digit

[RAMROCK]

Hello Jack,

Ok. You say, look for all combs in the lotto file on the last digit part and select those who match the sequence 1,2,3 don't metter the position of the digit. This time I don't have a clue... i am helping you with the magic mirror...i am working in a depth filter by column trying to find out at least 4 digits in most cases... but i am on this.

Hello Ram, this is so for the last digit 0-9 or inside the suits of the last digit See if the position has mennos UMSA sequential example = 1,2,3 Look at the test 14,32,33,40,42 = 51 = last digit = 4,2,3,0,2,1 = 1,2,3 we have the sequence It does not matter ordemo this example the sequence of three digits 1,2,3 is the position 2nd, 3rd, and 6th = 2,3,1, no matter the position, but the sequence of 1,2,3, see how many threads Are possible 0-9 and their statistical, 1,2,3 5,6,7 ... etc. results is that the pair It is sequential, one can crial a sequential filter all together even and odd pairs and sequential single sequence x = 0,2,4 only odd, 1,3,5, all together (even and odd) 1,2,3 3,4,5 etc. The position may be the biggest in the 1st place!! says you understand?

Ramrock!

 

[jack]

Hello, ram is easy, below is a list of possible training sequences in the last digit
There are study sequences of numbers like 12,13 15,16 etc, but my study is suits
Sequential, eg 7,8,9, no matter the order in 20 terms may suit
Be in position 9,7,8, ok formations below the sequences
0,1,2
1,2,3
2,3,4
3,4,5
4,5,6
5,6,7
6,7,8
7,8,9
8,9,0
They are in training in 9 threads for last digit 0-9 ok, and as sequential numbers
But it is the third digit, look at this example = 4,5,9, this example is not sequential ok

 

[jack]

Hello, look at this last digit result = 05,19,26,47,50,58 = 5,9,6,7,0,8 What are the last digit in sequence? Answer = 6,7,8, also has the same draw e7 = 5,6,7, 8.9 There are nine groups or formations of sequence ok this filter only last digit

 

[jack]

Hello, ram the dismemberment of pairs of 60/6 are 15 positions, it also helps
Herein, when 99% of the lottery has a pair followed (in sequences)
Are the following pairs of sequences =
0.1
2.3
3.4
4.5
5.6
6.7
7.8
8.9
1.0
0.9
But within the 15 rotating positions, par = 4.8 sample 4 may be in 2nd position and 8 in the 5th position, also sequences of pairs of digits only pairs = 0.2 or 1.3 only odd
The dismemberment of the 15 positions of the pairs will give clues to the formation of the suit

 

[jack]

Hello In the example above is the last pair of digits, but You Can See ram threads Pairs covering or joining the first digit and last digit example Resulatado = 12,15,25,42,56 = the number 12:56 has the double-digit sequences Both are right or left, It Seems a good filter then separated from the last digit The sequence together the two digit = first digit and

 

[jack]

Hello, ram in a lottery 60/6 have a total of 34,220 possible suits The table below the sum of each number = 1,10,19,28,37,46,55 = 1 for the sum 2,11,20,29,38,47,56 = 2 for the sum = 3,12,21,30,39,48,57 to sum ​​3 4,13,22,31,40,49,58 = 4 for the sum 5,14,23,32,41,50,59 = 5 for the sum 6,15,24,33,42,51,60= to the sum 6 7,16,25,34.43,52 = 7 for the sum 8,17,26,35,44,53 = 8 for the sum 9,18,27,36,45,54 to the sum = 9 Ram 20 to the sum of a result of such suits = Sum = 12,15,26 = 3,6,8 do with the other 19 suits each outcome of sweet sena Through the sum, if you can see if there is repetition of the sum by position, or you have something better in your studies,

 

[jack]

Hello Ram goal, and see how many suits have in the example = sum = sum = 4,2,6 suit how many suits we have within this sum? = 4,2,6,
Below the list of all possible sums of all suits =
Each result has 20 of these sums, you can create a filter
So within the 34.220 suits have fit within the 84 combinations of sums =
123
124
125
126
127
128
129
134
135
136
137
138
139
145
146
147
148
149
156
157
158
159
167
168
169
178
179
189
234
235
236
237
238
239
245
246
247
248
249
256
257
258
259
267
268
269
278
279
289
345
346
347
348
349
356
357
358
359
367
368
369
378
379
389
456
457
458
459
467
468
469
478
479
489
567
568
569
578
579
589
678
679
689
789

 

[jack]

Hello, ram is an example of a lottery of five numbers but you can use the concept to A lottery 60/6, the filters are the same k = NS = 6 (Numbers Sum). The Digit Numbers Sum Filter is Calculated by adding all five numbers in the drawing set and apply the Following formula: CINT (((HIT) ^ .5) / (3.14159 / 2) -2.51) For example, the drawing on 24 27 31 45 52 Formula: 24 + 27 + 31 + 45 + 52 = 179 179 ^ .5 = 13.37908816 (Root of 179) Divided by 2 = 3.14159 1.570795 Divided by 1.570795 = 13.37908816 8.517399253 Minus 2:51 = 8.517399253 6.007399253 Rounded = 6 So filter NS = 6 DM = 3 (Digit Mirrors). The Digit Mirrors are Calculated by taking the digits 1 through 9 and Creating a list of Mirrors Sets Mirror Set Listing: Digits 1 & 6 Digits 2 & 7 Digits 3 & 8 Digits 4 & 9 Digits 5 & 0 We DO NOT count the 0 (zero) in any of the number 01, 02, 03, 04, 05, 06, 07, 08, 09 For example, the drawing on 24 27 31 45 52 we have the digits 2,4,2,7,3,1,4,5,5,2 we have the Mirrors 2 & 7. The total equals 1 Digit Mirrors So filter DM = 1 Another example, drawing 03 15 29 35 54 w have the digits 3, 1, 5, 2, 9, 3, 5, 5, 4. The Mirror is the only Digit 4 & 9. So in this example the Scan Mirror or DM = 1. OD = 4 (Odd 2nd Digit). The OD Odd or 2nd Digit Filter is Calculated by taking all five numbers drawn from The Set then using the right digit from all five numbers drawn and count the total number of Odd Digits For example, the drawing on 03 15 29 35 54 Odd the Digits would be 3, 5, 9, 5. This would equal a total of 4 Odd Digits. So filter OD = 4

 

[jack]

Hello, the above example, you have to do for the 20 suits ok, will have to adapt The even and odd for the suits, you can use to filter your idea too.

 

[jack]

Hello, ram you have, do, is to exclude a mirror equivalent, then the mirror is to filter or delete bets columns lottery example = termination (last digit) of the outcome of the 1st position is a mirror is 6 then the next draw the column 6 is deleted, but in the 2nd position 6 can enter the game

 

[RAMROCK]

Hi Jack,

what I understand all features around the lottery in the statistical frame belongs to the normal curve, it is a multi-dimensional event, so you can get the best statistical filters under the nomal curve, for example you can get 6 filters under the normal curve (first SD arround the mean) and those filters will normally distributed also. keep this in mind!!

Ramrock.!!

 

[jack]

Hello, perfect, hard work, because we have to deal with repetitions and repetitions do not
Number of intervals of past results, you can create with this weight, dimunindo or increasing the weight (in the presence of the number bet) the great advantage of the positional desbobramento in 20 suits, will help at least one term to hit any of the 20 suits can become (pivot or reference). When the mirror objective is to delete the mirror ok
ie if the number on the last draw of the 1st position the last digit
2 is the next in the same position not play 7

 

[RAMROCK]

Hello Jack,

The mirror filter works as you describe, What i am doing is to find out the way to forecast 4 numbers at least based on the last 4-5 draws. Example: Column#1,Column#2,Column#3...and so on!!
03,15,16
11,22,25
05,21,32
06,17,33
01,17,19

Last Digit in Column#1 (3,1,5,6,1) --->sorting (1,3,5,6)
Wich numbers are missing for last digit?: 2,4,7,8,9,0 (6 digits missing)

Did come one of the 6 digits missing in that column? and now, here and forever this follows the law of the combinations and they are normally distributed of course!!

Ramrock!!

 

[jack]

Hello, ram, it is easy you can put the last digit, all even or all odd, has the table in the post above, then one of 20 positions, will hit all playing in the last last digit even or odd in suits, good for encontar 4 digits in each vertical position using the three frequencies, looking about 10 drawings in each of the vertical positions of each three suits (20 terms then we have 60 columns, the three frequencies (medium, hot and cold, use two digits warm , a medium and a cold) so one of the 20 terms will hit, ram each of the 20 suits can be the pivot,

 

[jack]

endings is the vertical caption = digits The odd = E = pairs L = low (0,1,2,3,4) H = high (5,6,7,8,9) = 1,7,2,3,4 LHLLL 00EOE = 3,1,4,8,2 LLLHL 00EEE = 5,6,2,4,9 HHLLH OEEEO = 7,2,3,4,2 HLLLL OEOEE

 

[jack]

0 0 2
0 0 4
0 0 6
0 0 8
0 2 0
0 2 2
0 2 4
6 0 2
0 2 8
0 4 0
0 4 2
0 4 4
0 4 6
0 4 8
0 6 0
0 6 2
0 6 4
0 6 6
6 0 8
0 8 0
0 8 2
0 8 4
0 8 6
0 8 8
2 0 2
2 2 2
2 2 4
2 2 6
2 2 8
2 4 2
2 4 4
2 4 6
2 4 8
2 6 2
2 6 4
2 6 6
2 6 8
2 8 2
2 8 4
2 8 6
2 8 8
4 0 4
4 2 4
4 4 4
4 4 6
4 4 8
4 6 4
4 6 6
4 6 8
4 8 4
4 8 6
4 8 8
6 0 6
6 2 6
6 4 6
6 6 6
6 6 8
6 8 6
6 8 8
8 8 0
8 2 8
8 4 8
8 6 8
8 8 8

 

[jack]

hello, ram also the last digit of each of the suits can only be odd, and also ram, you can use each of the 20 suits like a pick3 because exatemente as a pick3 repeats a digit in a draw for another 65 %, then the last two pairings can form a pair of 3x3 are nine pairs, eg Position = 1st suit 457 168 9 pairs = 41,46,48,56,58,71,76,78,51 = 562 in the next gave the pair was 56 ram, also here you can use table mirrors dimunir the amount of 9, the third digit, being random, or three frequencies, so here is where the prediction is born see the pair is the last digit of the suit of each of the 20 positions

 

[jack]

Hello, ram, the system is above 92% at least one digit of the next When joining the last two draws last digit of the pair to find the pair So, you can have the pair in 60% and 92% in just one four left you go looking for this So you have two option of the last six digits of the two results, to form the sea of the other four digits

 

[jack]

Hello, ram I noticed this pattern, you can also use the 60/6 and 49/6 example = Two recent results 12,15,26,36,37,52 01,15,27,37,42,56 Divide the six positions in two groups the first group position = 1 st 2 nd 3 rd and the other group 4th 5th and 6th place, then we have two groups: group A (1,2,3) and group B (4,5,6) The pattern is so when the group A is the pair with 9 pairs in another segment or group the last digit, are four different, and when the pair is in grupoB in group A 4 digits are different from the last two of six draws, then either this pair in six in group A or group B in this, the third digit, we see ok. As we do not know which group is the pair has to make the two conditions are 9 pairs of group A and 9 pairs of group B. DIS IS UNDERSTOOD THIS PART!

 

[jack]

Hello, ram, the hard hit are the last digit (the terminals 0-9) Particularly as the first interval is short digits 0-5, and has the positions, it is easier to set Therefore the focus in the study are the endings (last digist), ram another filter are adjancente (delta) of the last digit, example = the last digit of a sweepstakes are = 2,4,6,5,9,1 Adjacent left and right of each digit = 123 345 567 456 890 012 Then in column 1 = 0s adjacent ous delta = last digit for not playing in 1st position is 1.3 Or example in the 5th position does not play the last digit = 8:00 with this if it takes two columns, and so with the other four positions, good filter