive powers of 10, in which case
L(n) = L(n+1) = L(n+2) = m+1
.
Suppose 10^m < a < 3a < 9a < 10^(m+1)
This implies 10^m < a < (10/9)x10^m.
With a = 3^n, taking logarithms (to base 10) leads to
0 < n log3 – m < 1 – 2 log3 = 0.04575749056...
We seek solutions (m, n) to this inequality.

The denominator sequence of the continued fraction for log3 = 0.477121254719... is
[0; 2, 10, 2, 2, 1, 13, 1, 7, 18, 2, ...]
so the best rational approximations (convergents) to log3 are
0/1, 1/2, 10/21, 21/44, 52/109, 73/153, 1001/2098, 1074/2251, 8519/17855, 154416/323641, ...
These convergents alternate below and above log3:
0/1 < 10/21 < 52/109 < 1001/2098 < ... < log3 < ... < 1074/2251 < 73/153 < 21/44 < 1/2
All of these are "useful", but the following calculations will only make use of 10/21 and 1/2.

• The first solution to 0 < n log3 – m < 0.04575749056... is
(m, n) = (21, 10): then 21 x log3 – 10 = 0.019546349...
This calculation predicts L(21) = L(22) = L(23) = 11.
The three consecutive powers of 3 are the last three given explicitly at the start of this post:
3^21 = 10460353203
3^22 = 31381059609
3^23 = 94143178827
• The next (second) solution to 0 < n log3 – m < 0.04575749056... is double the first:
(m, n) = (42, 20): then 42 x log3 – 20 = 0.039092698...
This calculation predicts L(42) = L(43) = L(44) = 21.
The three consecutive powers of 3 are:
3^42 = 109418989131512359209
3^43 = 328256967394537077627
3^44 = 984770902183611232881
• Tripling the first solution to 0 < n log3 – m < 0.04575749056... does not give a solution,
because the difference in the first solution is larger than one third of the upper bound.
However (m, n) = (2, 1) gives us 2 x log3 – 1 = -0.04575749...
Adding this to triple the first solution yields a difference that falls into the constrained range:
(m, n) = (63, 30)+(2, 1) = (65, 31): then 65 x log3 – 31 = 0.012881557...
This calculation predicts L(65) = L(66) = L(67) = 32.
The three consecutive powers of 3 are:
3^65 = 10301051460877537453973547267843
3^66 = 30903154382632612361920641803529
3^67 = 92709463147897837085761925410587
And so on.
Once again, the algorithmic method implicit in these calculations is sufficient to prove
Fact: There are infinitely many positive integers n such that L(n) = L(n+1) = L(n+2).

/RogerE :D

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Re: Patterns of Prediction

Post by OldDuffer1 »

Indeed! :)

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Re: Patterns of Prediction

Post by RogerE »

Powers of 3 — a correction ;)

My post at Fri May 21, 2021 03:06:10 am included:

Number of digits in the powers of 3

Let L(n) denote the length of the decimal representation in 3 to the power n (= number of digits in the decimal representation of 3^n). The sequence of lengths begins
L(0) = 1, L(1) = 1, L(2) = 1, L(3) = 2, L(4) = 2, L(5) = 3, L(6) = 3, L(7) = 4, L(8) = 4, L(9) = 5, L(10) = 5, ....
Because 1 < 3 < 9 < 10, there will be at least two powers of 3 between any two consecutive powers of 10. Then
0 ≤ L(n+1) – L(n) ≤ 1
.
So far, so good. However, the very last calculation step in the following "proof" is a simple mistake:
Fact: L(n+3) = L(n) + 1 holds for every n ≥ 0.
Proof: If 10^m < 3^n < 10^(m+1) then
10^(m+1) < 27x10^m < 3^(n+3) < 27x10^(m+1) < 10^(m+2). [End of proof]
That last step should be 27x10^(m+1) < 10^(m+3), and the correct conclusion should be
Fact: L(n) + 1 ≤ L(n+3) ≤ L(n) + 2 holds for every n ≥ 0.
.
The corrected statement is not so elegant, but it's true, and the earlier mistaken version was false!
For example, L(2) = 1 and L(5) = 3
3
9
27
81
243
show us that
• L(1) = 1 and L(4) = 2
• L(2) = 1 and L(5) = 3
Sometimes the light of day reveals things not noticed in the depths of night ;)

/RogerE :D

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Re: Patterns of Prediction

Post by RogerE »

There is a risk/danger involved in revealing one's predictions while they are
still operative — they might influence the behaviour of their subjects.
I wonder whether that played a role after I posted this prediction:

Screen Shot 2021-05-24 at 2.34.57 am.png
.
In fact, the Puffin has reached 41986 and has now retired to bed (we assume).
We confidently expect him to reach 42K posts today (24 May 2021), but we
wonder whether this might have happened yesterday if the above prediction
had not been published openly ;)

/RogerE :D

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Re: Patterns of Prediction

Post by Ubobo.R.O. »

NO. You just simply stuffed up in your calculations !

My prediction for today: The Philatelic Bus will arrive in the charming little villiage of The Summit.
Full time horse non-whisperer, post box searcher and lichen covered granite rock percher. Gee I'm handsome !
You gottem birds, butterflies, shells, maps, flags and heads on stamps ? Me wantem !

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Re: Patterns of Prediction

Post by RogerE »

Well, the witness for the defence has spoken. (*)
The 42K post has now been passed by Mr Puffin, and celebrated in the Posting Post. :D
Shall we enjoy looking at some properties of 42? Here we go:

Perfect, deficient, and abundant numbers

A recent post in this thread looked at classical Greek mathematics, noting that lengths of unit-width rectangles were visually linked to divisors/factors and perfect numbers.

Visual/geometrical demonstration that<br />1, 2, and 3 are aliquot parts/divisors of 6<br />and 6 is perfect because it is equal to the sum <br />of its aliquot parts/divisors: 1 + 2 + 3 = 6
Visual/geometrical demonstration that
1, 2, and 3 are aliquot parts/divisors of 6
and 6 is perfect because it is equal to the sum
of its aliquot parts/divisors: 1 + 2 + 3 = 6


From that geometrical viewpoint, a divisor/factor of a given number is a small rectangle with the property that aligning multiple copies of it will match the length of the long rectangle representing the given number.
The aliquot parts/aliquot divisors of a given number are its divisors which are smaller than the number itself. A perfect number is such that one copy of each of the aliquot divisor rectangles aligns to match the length of the long rectangle representing the given number.

For instance, the aliquot parts/divisors of 10 are 1, 2, and 5: multiple copies of identical small rectangles of any one of those lengths align to match the length of a large rectangle of length 10:
10 = 1+1+1+1+1+1+1+1+1+1 = 2+2+2+2+2 = 5+5
If we align one of each aliquot divisor, the total length falls short of 10:
1 + 2 + 5 = 8 < 10
In classical Greek mathematics, this classifies 10 as a deficient number — its aliquot parts/divisors are too few to add up to the size/length of the number 10 itself.

On the other hand, the aliquot divisors of 12 are 1, 2, 3, 4, and 6. When rectangles of these lengths are aligned, the total length exceeds 12:
1 + 2 + 3 + 4 + 6 = 16 > 12
In classical Greek mathematics, this classifies 12 as an abundant number — its aliquot parts/divisors are more than sufficient to add up to the size/length of the number 12 itself. (In this case, if we dispensed with the divisor 4, the other aliquot parts/divisors happen to align so that they exactly match the length 12 — an unexpected bonus!)

Now we come to 42. The aliquot divisors of 42 are 1, 2, 3, 6, 7, 14, 21. Their sum is
1 + 2 + 3 + 6 + 7 + 14 + 21 = 54 > 42
Therefore 42 is an abundant number. In fact, the aliquot divisors 7, 14, and 21 already add up to 42 (another unexpected bonus‽)

How are the abundant numbers distributed?

The discussion so far readily leads us to wonder how common abundant numbers might be, and more specifically, to wonder how they are distributed within the sequence of whole numbers = positive integers.

Prime numbers, such as 2, 3, 5, 7, 11, 13, ... have only one aliquot divisor: 1. Therefore, all prime numbers are deficient.

Squares of prime numbers have only two aliquot divisors: the sum of those aliquot divisors is less than the square, so they are also deficient. For example
1 + 2 = 3 < 2^2 = 4; 1 + 3 = 4 < 3^2 = 9; 1 + 5 = 6 < 5^2 = 25; ...
In fact 1 + p = (p^2 – 1)/(p – 1) < p^2 for any prime p.
.
Cubes of prime numbers have only three aliquot divisors, and their sum is less than the cube, so they are also deficient. For example
1 + 2 + 4 = 7 < 2^3 = 8; 1 + 3 + 9 = 13 < 3^3 = 27; 1 + 5 + 25 = 31 < 5^3 = 125; ...
In fact 1 + p + p^2 = (p^3 – 1)/(p – 1) < p^3 for any prime p
You can see the pattern emerging here, and it will allow you to correctly predict that any power of a prime number is a deficient number.

That tells us: Every abundant number has at least two different prime factors. In slightly different words: The aliquot divisors of an abundant number must include at least two different prime numbers.

So, amongst the earliest positive integers the candidates for being abundant are
6, 10, 12, 15, 18, 20, 21, 22, 24, 26, 28, 30, ...
We already know that 6 is perfect (so not abundant), 10 is deficient, and 12 is abundant. Therefore 12 is the smallest abundant number. You can easily check that 15, 20, 21, 22, and 26 are deficient. More pleasantly, you can check that 18, 24, and 30 are abundant, and 28 is perfect (the second perfect number!). If we use s(n) to denote the sum of aliquot divisors of n, then:
s(12) = 16, s(18) = 21, s(24) = 36, s(28) = 28, s(30) = 42
.
Do you see any patterns emerging, that will allow you to predict which numbers are abundant?
.
As a related, intriguing possibility, notice that the abundant numbers 12, 18, 24, 30, 42 all have the property that a subset of their aliquot divisors has sum exactly equal to the number itself! For example, 5 + 10 + 15 = 30. Let us say that a number with this property is subperfect, to suggest that it has "a proper subset of aliquot divisors with sum perfectly equal to the number itself". Clearly, every subperfect number is abundant. Is every abundant number subperfect?

/RogerE :D

(*) Footnote: defence or defense? Check out https://tinyurl.com/2bfdxazb

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Re: Patterns of Prediction

Post by RogerE »

Reviewing that prediction about Puffin reaching 42K posts.

plLcmLa.jpg


I made my prediction on Sun 24 May in the post

Image
.
A count back shows that at 10:31:06 on Sun 23 May Puffin's total number of posts stood at 41971.

Despite Puffin's thought that I had calculated for that prediction, in fact it was a pure guess, based only on my impression of how many posts Puffin adds each day.

Since then Puffin has proclaimed complete innocence, indicating that he was totally uninfluenced by my prediction (probably he didn't even notice it on Sunday). The evidence is entirely in his favour.

Puffin's earliest post on Sun 23 May was at 00:06:28, his post number 41933.
At the time I made my prediction he had already made 39 posts on Sun 23 May.
By the time the day was over, Puffin had added post number 41984, at 22:35:19. That amounts to a total of 53 posts for the day, well above his "lifetime" daily average, which is currently at 33.80.

If I had made an actual calculation, rather than an impressionistic guess, I might have calculated
something like this:
At "close of business" (midnight) on Sat 22 May Puffin had made 41932 posts.
His long term average is almost 34 posts per day.
He needs 42000 - 41932 = 68 posts to reach 42000.
At his long term average rate of posting, that's exactly 2 days' worth.
Therefore, assuming he keeps to his long term average, we expect him to reach 42000 posts at around midnight on Mon 24 May.


In practice:
Puffin added exactly 34 posts on Sat 22 May (as well as attending to other business between 10:30am and 5:30pm), but on Sun 23 May he was very prolific on Stampboards, and added 53 posts for the day, reaching 41985 posts at "close of business" (midnight). The next 15 posts were added today, Mon 24 May, by 10:11:45am, when post number 42000 = 42K was added. By noon today he had already added post number 42007.

Improving such predictions:
Any lessons to be learnt? Probably that long term average number of posts per day is likely to underestimate current rate of posting per day, because accumulated experience as a poster probably results in a gradual increase in the number of posts per day by the individual. Also, an individual's daily routine depends on the day of the week, so seven average daily posting numbers would give finer tuning than an average over all days: the average number of posts on Sun, on Mon, on Tues, ..., on Sat, would give us seven more sensitive numbers to work with. Hence, if we knew Puffin's Sunday average, and Monday average, those two would be expected to yield a better prediction. On the other hand, it should always be kept in mind that averages are better predictors when more instances of the "sample set" are involved. In other words, predicting results after many days using known past averages is likely to produce more reliable results than predicting the outcome after just a few days.

/RogerE :D

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Re: Patterns of Prediction

Post by RogerE »

More on perfect, deficient, and abundant numbers

In my post of Mon May 24, 2021 14:14:27 pm I wrote:
"You can easily check that 15, 20, 21, 22, and 26 are deficient."
If a diligent reader follows this post, this claim would be checked.
Let s(n) denote the sum of aliquot divisors of n. Then
s(15) = 9, s(20) = 22, s(21) = 11, s(22) = 13, s(26) = 16
so one of my claims was mistaken! In fact 20 is abundant, not deficient.

When is n = pq deficient?
Let p and q be distinct primes, such that p < q. If n = pq, then
s(n) = s(pq) = 1 + p + q ≤ q + q = 2q ≤ pq = n
The step using 1 + p ≤ q is an equality precisely when p and q are primes differing by 1, so p = 2, q = 3.
In that case, n = 6 and s(6) = s(2x3) = 1 + 2 + 3 = 3 + 3 = 2x3 = 6, reminding us that 6 is perfect.
If q > 3 then p and q differ by at least 2, so 1 + p < q, and then
s(n) = s(pq) = 1 + p + q < q + q = 2q ≤ pq = n
In short, s(n) < n.
Fact: If p, q are distinct primes, p < q and q > 3, then n = pq is deficient.

It follows that 10, 14, 15, 21, 22, 26, 33, 34 35, 38, ... are all deficient.


When is n = 4q deficient?
Let q ≥ 7 be prime. If n = 4q, then
s(n) = s(4q) = 1 + 2 + 4 + q + 2q = 7 + 3q ≤ 4q = n
The step using 7 ≤ q is an equality precisely when q = 7.
In that case, n = 28 and s(28) = s(4x7) = 1 + 2 + 4 + 7 + 14 = 7 + 3x7 = 4x7 = 28,
reminding us that 28 is perfect.
If q > 7 then 1 + 2 + 4 < q, so
s(n) = s(4q) = 7 + 3q < q + 3q = 4q = n
In short, s(n) < n.
Fact: If q > 7 is prime, then n = 4q is deficient.
Note that q = 3, 5, 7 produce instances of n = 4q which are not deficient:
q = 3: s(12) = s(4x3) = 1 + 2 + 4 + 3 + 6 > 3 + 3x3 = 4x3 = 12, so s(12) > 12.
q = 5: s(20) = s(4x5) = 1 + 2 + 4 + 5 + 10 > 5 + 3x5 = 4x5 = 20, so s(20) > 20.
q = 7: s(28) = s(4x7) = 1 + 2 + 4 + 7 + 14 = 7 + 3x7 = 4x7 = 28, so s(28) = 28.


When is n = 9q deficient?
Let q > 3 be prime. If n = 9q, then
s(n) = s(9q) = 1 + 3 + 9 + q + 3q = 13 + 4q < 9q = n
The step using 13 < 5q follows directly from q > 3.
In short, s(n) < n.
Fact: If q > 3 is prime, then n = 9q is deficient.
Note that q = 2 produces an instance of n = 9q which is not deficient:
q = 2: s(18) = s(9x2) = 1 + 3 + 9 + 2 + 6 = 13 + 4x2 > 9x2 = 18, so s(18) > 18.


If a = p^2, when is n = aq deficient?
Let p, q be distinct primes. If a = p^2 and n = aq, then
s(n) = s(aq) = 1 + p + p^2 + q + pq = (1 + p + p^2) + (p + 1)q.
Then s(aq) < aq precisely when (A): 1 + p + p^2 < (p^2 – p – 1)q.
Condition (A) is equivalent to (p + 1)(q + 1) < (q – 1)p^2,
which is equivalent to
(A') 1 + 1/p < p[1 – 2/(q+1)].
p = 2: (A') requires 3/4 < 1 – 2/(q+1), so q ≥ 11.
p = 3: (A') requires 4/9 < 1 – 2/(q+1), so q ≥ 3.
p ≥ 5: (A') requires (1 + 1/p)/p ≤ 6/25 < 1 – 2/(q+1), so q ≥ 2.
Earlier discussions already covered the cases p = 2 (a = 4) and p = 3 (a = 9).
When p ≥ 5, the above argument shows that (A) holds for all primes q distinct from p.
Hence, summarising all cases:
Fact: If p and q are distinct primes and a = p^2, then n = aq is deficient
when (1) p = 2, q ≥ 11; (2) p = 3, q ≥ 5; (3) p ≥ 5, q ≥ 2.


/RogerE :D

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Re: Patterns of Prediction

Post by RogerE »

More on abundant numbers

Multiples of 6

Let 6a be any multiple of 6.
Clearly a, 2a, 3a are aliquot divisors of 6a.
Their sum is 6a.
Hence s(6a) ≥ a + 2a + 3a = 6a,
so if there are any other aliquot divisors of 6a then 6a is abundant.
If a > 1, then 1 is a further aliquot divisor of 6a, so s(6a) ≥ 6a +1 > 6a when a > 1.
If a = 1, there are no other aliquot divisors, so in this case s(6a) = s(6) = 6.
Hence
Fact: If a is any positive integer, then 6a is perfect if a = 1 and is abundant if a > 1.
So, from 12 onwards, every sixth number (at least) is abundant.

Multiples of 28

The same idea will work nicely for multiples of 28:
Let 28a be any multiple of 28.
Clearly a, 2a, 4a, 7a, 14a are aliquot divisors of 28a.
Their sum is 28a.
Hence s(28a) ≥ 28a,
so if there are any other aliquot divisors of 28a then 28a is abundant.
If a > 1, then 1 is a further aliquot divisor of 28a, so s(28a) ≥ 28a +1 > 28a when a > 1.
If a = 1, there are no other aliquot divisors, so in this case s(28a) = s(28) = 28.
Hence
Fact: If a is any positive integer, then 28a is perfect if a = 1 and is abundant if a > 1.

Multiples of 20

We have also noted that s(20) = 1 + 2 + 4 + 5 + 10 = 22 > 20, so 20 is abundant.
Fact: 20 is the smallest abundant number that is not a multiple of 6.

Let 20a be any multiple of 20.
Clearly a, 2a, 4a, 5a, 10a are aliquot divisors of 20a, and their sum is 22a so s(20a) > 20a.
Hence
Fact: If a is any positive integer, then 20a is abundant.
By the same reasoning,
Fact: Every positive multiple of an abundant number is abundant.


Other multiples of 10

s(50) = 1 + 2 + 5 + 10 + 25 = 43 < 50, so 50 is deficient.
s(70) = 1 + 2 + 5 +10 + 7 + 14 + 35 = 74 > 70, so 70 is abundant.
Hence
Fact: If a is any positive integer, then 70a is abundant.

Challenge question

So far each family of abundant numbers we have found contains only even numbers.
Q: Are there any odd numbers which are abundant?

/RogerE :D

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Re: Patterns of Prediction

Post by RogerE »

Odd abundant numbers

Time to answer my latest challenge question. (Some day a diligent reader will come along and be happy to find the follow up discussion).

Products of odd primes

Systematic examination of the product of the first few odd primes gives a pattern of growth is s(n)/n:
n = 3: s(n) = 1 < 3, so 3 is deficient, and 1/3 = 0.33333...
n = 3x5 = 15: s(n) = 9 < 15, so 15 is deficient, and 9/15 = 0.60000...
n = 3x5x7 = 105: s(n) = 87 < 105, so 105 is deficient, and 87/105 = 0.82857...
n = 3x5x7x11 = 1155: s(n) = 1149 < 1155, so 1155 is deficient, and 1149/1155 = 0.99480...
n = 3x5x7x11x13 = 15015: s(n) = 17241 > 15015, so 15015 is abundant, and 17241/15015 = 1.14825...
n = 3x5x7x11x13x17 = 255255: s(n) = 325353 > 255255, so 255255 is abundant, and 315353/255255 = 1.27461...
n = 3x5x7x11x13x17x19 = 4849845: s(n) = 6762315 > 4849845, so 4849845 is abundant, and 6762315/4849845 = 1.39436...

This establishes:
Fact: Every odd multiple of 15015 is an odd abundant number.

[In fact Euler proved that the sum of reciprocals of primes is divergent, which here implies that if n is the product of the first m odd primes, then s(n)/n grows larger than any prescribed number as m grows, and is always greater than 1 when m ≥ 5.]

Tweaking the calculations here, we can find odd abundant numbers smaller than 15015:
n = 3x3x5x7x11 = 3465: s(n) = 4023 > 3465, so 3465 is abundant, and 4023/3465 = 1.16103...
n = 3x3x3x5x7 = 945: s(n) = 975 > 945 , so 945 is abundant, and 975/945 = 1.03174...

Note that 975 – 945 = 30 = 3 + 27, so when n = 945, the difference between s(n) and n is equal to the sum of two of the aliquot divisors of n. In total, n = 945 has 15 aliquot divisors, so the other 13 divisors must have sum equal to n:
1 + 5 + 7 + 9 + 15 + 21 + 35 + 45 + 63 + 105 + 135 + 189 + 315 = 945
If k is any positive integer, multiplying all 13 of those aliquot divisors by k yields a sum equal to 945k, so 945k is equal to the sum of 13 of its aliquot divisors.

This establishes:
Fact: Every multiple of 945 is abundant, and each one is subperfect..

Probably 945 is the smallest odd abundant number, though the argument above has not completely demonstrated that it is the smallest.

Footnote: the Online Encyclopedia of Integer Sequences = OEIS is full of all kinds of sequences to gladden the heart of the number enthusiast. It turns out that the sequence A005231 lists the odd abundant numbers, and that sequence begins with 945, confirming the above conjecture that 945 is the smallest. :D
https://oeis.org/A005231
.


/RogerE :D

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Re: Patterns of Prediction

Post by RogerE »

Absolute and relative abundance of abundant numbers

If n is an abundant number, the sum of its aliquot divisors is greater than n, that is, s(n) > n.
In that case, let us compare n and s(n).

Define the absolute abundance of n as
A(n) = s(n) – n
and the relative abundance of n as
R(n) = s(n)/n
.

For example:
s(12) = 16, so A(12) = 4 and R(12) = 16/12 = 4/3 = 1.33333...
s(18) = 21, so A(18) = 3 and R(18) = 21/18 = 7/6 = 1.16666...
s(20) = 22, so A(20) = 2 and R(20) = 22/20 = 11/10 = 1.10000...
s(24) = 36, so A(24) = 12 and R(24) = 36/24 = 3/2 = 1.50000...
s(30) = 42, so A(30) = 12 and R(30) = 42/30 = 7/5 = 1.40000...

Growth of abundance with additional prime factors

Suppose n is abundant, and p is a prime which is not a factor of n.
The aliquot divisors of pn are all the aliquot divisors of n, as well as p times each of those
aliquot divisors, and also n itself. Then
s(pn) = s(n) + ps(n) + n = (p + 1)s(n) + n
so A(pn) and R(pn) satisfy:
A(pn) = s(pn) – pn = p[s(n) – n] + s(n) + n = pA(n) + s(n) + n = (p + 1)A(n) + 2n;
R(pn) = s(pn)/pn = (1 + 1/p)s(n)/n + 1/p = (1 + 1/p)R(n) + 1/p.
.
For example, let n = 20 and p = 3. Then
s(60) = (1 + 2 + 4 + 5 + 10) + (3 + 6 + 12 + 15 + 30) + 20 = 4x22 + 20 = 4xs(20) + 20 = 108.
A(60) = 48 = 3x2 + s(20) + 20 = 4A(20) + 2x20;
R(60) = 108/60 = 9/5 = 1.80000... = (4/3)x(11/10) + 1/3 = (1 + 1/3)R(20) + 1/3.

These results show that when n is multiplied by a new prime p, both the absolute abundance
and the relative abundance increase. Quantifying:

Fact: If n is abundant and p is a prime that is not a factor of n, then
A(pn) = (p + 1)A(n) + 2n;
R(pn) = (1 + 1/p)R(n) + 1/p.

.
Abundant numbers with small absolute or relative abundance
.
For any given abundant number, the last fact shows it is easy to find abundant numbers with larger absolute abundance or relative abundance. So let's consider the more interesting challenge of finding numbers with very small absolute abundance or relative abundance.

Absolute abundance
We have seen that A(12) = 4, A(18) = 3, A(20) = 2. Also A(70) = 4 occurred earlier.
Some questions of interest:
• Is there an abundant number n that satisfies A(n) = 1 ?
• Are there infinitely many abundant numbers n with A(n) ≤ 4 ?
• Are there abundant numbers n that satisfy A(n) = n ?
• This last question is equivalent to asking for n such that R(n) = 2.

Relative abundance
We have seen that
R(36) = 1.50000...,
R(30) = 1.40000...,
R(12) = 1.33333...,
R(18) = 1.16666...,
R(20) = 1.10000...
R(70) = 1.05714...
• Given any positive integer k, is there an abundant number n such that 1 < R(n) < 1 + 1/10^k ?
• We have just seen that n = 70 is a solution when k = 1. Can we find a solution when k = 2 ?

Notice that if n is abundant, and the prime p is an aliquot divisor of n, but the prime q > p is not an aliquot divisor of n, then nq/p is an integer which need not be abundant, but its aliquot divisors are those of n/p, together with q times each of those divisors, and n/p itself. For example, if n = 12, p = 2 and q = 5, then the aliquot divisors of n/p = 6 are 1, 2, 3; q times those gives us 5, 10, 15; and n/p = 6. I claim that the seven numbers just listed are the aliquot divisors of nq/p = 12 x 5/2 = 30. Then
s(30) = s(6) + 5 x s(6) + 6
In general,
s(nq/p) = s(n/p) + q x s(n/p) + n/p,
so R(nq/p) = (1/q)s(n/p)/(n/p) + s(n/p)/(n/p) + 1/q = (1 + 1/q)R(n/p) + 1/q.
Notice that in fact n/p need not be abundant for this calculation to hold!
Pursuing the illustrative example, this gives us
R(30) = (1 + 1/5)R(6) + 1/5 = 7/5 = 1.4, since R(6) = 1.

Fact: If n is a multiple of the prime p, but not of the prime q, then
R(qn/p) = (1 + 1/q)R(n/p) + 1/q = R(n/p) + (R(n/p) + 1)/q
If n is abundant, nevertheless n/p might be deficient or perfect, and a suitable q might manage
to produce a relative abundance very close to 1.

For instance, take n = 58, p = 2, q yet to be decided. Then
R(n/p) = R(28) = 1, so R(qn/p) = R(28q) = 1 + 2/q.
If we choose q so that 2/q < 1/100 we expect to find a solution to the k = 2 challenge.
With q > 200, let's choose q = 211, the first prime greater than 200. Then
qn/p = 211 x 28 = 5908, and
s(5908) = (1 + 2 + 4 + 7 + 14) + 211 x (1 + 2 + 4 + 7 + 14) + 28 = 213 x 28 = 5964.
Then R(5908) = 213/211 = 1.0094786... < 1.01 = 1 + 1/10^2.
(Incidentally, this calculation shows that A(5908) = 2q = 2 x 28 = 56.)
Clearly, for any given k if we choose q to be any prime greater than 2 x 10^k, then
1 < s(28q) < 1 + 1/10^k.
.
Fact: There are abundant numbers with relative abundance arbitrarily close to 1.

/RogerE :D

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Re: Patterns of Prediction

Post by RogerE »

Prime numbers

The prime numbers ar the building blocks for the integers, under multiplication.
The sequence begins 2, 3, 5, 7, 11, 13, 17, 19, 23, ...
There are exactly 25 primes below 100.

The sequence of primes is notoriously "unpredictable" in detail (the fine scale), but is
remarkably regular on the very large scale, exhibiting a logarithmic rate of growth.

If we concatenate the primes to form an unending sequence of digits, the first 64 digits
of the sequence are
.
D = 2357111317192329313741434753596167717379838997101103107109113127...
.
Does every decimal digit occur in D?

Yes! First occurrences of the decimal digits in D are as follows:
f(2) = 1, f(3) = 2, f(5) = 3, f(7) = 4, f(1) = 5, f(9) = 12, f(4) = 21, f(6) = 31, f(8) = 41, f(0) = 48

Sorting, these first occurrences in D are
f(0) = 48, f(1) = 5, f(2) = 1, f(3) = 2, f(4) = 21, f(5) = 3, f(6) = 31, f(7) = 4, f(8) = 41, f(9) = 12.

Does every ordered pair of decimal digit occur in D?

The first occurrences begin:
f(23) = 1, f(35) = 2, f(57) = 3, f(71) = 4, f(11) = 5, f(13) = 7, f(31) = 8, f(17) = 9, f(19) = 11,
f(92) = 12, f(32) = 14, f(29) = 16, f(93) = 17, f(37) = 19, f(74) = 20, f(41) = 21, f(14) = 22,
f(43) = 23, f(34) = 24, f(47) = 25, f(75) = 26, f(53) = 27, f(59) = 29, f(96) = 30, f(61) = 31,
f(16) = 32, f(67) = 33, f(77) = 34, f(73) = 37, f(79) = 39, f(98) = 40, f(83) = 41, f(38) = 42,
f(89) = 43, f(99) = 44, f(97) = 45, f(10) = 47, f(01) = 48, f(03) = 51, f(07) = 54, f(09) = 57,
f(91) = 58, f(12) = 2, f(27) = 63, ...

Sorting, these first occurrences in D are

f(01) = 48, f(03) = 51, f(07) = 54, f(09) = 57,
f(10) = 47, f(11) = 5, f(12) = 2, f(13) = 7, f(14) = 22, f(16) = 32, f(17) = 9, f(19) = 11,
f(23) = 1, f(27) = 63, f(29) = 16,
f(31) = 8, f(32) = 14, f(34) = 24, f(35) = 2, f(37) = 19, f(38) = 42,
f(41) = 21, f(43) = 23, f(47) = 25,
f(53) = 27, f(57) = 3, f(59) = 29,
f(61) = 31, f(67) = 33,
f(71) = 4, f(73) = 37, f(74) = 20, f(75) = 26, f(77) = 34, f(79) = 39,
f(83) = 41, f(89) = 43,
f(91) = 58, f(92) = 12, f(93) = 17, f(96) = 30, f(97) = 45, f(98) = 40, f(99) = 44, ...
.
There are still numerous ordered pairs of digits which have not yet made their appearance,
including 00, 22, 33, 44, 55, 66 and 88.
Shall we optimistically predict that all remaining ordered pairs of digits will occur sooner or
later in the prime digit sequence D ?

/RogerE :D

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Re: Patterns of Prediction

Post by RogerE »

Prime numbers cont.

Mathematicians who have contributed to our understanding of prime numbers have
been honoured philatelically. Probably the earliest mathematician to leave us a
theorem about prime numbers is Euclid, who proved
Fact: There are infinitely many prime numbers.
This is equivalent to the fact that the list of prime digits D never comes to an end:
D = 2357111317192329313741434753596167717379838997101103107109113127...

Euclid
.
Ukraine, 2019: Euclid
Ukraine, 2019: Euclid
.
Maldives: Euclid
Maldives: Euclid
.
The logarithmic behaviour of the "large scale" distribution of the prime numbers was
established by Jacques Hadamard and by Charles Jean de la Valée Poussin in 1896:

Hadamard
.
Ukraine, 2017: Hadamard
Ukraine, 2017: Hadamard
.
Distribution of the primes: the Prime Number Theorem
.
Screen Shot 2021-06-01 at 11.35.30 am.png
.
https://en.wikipedia.org/wiki/Prime_number_theorem
.
/RogerE :D

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Re: Patterns of Prediction

Post by Malaya »

Brit-Col wrote:
13 Jan 2021 06:25

To make the discussion more interesting to this chat board apply it to stamps. Pareto principle: 80% of stamps are virtually worthless, 20% (being generous) hold 99% of the total value. So predict where the money will flow.
Here's a plot of the catalogue values (pooled data from four different catalogues) of the post-war Malaya coconut definitives against the quantity printed and the denomination, and for both explanatory variables, the response variable (catalogue price) has a few very high values and many low values.
3D scatter plot of relationship of catalogue values of post-war Malaya coconut definitive small heads issues to denomination and print quantity. The wireframe is a fitted generalized additive model to show the overall trend.
3D scatter plot of relationship of catalogue values of post-war Malaya coconut definitive small heads issues to denomination and print quantity. The wireframe is a fitted generalized additive model to show the overall trend.
The catalogue values are a "mint-used aggregate price" calculated by taking the arithmetic mean of the mint and used values for each listing, weighted by the inverse of the prices. This gives an overall idea of the value of the particular listing that takes into account both mint and used states.

Here are plots showing how the catalogue value and the "value for money" (calculated by dividing the mint-used aggregate price of a particular stamp by its print quantity) of the stamps are related to the quantities printed:
scatter plots of relationship of catalogue value and value for money to print quantity of the post-war Malaya coconut definitive small heads issues.
scatter plots of relationship of catalogue value and value for money to print quantity of the post-war Malaya coconut definitive small heads issues.

Even with log-log axes, the relationship is still bent, showing that it is even more powerful than a power law.

More details of how the data was collected and analysed can be found here.

More power laws/long-tailed distributions (detailed analysis here):
Abundance ranking of post-war Malaya coconut definitive issues based on quantities printed. Exponent of fitted power law = 1.78.
Abundance ranking of post-war Malaya coconut definitive issues based on quantities printed. Exponent of fitted power law = 1.78.

For each denomination of post-war Malaya coconut definitives, there were a few large orders and many small orders for stamps to be printed.
For each denomination of post-war Malaya coconut definitives, there were a few large orders and many small orders for stamps to be printed.

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Re: Patterns of Prediction

Post by Malaya »

RogerE wrote:
30 Jan 2021 14:18
Pattern Recognition

Pattern recognition is a topic relevant to this thread.
In fact, it is two topics, both relevant to this thread!

The Wikipedia article entitled Pattern recognition begins with the introductory note
This article is about pattern recognition as a branch of engineering. For the cognitive process, see Pattern recognition (psychology). For other uses, see Pattern recognition (disambiguation).
https://en.wikipedia.org/wiki/Pattern_recognition
.
So, here I want to discuss Pattern recognition as a branch of engineering/computer science. The Wikipedia link just cited is the source of the quotations included here.

What is Automated Pattern Recognition?
Wikipedia wrote:
Pattern recognition is the automated recognition of patterns and regularities in data.

It has applications in statistical data analysis, signal processing, image analysis, information retrieval, bioinformatics, data compression, computer graphics and machine learning.

Pattern recognition has its origins in statistics and engineering. Some modern approaches to pattern recognition include the use of machine learning, due to the increased availability of big data and a new abundance of processing power. However, these activities can be viewed as two facets of the same field of application, and together they have undergone substantial development over the past few decades.

A modern definition of pattern recognition is:

The field of pattern recognition is concerned with the automatic discovery of regularities in data through the use of computer algorithms and with the use of these regularities to take actions such as classifying the data into different categories. [Bishop, Christopher M. (2006). Pattern Recognition and Machine Learning. Springer]
.
Automated pattern recognition can be a powerful tool for analytical philately as well. For example, computer vision can be used to automatically detect and measure the shape, size and location of perfin holes:
Using computer vision to automatically detect perfin holes in Malayan perfins. The size, area and positions etc. of the holes can be measured accurately given high-resolution undistorted images.
Using computer vision to automatically detect perfin holes in Malayan perfins. The size, area and positions etc. of the holes can be measured accurately given high-resolution undistorted images.

The computer can do all kinds of measurements automatically, accurately and on large numbers of stamps in a very short time, which is impossible to do by eye.

Spatial analysis of Malayan perfin holes using average nearest neighbour statistics. This is a measure of how readable the perfin is.
Spatial analysis of Malayan perfin holes using average nearest neighbour statistics. This is a measure of how readable the perfin is.

Computerised alignment and quantification of HSBC perfin using principal component analysis and singular value decomposition. This helps to assess die differences or authenticity.
Computerised alignment and quantification of HSBC perfin using principal component analysis and singular value decomposition. This helps to assess die differences or authenticity.

Full details of the analysis can be found here, and the open-source computer code can be downloaded here.

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Re: Patterns of Prediction

Post by RogerE »

Prime numbers cont. (2)

The Prime Number Theorem, as proved by Hadamard and by de la Vallée Poussin
implies that P(n), the nth prime number, is about n.log(n), where log(n) is the
natural logarithm of n.

More precisely, the absolute size of the fraction
[P(n) – n.log(n)]/n
diminishes as n increases, and for any small fraction 1/m (e.g., 1/1000) there is a
corresponding bound b(m), depending on m, such that
-1/m < [P(n) – n.log(n)]/n < 1/m for every n ≥ b(m)
Admittedly, that is a more involved statement than: "P(n) is approximately n.log(n)".
But the more involved statement is more precise, because it introduces a measure of
how good the approximation is. What is still missing in that statement is an explicit,
formula for b(m).

A wonderful, permanently valuable paper in the field of computational mathematics,
dealing with this topic and many related matters, was published in 1962:
Rosser, J. Barkley; Schoenfeld, Lowell. Approximate formulas for some functions
of prime numbers
. Illinois J. Math. 6 (1962), no. 1, 64–94.
.
Screen Shot 2021-06-01 at 12.47.12 pm.png
.
Some key snapshots from page 69:
.
Screen Shot 2021-06-01 at 12.15.59 pm.png
Screen Shot 2021-06-01 at 12.46.16 pm.png
.
Note that the expressions are fully explicit, including the lower bounds, after which
the inequalities are guaranteed to hold.
.
Applying Rosser and Schoenfeld's Theorem 3

Let's look at some computations to see how the R&S Theorem 3 works :D

• A computational fact is that 9973 is the largest prime below 10,000, and it is prime number 1229.
If we substitute n = 1229 into R&S Theorem 3, we get: 9310.9 < P(1229) < 10539.9
That is an interval of 1229 numbers, guaranteed to contain the prime P(1229). The
midpoint of that interval is 9925.4, which is 48 less than the actual prime 9973.
The size of the interval guaranteed to contain P(1229) is less than 12.4% of the actual prime 9973
and the error of the midpoint is less than 0.5% of the actual prime.

• As a "bigger" example:
a computational fact is that 15,485,863 is the 1,000,000th prime number.
If we substitute n = 1,000,000 into R&S Theorem 3, we get:
14941392.4 < P(1,000,000) < 15941302.4
The midpoint of the interval is 15,441,302 which is 44560 less than the actual prime 15,485,863.
The size of the interval guaranteed to contain P(1,000,000) is less than 6.5% of the actual prime 15,485,863
and the error of the midpoint is less than 0.3% of the actual prime.

• As a "far bigger" example:
If we substitute n = 10^1000000 into R&S Theorem 3, we get:
2302598.2 x 10^1000000 < P(10^1000000) < 2302599.3 x 10^1000000
It follows that the prime P(10^1000000) has 1000007 digits, and its six leading digits are 230259.

This sequence of examples shows how to "sneak up" on prescribing an initial block of digits of a prime.

Prescribing an initial block of digits of a prime

Let us prove:
Fact: The digit block 8888888 occurs infinitely often in the prime digit sequence D
We apply R&S Theorem 3 to show that there is a prime with 8888888 as its initial block of digits.
The natural logarithm of 10 is 2.302585092994...
Then the natural logarithm of n = 10^k is k x 2.302585092994...
By choosing k = 386039494, it follows that the natural logarithm of n = 10^k is b = 888888784.19...
The natural logarithm of b is 20.605..., so b + ln(b) - 1.5 = 888888803.29...
Then R&S Theorem 3 shows P(n) falls in the interval
n.[b + ln(b) - 1.5] < P(n) < n.[b + ln(b) - 0.5]
which implies
888888803.29... x 10^386039494 < P(n) < 888888804.29... x 10^386039494
Therefore the prime P(n) = P(10^386039494) has 9 + 386039494 = 386039503 digits, and its leading
eight digits are 88888880. (We are not sure whether the ninth digit is 3 or 4.)

Notice that choosing k' = 386039536 and n' = 10^k' yields b' = ln(n') = 888888880.89...
Let n* = 10^(k'.10^m), for a positive integer m.
Then the natural logarithm of n* is b* = b' x 10^m where b' = 888888880.89..., so ln(b*) = m.ln(10) + ln(b').
For instance, if m = 10 then b* + ln(b*) = 888888880.89... x 10^10 + 43.6... ,
The last terms (43.6) affects the first term only in its 18th digit and beyond, so has not affect on the
leading nine digits. In fact, for each positive integer m it follows that the prime P(n*) has 88888888
as its leading block of digits. For each m the prime P(n*) has this initial block of digits, so D
has infinitely many occurrences of this block of digits. [End of proof]

This proof models the algorithm which guarantees
Fact: Any given finite block of digits occurs infinitely often in the prime digit sequence D

/RogerE :D

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Re: Patterns of Prediction

Post by RogerE »

Patterns of prediction suggest that BigSaint and Ubobo.R.O. will remain unposted and otherwise occupied for approximately the next three hours.
.
plLcmLa.jpg
.
Background for this prediction:

The most recent post by Ubobo.R.O.[/b] was at Sat Jun 05, 2021 10:56:14 am.
The most recent post by BigSaint[/b] was at
Sat Jun 05, 2021 12:25:48 pm

The prediction is made even though BigSaint has posting privileges that will no doubt result in completion of the following reserved places:

Screen Shot 2021-06-05 at 2.03.17 pm.png
Screen Shot 2021-06-05 at 2.03.00 pm.png
Screen Shot 2021-06-05 at 2.02.46 pm.png
.
/RogerE :D

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Re: Patterns of Prediction

Post by Ubobo.R.O. »

UBOBO.R.O and BIGSAINT
equals
SATURDAY AFTERNOON
equals
HORSE RACING
equals
YELLING AT TV
equals
NO POSTING.
Full time horse non-whisperer, post box searcher and lichen covered granite rock percher. Gee I'm handsome !
You gottem birds, butterflies, shells, maps, flags and heads on stamps ? Me wantem !

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Re: Patterns of Prediction

Post by RogerE »

That was my working assumption, based on past posting patterns, leading to a satisfactory prediction:

Screen Shot 2021-06-05 at 11.36.53 pm.png
.
;)

/RogerE

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Re: Patterns of Prediction

Post by RogerE »

A Stampboards Prediction

By accident, I noticed the following current statistics at the foot of the Introduction page:

Screen Shot 2021-06-06 at 12.45.34 am.png
I predict that during the next hour the total number of posts will reach
.
72K** = 7,200,000 (*)
.
:D :D :D
.
(*) The scale factors K = 10^3, K* = 10^4, K** = 10^5 provide a convenient set of measures.
K** = 10^5 has a special name in india: it is the lakh.
Wikipedia wrote:A lakh /læk, lɑːk/; abbreviated L; sometimes written lac, is a unit in the Indian numbering system equal to one hundred thousand (100,000; scientific notation: 10^5). In the Indian 2,2,3 convention of digit grouping, it is written as 1,00,000.
Thus, in the conventional parsing notation in use in India, the predicted total is 72,00,000
while the British and US conventions have it as 7,200,000 and some European conventions have
it as 72.000.000 or 7 200 000.
.
/RogerE

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Re: Patterns of Prediction

Post by RogerE »

A Stampboards Prediction cont.
RogerE wrote:
06 Jun 2021 01:56
I predict that during the next hour the total number of posts will reach
.
72K** = 7,200,000 (*)
.
:D :D :D
Follow up:
Screen Shot 2021-06-06 at 3.09.16 am.png
.
Well, that was not a surprising prediction! Made about two and a quarter hours ago,
there have been 23 posts in that period, taking us to 72K** + 4.
If anything, the time it took to reach 72K** from when I made the prediction was a bit longer than I expected.
If the new posts were added at a uniform rate, then they would have required (19/23)x136 = 112 min
to reach 72K**, so that would have occurred at 2:48 am.

/RogerE :D

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Re: Patterns of Prediction

Post by RogerE »

Subjective age
.
A piece by Graham Lauton, "Don't act your age!" was the cover-featured article in New Scientist, 20 Mar 2021. It appears on pp.36-40.

The theme is the distinction between chronological age (how many birthdays you have passed — measured by the calendar) and subjective age (a sociological/psychological attitude measure). In between these two there is also biological age (physiological/physical fitness — as measured by "biomarkers").

It is thought that Mark Twain was the author of the quip: Age is an issue of mind over matter. If you don't mind, it doesn't matter.

The scientific underpinning for Lauton's article is a longitudinal study of some 17,000 people followed/studied over 20 years, conducted by FSU's Antonio Terracciano and colleagues. They summarised their findings thus: People who feel younger live longer. Those who feel older have a shorter lifespan.

Two specific statements from Lauton's article:
Positive attitude is more strongly associated with long life than any biomarker.
A lower subjective age is associated with better health and well-being.

Terracciano links lower subjective age to more friendship, better sex, new experiences and healthier attitude to ageing.

That's a Pattern of Prediction worth noticing, don't you think?

/RogerE :D

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Re: Patterns of Prediction

Post by RogerE »

Subjective age — a sequel
.
A personal comment: my previous post was added on the day before my wife's 80th Birthday. :D

I think we both practise the advice in Graham Lauton's article "Don't act your age!" We are happy to tell others that we have reached the quintessential "fourscore" marker of longevity.

In fact we found multiple ways on multiple days to celebrate my wife's birthday, beginning on 10 June.

Our celebratory activities not only made full use of the Queen's Birthday long weekend, but have continued on numerous days since then, including the past weekend (26–27 June), so my wife has been "Queening" for much of this memorable month. These celebrations were first "reported" here at
https://www.stampboards.com/viewtopic.php?f=6&t=82193&start=9021
and two relevant later posts are at
https://www.stampboards.com/viewtopic.php?f=6&t=82193&start=9052
https://www.stampboards.com/viewtopic.php?f=6&t=82193&start=9214
These note numerous occasions in which we have happily shared our celebratory mode with family and friends. Recall this positive conclusion from the 20-year-long study conducted by Antonio Terracciano and colleagues: Positive attitude is more strongly associated with long life than any biomarker.

"Fourscore" connotation in our culture

In reference to longevity, our culture inherits the term "fourscore" from Biblical references, in the phrasing of King James English. Two relevant instances:
I am this day fourscore years old: and can I discern between good and evil? [2 Sam 19:35]
The days of our years are threescore years and ten; and if by reason of strength they be fourscore years, yet is their strength labour and sorrow; for it is soon cut off, and we fly away. [Psa 19:10]

The use of "score" for the quantity "twenty" is related to cutting or scratching a recording mark (for example, noting a count using up all the fingers and all the toes):
["Score" comes] from Old Norse skor "mark, notch, incision; a rift in rock".

The connecting notion probably is counting large numbers (of sheep, etc.) with a notch in a stick for each 20. That way of counting, called vigesimalism, also exists in French. In Old French, "twenty" (vint) or a multiple of it could be used as a base, as in vint et doze("32"), dous vinz et diz ("50"). Vigesimalism was or is a feature of Welsh, Irish, Gaelic and Breton (as well as non-IE Basque), and it is speculated that the English and the French picked it up from the Celts.
https://www.quora.com/Why-is-the-number-20-called-a-score
/RogerE :D

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Re: Patterns of Prediction

Post by RogerE »

An amplified version of the previous post, concentrating on the linguistic aspects, is now included in the Stamps and Languages thread, at
.
https://www.stampboards.com/viewtopic.php?f=13&t=90529&start=828
.
/RogerE :D

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Re: Patterns of Prediction

Post by RogerE »

Can the abacus help us master computational skills?

In another thread, Ubobo.R.O. has shown us a street abacus in Broome, Western Australia:
https://stampboards.com/viewtopic.php?f=6&t=82193&start=9348

thumbnail_No.11 street abacus in Chinatown Broome.jpg

This has prompted me to read about the abacus. I have found a very interesting post in Vox, which is a website specialising in explanatory journalism.

The writer presents ideas that indicate learning the abacus can develop one's "mental arithmetic" skills, and includes some interesting remarks about learning more generally. Since these subjects concern prediction of results/benefits expected to evolve from certain activities, it seems appropriate to share them in this Patterns of Prediction thread.
I will split it into two posts, because it has a rather natural transition from initial narrower focus to wider focus in its later part...

I learned how to do math with the ancient abacus
— and it changed my life

By Ulrich Boser – 6 Mar 2017, 8:00am EST
https://www.vox.com/first-person/2017/3/6/14766970/abacus-math-learning-lessons
.
Credit: Getty Images
Credit: Getty Images
.
Ulrich Boser wrote:
.
Screen Shot 2021-07-06 at 2.52.09 pm.png
Screen Shot 2021-07-06 at 2.52.53 pm.png
Screen Shot 2021-07-06 at 2.53.28 pm.png
Screen Shot 2021-07-06 at 3.02.56 pm.png
.
This article continues in the next post.
/RogerE :D

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Re: Patterns of Prediction

Post by RogerE »

Can the abacus help us master computational skills? cont.


I learned how to do math with the ancient abacus
— and it changed my life

By Ulrich Boser – 6 Mar 2017, 8:00am EST
.
(Continued from previous post: Tue Jul 06, 2021 14:49:31 pm)
.
Screen Shot 2021-07-06 at 3.04.39 pm.png
(Continuing with inset advertisement suppressed)
Screen Shot 2021-07-06 at 3.05.10 pm.png
Screen Shot 2021-07-06 at 3.05.27 pm.png
Screen Shot 2021-07-06 at 3.06.04 pm.png
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(Continuing with inset advertisement suppressed)
Screen Shot 2021-07-06 at 3.08.03 pm.png
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(Continuing with inset advertisement suppressed)
Screen Shot 2021-07-06 at 3.09.09 pm.png
Screen Shot 2021-07-06 at 3.09.36 pm.png
.
https://www.vox.com/first-person/2017/3/6/14766970/abacus-math-learning-lessons
.
Note that the children using their abacuses have them flat, so after the beads are moved along the rods they stay in their new positions, until the user wants to move them again. Similarly, the rods in Broome's street abacus are horizontal, so the beads can be moved along the rods and will stay in their new positions until the user wants to move them again. On the other hand, the abacus at the start of Ulrich Boser's article (in the previous post) is shown with its rods vertical. This shows us how elegant the computing instrument is, but it is not in computational orientation — for that, it needs to be rotated so that its rods are horizontal.

If you would like to see a very basic introduction to how to use the abacus, here is a 3min video I recommend:
https://www.youtube.com/watch?v=FTVXUG_PngE
.
/RogerE :D

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Re: Patterns of Prediction

Post by RogerE »

How WEIRD are you?

That is the title of an article by Dan Jones, in which he interviews Joe Henrich. It was published in New Scientist, 5 Sep 2020, pp46-49.

I will share here a few extracts from the article.
How does the culture we live in influence our psychology, motivation and decision making? Joe Henrich was a cultural anthropologist working in the Amazon when he first tried to find out...
He realised that his findings have big implications for psychological research, which tends to focus on students from Western backgrounds. In 2010, he introduced the "WEIRD" concept to describe the unusual psychology of the subjects in the majority of these studies.
Now professor of human evolutionary biology at Harvard University, he tells New Scientist about the origins of WEIRDness, its impact on history and its role in the modern world.
Henrich coined the acronym
WEIRD = Western, Educated, Industrialised, Rich and Democratic
to characterise the backgrounds of the subjects often studied in psychological research.

Henrich says:
[Two cultural psychologists and I] noticed that in the behavioural sciences, and psychology in particular, about 96 per cent of study participants were from Western, Educated, Industrialised, Rich and Democratic societies — and that they were often psychological outliers in comparison with other populations.
WEIRD people tend to show greater trust in strangers and fairness towards anonymous others; think more analytically rather than holistically; make more use of intentions in moral judgements; are more concerned with personality, the self and the cultivation of personal attributes; they are more individualistic and less loyal to their group; and they are more likely to judge the behaviour of others as reflecting some enduring disposition rather than temporary situational factors.
On the downsides of WIERDness, Henrich says:
In societies where there is a strong sense of kinship, like Fiji where I have done fieldwork, there is a sense of security, community, oneness — a kind of comfort that comes from the warm embrace of knowing you are at the centre of a tight web of relations who will always have your back. They aren't tied to you because you are a convenient contact or are currently smart or successful, they are tied to you in a deep way...
People living in tribal or clan-based societies also tend to see themselves as links in a chain connecting past to future, creating a sense of continuity that gives people a real sense of meaning and security.
A key summary statement from Henrich:
The picture of "human psychology" portrayed in the textbooks, and still in many journal articles, doesn't represent the psychology of Homo sapiens at all...
This bias hampers our efforts to understand the origins and nature of psychological processes and brain development. Much of what looks like reliably developing features of minds, with clear developmental trajectories over childhood, turns out to be the result of cultural products, like the institutions, values, technologies or languages individuals confront and must learn, internalise and navigate to make their way in the world. This applies not only to psychology and neuroscience... but also to aspects of human physiology, anatomy and health.
.

Our expectations of ourselves and others are in fact predictions, based on patterns of experience or learning. We can now reflect that the WEIRD society most of us inhabit has pervasive influence on the patterns we use to make those predictions.

/RogerE :D

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Re: Patterns of Prediction

Post by satsuma »

There is often another bias in Weird subjects.

They tend to be of college age and many can get course credit for participating.

So for example, one bias could be that they are less risk adverse than a truly representative sample of their culture.

Another potential inherent bias is the problem of self selection. For example it's likely far more extroverts would volunteer as subjects than would comparative introverts.

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Re: Patterns of Prediction

Post by RogerE »

Quite right, thanks satsuma. The WEIRD category serves as a nice flag to remind us of multiple sources of selective bias in common designs of psychological experiments.

/RogerE :D

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Re: Patterns of Prediction

Post by RogerE »

Gambling

Turning predictions into occasions for gambling is social behaviour widely exploited by money-makers (governments in particular, but casinos, totalisators, loto operations, etc). Some forms of gambling have evident skill-based components, but others are purely "chance" (random?) processes.

Australia has an unenviable rank in the gambling culture:

Screen Shot 2021-07-12 at 12.18.09 pm.png
.
https://www.worldatlas.com/articles/countries-that-gamble-the-most.html
.
Note that the rightmost column is not headed "Gaming Profits per Adult".

/RogerE :o

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Re: Patterns of Prediction

Post by RogerE »

Gambling cont.

What is gambling? Well, if you consult dictionaries you will find a range of meanings, a whole fuzzy set of meanings, as is the case with many words. When a word is used, we learn to deduce from its context the most likely meaning it is intended to have from within its fuzzy set of potential meanings.

We could get distracted by semantic questions or misdirected into focussing on exploring the full range of meanings of the word "gambling", when the main concern is with gambling as a real problem which causes individuals to lose more money than they can afford to lose, and perhaps even more money than they have...

The previous post, ranking countries by their annual average gambling losses per adult, leaves no doubt that it is talking about official figures tracking money lost in organised public gambling activities. Those who have a gambling problem, and those who seek to help them, don't have any doubts about which part of the fuzzy set of meanings of the word "gambling" they are concerned about. Lifeline says

Screen Shot 2021-07-13 at 12.24.03 am.png
https://www.lifeline.org.au/get-help/information-and-support/problem-gambling/
.
The side of gambling that I find particularly objectionable is the money-making side of the "industry", where individuals seek to exploit and profit from those with a tendency to gamble too much, especially those with a serious gambling problem.

By the way, notice that the average annual gambling loss per adult Australian translates to more than $100 per month. I think about 30% of Australians don't gamble, and many more Australians reading this $100 figure would note that it is far more than they personally lose to gambling in any form. The significance of the average must be that there are some people who lose far more than $100 per month. That's where the problem lies...

/RogerE :(

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