Braselton Ga United States
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January 14, 2012
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I had giving up on cash 3 & 4 till i got the HT program and The Slider i take what i think is best i only play 1 to 3 lines and i play those for 7 days. HT color history tell me what to look for and the Slider tell me what to play. Frankenstein is all i got they have been good to me Frankenstein HT and Frankenstein Slider. I guess i will call the new program Franken Thing. I don't think your old programs will never Die either.
United States
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August 26, 2012
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Quote: Originally posted by RL-RANDOMLOGIC on Jun 4, 2017
Guys
Nothing special about rogues and I suppose one could use just about any method that breaks the game down
into defined blocks. The idea behind it is to shrink the pool of choices. In pick-3 we pick 3 digits each from a
pool of 0 to 9. Using rogues requires picking 5 digits from a pool of 0 to 3. If we select a value for any 4 of the
5 rogue positions and let the 5th run wild then a setup will produce 4 lines. If we let 2 of the 5 rogues run wild
then we have 16 lines. Which is easier, picking 3 digits from 0 to 9 or picking 5 digits from 0 to 3. We can't shorten
the odds for the game and like JW said the rogues for pick-3 - 1024. A pick-3 has 1000 total combinations so what
happens to the extra 24. These are the rogue values, ghost values that only exist so that any matrix can be divided
equally by 4. To get the total rogue positions for a matrix we multiply by 4 until the value is >= to the total lines in
the matrix.
5-39 = 575757 total combinations
rogue position A = 4
rogue position B = 4*4 = 16
rogue position C = 16*4 = 64
rogue position D = 64*4=256
rogue position E = 256*4=1024
rogue position F = 1024*4=4096
rogue position G = 4096*4=16384
rogue position H = 16384*4=65536
rogue position I = 65536*4=262144
rogue position J = 262144*4=1048576
All the so called rogue values, ie values greater than the total lines in the matrix will end up in block-3 in rogue-A
This is why with some games Rogue-A is limited to 0 to 2. Here is a picture of the old rogue program where I set
the first 8 of the 10 rogue values. Eight rogues will reduce a 5-39 to 16 lines and every line will match at least 3
numbers. The big question is, is it easier to pick 5 to 6 numbers to play from a pool of 39 or choose 8 rogues from
a pool of 0 to 3? Look at the history for Rogue A. 13 of the last 18 games it was a (1). The thing tries to predict
the rogue value for each position.
We start with dividing the total rogues into 4 groups where each group for my 5-39 will have 262144 rogue values.
inside each of those groups the we have 4 blocks where each block has 65536 rogues and inside each of those groups
we have 4 more blocks each have 16384. Each block is divided by 4 until each block only contains 1 rogue. Why do it
this way you might ask, lets take pick3 where we have 1000 total sets. If we divide 1000 by 4 we get 4 blocks of 250
lines. So far so good but when we divide 250 by 4 we get 62.5. Hmmmmm. now if we divide 62.5 by 4 we get 15.625.
Do we round up or round down? Each block needs to be divisible by 4 because the rogue as I call them are a address
of sorts.
RL
Well, your way works for you and that is all that counts.
In the old filter way we could try something like this:
California (CA) Daily 3 Midday Sat, Jun 03, 2017 2-9-1 Fri, Jun 02, 2017 7-2-6 Thu, Jun 01, 2017 5-3-1 Wed, May 31, 2017 3-1-1 Tue, May 30, 2017 7-2-8 Mon, May 29, 2017 8-8-2 Sun, May 28, 2017 9-7-0 Sat, May 27, 2017 8-4-5 Fri, May 26, 2017 1-4-4 Thu, May 25, 2017 7-5-6 Wed, May 24, 2017 7-4-1 Tue, May 23, 2017 5-7-8 Mon, May 22, 2017 5-4-3 Sun, May 21, 2017 2-1-8 Sat, May 20, 2017 1-8-5 Fri, May 19, 2017 4-0-9 Thu, May 18, 2017 7-7-8 Wed, May 17, 2017 9-8-5
What I am going to show is "after the fact" so there is no real prediction done here, it is just an example.
Let us say that it was Jun 1, 2017 and we wanted to predict right the next winning number on Jun 2, 2017.
Fri, Jun 02, 2017 7-2-6
For position 1 a 7 would come out next so we would filter out Even Digits (0 2 4 6 8), Low Digits (0 1 2 3 4), Out Digits (0 1 2 8 9).
On position 1 we started with all 10 Digits= 0 1 2 3 4 5 6 7 8 9
Filter Out:
Even Digits, so now we have 1 3 5 7 9 Digits left.
Low Digits, so now we have 5 7 9 Digits left.
Out Digits, so now we have 5 and 7 Digits left.
Now let us look at my:
Bala-Nced Filter
-------------
1 2 4 5 8 = Bala = B
0 3 6 7 9 = Nced = N
So we have a 5 and a 7 left on position # 1
If we filter out the Bala Digits, we now have a 7 left on position # 1
And the next digit for position # 1 was a 7.
As you can see, this is one more reason why we also need the Bala-Nced filter.
All of this was done after the fact, we had to be right 4 times on a sort of heads and tails guessing game, but with the lottery games we have a past history of numbers-patterns.
Here I showed 2 patterns digits filters only.
You have to pick right 4 times for each position so 4 X 3 = 12, you would need to be right 12 times, to filter down to just 1 pick 3 straight, if you don't chose right you might end-up with either:
One or more non-winning number(s) or no predicted number at all.
Or maybe too many predicted numbers, winning or not.
RL's software program probably beats this old way of predicting.
It is very likely, almost for sure that I would not have been able to predict the 726 winning number with this particular method and with only 1 predicted straight number.
But it might have been possible with other prediction method and with very many more produced and predicted numbers.
Not all prediction methods are equaly accurate, filter too much and you might just filter out the next winning number also.
COLUMBUS,GA. United States
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June 3, 2004
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Quote: Originally posted by RL-RANDOMLOGIC on Jun 5, 2017
You many have noticed in the pick-3 history there are 5ea games where the first 2 digits were 2-2 out
of the last 19 games. 19*32=$608 and $300* 5 = $1500.00 - 608 = $892 in profits.
A quick search for 2-2 showing in rogue positions A and B shows 109 hits in the last 1000 games. Not much
profit but playing 22 every game would have covered ticket cost and made a small profit.
P.S. Notice the max skip for 22 is 80 games, playing $32.00 for 80 games with no hits would be hard on the
moral.
RL
RL,
If I lived in Mo, I would play these 5 combos in Eve draw, until the root 5 hit. All of the combos below are all over 1000 skips out. The root 5 is at max skips.
197
205
213
217
225
249
461
469
473
481
489
493
509
705
717
721
729
733
753
757
761
969
981
985
These are root 5 combos.
473
509
761
Good luck!! Hope root 5 hits tonight. If i only had $1!!