your 'statistics' are horribly flawed - try again
a REAL qproof
edited from >>2364830 (Bread #2980)
What are the odds that Q posts at the same time as POTUS?
For simplicity lets assume that POTUS tweets 10 times per day, at roughly 1 post per hour, in a 10 hour window.
In other words:
Your chance of posting 1 minute before trump is 1 in 60 or 1.67% chance
Your chance of posting 20 seconds before trump is 20 in 3600 or 0.54% chance
Your chance of posting 10 seconds before trump is 10 in 3600 or 0.27% chance
what about multiple times in a row? Just multiply the values together:
two times in a row at 1 minute before? 1/60 * 1/60 or 0.027% chance
two times in a row at 10 seconds before? 10/3600 * 10/3600 or 0.00077% chance
three times in a row at 1 minute before? 0.00046% chance
three times in a row at 10 seconds before? 0.0000021% chance
If you factory workers or engineers are familiar with 6 sigma process, 6 sigma is used to ensure no defects ever occur. 6sigma is equal to 99.99966%, which assumes success in all conditions → That means, the chance of failure (to be considered success) is less than 0.00033%
In other words, it only takes a few times in a row to make it statistically impossible to be a coincidence
(i'm continuing to refine it)
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
>BUT IM A SHILL AND THAT ISNT GOOD ENOUGH FOR ME, HOW ELSE CAN YOU CONVINCE ME?
WELL, KEEP READING FELLOW NERD AND LETS TRY A DIFFERENT ANGLE
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.