statistically IMPOSSIBLE long ago

 

QPROOF

edited from >>2364830 (Bread #2980) with Today's information

 

What I am about to share is not 100% completed work (I made rough assumptions on word classifications, and did not conduct an actual machine learning based word or sentence embeddings on trump's & Q's posts), however, should be factually based with enough information to quell even the toppest of kek anons.

 

What are the odds that Q uses the same two words as POTUS in a 240 character twitter post, within an hour of POTUS posting?

 

If you factory workers or engineers are familiar with 6 sigma process, 6 sigma is used to ensure no defects ever occur.

6sigma is 99.99966%, which assumes success

That means, the chance of failure (to be considered success) is 0.00033%

 

Time Based Chance

Now, lets consider 3 posts that happen from Q within 1 minute before POTUS in a row.

 

For simplicity lets assume the following:

There are 24 hours in the day, yet assume it is known that POTUS only posts in 10 hour window

Assume POTUS averages about 10 tweets per day, or 1 post per hour.

Your chance of posting 1 minute before trump is 1 in 60 or 1.67% chance

Your chance of posting 1 minute before trump, two times in a row is 1 in 3600 or 0.028% chance

Your chance of posting 1 minute before trump, three times in a row is 1 in 216000 or 0.0004630% chance -just about equal to 6sigma

 

IF THAT DIDNT CONVINCE YOU… KEEP READING FELLOW NERD

 

Word Choice Chance

240 possible characters

 

Assuming that within the English dictionary that…

2000 words are common words

2000-10000 are rare words

10000-40000 are exotic words (such as names)

 

common words average about 4 characters and 1 space

rare words average about 6 characters and 1 space

exotic words average about 7 characters and 1 space

 

For any given post, of 240 characters, lets assume:

 

80% common words

15% rare words

5% exotic words

 

Let us further assume, on average, only 60% of the characters are used in a tweet, which gives us:

 

2400.80.6 / 5 = 23.04 common words

2400.150.6 / 7 = 3.08 rare words

2400.050.6 / 8 = 0.9 exotic words

 

Lets assume that probability of matching a common word is 0.01137416192 (23 possible words as N, 2000 possible word choices as probability of success)

Lets assume that probability of matching an rare word is 0.0003749062559 (3 possible words as N, 8000 possible word choices as probability of success)

Lets assume that probability of matching an exotic word is 0.0000333333 (1 possible words as N, 30000 possible word choices as probability of success)

 

Lets take the rogue example

The probability of the 'rogue' example, assuming that 'rogue' is a rare word, is about 0.037% chance to occur

 

At the moment, there are over 30 examples like this, given 30+ examples, the probability just using average words is 0.0000000000000000000000000000000000000000000000003%

 

To give this a further proof, only 3 examples of coincidence would provide about 0.00000565%

 

The probability of Q being just a COINCIDENCE is stastically impossible.