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Gap Strategy - Texas Lotto -0828

Topic closed. 7 replies. Last post 6 years ago by LANTERN.

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bobby623's avatar - abstract
San Angelo, Texas
United States
Member #1097
January 31, 2003
1394 Posts
Offline
Posted: August 28, 2010, 8:23 pm - IP Logged


http://www.lotterypost.com/thread/219231

 

Pumpi76 initiated a thread where he suggested that jackpots could be won by playing numbers suggested by the entire LP membership.
Since no one came forward with a plausible idea as to how the 96,000+ numbers could be brought together in a usable way, due mainly to an apparent lack of interest, the discussion shifted to systems, strategies, workouts, etc.

One member, RJOH, demonstrated his playing methods by generating 200 combinations for the 0825 PB drawing. These were posted in an
open forum. His evaluation is available.

Another member, Luminus, stressed his method, which is a variation of a popular commercial system.
 
This post is a follow-up to my promise to demonstrate Gap Strategy for the 0828 Texas Lotto drawing.

I was going to generate 100 combinations, but I decided 50 would be adequate. Besides, my pockets are shallow.

Texas Lotto is a Pick 6 of 54 numbers game.

Using various data streams, the first step in using Gap Strategy is choosing one or more 'structures' to guide number
selections.

Since the presence of all 6 numbers in a winning combination is rare, the combinations played must have
2, 3, 4 or 5 of the 6 number groups, A, B, C, D, E and F.

For example: Winning combination 1.10.23.31.43.50 becomes ABCDEF. There is only one combination having this structure, which
is called a 6-point structure.

There are 6 1-point structures, 15 2-point structures, 20 3-point structures, 15 4-point structures and 6 5-point structures.

A current structure inventory shows that 4-point structures come up more often than any other structure, followed by 5-point and
3-point.

Positional analysis inventory for Postions 1 and 2 shows that 'AB' has come up 131 times, 'BC' 56 times, 'AC' 37 times,
'BD' 11 times.

Simular inventories for Positions 3 and 4 shows 'CD' 42 times, 'CD' 36 times, 'CE' 20 times, 'E.' 20 times, 'EF' 19 times,
'DF' 16 times, and 'D.' 12 times. (D. is correct)

The inventory for Positions 4 and 5, yes, 4 and 5, shows 'EF' 61 times, 'DE' 59  times, 'DF' 31 times, 'E.' 21 times. These
are mainly from 5-point structures. (E. is correct)

Using other data, I've decided (guessed) that the winning combination for 0828 will be either a 3-point or a 4-point structure.
Specifically ABCE, ABDE, BCEF, BCD or BDE.

Further, I decided to play 10 combinations for each of the possibilities.

Wihout going into the nitty gritty, here are worksheet headers and the combinations for each structure.

BCD - 3 numbers each letter, 9 total numbers, 10 sets, 60 spots to fill, All numbers used 6 times, 6 numbers used 7 times.

Numbers to be wheeled:  11.12.16.23.25.29.31.33.39

Using LS2004 wheeler, the 10 combinations are:

11.23.25.29.33.39; 11.12.23.29.31.39; 11.16.25.29.31.39; 11.16.23.29.31.33; 12.16.23.25.31.33;
11.16.25.31.33.39; 12.23.25.29.31.33; 12.16.23.25.33.39; 11.12.16.23.29.39; 12.16.25.29.31.33

BDE - 3 numbers each letter, 9 total numbers, 10 sets, 60 spots to fill, All numbers used 6 times, 6 numbers used 7 times.

Numbers in wheel: 12.15.17.30.32.33.40.42.46

Combinations to play are:

12.15.17.30.40.42; 12.15.17.32.40.42; 12.15.30.32.33.46; 12.15.30.33.40.46; 12.17.30.33.42.46;
12.17.32.33.40.46; 12.17.32.33.42.46; 15.17.30.32.33.40; 15.30.32.40.42.46; 17.30.33.40.42.46

ACDE - 3 numbers each letter, 12 total numbers, 10 sets, 60 spots, All numbers 5 times.

Numbers in wheel: 6.8.9.23.25.29.31.33.39.43.44.46

Combinations to play are:

6.8.9.25.39.43; 9.25.29.31.44.46; 8.25.31.33.44.46; 8.23.29.31.39.44; 8.9.23.29.33.44;
6.9.23.31.33.46; 6.9.23.39.43.46; 6.23.25.33.39.43; 8.29.31.43.44.46; 6.25.29.33.39.43

ABCE - 3 numbers for each letter, 12 total numbers, 10 sets, 60 spots to fill, All numbers 5 times.

Numbers in wheel: 3.5.6.11.12.16.21.22.25.40.42.46

Combinations to play are:

3.5.6.12.25.40; 5.11.21.22.25.42; 6.16.25.40.42.46; 3.12.16.21.40.46; 3.5.11.16.22.46;
6.16.21.22.42.46; 5.6.12.21.22.42; 3.11.12.22.25.40; 5.11.12.16.40.42; 3.6.11.21.25.46

BCEF - 3 numbers for each letter, 12 total numbers, 10 sets, 60 spots, All numbers 5 times.

Numbers in wheel: 11.12.17.21.23.25.40.44.46.51.53.54

Combinations to play:

11.23.44.46.51.53; 17.21.23.25.46.53; 12.25.40.44.51.54; 11.12.23.46.51.54; 11.12.17.21.40.53;
12.17.25.40.44.53; 17.21.25.44.51.53; 11.21.40.46.51.54; 11.23.25.40.44.54; 11.17.21.23.46.54

These are my 'best guesses' based on the Gap Strategy data streams.
Other players using the workouts could very well reach other conclusions.

Gap Strategy is versatile. That is, if I only wanted to play 5 sets for a pick 6 game, I'd select a scheme that
would provide a minimum of 3 numbers for each letter, 9 numbers overall, 30 spots to fill and a wheel setting
to generate 5 sets. I'd then have to adjust the recommended combinations to ensure that
all combinations have at least one of the nine numbers in the wheel.

I could, of course, adopt a scheme where I would generate 5 or more numbers for each
letter in the structure to be played. However, there are limits. Playing 5 numbers for
4 letters would require more than 10 combinations. I can't say exactly how many because
I've never contemplated working with such a scheme.


Thanks for your interest.

Bobby623

 

 


-

    bobby623's avatar - abstract
    San Angelo, Texas
    United States
    Member #1097
    January 31, 2003
    1394 Posts
    Offline
    Posted: August 28, 2010, 11:42 pm - IP Logged

    Winning numbers: 1.12.25.36.39.45 = ABCDDE = ABCDE = 5-point structure.

    One $3 winner.

    Several 'missed by one.'

    The $8 million jackpot was won in a city to be announced.

    Thanks for your interest.

      rcbbuckeye's avatar - Lottery-043.jpg
      Texas
      United States
      Member #55889
      October 23, 2007
      5615 Posts
      Offline
      Posted: August 29, 2010, 2:47 pm - IP Logged

      Winning numbers: 1.12.25.36.39.45 = ABCDDE = ABCDE = 5-point structure.

      One $3 winner.

      Several 'missed by one.'

      The $8 million jackpot was won in a city to be announced.

      Thanks for your interest.

      Winning tic purchased in Hooks, Tx, wherever that is . Tx IS a big state.

      A couple weeks ago there was a story about the Tx Lottery Comm being short of money due to 2 jackpot wins within 2 weeks, now we have another win about 2 weeks from the last win.

      bobby623, that six decade combo that was drawn is pretty rare. I did that kind of analysis recently, and you are correct that 4 decades play the most.

      However, the person that won last night apparently didn't know or didn't care, he picked his own numbers.

      CAN'T WIN IF YOU'RE NOT IN

      A DOLLAR AND A DREAM (OR $2)

        rcbbuckeye's avatar - Lottery-043.jpg
        Texas
        United States
        Member #55889
        October 23, 2007
        5615 Posts
        Offline
        Posted: August 29, 2010, 3:15 pm - IP Logged

        Correction.... 5 decades drawn last night.

        CAN'T WIN IF YOU'RE NOT IN

        A DOLLAR AND A DREAM (OR $2)

          bobby623's avatar - abstract
          San Angelo, Texas
          United States
          Member #1097
          January 31, 2003
          1394 Posts
          Offline
          Posted: August 29, 2010, 4:55 pm - IP Logged

          rcbbuckeye

          The winning numbers could fit birthdays or other personal dates.

          Or, maybe the winner is using a similar strategy and decided to play the 5-point structure.

          I thought about dropping one 4-point and going with a 5-point, but none of the other data streams
          supported a change.

          That's the basic lottery story - you never know.

          By the way - Hooks is in Bowie County in far, far northeast Texas. It's within driving range of Arkansas.

            LANTERN's avatar - kilroy 28_173_reasonably_small.jpg
            Tx
            United States
            Member #4570
            May 4, 2004
            5180 Posts
            Offline
            Posted: September 5, 2010, 1:16 pm - IP Logged


            http://www.lotterypost.com/thread/219231

             

            Pumpi76 initiated a thread where he suggested that jackpots could be won by playing numbers suggested by the entire LP membership.
            Since no one came forward with a plausible idea as to how the 96,000+ numbers could be brought together in a usable way, due mainly to an apparent lack of interest, the discussion shifted to systems, strategies, workouts, etc.

            One member, RJOH, demonstrated his playing methods by generating 200 combinations for the 0825 PB drawing. These were posted in an
            open forum. His evaluation is available.

            Another member, Luminus, stressed his method, which is a variation of a popular commercial system.
             
            This post is a follow-up to my promise to demonstrate Gap Strategy for the 0828 Texas Lotto drawing.

            I was going to generate 100 combinations, but I decided 50 would be adequate. Besides, my pockets are shallow.

            Texas Lotto is a Pick 6 of 54 numbers game.

            Using various data streams, the first step in using Gap Strategy is choosing one or more 'structures' to guide number
            selections.

            Since the presence of all 6 numbers in a winning combination is rare, the combinations played must have
            2, 3, 4 or 5 of the 6 number groups, A, B, C, D, E and F.

            For example: Winning combination 1.10.23.31.43.50 becomes ABCDEF. There is only one combination having this structure, which
            is called a 6-point structure.

            There are 6 1-point structures, 15 2-point structures, 20 3-point structures, 15 4-point structures and 6 5-point structures.

            A current structure inventory shows that 4-point structures come up more often than any other structure, followed by 5-point and
            3-point.

            Positional analysis inventory for Postions 1 and 2 shows that 'AB' has come up 131 times, 'BC' 56 times, 'AC' 37 times,
            'BD' 11 times.

            Simular inventories for Positions 3 and 4 shows 'CD' 42 times, 'CD' 36 times, 'CE' 20 times, 'E.' 20 times, 'EF' 19 times,
            'DF' 16 times, and 'D.' 12 times. (D. is correct)

            The inventory for Positions 4 and 5, yes, 4 and 5, shows 'EF' 61 times, 'DE' 59  times, 'DF' 31 times, 'E.' 21 times. These
            are mainly from 5-point structures. (E. is correct)

            Using other data, I've decided (guessed) that the winning combination for 0828 will be either a 3-point or a 4-point structure.
            Specifically ABCE, ABDE, BCEF, BCD or BDE.

            Further, I decided to play 10 combinations for each of the possibilities.

            Wihout going into the nitty gritty, here are worksheet headers and the combinations for each structure.

            BCD - 3 numbers each letter, 9 total numbers, 10 sets, 60 spots to fill, All numbers used 6 times, 6 numbers used 7 times.

            Numbers to be wheeled:  11.12.16.23.25.29.31.33.39

            Using LS2004 wheeler, the 10 combinations are:

            11.23.25.29.33.39; 11.12.23.29.31.39; 11.16.25.29.31.39; 11.16.23.29.31.33; 12.16.23.25.31.33;
            11.16.25.31.33.39; 12.23.25.29.31.33; 12.16.23.25.33.39; 11.12.16.23.29.39; 12.16.25.29.31.33

            BDE - 3 numbers each letter, 9 total numbers, 10 sets, 60 spots to fill, All numbers used 6 times, 6 numbers used 7 times.

            Numbers in wheel: 12.15.17.30.32.33.40.42.46

            Combinations to play are:

            12.15.17.30.40.42; 12.15.17.32.40.42; 12.15.30.32.33.46; 12.15.30.33.40.46; 12.17.30.33.42.46;
            12.17.32.33.40.46; 12.17.32.33.42.46; 15.17.30.32.33.40; 15.30.32.40.42.46; 17.30.33.40.42.46

            ACDE - 3 numbers each letter, 12 total numbers, 10 sets, 60 spots, All numbers 5 times.

            Numbers in wheel: 6.8.9.23.25.29.31.33.39.43.44.46

            Combinations to play are:

            6.8.9.25.39.43; 9.25.29.31.44.46; 8.25.31.33.44.46; 8.23.29.31.39.44; 8.9.23.29.33.44;
            6.9.23.31.33.46; 6.9.23.39.43.46; 6.23.25.33.39.43; 8.29.31.43.44.46; 6.25.29.33.39.43

            ABCE - 3 numbers for each letter, 12 total numbers, 10 sets, 60 spots to fill, All numbers 5 times.

            Numbers in wheel: 3.5.6.11.12.16.21.22.25.40.42.46

            Combinations to play are:

            3.5.6.12.25.40; 5.11.21.22.25.42; 6.16.25.40.42.46; 3.12.16.21.40.46; 3.5.11.16.22.46;
            6.16.21.22.42.46; 5.6.12.21.22.42; 3.11.12.22.25.40; 5.11.12.16.40.42; 3.6.11.21.25.46

            BCEF - 3 numbers for each letter, 12 total numbers, 10 sets, 60 spots, All numbers 5 times.

            Numbers in wheel: 11.12.17.21.23.25.40.44.46.51.53.54

            Combinations to play:

            11.23.44.46.51.53; 17.21.23.25.46.53; 12.25.40.44.51.54; 11.12.23.46.51.54; 11.12.17.21.40.53;
            12.17.25.40.44.53; 17.21.25.44.51.53; 11.21.40.46.51.54; 11.23.25.40.44.54; 11.17.21.23.46.54

            These are my 'best guesses' based on the Gap Strategy data streams.
            Other players using the workouts could very well reach other conclusions.

            Gap Strategy is versatile. That is, if I only wanted to play 5 sets for a pick 6 game, I'd select a scheme that
            would provide a minimum of 3 numbers for each letter, 9 numbers overall, 30 spots to fill and a wheel setting
            to generate 5 sets. I'd then have to adjust the recommended combinations to ensure that
            all combinations have at least one of the nine numbers in the wheel.

            I could, of course, adopt a scheme where I would generate 5 or more numbers for each
            letter in the structure to be played. However, there are limits. Playing 5 numbers for
            4 letters would require more than 10 combinations. I can't say exactly how many because
            I've never contemplated working with such a scheme.


            Thanks for your interest.

            Bobby623

             

             


            -

            bobby

            You said that:

            There are 6 1-point structures, 15 2-point structures, 20 3-point structures, 15 4-point structures and 6 5-point structures.

            A current structure inventory shows that 4-point structures come up more often than any other structure, followed by 5-point and
            3-point.

            Positional analysis inventory for Postions 1 and 2 shows that 'AB' has come up 131 times, 'BC' 56 times, 'AC' 37 times,
            'BD' 11 times.

            Simular inventories for Positions 3 and 4 shows 'CD' 42 times, 'CD' 36 times, 'CE' 20 times, 'E.' 20 times, 'EF' 19 times,
            'DF' 16 times, and 'D.' 12 times. (D. is correct)

            The inventory for Positions 4 and 5, yes, 4 and 5, shows 'EF' 61 times, 'DE' 59  times, 'DF' 31 times, 'E.' 21 times. These
            are mainly from 5-point structures. (E. is correct)

            Can you please do the same for the TxCash5?

            Thanks!

            Fernando.

            BibleOnline  ParishesOnline  ChristianRadioOnline   MassOnline   Mass

            "Ten measures of beauty descended to the world, nine were taken by Jerusalem."

              bobby623's avatar - abstract
              San Angelo, Texas
              United States
              Member #1097
              January 31, 2003
              1394 Posts
              Offline
              Posted: September 5, 2010, 3:53 pm - IP Logged

              Lantern

              I quit playing Tx Cash 5 when the jackpot dropped below $25,000.
              A lot of work with a good possibility that there would be more than one winner.

              You can obtain the data you want yourself by doing the following.

              1. Using graph paper, log the winning combinations in numerical order. Go back 30 more drawings.

              2. Convert numerical combinations to what I call an 'alpha sequence' by substituting a letter
              for the numerals  in a separate column. Divide into P1p2 P3 P4P5.

              Key: 1-9 = A, 10-19 =B, 20-29 = C, 30-37 = D.

              3. Convert the alpha sequences to 'structures' by eliminating duplicates. For example,
              ABBCD becomes ABCD, which is a 4-point structure.

              To this point, you should have"

              Date 01 10 20 30 37       AB C DD   ABCD 4
              date 14 23 33 34 35       BC D DD   BCD. 3
              date 07 08 09 11 16       AA A BB   AB.. 2

              etc

              You can number the structures individually on same sheet, or use a separate sheet maintain
              an inventory.

              I use a separate sheet.

              There are 4 1-point, 6 2-point, 4 3 point and 1 4-point structures in every 5/3X game.

              Update the totals after each drawing.

              On a 3rd sheet, log the P1P2 pairs. Number them 1-up.  I use Gap Strategy, therefore, I enter
              the gap number in a 4th column

              E. Heading  #  P1P2 #
                          1   AB  -
                          2   AB  1

              Strike through each pair as they are tabulated. Don't obliterate, just a mark to show it shouldnt
              be counted a second time.

              Do the same for P2P3, and P3P4.
              Keep in mind that some pairs will be '..'

              When you are done, you will have an exact total for each pair, and make play choices accordingly.

              Construct a structure follower chart.

              Prepare a 4x4 square with rows and columns marked 1 thru 4. Count the number of times one structure
              follows another and put the total in appropriate box.

              When you are done you will know exactly how many times the structures follow one another. If the last
              structure was a 3, you can refer to chart and decide if the next will be a 3, a 4, a 2 or a 1, depending
              on the totals and current trends.

              You can keep the same type of stats for the alpha sequences.
              Some sequences come up way more often than others.

              As far as the numbers are concerned, you will have to adopt appropriate inventory streams and followers.

              Good luck!

                LANTERN's avatar - kilroy 28_173_reasonably_small.jpg
                Tx
                United States
                Member #4570
                May 4, 2004
                5180 Posts
                Offline
                Posted: September 5, 2010, 4:27 pm - IP Logged

                Lantern

                I quit playing Tx Cash 5 when the jackpot dropped below $25,000.
                A lot of work with a good possibility that there would be more than one winner.

                You can obtain the data you want yourself by doing the following.

                1. Using graph paper, log the winning combinations in numerical order. Go back 30 more drawings.

                2. Convert numerical combinations to what I call an 'alpha sequence' by substituting a letter
                for the numerals  in a separate column. Divide into P1p2 P3 P4P5.

                Key: 1-9 = A, 10-19 =B, 20-29 = C, 30-37 = D.

                3. Convert the alpha sequences to 'structures' by eliminating duplicates. For example,
                ABBCD becomes ABCD, which is a 4-point structure.

                To this point, you should have"

                Date 01 10 20 30 37       AB C DD   ABCD 4
                date 14 23 33 34 35       BC D DD   BCD. 3
                date 07 08 09 11 16       AA A BB   AB.. 2

                etc

                You can number the structures individually on same sheet, or use a separate sheet maintain
                an inventory.

                I use a separate sheet.

                There are 4 1-point, 6 2-point, 4 3 point and 1 4-point structures in every 5/3X game.

                Update the totals after each drawing.

                On a 3rd sheet, log the P1P2 pairs. Number them 1-up.  I use Gap Strategy, therefore, I enter
                the gap number in a 4th column

                E. Heading  #  P1P2 #
                            1   AB  -
                            2   AB  1

                Strike through each pair as they are tabulated. Don't obliterate, just a mark to show it shouldnt
                be counted a second time.

                Do the same for P2P3, and P3P4.
                Keep in mind that some pairs will be '..'

                When you are done, you will have an exact total for each pair, and make play choices accordingly.

                Construct a structure follower chart.

                Prepare a 4x4 square with rows and columns marked 1 thru 4. Count the number of times one structure
                follows another and put the total in appropriate box.

                When you are done you will know exactly how many times the structures follow one another. If the last
                structure was a 3, you can refer to chart and decide if the next will be a 3, a 4, a 2 or a 1, depending
                on the totals and current trends.

                You can keep the same type of stats for the alpha sequences.
                Some sequences come up way more often than others.

                As far as the numbers are concerned, you will have to adopt appropriate inventory streams and followers.

                Good luck!

                bobby

                O.K.

                Thanks you much!

                Fernando.

                BibleOnline  ParishesOnline  ChristianRadioOnline   MassOnline   Mass

                "Ten measures of beauty descended to the world, nine were taken by Jerusalem."