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Next step help

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

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Ramijami's avatar - wired shell.jpg
Cape Town
South Africa
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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

    bootleg233's avatar - Lottery-034.jpg
    Tn
    United States
    Member #54963
    September 4, 2007
    1166 Posts
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    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 Big Smile

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

      Ramijami's avatar - wired shell.jpg
      Cape Town
      South Africa
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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


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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.

          Ramijami's avatar - wired shell.jpg
          Cape Town
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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.

            Ramijami's avatar - wired shell.jpg
            Cape Town
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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.

              Ramijami's avatar - wired shell.jpg
              Cape Town
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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.

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                bgonçalves
                Brasil
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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-
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                  Bahamas
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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

                    Ramijami's avatar - wired shell.jpg
                    Cape Town
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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 :-)

                      Ramijami's avatar - wired shell.jpg
                      Cape Town
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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 Smile

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                        bgonçalves
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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

                          Ramijami's avatar - wired shell.jpg
                          Cape Town
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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?

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                            bgonçalves
                            Brasil
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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

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                              bgonçalves
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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