Hello:
Here's a very consistent system that will often identify at least 3 out of the 4 numbers in at least one of the next 10 upcoming draws.
If you look at the frequency statistics of consecutive pairs(01, 12, 23, 45, 56, 67, 78, 89) for the P4, you'll find that they appear somewhere between 60% - 80% of the time within a draw. This system is based on that statistic.
Here are the rules:
1. Consider the last 20 draws of the P4 you are playing.
2. Find 2 consectuive pairs from the list (preferably all different digits, for example don't choose 1223), where no three out of the 4 numbers, in those pairs, appear in any of the any of the last 20 draws.
3. In the next 10 draws, there's a good chance that at least three of the four numbers you chose will appear in at least one of the 10 future draws.
Examples:
For the test, we chose the P4 in Massachusetts, Georgia, and Connecticut. For the 20 day period we chose
Nov. 29, '08 to Dec. 18, '08.
The past 20 draws for Mass. were:
2148
9693
2613
5622
4715
7559
4083
7565
5458
9185
0858
1962
7588
1810
9094
7329
4286
2130
9006
0099
For the two consecutive pairs we chose 67 and 89, since no three numbers belonging to these pairs showed up in the last 20 draws. Eight days later(Dec. 26,'08), the number 7688 was drawn.
Georgia
The past 20 draws for the test period were:
7824
2168
2340
7813
2459
3133
7490
7379
6219
7648
6836
3731
9564
6167
9221
5159
2290
0594
4976
6106
We chose the same pair sets 67 and 89 as in the Mass. case, and on Dec.25,' the number 8729 was drawn.
Ct.
This one is very interesting. The past draws for the test period were:
7053
0462
6803
0289
1958
6997
2011
7808
5152
4279
1042
9779
7400
4470
7753
0142
2101
1902
3397
8057
In this scenario we noticed that were 3 sets of pairs that could be chosen to satisfy the rules:
(12-67), (45-67), and (01-67). On Dec. 19, 21, 23, and 24 the following numbers were drawn: 0546, 6410, 4456, and 2796.
The question that begs to be answered now is: How do we get the fourth digit to play? One way is to consider for the fourth digit the other 6 that are left(for non-double box). This would result in 24 numbers for each set of pairs selected. Another strategy would be to limit the plays to doubles. Notice that in two of the three tests a double was drawn. In this scenario, there would be 12 boxed plays per pair set chosen. Since doubles usually pay more at the local candy store, this might be a profitable strategy.
Comments welcomed.
jayemmar
