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The Theory of
Line Patterns 		No figure out of the possible 30^49 (2E72)
is the square root or greater as improbable by design
using prime digits 2 and 5
as all of the 2^23 * 29! / 18! (1E22) figures shown
using prime digits 3 and 7 by the noted sub-pattern. 		math on 37 patterns
scholastically
verified
List of pledges to the first person who successfully disproves "The Theory of patterns" by the conditions set out on this website https://www.thetheoryofpatterns.com/rules.php. (No money changes hands unless and until someone actually disproves this theory)

For anyone concerned about pledging, remember that the odds of finding any string of characters that match all 37 primary patterns of the pseudo sentence, that are not reverse engineered from the pseudo sentence, is 1 in 5E39. There are only 22^28, which equals about 1E37, possible combinations of 28 pseudo letters, which is 500 times less than 5E39. That is just to get a set of characters. To get actual words would be considerably more difficult. To get a legitimate sentence would be considerably more difficult than that. To create a sentence that clearly and articulately describes the same event would be even considerably more difficult than that. To do this without incredible luck or an alternate method is impossible. Rest assured, your money is safe.

1 in 5E39 is like flipping 132 normal coins that are not rigged and all of them coming up heads at the same time. If everyone on earth flipped 132 coins at a time 1,000,000 times per second each for 30,000,000 seconds per year without ever eating, sleeping, or taking a coffee break, for 1,000,000 years, there is less than one in 24,000,000,000 chance that anyone would ever flip all 132 coins as heads at one time, assuming that they did not use trick coins.

Assuming that you did have a random set of characters that met all 37 primary patterns of the pseudo sentence and lets assume that you are doing this in English, Then the probability of any group of characters being a real English word is about one over the number of characters in the alphabet to the power of the number of letters in the word all times the number of words in the English language with that many characters.

Let's take a list of 58,000 words and count the number of each word size. We then take 26, the number of letters in the English alphabet, to the power of the number of letters in the word, divided by this count. The results are as follows:

letterscount(26 ^ letters) / count
24714
358929
42294199
542662785
6693644538


We determine below that this will take about 70 characters so lets assume the we have 14 words with the same distribution of word lengths as the pseudo sentence.

14^2 * 29^4 * 199^2 * 2785^4 * 44538^2 = 1E36 inverse probability of the random characters being all words.

The inverse of the probability of all of the words being real words and making a legitimate sentence is about the number of possible word combinations, 1E5^14, divided by the number of legitimate sentences that can be made, 1E48, times this 1E36, which is about 1E58. The inverse of the probability that this sentence clearly articulates a belief in the event of the pseudo sentence is the number of possible sentences, 1E48, divided by the number of clear descriptions, 1E24, times this 1E58, which make the inverse of the probabiity about 1E82.

Now lets divide this number by a billion trillion for no reason whatsoever, just because the number has become so outrageously large. Remember that we have already assumed that their are a trillion trillion possible ways to describe the event of the pseudo sentence which is probably a little high to start with.

If we take this times our 1 in 5E39, we get an overall probability of meeting all of these conditions at about 1 in 1E100 which is one in a googol. 26^70 = 1.568E100 so this would take at least seventy characters. The actual probability is closer to one in a googol squared, but for now we will stick to one in a google.

The probability that the author wrote the pseudo sentence with all 37 primary patterns of the pseudo sentence without incredible luck or an alternate method is about 1 in a googol by this theory.

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