- Previous Private Message -
Sent By : CWILSON
Hi CWILSON,
The intent of the Offset9 system is to isolate 5 of the possible 10 digits and to play those 5 digits in all combinations, giving the player 10 picks.
Example: List the last 9 winning draws for the Pick 3 game you wish to test it on.
730
250
003
446
781
349
261
315
117
Now, figure the offsets for each pick by listing the digits (plus 1 and minus 1) for each individual digit.
Example: Offsets for 730 > : The first digit 7 > offsets are 6 (7 - 1) and 8 (7 + 1) : The second digit 3 > offsets are 2 (3 - 1) and 4 (3 + 1) : The third digit 0 > offsets are 9 (0 - 1) and 1 (0 + 1)
I realize that 0 - 1 is actually -1 but in this system, the numbers are put into a loop where 1 comes after 0 and 9 comes before 0, so 0 - 1 will always = 9.
Here are the offsets for the above picks...
730 > offsets > 629 and 841
250 > offsets > 149 and 361
003 > offsets > 992 and 114
446 > offsets > 335 and 557
781 > offsets > 670 and 892
349 > offsets > 238 and 450
261 > offsets > 150 and 372
315 > offsets > 204 and 426
117 > offsets > 006 and 228
Now, just focusing on the offset numbers, you want to add up all of the 0's for a total. Then the 1's, 2's, 3's, 4's, ect... to get the offset occurance totals as follows...
These are the total number of occurances from both columns of the offset numbers...
0 = 6
1 = 6
2 = 8
3 = 5
4 = 6
5 = 5
6 = 5
7 = 3
8 = 4
9 = 5
Do you see how I did that?
Now, you want to add up the occurance totals from the actual winning draws. That's the first column of numbers I have listed starting with the winning pick > 730 and ending with 117. We want to know how many 0's are there. Then the 1's, 2's, 3's, ect...
0 = 4
1 = 5
2 = 2
3 = 4
4 = 3
5 = 2
6 = 2
7 = 3
8 = 1
9 = 1
Now, we want to subtract the actual winning pick occurances I have just listed, from the offset occurances, so next you will see the combined occurance lists. The offset values are listed first, followed by a minus sign, then the actual winning pick values are listed to form a simple mathematical expression...
0 > 6 - 4 = +2
1 > 6 - 5 = +1
2 > 8 - 2 = +6
3 > 5 - 4 = +1
4 > 6 - 3 = +3
5 > 5 - 2 = +3
6 > 5 - 2 = +3
7 > 3 - 3 = 0
8 > 4 - 1 = +3
9 > 5 - 1 = +4
Therefore...
0 = +2
1 = +1
2 = +6
3 = +1
4 = +3
5 = +3
6 = +3
7 = 0
8 = +3
9 = +4
At this point you can see what numbers are hot by their value. The 2 and the 9 are hottest and the 1 ,3 and the 7 are coldest. The 4, 5, 6, and 8 are hot as well. BTW, the value after subtracting actuals from offsets CAN be a negative number... Example, 3 - 5 = -2. It just so happens in the above example that all values are 0 or positive numbers.
Let me just state right now that the next draw for this example was 657 and the Offset9 system did not produce that winner, because the 7 wouldn't have been chosen to play if the player were only selecting the hottest digits. But you can think about the results this system produces and come up with a method of selecting numbers that may produce above average winners.
If you do these calculations on a daily basis, using the most recent 9 draws, you will start to see increases and decreases in the frequency of each of the digits. Then you can graph those results and bet on the upward trending lines those results would produce. A daily procedure such as graphing is quite an eye-opener!
I know it's a lot of work, but it may be worth it.
Like I said in the original post, I haven't tested this system out and only interpretted the data in a somewhat popular way by playing the hottest digits only. But you may come up with another way to interpret the results and improve it greatly!
Like they say in Meteorology - "Warm air finds a colder place." So with hot and cold digits - sooner or later the cold ones will heat up and the hot ones will cool off. It's just a matter of how a person looks at it combined with timing and patience.
I'll go ahead and post this in the Lottery System forum so those interested can better understand how it works.
Good luck!