My Q Proof ultimate weapon! (Doubters and shills: Try and debunk this, I dare you.)
(media.greatawakening.win)
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You are missing the point Pede. I am not saying you are a bad person. What you are doing is a version of Apophenia, the tendency to perceive meaningful connections between unrelated things.
The fact that there was such a large amount of letters just about anything could be spelled out. One thing is certain about coincidence. The phenomenon fascinates believers and skeptics alike.
So, back up your position. Find me ONE tweet that has that exact match (every letter, specifically by upper case or lower case) matching AND every punctuating character matching. Once you fail at that, try and wrap your head around having one posted at the same time as Q and/or POTUS.
Entertain me....see I& you can do it.
(Spoiler alert: You can’t) The chance if that happening even once is 312 million to one.
What you are displaying is a textbook case of Dunnings-Kruger.
It's Dunning-Kruger and you still missed the point.
Not saying I agree with him, I actually believe your post is right. Just saying you still definitely missed his point.
Looks like you couldn't answer the challenge and lost the debate.
Congrats, you caught a typo.
And the POINT is mathematics.
These are non-debatable and not subject to opinion.
The “point” that you my friend are missing is simply that facts don’t give a single fuck about your feelings ??♂️
But do go on...I find discussions like this entertaining....especially when the provocateur is clearly out of their depth on the topic. ?
If you want to be taken seriously you're really going to need to do this analysis yourself.
No, I REALLY am smarter then you. 127 IQ in 1960 in High School, didn't get dumber for sure :-)))
Good bye.
Lol...by your own statement I can assure you: no, no you really aren’t.
op is correct that statistically this is impossible. Even if you only look at the digits in the tripcode, 1, 1, 3, and 6, the probability that every single one of 4 random digits would show up in a tweet which has around 10 (relatively random) digits is pretty low: (1-(1-1/10)^10)^4 = ~0.18 = 18%
Then there are a total of 13 capital letters in the tweet, 3 capitol letters in the trip code (the following calculation will ignore letter frequencies for simplicity): (1-(1-1/26)^13)^3 = ~0.06 = 6%
together .06*.18 = ~1% And this number will continue to shrink the more factors you take into account
These are absolutely not methods for calculating the exact probability of this, but will generally correlate with the actual answer, and should at least make a point. Anyone feel free to correct me if i made any mistakes.
Blocked
It was a Q proof, posted immediately after Q...it’s not like he missed a W. That’s called confirmation.