Next step help

Reply #17

crissie — Thu, 24 Nov 2011 00:56:57 GMT

hello thanks for the info sure looks like a lot of work lol. well continue to do well and best wishes with your system

Reply #16

dr san — Wed, 23 Nov 2011 23:31:24 GMT

hello, ramie, you can also use a lottery two pick3 49 / 6

That is the 1st 2nd 3rd place = a pick3 and 4 th 5 th 6 th = other pick3

Vertically, you can use the pick3 studies, using the filters of the study pick3 forum

Loery post in the final digit lottery 6 49 / 6 0-9

Rami. One factor of the repetition of the group so he's turning down of the last result positions = 15 pairs and trios

Reply #15

Ramijami — Wed, 23 Nov 2011 18:24:34 GMT

Thanks dr san, will have a look at your suggestions

Reply #14

dr san — Tue, 22 Nov 2011 20:25:39 GMT

Hello, ramie in a game 49 / 6 have 6 positions

Giving a trio, or three locations where the filter worked

The rest is played the same group of the previous result, goal is to get pairs and trios

(Three positions) of the six possible as it is not known which three positions

They are 6x3 = 15 positions, where the filter will use as reference works, the group set

You have to study the positions, and match the pairs or trios reference

Reply #13

dr san — Tue, 22 Nov 2011 20:05:52 GMT

Ok ramie

After separating the last digit will classify the digits into low and high

2, 7, 5, 6, 8, 6

L, H, L, H, H, H

Next, DO NOT PLAY THE SAME GROUP = EXAMPLE

1st place DI WAS LAST DRAW DOWN THE NEXT PLAY HIGH,

CLEAR AS IS 6 digits ranging from 0 to 9 two or more digits REPEAT OF PREVIOUS RESULT IS GOING TO FIND THAT POSITION AGAIN, THE REST OF CHANGE FOR THE OTHER PARTY. EASY WELL, THEN HAS THE SUBGROUPS

Reply #12

Ramijami — Tue, 22 Nov 2011 19:41:43 GMT

dr san, how would you then use this info to chose which final digit to leave out

Reply #11

dr san — Tue, 22 Nov 2011 19:27:56 GMT

hello ramie, you can use both forms of lottery, ok

example in ascending order ofa lottery 49 / 6

12,17,25,36,38,46

random order = 46,17,36,25,12,38

12,17,25,36,38,46

only analyzed the final digit = 2,7,5,6,8,6

Reply #10

Ramijami — Tue, 22 Nov 2011 19:16:11 GMT

Hi crissie,

The formulae are based on the article mentioned in my first post. The author uses 10, 20, and 50 days to select stocks, and I

use 9, 18, and 45 draws to select the numbers. In the following explanation Ave9 will refer to the average apperance of the

number in the last 9 draws, Ave18 average apperance of number in last 18 draws and Ave45 the average apperance of the number

in the last 45 draws. You have to count the occurence of each number in 9, 18 and 45 draw... [ More ]

Reply #9

Ramijami — Tue, 22 Nov 2011 19:13:31 GMT

Hi dr san,

Would you use it based on the way the numbers are drawn, or from lowest to highest to chose? Can you possibly give an example please

Reply #8

crissie — Tue, 22 Nov 2011 11:04:18 GMT

how did you choose your numbers its interesting would like to try out florida 5 ball/ new york

Reply #7

dr san — Mon, 21 Nov 2011 14:28:11 GMT

Hello, ramie, a way to eliminate the final digit from 0 to 9

Classified in two groups =

Low = 1,2,3,4,5

High = 6,7,8,9,0

Look at the last draw of each vertical position, eg

If the last draw in the 1st position gave the final digit down

On the next play No. 1 spot in the high group, is the subdivision of even and odd

In the low-and high

Reply #6

Ramijami — Mon, 21 Nov 2011 14:11:03 GMT

Formula is still holding up providing at least one 5 numbers match in the six sets. Powerball doing slightly better with two sets having all 5 of the winning numbers in the last two draws, but still between 25 to 30 numbers per set. For Lotto it's moved away from the 3/4/6 sets which occurred most often to the 1/2/5 sets.

Does anybody have an accurate way of eliminating at least 1 end digit (one of 0-9) and one numerology digit (one of 1-9) that I can look at as a way to eliminate some mor... [ More ]

Reply #5

Ramijami — Sun, 13 Nov 2011 17:34:37 GMT

Draw was 2 8 14 33 41 46 Bonus 4

Set 2 had 5 numbers amongst the 25.

Reply #4

Ramijami — Sat, 12 Nov 2011 08:55:05 GMT

Hi ACPutz,

I had been counting the occurences as a way of finding a solution, but hadn't thought about two master sets, so thanks for the suggestion. Based on that suggestion, these are the two sets for the next draw with the first set containing the 5/6 occurences:

2, 7, 9, 10, 13, 19, 34, 36, 37, 42, 44,

1, 3, 4, 5, 6, 8, 11, 12, 14, 15, 16, 17, 18, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 35, 38, 39, 40, 41, 43, 45, 46, 47, 48, 49

In the previous draw, the... [ More ]

Reply #3

ACPutz — Fri, 11 Nov 2011 19:57:43 GMT

Hey Ram -

You might want to make a master set from these six. If five of the six sets contain at least five of the numbers, than I would go through and see how many times each number appears (ex. 4 appears in sets 1,4,5) so I would put that aside in a different set while the #2 appears in sets #1,2,3,4,6 (5 sets). I would make a set that has every number that appears five times and use that to come up with five numbers for my combinations. Then the sixth number I would just pick from tha... [ More ]

Reply #2

Ramijami — Fri, 11 Nov 2011 18:57:03 GMT

Thanks bootleg233, will post the draw and see how well it does

Reply #1

bootleg233 — Fri, 11 Nov 2011 17:24:40 GMT

If it were me I would do away with any doubles of numbers so the row 6 would be completely gone because all those are in 3 and 4. Then just keep doing away with them until no duplicates and see whats left to work with might help not sure! But it does cut the numbers down. Lol good luck!

A Few less after getting rid of Dbles.

(ROW3) 1, 2, 3, 6, 7, 9, 10, 11, 13, 14, 17, 18, 19, 20, 22, 23, 27, 29, 33, 34, 35, 36, 37, 39, 40, 41, 42, 44, 45,

(ROW4) 4, 5, 8, 12, 15, 16, 21, 24, 25, 26,

Next step help

Ramijami — Fri, 11 Nov 2011 10:52:09 GMT

Hi everybody,

Need some help with a next step. Using a formula to help me select numbers for the next draw of my 6/49 lottery based on the first component of a stock selection strategy. Not sure about posting links but you can google Beyond Just a stock Chart to get an idea of what I am using. I have adapted it and currently use it to get six sets of between 21 and 35 numbers which contain 5 numbers in at least one set. Out of the 10 draws I backtested two had all 6 numbers in a set, and t... [ More ]