Q answered yes to the JFK Jr. question. Check the ID's and do the odds. 1: 2-Trillion chance of ID's appearing as such.
Q answered yes to the JFK Jr. question. It is a MIRROR.
The QUESTION AND ANSWER WERE FLIPPED. ANSWER, THEN QUESTION.
SEE SEE the IDs, then fill in the blanks…
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Odds of cc on both "randomly-generated" IDs:_______
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Odds of first c being second-to-last in both instances:_______
3.Odds of second c being last in both instances:_____
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Odds of both c's appearing last and in succession: ___
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Odds of interrogator's randomly-generated id adding up to 23 (PAIN/JFK Jr.):_______
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Odds of Q's randomly-generated id adding up to 17 (Q): ___
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Odds of a flipped question/answer (mirror):___
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Odds of 2 of the above happening together:___
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Odds of 3 of the above happening together:___
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Odds of 4 of the above happening together:____
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Odds of 5 of the above happening together:_____
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Odds of 6 of the above happening together:_____
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Odds of 7 of the above happening together:_____
How many coincidences before mathematically impossible? -Q
Q !CbboFOtcZs ID: 07b0b9 No.2263683 📁
Jul 24 2018 11:51:59 (EST)
Q !CbboFOtcZs ID: 07b0b9 No.2263659 📁
Jul 24 2018 11:50:26 (EST)
https://www.huffingtonpost.com/entry/qanon-conspiracy-real-life_us_5b54bbafe4b0b15aba8fe484📁
PANIC!
Q
Now do you understand why mirrors and disinformation is necessary?
Logical thinking.
[20]
Q
No, there are 35 possibilities per space in the ID. Go back and do the odds properly, retarded boy.
Garbage. Go back and do the odds properly. Even in the lottery you don't have to match the number to the space it occupies.
Here, since you are mentally retarded I will get you started. There is a 1 : 216 chance that a given letter or number will appear in a given space.
That's wrong. But let's say that's right. What are the odds that the last c will appear in the last space in both ID's?
So you are saying it is impossible that the letter c would appear. That is actually what you just said, retard.
There are 26 possible letters and 10 digits possible per space in the 6-space ID: 1:35 chance of c appearing in a given space IN JUST ONE OF THE IDs. Now, what are the odd of a c appearing in the same space in both IDs?
What don't you understand? In any given space there are 26 possible letters and 10 possible numerical digits. 1:36 chance per space for JUST ONE ID.
OK. It is 16 for one ID. What are the odds that the same letter will appear in the very next slot?
Yes, you are right. My bad So it is 1:16 for one ID. what are the odds that the same character will appear in the very next slot?
OK. What are the odds that both c's will match both c's and match the same slots in both IDs?
Now for just Q's ID, what are the odds that his digits will add up to 17 AND have the matching Cs?
Q's numerical digits add up to 17 (Q).
The interrogator's digits ad up to 23 (PAIN [JFK Jr])
The cc appears in the same spot in both IDs
Q uses mirrors - i.e., reversing things. Q&A could very well be A&Q in this case.