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# Next step help

Topic closed. 17 replies. Last post 5 years ago by crissie.

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Cape Town
South Africa
Member #66343
October 23, 2008
115 Posts
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 Posted: November 11, 2011, 5:52 am - IP Logged

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 the other 8 draws had at least 5 in a set. I have narrowed it down to three sets that will likely contain the 5 numbers, and this was confirmed in the "forward test" on Wednesday with a set containing 5 of the winning numbers in 30. I need help with a next step to either reduce the numbers in the 3 "promising" sets further, or a way to select the best set to play, or BONUS a combination of the two.

This example of the six sets is for the South Africa Lotto Draw tomorrow (12/11/2011):

1.)  2,  4,  6,  7,  9, 10, 11, 13, 14, 19, 21, 22, 25, 27, 28, 29, 31, 32, 34, 35, 36, 37, 41, 42, 43, 44, 47, 48, 49,
2.)  2,  5,  6,  8, 10, 11, 12, 13, 14, 15, 16, 19, 22, 24, 26, 27, 29, 30, 34, 35, 36, 37, 38, 41, 46,
3.)  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,
4.)  2,  4,  5,  7,  8,  9, 10, 12, 13, 15, 16, 19, 21, 24, 25, 26, 28, 30, 31, 32, 36, 37, 38, 42, 43, 44, 46, 47, 48, 49,
5.)  1,  3,  4,  7,  9, 17, 18, 20, 21, 23, 25, 28, 31, 32, 33, 34, 37, 39, 40, 42, 43, 44, 45, 47, 48, 49,
6.)  1,  2,  3,  5,  7,  8,  9, 10, 12, 13, 15, 16, 17, 18, 19, 20, 23, 24, 26, 30, 33, 34, 36, 37, 38, 39, 40, 42, 44, 45, 46,

The 3 sets that usually produce the results are 3,4 and 6. Any suggestions or ideas will be appreciated.

Thanks

Tn
United States
Member #54963
September 4, 2007
1166 Posts
Online
 Posted: November 11, 2011, 12:24 pm - IP Logged

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 the other 8 draws had at least 5 in a set. I have narrowed it down to three sets that will likely contain the 5 numbers, and this was confirmed in the "forward test" on Wednesday with a set containing 5 of the winning numbers in 30. I need help with a next step to either reduce the numbers in the 3 "promising" sets further, or a way to select the best set to play, or BONUS a combination of the two.

This example of the six sets is for the South Africa Lotto Draw tomorrow (12/11/2011):

1.)  2,  4,  6,  7,  9, 10, 11, 13, 14, 19, 21, 22, 25, 27, 28, 29, 31, 32, 34, 35, 36, 37, 41, 42, 43, 44, 47, 48, 49,
2.)  2,  5,  6,  8, 10, 11, 12, 13, 14, 15, 16, 19, 22, 24, 26, 27, 29, 30, 34, 35, 36, 37, 38, 41, 46,
3.)  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,
4.)  2,  4,  5,  7,  8,  9, 10, 12, 13, 15, 16, 19, 21, 24, 25, 26, 28, 30, 31, 32, 36, 37, 38, 42, 43, 44, 46, 47, 48, 49,
5.)  1,  3,  4,  7,  9, 17, 18, 20, 21, 23, 25, 28, 31, 32, 33, 34, 37, 39, 40, 42, 43, 44, 45, 47, 48, 49,
6.)  1,  2,  3,  5,  7,  8,  9, 10, 12, 13, 15, 16, 17, 18, 19, 20, 23, 24, 26, 30, 33, 34, 36, 37, 38, 39, 40, 42, 44, 45, 46,

The 3 sets that usually produce the results are 3,4 and 6. Any suggestions or ideas will be appreciated.

Thanks

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, 28, 30, 31, 32, 38, 43,  46, 47, 48, 49,

WHEN IT FEELS THE WHOLE WORLD SUCKS!

RELAX.........IT'S ONLY GRAVITY

I think I can I think I can!!!!

Cape Town
South Africa
Member #66343
October 23, 2008
115 Posts
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 Posted: November 11, 2011, 1:57 pm - IP Logged

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

United States
Member #79057
August 26, 2009
70 Posts
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 Posted: November 11, 2011, 2:57 pm - IP Logged

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 that other set, the one that holds all the numbers that don't appear 5 times.   Then I would settle on a range of sums and number of odd or even digits.  I would also limit the number of unique digits to 4 to 7.  Hope that's a help.

Cape Town
South Africa
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October 23, 2008
115 Posts
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 Posted: November 12, 2011, 3:55 am - IP Logged

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 5/6 set had 3 numbers amongst 14, and the other 3 came from the bigger set.

Cape Town
South Africa
Member #66343
October 23, 2008
115 Posts
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 Posted: November 13, 2011, 12:34 pm - IP Logged

Draw was 2 8 14 33 41 46 Bonus 4

Set 2 had 5 numbers amongst the 25.

Cape Town
South Africa
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October 23, 2008
115 Posts
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 Posted: November 21, 2011, 9:11 am - IP Logged

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 more numbers? Since I can't accurately select the best sets I'm looking at reducing the numbers and possibly using all 6 sets with filters.

bgonÃ§alves
Brasil
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June 9, 2010
2133 Posts
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 Posted: November 21, 2011, 9:28 am - IP Logged
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-
New Member

Bahamas
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September 10, 2011
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 Posted: November 22, 2011, 6:04 am - IP Logged

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

Cape Town
South Africa
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October 23, 2008
115 Posts
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 Posted: November 22, 2011, 2:13 pm - IP Logged

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 :-)

Cape Town
South Africa
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October 23, 2008
115 Posts
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 Posted: November 22, 2011, 2:16 pm - IP Logged

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 draws. So for example say the number one

has appeared 1 time in the last 9 draws, 3 times in 18 and 4 times in 45, you would divide 1 by 9 to get the Ave9, 3 divided

by 18 to get Ave18 and 4 divided by 45 for Ave45. You would do this for every number in your game. Then you apply the

following formulae to each number to select the numbers for the sets:

Set1
if (Ave9>=Ave18 and Ave18>=Ave45) OR (Ave9<=Ave18 and Ave18<=Ave45) then number selected (if it meets the first part OR the

second part then it is selected, if it doesn't meet any of the two criteria then the number is rejected)

Set2
if (Ave9>=Ave18 and Ave18>=Ave45) OR (Ave9<=Ave18 and Ave18>=Ave45) then number selected

Set3
if (Ave9>=Ave18 and Ave18>=Ave45) OR (Ave9>=Ave18 and Ave18<=Ave45) then number selected

Set4
if (Ave9<=Ave18 and Ave18<=Ave45) OR (Ave9<=Ave18 and Ave18>=Ave45) then number selected

Set5
if (Ave9<=Ave18 and Ave18<=Ave45) OR (Ave9>=Ave18 and Ave18<=Ave45) then number selected

Set6
if (Ave9<=Ave18 and Ave18>=Ave45) OR (Ave9>=Ave18 and Ave18<=Ave45) then number selected

A bit complicated but this gives the six sets, one of which should have 4/5 or 5/6, and sometimes 5/5 or 6/6 depending on

your game. I then do this process for every draw to generate the six sets. I am still only testing and looking for ways to

further reduce the numbers in the sets. If you give me your most recent 45 draws then I can generate sets for you for the

next draw to test, would be interesting to see if it produces similar results

bgonÃ§alves
Brasil
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June 9, 2010
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 Posted: November 22, 2011, 2:27 pm - IP Logged

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

Cape Town
South Africa
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October 23, 2008
115 Posts
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 Posted: November 22, 2011, 2:41 pm - IP Logged

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

bgonÃ§alves
Brasil
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June 9, 2010
2133 Posts
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 Posted: November 22, 2011, 3:05 pm - IP Logged

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

bgonÃ§alves
Brasil
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June 9, 2010
2133 Posts
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 Posted: November 22, 2011, 3:25 pm - IP Logged

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

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