dChan
Anonymous ID: 0d9b0d July 30, 2018, 7:49 p.m. No.2364830   🗄️.is 🔗kun   >>4844 >>4850 >>4851 >>4856 >>4860 >>4873 >>4886 >>4902 >>4915 >>4923 >>4926 >>4952 >>4969 >>4989 >>5026 >>5037 >>5056 >>5165 >>5266 >>5320 >>5340 >>5545

QPROOF

What I am about to prove is not 100% completed (I made rough assumptions on average vs rare vs exotic, and on trump's twitter breakdown), however, is still factually based

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

240 possible characters

2000 words are average words

2000-10000 are rare words

10000-40000 are exotic words (such as names)

 

average 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% average 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 average words

2400.150.6 / 7 = 3.08 rare words

2400.050.6 / 8 = 0.9 exotic words

 

Lets assume that probability of matching an average 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 15 examples like this, given 15 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%

 

If you workers 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%

 

The probability of Q being a Larp based on just a few examples is stastically impossible.

The probability of being a conspiracy theory is 0.

The probability of being a conspiracy theory is 0.

Anonymous ID: 0d9b0d July 30, 2018, 8:11 p.m. No.2365210   🗄️.is 🔗kun

>>2365165

yeah, framework for someone to run it on his twitter vs global twitter to identify average/rare/common and his average metric on twitter length

 

i didn't even apply the "within an hour" to the probability, because I wasn't sure how to account for that… Would probably need to look at the frequency of Trump's posts, and then point out the probability of knowing when he is going to post.