Hello Folks. This is my small contribution to all the great people on this site.This is for the Pick 3 system. Pick 4 will be done later. The workup takes 15 minutes. With a small program, it'll take seconds.
My method is called DSUM as in 'dee-some'. No kidding-It works 100% of the time. Let me repeat, **100%**. I have used it and I whave won on Betslips everytime I have played. Its been $37.50, $75.00, $300.00, etc...everytime I have played. The only time I don't win or don't win consistently is when I try other systems and start experimenting.
I win every draw that is held. The small problem is filtering just a little bit more. Just a little. Maybe you guys can help refine. Its not complicated. Here it goes:
The DSUM method starts out with the base 10, which is the source from which all numbers spring. The base is the wrap method.ou know, you take the last draw and add 1 to each number until the number wraps around to itself. For example take the number 472--
"D" is for down. You only ever need go down by 1 digit only. Example is 7 goes to 6, or 9 goes to 8.
"S" is for Same. The number stays the same.
"U" is for Up. You only ever need go up by 1 digit only. Example is 7 goes to 8 or 9 goes to 0.
"M" is for Mirror. You put in the mirror of a number. Example is 9 goes to 4 or 6 goes to 1.
You only have to go up and down by 1 digit only.
Now each 3 digit number in a draw has only 64 possible permutations. 16 for "D", 16 for "S", 16 for "U" and 16 for "S". Using the draw 472. You just have to permutate 472 through D-S-U-M and then wrap down that permutation. Take a look and notice how you will get many duplicates. All but one can be eliminated. Again, its not complicated. Stay with it. Here's the formula
DSUM
of 4 7 2
The D's
1 2 3 4 5 6 7 8
DDD DDS DDU DDM DSD DSS DSU DSM
361 362 363 367 371 372 373 377
472 473 474 478 482 483 484 488
583 584 585 589 593 594 595 599
694 etc... etc... etc... 604 605 etc... etc...
705 Wrap each of these 8 columns down.
816
927 ETC
038
149
250 251 252 256 260 261 262 266
9 10 11 12 13 14 15 16
DUD DUS DUU DUM DMD DMS DMU DMM
381 382 383 387 321 322 323 327
Wrap each of these 8 columns down.
Now we do the S's for "472"
17 18 19 20 21 22 23 24
SDD SDS SDU SDM SSD SSS SSU SSM
461 462 463 467 471 472 473 477
572 573 574 578 582 583 584 588
Wrap each of these 8 columns down.
25 26 27 28 29 30 31 32
SUD SUS SUU SUM SMD SMS SMU SMM
481 482 483 487 421 422 423 427
Wrap each of these 8 columns down.
Now the U's for 472
33 34 35 36 37 38 39 40
UDD UDS UDU UDM USD USS USU USM
561 562 563 567 571 572 573 577
Wrap each of these 8 columns down.
41 42 43 44 45 46 47 48
UUD UUS UUU UUM UMD UMS UMU UMM
581 582 583 587 521 522 523 527
Wrap each of these 16 columns down.
Now the M's for 472
49 50 51 52 53 54 55 56
MDD MDS MDU MDM MSD MSS MSU MSM
961 962 963 967 971 972 973 977
Wrap each of these 16 columns down.
57 58 59 60 61 62 63 64
MUD MUS MUU MUM MMD MMS MMU MMM
981 982 983 987 927 922 923 927
Wrap each of these 16 columns down.
The number of the next draw will absolutely be found in the above wraps.
There will be about 38 - 40 columns that will be duplicates and those can
eliminated. You will have 20 columns of the 64 that will used to find the next
draw. You can look at the numbers by columns or by rows. Rows are
interesting , because one number in the original base 10 can always be
permutated by one of the DSUM manipulatins to find the winning answer.
Again I have laid out all the possible permuations. Since you don't know how
to exactly tweak or permutate the number you play the whole string across.
You will end up with 20 numbers. The problem I'm having is finding a good
strong pointer in one row of the original base 10 column. Remember, the
original base 10 is to take the last draw and do the wrap. Perform DSUM on
that number, and play the string of numbers in that row. If you're not sure
whcih number to pick, like I often am you will have to pick ad just not from
only 1 row, but all ten rows of 20 colums.
At this point I look at the last five draws to help me elimante the numbers
that will not probably come out. If playing by betslips it becomes very
profitable. I'll eventually try to end up with 50.
Anyone can apply any other method to help pull out the winning
number. You can use any other valid method on this site to filter out the
unlikely numbers, and come up with the winning draw. Remember the
winning number is there 100% of the time. The only problem is finding the
valid pointer. Another strong technique I combine with the above method is
to help find the next draw is to take the previous draw, and find the six
combos. For example, take 472 again. The combos are:
472
427
274
247
742
724
In the case of doubles you would have 3 combos. Now you take the
combos and divide each by pi - you know 3.14. Put the result in 4 columns.
In the 1st column is just the result from dividing by pi. In the 2nd column
you add 111 to the original result. In the 3rd column you add 666 to the
original result, and in the 4th and last column you add 888 to the original result.
I have found that you don't really need the 888 column. I just like to. Often
times you will see the numbers converging in a pattern, they will
converge on a certain number. Even if they don't I will merge this second
technique with DSUM. I always come out with the right number.
If you back-test DSUM you will find the winner is there
everytime. The system is very profitable online with betslips. I made about
$600 bucks in several days, by winniing incrementally in every draw. I just
don'y like playing many numbers. Can any brilliant mind find a somewhat
perfect way to filter? or simply or I guess rather not simply, find the way to
the row in the original base 10. If you can, oh boy--you will win everytime,
beacsue the permutaition willbe in that row. If you were to play all 64
'tweaks' you would hit STRAIGHT WITHOUT FAIL!!!
My pick 4 method is like the one above. Its just that there are a lot more
columns. It works just as well. I have done the Pick 5 work through as well.
Only thing is that I have to make sure that there is the same resonance with
the numbers that exist in Pick 3 and Pick 4. I'm trying to see how my DSUM
would work on the numbers. For example to go down a digit in Pick 5 would
I do 23 to 22 or 23 to 12. The same goes for the mirrors of these numbers.
I've beenavoiding it beacuse it will be a herculean effort. I needto leran
some programming.it wille done though, because the implications, at least
for me, is mouth-watering. This DSUM method works for me.
Oh by the way,
another much simpler way to always get the final draw it to forget the DSUM
Wraps, and just take the 6 combos of the last draw divided by Pi with 111,
666, and/or 888 added to them and perform a DSUM string for eah
number without the Wraps on them. Then filter out the duplicates and the
unlikely numbers. The winning draw will always be found there. I hope I
was somewhat coherent in my explanation. and I hope some people find
this very useful. I'm working on finding the strong pointer. Oncei get it. It
will be shown. Wouldn't it be great to play just 20 numbers, and know that
wiithout a doubt your winning number will always be found there. Until
later....Ciao.
Kola
Rookie Time TravelerMilky Way Spiral
