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# Probabilities of Primes and Composites

Phaethon

Posts : 146
Points : 237
Join date : 2019-08-05
Preamble
The aim here is to discover the frequencies of numbers for the purpose of guiding the selection of a numerical base that best represents the most frequent numbers.

Proposition 1:
Suppose that all natural whole numbers are equally probable.
Caveat:
It is not actually believed that all whole numbers have an equal probability of being selected.
Corollary:
Then, if a number from the set of natural whole numbers be selected at random, each whole number has an equal probability of being selected.

Expected Frequencies of Primes
Query:
What is the number of times an $$i$$th prime number $$p_i$$ is expected to occur in such a randomly selected whole number?
Response:
Consider the prime number 2. It occurs at least once in half of all whole numbers. In some of these even numbers, the prime number two as a factor occurs more than once. For example, in the composite number four that is formed by a multiple of prime numbers as two to the power of two, the prime number two as a factor appears twice. In the composite number eight, the prime number two appears three times as a factor.

Caveat:
By the number of times a prime factor appears as a factor in a composite number is meant not the number of times that prime number appears added with itself in a sum to make up the composite number, because this number of times is the composite number divided by the prime number, but rather the exponent of the prime number in the prime factorisation of the composite number.

The prime number two appears as a factor at least once in half of all whole numbers, so, given the first Proposition, the frequency of the prime number two appearing as a factor at least once is 1/2.

The prime number two as a factor appears at least twice in a quarter of all numbers. Given Proposition 1, the frequency of the prime number two appearing as a factor at least twice in whole numbers is 1/4.

In general, the frequency of the prime number $$p$$ appearing as a factor at least $$n$$ times is 1/p^n.

The expected frequency $$F(p)$$ of a prime number $$p$$ appearing as a factor, taking into account how often that prime number appears within a number, in a randomly selected whole number, given Proposition 1, is the sum $$S_{\infty} = \sum_{n=1}^{n=\infty}\frac{1}{p^{n}}$$ of the frequencies $$\frac{1}{p^{n}}$$ of that prime number $$p$$ appearing as a factor at least $$n$$ times. This summation is a geometric series with initial term 1/p and common ratio $$\frac{\frac{1}{p^{n+1}} }{\frac{1}{p^{n}}} = \frac{1}{p}$$ between adjacent terms. The sum to infinity $$S_{\infty}$$ is therefore $$\frac{\frac{1}{p}}{1-\frac{1}{p} } = \frac{1}{p-1} = F(p)\text{.}$$

Caveat:
Since the expected frequencies were derived assuming Proposition 1, but Proposition 1 is not believed, the expected frequencies so formulated are not believed to be a realistic model.

The expected frequencies $$F(p)$$ for the smallest prime numbers can be compared in a table.

Table 1. Expected Frequencies for Primes
 Prime, $$p$$ Expected Frequency, F(p) = $$\frac{1}{p-1}$$ 2 1 3 1/2 5 1/4 7 1/6 ⑪ 1/⑩

It can be seen from this derivation that the prime number two is twice as frequent as the prime number three, and that the prime number three is twice as frequent as the prime number five. For example, considering the numbers up to eleven, the prime factor two occurs once in the number two, twice in the number four, once in the number six, three times in the number eight, and once in the number ten, or a total of eight times. In the same span of numbers, the prime factor three occurs once in the number three, once in the number six, and twice in the number nine, or a total of four times, which is half as often as for the prime number two. The prime factor five occurs once in the number five and once in the number ten, or a total of twice, which is half the frequency of the prime number three.

For ranking the prime numbers for importance by their frequencies of occurrence, the principles of the preceding argument have been described previously elsewhere.

For the purpose of ascertaining the base most likely to produce rounding in numbers formed from randomly selected prime numbers, yesterday I proceeded with the following reasoning.

Probabilities of Primes
Proposition 2:
Suppose that the probability $$P(p)$$ of a prime number being selected is proportional to its expected frequency $$F(p)$$.
Caveat:
Since the Expected Frequency as formulated above is not believed because it was derived from Proposition 1 which is not believed, the probabilities of primes are not believed to be proportional to the expected frequencies so formulated, but may nevertheless be proportional to the actual expected frequencies of another formulation.
Discussion:
The expected frequency of the prime number two was one. However, the probability of the prime number two appearing in a randomly selected number is less than one, because half of all numbers do not contain the prime number two as a factor. The probability for the prime number two need not be a half, however. All numbers are considered to be broken into their constituent prime numbers, which are then placed into a bag and shuffled. A prime number is then withdrawn at random from this mix. The probability sought is that of the selected prime number being a certain prime number. For example, what is the probability of the withdrawn prime number being the prime number two? Let this probability be called P(2). Then, given the second Proposition, the probabilities of all other prime numbers can be expressed in terms of P(2), as the multiple P(2)F(p) of P(2) and the expected frequency F(p).

Probabilities of Composites
Proposition 3:
The probability of selecting a composite number is the product of the probabilities that are raised to the exponents of the primes for the prime numbers in the prime factorisation of the composite number.
Discussion:
By one of the Fundamental Principles of Counting, the probability of selecting a combination of prime numbers, such as a prime number $$p_a$$ and a prime number $$p_b$$, where $$a$$ and $$b$$ may be the same or different, and forming a composite number out of them, shall be the multiple P($$p_{a}$$)P($$p_{b}$$) of the probabilities P($$p_{a}$$) and P($$p_{b}$$) of selecting the prime numbers individually.

Since P(3) = 2P(5), multiplying both sides of the equation by [P(2)]^2 gives
[P(2)]^2 P(3) = 2[P(2)]^2 P(5),
thus:
P(①⓪) = 2P(2) P(⑩)
and the ratio of the probabilities for the numbers twelve and ten is:
P(①⓪)/P(⑩) = 2P(2).
If the numbers twelve and ten were equally probable, then P(①⓪)/P(⑩) = 1 and 2P(2) = 1, so P(2) = 1/2.

Suppose the probability P(2) to be a half. Then, probabilities of numbers being selected from a pool of prime numbers could be tabulated as follows.

Table 2. Probabilities for Whole Numbers
 Whole Numbers, n Probability, P(n) 2 1/2 3, 4 1/4 5, 6, 8 1/8 7 1/①⓪ 9, ⑩, ①⓪, ①④ 1/①④ ⑪ 1/①⑧

Regardless of the value of P(2), the sum of the probabilities of the prime numbers would be $$P(2)\sum_{i=1}^{i=\infty} \frac{1}{p_i - 1}\text{.}$$ However, if the summation does not converge then this scheme of probabilities cannot be the correct one. The series diverges because the sum of the reciprocals of the primes diverges. This means that as the sizes of the prime numbers being considered in the sum increase, the value of P(2) must decrease towards zero in order for the sum of probabilities to be one, with the implication that the probability of the prime number two being selected decreases because of the increasingly large number of other prime numbers expected to be found in the large composite numbers.

Hence, and because the larger numbers approaching infinity are less likely to be encountered than the smaller numbers, it is necessary to adjust the probabilities of numbers being selected at random dependent on their sizes, contrary to Proposition 1.

It is expected that the effect of such an essential adjustment would be to make the smaller prime numbers, such as two and three, more probable and the larger prime numbers, such as five and seven, less probable.

It may be that the adjustment can be done in such a way that maintains the equally of the probabilities of the numbers three and four as seen in Table 2, if P(3) = [P(2)]^2. On the other hand, the prime number five can be expected to move down the rows of the table such that it would become less probable than six or eight. Consequently, the composite number ten would move down to being less probable than the numbers nine, twelve, and ①④. Similarly, the prime number seven would be expected to become less probable, such that it might end up below the numbers nine, twelve, ①④, or being even less probable than ten.

Consider a ranking of the numbers that has been proposed previously elsewhere which places the numbers in an order that fulfils these expectations.

Unimportance Rank of Numbers
Let the unimportance, say U(n), of a number $$n$$ be calculated by subtracting one from each prime factor in the prime factorisation of a number and adding up these subtractions. This produces the following table ranking the first few numbers.

Table 3. Unimportances of Numbers
 Whole Numbers, $$n$$ Unimportance Score 2 1 3, 4 2 6, 8 3 5, 9, ①⓪, ①④ 4 ⑩ 5 7, ①③, ①⑧ 6

Probabilities from Unimportances
Since the unimportance score of a composite number is calculated by the addition of the unimportance scores of its prime factors, but for probabilities a multiplication or product rather than addition of scores is required, to convert the unimportance to a probability, exponentiate the unimportances by raising the unimportance score as the exponent of the base two. Suppose that the resulting scores are proportional to the probabilities for the numbers. Summing the probability scores for all the prime numbers should be equal to one, so to achieve this the sum can be normalised by multiplying it by a constant $$k$$ such that this sum of all probabilities is one. This constant then distributes into the sum, multiplying by each term to produce the probabilities of the primes.

Table 4. Prime Probabilities from their Unimportances
 Prime, $$p$$ Probability, $$P(p) = \frac{k}{2^{p-1}}$$ 2 k/2 3 k/4 5 k/①④ 7 k/⑤④

$$\sum P(p_{i}) = k\sum_{i=1}^{i=\infty} \frac{1}{2^{p_{i}-1}} = 1$$
This series converges by the ratio test. The sum is found to converge at ≈0.829365019, about ⓪⁏⑨⑪⑤①⑧⑥. The reciprocal ≈1.205741715 of this is $$k$$. Then the probability P(2) for the prime number two being selected is k/2 ≈ 0.602870857. From this probability, the probabilities for the other prime numbers and composite numbers can be calculated and tabulated.

Discussion:
From these results of probabilities from unimportances, it seems that the probability for the prime number five is far too low. This implies that the unimportance scores and rank that others started advocating produce unrealistic probabilities. It seems that a more realistic scheme would be intermediate between those of Tables 2 and 3. Thus, it is expected that the prime number five would appear less probable than the numbers six and eight, but more probable than the numbers nine, twelve, and ①④.

Reference
https://www.tapatalk.com/groups/dozensonline/viewtopic.php?p=40010332#p40010332
https://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes

Phaethon

Posts : 146
Points : 237
Join date : 2019-08-05
Phaethon wrote:to convert the unimportance to a probability, exponentiate the unimportances by raising the unimportance score as the exponent of the base two.

Actually, instead of base two, use the base ≈ 1.79256877 ≈ ①⁏⑨⑥①⑥⑧⑤⑧ approximately. Then k is one and the probability of a prime $$p$$ becomes $$P(p)≈0.55786^{p-1}$$ ≈ ⓪⁏⑥⑧④$$^{p-1}$$. These probabilities produce the same rank for primes and composites as the unimportance score.

Phaethon

Posts : 146
Points : 237
Join date : 2019-08-05
When the unimportance scores were being used to form probabilities, the aim was not to replicate the ranks of the prime and composite numbers by that score, but rather to produce a sum of probabilities that would converge while the rank would resemble as much as possible the order from the expected frequencies. The unimportance score was selected because it had the property of reduced ranks for larger primes relative to the expected frequencies, and it had been guessed that a suitable convergence would cause the ranks of the prime numbers five and seven to fall relative to their ranks from those expected frequencies. The investigation was to devise some function into which the unimportance scores could be imputed such that probabilities would result that would converge when summed. The rank of primes and composites from the probabilities might differ from the rank by unimportance scores, just as the rank by the expected frequencies differs from the rank by the unimportances.

In order for the combination of scores to be multiplicative instead of additive, the unimportance scores were raised as the exponent of some base. In order for these powers to be proportional to probabilities, their reciprocals are taken for the resulting scores. The base on which the negative unimportance score is raised to be proportional to the probability can be varied. The base two was chosen in the discourse so that the probability for the prime number three would be half that for the prime number two, just as the expected frequency for the prime number three was half that of the prime number two, although this relationship may no longer hold after the necessary adjustment to the expected frequencies for convergence. However, the resulting series of probabilities appeared to converge too quickly because the probabilities for larger prime numbers were vanishingly small.

The probabilities P(p) of primes $$p$$ could be expressed by the equation $$P(p) = b^{p + a}\text{,}$$ where some base $$b$$ and a constant $$a$$ are chosen such that the series $$\sum_{i=1}^{i=\infty} b^{p_{i} + a}$$ from the sum of the probabilities for all primes converges to one.

As mentioned, $$a$$ is minus one when the base $$b$$ is about ⓪⁏⑥⑧④, and the rank of primes and composites by the resulting probabilities is the same as by the unimportance scores. This has the benefit that the ranks for the numbers three and four are the same, just as they are from the unadjusted expected frequencies. However, several authors complained about the rank for the powers of the prime number three being too high relative to the rank for powers of the prime number two by the unimportance scores. In order to increase the importance of binary relative to ternary powers, the reciprocal of the base $$b$$ would have to be increased from this value 1/⓪⁏⑥⑧④ and would end up being closer to base two. However, this causes the probability for the prime number five and larger primes to decrease too dramatically. That is, the summation series of probabilities converges too rapidly. When $$a$$ < -1, the prime number five is less probable than ①④, ①⓪, and nine. When $$a < -1$$, four squared is more probable than twelve. Surely, dozenists would rather not imply that four squared is more important than twelve.

If anything, the series should be made to converge more slowly than implied by the unimportance scores in order to resemble the expected frequencies better. When -1 < a < -⓪⁏⑥, the probability of the prime number five is under the probability of the number eight but still more than the probabilities of the numbers nine, twelve, and four squared. When $$-1 < a$$, three is more probable than four, with the result that twelve is more probable than four squared. The absolute probability of the prime number three has a maximum near about $$b \approx$$ ⓪⁏⑦⑤ and $$a \approx$$ -⓪⁏⑦②⑧. The absolute values of these two parameters are equal near ⓪⁏⑦④⑧. When -1 < a < -⓪⁏⑥, the probabilities of numbers composed only of the first two or three prime numbers happen to be near their reciprocals. This means that even after dozens of trials, the prime number two would appear to happen about half the time, the prime number three would appear to have happened about a third of the time, four would appear about a quarter of the time, six would appear about a sixth of the time, eight would appear about an eighth of the time, nine would appear about a ninth of the time, and twelve would appear about a twelfth of the time.

I think -⓪⁏⑥ << a < 0 is unlikely because then the ratio of the probabilities of the prime numbers five and two becomes less than the ratio from the expected frequencies, which is four to one. This could be giving too much importance to the prime number five. In this range, fifths become more important than eighths, and consequently ten becomes more important than four-squared.

The simplest equation seems to be when $$a = 0$$ and $$b\approx$$ 0.6774 $$\approx$$ ⓪⁏⑧①⑥⑥⑦②⑩④⑥. Then, the importances of primes are ranked merely according to their sizes without consideration of their expected frequencies. It is a nice idea, but it is a different kind of problem. Nevertheless, the resulting rank is not extraordinarily different from that by the expected frequencies. Five and six would have the same rank, and consequently ten and twelve would be equals. Eight and nine would have the same rank, whereas with $$a < 0$$ eight would be more probable than nine, which most people prefer.

It is highly implausible that $$a$$ would be greater than zero. If this were to happen, four squared would be less probable than twice nine, which would be too incredible for many people. Thus ten is unlikely to ever be more important than twelve.

Phaethon

Posts : 146
Points : 237
Join date : 2019-08-05
I have come to the conclusion that the best exponential probability function $$P(p)$$ of prime numbers in the form given above has parameters $$b \approx$$ 0⁏768 and $$a \approx$$ $$-$$0⁏6. This function for probabilities of primes has a number of favorable properties:
• The probability of the prime number two is very nearly a half. This is in agreement with the rough trend of its probability among the first dozen or so natural numbers.
• The probabilities of the most important numbers are also near their reciprocals or units fractions.
• The probability of the prime number three is a good compromise between its reciprocal and its expected frequency when the probability of the prime number two is a half.
• The probability of the prime number five is very nearly a quarter of the probability of the prime number two, in accordance with the expected frequencies.
• The characteristic of this grading is that the numbers five and eight are equally important, in accordance with the expected frequencies when the probability of the prime number two is a half.
• The rank of the more important numbers is similar to that by the expected frequencies in conjunction with a probability of half for the prime number two.
• Larger prime numbers are less important than predicted by the expected frequencies, in accordance with the principle of larger natural numbers being less probable than smaller ones, and desirable for the sum of the probabilities to converge.
• By taking the logarithm of the function to its base, scores for ranking primes and composites may be easily computed because of the simple rational value of the parameter $$a$$.
• The value of the parameter $$a$$ is the good halfway compromise between its values in the probability functions for the unimportance algorithm and the rank by size alone.

Last edited by Phaethon on Mon Jul 25, 2022 9:47 am; edited 1 time in total (Reason for editing : Added: ", in accordance with the expected frequencies when the probability of the prime number two is a half")

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