Quote:
Originally Posted by MDawg
UNKewlJ has been panicking on two fronts. One is that he’s been outed as a small time red green chipper who plays a lil blackjack only as a cover for sugar daddy hustling.
The other is that he’s been ID’ed as a Pac Islander looking fella.
As MaxPen would say...RIP.
The things he denies the most...are the ones that hit home the closest in truth.
Quote:
Originally Posted by MaxPen
Dude, if I really cared, I would have left this forum several years ago. And if I couldn't live without posting I would have returned with a new identity, like .....well Mdawg.....your reincarnated identity of a formerly banned member. :DQuote:
Originally Posted by MDawg
And I don't keep posting about the lawsuit. It is always you who brings it up, as you just did. ;)
There was no lawsuit. But there was a back-room incident and several months later a settlement. I don't really need to bring up either. I shared the backroom incident as a sort of warning to players, that this sort of thing that most of us thought were a thing of the past (mid 2000's) is happening again with more frequencies. And not just to blackjack players, although we bare the brunt of it. Seems to be a whole new group of people in charge or that have risen to positions of power in casinos that think they don't have to play by the rules (and laws). They need a refresher course.
The reason you don't leave is because you have nothing to lose, and nothing else. This! is all you have.
And as far as the healing of wounds, you continue with your nonsense you still post about threatened violence against forum members with pistol and sword, wished death upon forum members, and expressed glee when forum members have died. Also you have been caught lying about most everything and anything about yourself, a proven and outed pathological liar, with the latest being that nonsense about how oh yeah, actually - my backrooming incident was on the news, twice! :rolleyes:Quote:
Originally Posted by MaxPen
Nothing has changed with you, in fact, you've gotten worse. And THAT'S :D the difference between you and most people.
I most certainly have not expressed glee when forum members have passed away or even announced they were ill. No matter how shitty someone has treated me I always with them the best and express condolences at their passing.
At least until now. But I will tell you what. How about you die and we will see how I feel about it. :cool:
I'm not sure what is more deluded thinking - you believing anyone believes your stories or you telling everyone you don't really care.Quote:
Originally Posted by kewlJDude, if I really cared, I would have left this forum several years ago. And if I couldn't live without posting I would have returned with a new identity, like .....well Mdawg.....your reincarnated identity of a formerly banned member. :DQuote:
Originally Posted by MDawg
And I don't keep posting about the lawsuit. It is always you who brings it up, as you just did. ;)
There was no lawsuit. But there was a back-room incident and several months later a settlement. I don't really need to bring up either. I shared the backroom incident as a sort of warning to players, that this sort of thing that most of us thought were a thing of the past (mid 2000's) is happening again with more frequencies. And not just to blackjack players, although we bare the brunt of it. Seems to be a whole new group of people in charge or that have risen to positions of power in casinos that think they don't have to play by the rules (and laws). They need a refresher course.
Quote:
Originally Posted by MDawg
As Norm said,
with reference to your threatening forum members and saying that you will badmouth dead admin for fifty years after their death (in your case, more like ten because withyou'll be lucky to outlive anyone),Quote:
Originally Posted by soxfan
"Who does that?"
Einstein had a notoriously terrible memory, so he let libraries maintain the trivia. But, once, on a train, unable to find his ticket, the conductor told him that he may ride for free. Einstein noted that he needed his ticked to find out where to get off the train.Quote:
Originally Posted by Dan DruffThey had some kind of contest like this in my son's school.Quote:
Originally Posted by 1Hit1der
He rattled it off to 56 places, which is more than I ever had known. I think the most pi places I knew was like 15. All I remember now is 3.141596254.
He then told me that the kid who won knew it to like 180 places. Useless info.
Anyway, I'm sure that the precise digits of pi are numerologically predetermined. One way may be by empirically (by trial and error) finding its spot on the mirrored part of the graph of the fine-structure constants.
(-2*9.01222434314022^2 + 2*9.01222434314022 + 113) = -31.41592653589791 ---> 3.141592653589791, numerologically speaking.
And, as with determining the fine-structure constants, themselves, it's sufficient to work with only the first nine digits of the value 9.0122243[4]314022, namely, 9.0122243[4], and, then, to reflect, and next double them up, to a repeating length of seventeen digits, namely, [9].0122243[4]3422210[9] repeat, which only twice repeated in the equation above, becomes something like (-2*9.01222434342221090122243434222109^2+2*9.0122243 4342221090122243434222109 + 113) = -31.41592654549941 ---> 3.141592654549941, the physical analog of the (exact) mathematical value of pi.
The mirrored and repeated part, 9.0122243[4] = (9.01 + 0.00222 +0.00000434) = [9.01 + 0.00222 + (0.00000343 + 0.00000090 + 0.00000001)] = {9.01 + 0.00222 + [(0.007 * 0.007 * 0.07) + (0.00000030 + 0.00000060 + 0.00000001)]} = {(0.01 + 6 + √9) + 10*0.000222 + [(0.007 * 0.007 * 0.007*10) + (√(0.00000000000009) + 0.00000060 + 0.00000001 )]} ---> 1691__2/7_7/2__1961, by expanding the outside terms by two digits as contract the inside terms by two digits on each side of it. Note that the same things may be done to end up with 6_9__1271_1721__9_6 instead.
More thus perfectly constructed outcomes similar to the ones at https://vegascasinotalk.com/forum/sh...l=1#post171482 .
https://youtu.be/p_HkvlO8r6E
Now if Druff doesn't mind, too much, I will finish off my post above, about how pi relates to the physical fine-structure constants. All of the known mathematical constants do.
Note as well that things above might have ended up with 1691__1271_1721__1961 instead, which is even more in line with the results at
https://vegascasinotalk.com/forum/sh...l=1#post171482
And, that (-2*9.01222434314022^2 + 2*9.01222434314022 + 113) = -31.41592653589791 ---> 3.141592653589791 was chosen because it's the spot in the thus equations that naturally begins with the first digits of pi, namely, -31 ---> 3.1, given that (-2*9^2 + 2*9 + 113) = -31, on the 113-side, which has to do with the fine-structure constants 137, and 142.
Now to finish things off by bringing the numeral, 113, from the 9's, and 4's, of [9].0122243[4]3422210[9], which is the usual way of numerologically expanding the digits in those positions. Interestingly, there's 9 = [3^2 + (1 - 1)], 0122243 = 11*11113, and, [³√(11 - 3)]^2, which together is 311__11_11113_3113 ---> 31111_11113__3113 ---> 311_113__311_113, by overlapping the sets of four 1's, on the left-hand side, and, splitting off the sets of two 1's, on the right right. Note that exponents under other exponents, here 3 under 2, become part of the result.
What is the basic relation between pi, and the numeral, 113? "335/113 is the best rational approximation of pi with a denominator of four digits or fewer, being accurate to six decimal places. It is within 0.000009% of the value of pi, or in terms of common fractions overestimates π by less than 1/3748629. The next rational number (ordered by size of denominator) that is a better rational approximation of pi is 52163/16604, though it is still only correct to six decimal places. To be accurate to seven decimal places, one needs to go as far as 86953/27678. For eight, 102928/32763 is needed."
And, so, another very curious set of equations with their own corresponding forms, to do with the digits of 1961, and 1721, again, matter/antimatter pairings of the results.
#1. [(19 + 0^0=0)^2 - 6 + 1 - 0^0] = 355, or, 355 + 1,
#2. {[12 - (antilog(-0^0)=1)^-1] + 7^3 + 1 - 0^0} = 355, or, 355 - 1,
#3. {[(0100 + ²√(900) + antilog(0^0)=10] + 6^3 - 1 + 0^0} = 355, or, 355 + 1,
#4. {[0010^(2^0) + 2^(²√(009)) + 0^0=01]^2 - 7 + 1 + 0^0} = 355, or, 355 + 1. And,
#5. [(-1 - 9 - 0^0=01)^2 - 6 - 1 - 0^0] = 113, or, 113 + 1,
#6. {[-(1 - 2) + (antilog(-0^0)=10)^-1]^2 - 7 - 1 + 0^0} = 113, or, 113 + 1,
#7. [(-0010^(2^1) - ²√9 - 0^0=0) + 6^3 + 1 - 0^0] = 113, or, 113 - 1,
#8. {[-0100 - 2^(9 - ²) - antilog(0^0)=1] + 7^3 - 1 + 0^0} = 113, or, 113 + 1.
Note that, in equation #3, the 0^0 = 01 reverses to 10 = anti-log(0^0). And, that, in equations #3, #4, and #6, #8, respectively, that the 009 reverses to 900, and, the ²√9 reverses to (9 - ²). The latter reversal involves being able to reverse a numeral, but, not a square root (in the exponent on 2), versus, not being able to reverse a single-digit numeral, but, being able to reverse the square root, and the thus numeral, by writing both as numerals (in the exponent on 2).
Again, I had to resort to the anti-matters numerals of 113, and 355, in terms of symbolically creating them from the left-over amounts, by 0^0 for 0, or 1. To write the above equations with 113_311, and 355_553, on their respective right-hand sides. Then (10 - 1 - 3^2) = [3 - 1 - log(100)] = 0 ---> 113, or, (3 - 1 - 1) = 1 ---> 113, and, [3 - log(500 + 500)] = 0 ---> 355, or, (-3^2 + 5 + 5) = 1 ---> 355.
As well, the 012243 part of [9].0122243[4] = [(-3 + 5 + 5) + (5 - 3)^2 * (30550 + 3^2)] ---> 355_553__355_553, which corresponds to 311_113__311_113, by stripping a second 5 out of the (5-3)^2 term, because it's squared, and, by splitting off, or shedding of, the two 5's overlapped in the (30550 + 3^2) term. As well, 0122243 ---> 342221 = 11*311110 .
And, oh, I had to go with ²√ for the square root of, and, ³√ for the cube root of, because the formatting here doesn't allow for as compact a thus sign for square root, as it does for cube root as ∛.
I realized, later on, that if 0122243 = 11*11113 can reverse to 3422210 = 11*311110, then the same may hold for 0122243 = [(-3 + 5 + 5) + (5 - 3)^2 * (30550 + 3^2)]. Let's take a quick look.Quote:
Originally Posted by 1Hit1der
3422210 = [-(5 - 3)^2 * antilog(0^0=1) + (3 + 5 + 5)^2 * (3^2 * 05*50 * 3^2)] ---> 355_553__355_553, as before, with another 5 upfront pulled from the (5 - 3)^2 term, in the same way as before. Very amazing. Along with noticing that the days are getting longer, again, with some early morning sunshine. Ha.
Ah, more one interesting, but nutty, thing. The whole numeral, 901222434, without breaking it up into its thus logical components, becomes 901222434 = 2 * 3^2 * 7 * 7152559 = {2 * 7 * 1 * √9*√9 * [(660 - 101) + 7152000} ---> 2/7 of 1961 with 1/7152, by overlapping the twin 9's, and 6's, together, over my randomly generated, and often repeated, user-numeral, 7152, from the gematria forums. Numerals, as the "gift that keeps giving". Wow!
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Well, I knew I was good, but, I didn't think that I was this good. Ha.Quote:
Originally Posted by 1Hit1der
Turns out that if reverse the whole numeral, 901222434, without breaking it up into its thus logical components, 434222109 = [(100000 + 77 + 2*10000 + 1000*7) * (7 + 2 + 1*7 + 1) * (5 + 2 + 60) *√9] ---> 1721_1721 ---> 1721, by overlapping each of the twin 7's, and, next, the twin 1721's, with again 7152, but no longer inverted, and, 6/9 as June 30 by 30 = √(900) ---> 9, for the middle of the year, instead of July 2.