Substitution works!
Just about every lottery player using a system or workout to guide their activities encounters
'n-sequence' data strings of one form or another.
Usually, the challenge is to evaluate the 'endings' and try to use them to guess what the 'next'
digit might be.
The next digit being the lottery number to play.
While I'm certain there is a mathmatical solution, I've yet to find a formula or process that
will provide correct answers in a majority of cases.
The best solution and the one I follow is to organize the facts and then use intuition, experience, and other human (brain matter) qualities to choose specific numbers to play.
My assumptions are often correct, but my win/loss record is nothing to brag about.
The workout begins by substituting winning lottery numbers with other numbers.
For example, in my Texas Daily 4 game, lottery digit '5' is '1A', '9' is '2A' and '1' is '3A'
This is a fixed conversion governed by a 'key.'
Substitution table:
5, 9, 1 = 1A, 2A, 3A
6, 8, 7 = 4B, 5B, 6B
4, 3, 2, 0 = 7C, 8C, 9C, 0R
Here is the data arrangement used for choosing numbers to play for the 062313 Daily 4 drawing.
a b c d
41 1A 7 5 15 4B 1 6 19 7C 2 4 30 0R 29 0
2 2A 19 9 25 5B 5 8 2 8C 16 3
43 3A * 1 55 6B 6 7 11 9C 3 2
a - from tracking charts showing the number of times an item has come up in the
workout since 010612.
b - key assignment for lottery number.
c - from tracking charts showing the gap/skip for the key assignment.
Example: 1A has missed 7 consecutive opportunities, 2A has missed 19 opportunties,
3A was the last 'A' logged, as indicated by an asterisk.
d - lottery number assignment according to a key.
USAGE
According to 'n-sequences' in other tracking charts, the 'next' winning 'structure' could be
a 'triple', or, specifically, 'BBB'.
Note: Triple in this instance is NOT repeat NOT a lottery combination with same digit
repeated 3 times.
The task at hand is to choose three lottery numbers from the "B" group, which is an automatic
choice given that there are only three choices - 6, 8, 7.
Other tracking charts suggest that of the three possibilities, structure
"A687" would be the best choice.
Or, depending on how much money a player wants to spend, all three structures (A687,
C687, R687) could be played.
The task at this point is choosing a lottery number from the 'A' group.
There are 3 possibilties.
3A has been the most popular and was the last 'A' in the tracking chart. Will it repeat?
2A has missed 19 opportunities and is 'due'
1A has been popular.
Referring to the 'C' group.
'0R' is the most popular, followed by '7C'
'8C' has been 'out' the longest, missing 16 consecutive opportunities.
In this instance, the fourth winning digit came from the 'A' group.
The exact sequence would be based on 'other' tracking charts.
5687, 9687, 1687 or other combinations.
What is your choice??
While using inventory totals and 'times out' data is sufficient in choosing winning
lottery numbers, additional analysis is helpful.
This about to get a little more complex.
FACTS
GA
313 - 3322133223
13 - 3121331123222212123323332133223233233332221221233
3 - 1/67 2/62 3/58
FA
122 - 322322131231
22 - 313233231333311221323311333122231331
2 - 1/51 2/52 3/68
F/G
3.2
2.1
1.3
Note: The next digits in the sub-strings were determined by using a search tool
developed by Jimmy.
I entered a complete string and then asked the tool to scan for specific n-sequences.
The results are as shown.
EXPLANATIONS
GA - Group A
313 - last 3 digits in a string generated without regard to lottery number position in winning
combinations since 010612. There are several hundred digits in the overall string.
(N-sequences for strings generated by lottery number position are different)
13 - last 2 digits in the string.
3 - the last digit in the string.
OBSERVATIONS
1. N-sequence 313 has come up ten times with the 'next' digits as shown.
2. N-sequence 13 has come up 50 times with the next or third digits as shown.
3. The last digit in the whole string is '3'
A breakdown in a 3x3 matrix shows 3.1 has come up 67 times, 3.2 62 times and 3.3 58 times.
FA - Followers for GA
OBSERVATIONS
1. n-sequence 122 has come up 12 times with the next digits as shown.
2. n-sequence 22 has come up 36 times with the next digits as shown.
3. The last digit in the string is '2'.
A breadown in a 3x3 matrix shows 2.1 has come up 51 times, 2.2 52 times and 2.3 68 times.
F/G is a scale.
For example, if the next digit chosen in the FA group is '2', the next GA digit is '1'
Thus FA 2 equals GA 1, which is 1A.
1A equals lottery number '5'
Thus the final combination based on structure ABBB is "5687"
The winning combination was "8765", which showed up on June 29 Night drawing.
To answer the obvious question - When I decide to base my lottery combination
on a 'triple structure,' I usually keep it in play until that particular structure arrives.
In this instance, I chose '2A' , lottery number '9' and missed winning $100 by one number!
HELP NEEDED
I sure would appreciate it if someone could assist me in finding a good method for
determining the 'next' digit in those 'last 3' and 'last 2' sub-strings.
I usually rely on visual analysis, or construct additional charts.
This is important because the next digits selected from the sub-strings determines
which lottery numbers will be played.
I realize that the absence of "other strings" distorts this posting, but this is not the
time nor place to describe the complete workout.
Thanks for your interest.