Welcome Guest
Log In | Register )
You last visited December 8, 2016, 8:49 pm
All times shown are
Eastern Time (GMT-5:00)

Breaking down Daily 4 a Preliminary Report

Topic closed. 17 replies. Last post 11 months ago by lakerben.

Page 1 of 2
51
PrintE-mailLink
Avatar
Lincoln, California
United States
Member #167130
June 27, 2015
256 Posts
Offline
Posted: January 5, 2016, 7:09 pm - IP Logged

I have posted a little bit about this process previously in the Mathematics Forum under the Topic “Predictions from a Sequence”.

Divide and Conquer is the General idea behind the approach.

Playing Daily 4 only requires 4 choices, one for each number. 

Decision 1 (Pick First Number) has 10 possible options as do each succeeding Decision.  I replace the first decision with an option that has only 5 options.  The 5 options are the number of hits expected from 1 of 2 groups of numbers.  Generally the Number Groups could be set in several ways, (0-4 & 5-9; Odd & Even; Alternate Numbers; etc.)  My groups are formed by taking the numbers from the 4 Strings Broken on the previous Draw, placing those in the “R” Group with the remaining numbers placed in the “N” Group.  For each draw there will be from 0 to 4 numbers selected from the “R” Group.    The number of digits in each group varies from 6 to 8 in the “R” Group based on the number of broken strings.  An all Single Draw breaks 15 Strings and places 6 digits in the “R” Group.  A Double Draw breaks 35 Strings and places 7 Digits in the “R” Group.  A Double-Double Draw breaks 70 Strings and places 8 digits in the “R” Group. A Triple breaks 70 Strings and places 8 Digits in the “R” Group.  A Quadruple draw breaks 126 Strings and places 9 digits in the “R” Group (Note his has happened only 3 times in the 2743 games in the data). The following is a summary of the 5 options for Ca. Daily 4 as of Game 2777 (Dec 25, 2015).

 Games in Data   2743      Last Game           2777     

"R" Hits      Hits     Frequency         Games Out            GO/F

1                  305            9                         21               234%

4                  544            5                           7               139%

3                 1050           2.6                       3                115%

2                  802            3.4                       1                 29%

0                  49             56                         2                 4%

Looking at the options, 0 can be eliminated for a few draws.  I should noted that there are not that many combinations created from this option so it can be played for less than the other 4 options.  Monitoring the Games out Divided by the Frequency shows which options are due or overdue.

With this decision there is a 1:5 Chance of being right (1:4 if 0 is excluded)

Decision 2 is what positions will the R and N Groups occupy in the Draw.  The number of options is determined by the first decision as follows: 0=NNNN; 1= NNNR, NNRN, NRNN, RNNN; 2= NNRR, NRNR, NRRN, RNNR, RNRN, RRNN; 3= NRRR, RNRR, RRNR, RRRN; 4= RRRR.   The odds of getting this right (assuming decision 1 is correct) vary from 1:1 for 1:6.

Looking at Game 2778 there were 7 numbers in the “R” Group.  (Draw 2777 was a Double) The “R” Numbers 0135689; The “N” Numbers 247.  There were 2 numbers from the “R” Group drawn  The position of the group Numbers was RRNN. 

At this point My Wheel generates all of the Single, Double, Double-Double, Triple and Quadruple Combinations from the First 2 Decisions.  There are 146 Single Combinations, 168 Double Combinations, and 12 Double-Double Combinations. 

At this point there are 255 Combinations that could be played from the First 2 decisions.   Playing all combinations as a Straight would have produced 1 winning combination.  There was 1 winning Straight ticket that paid $5925 so dividing by 2 Win $2962.50.  Playing all combinations as Boxes would have Won $826.00. (there were 28 Box winning tickets at $236.00 each=$6608.00 total prize. Divided by 32=$206.5 each). There were 4 Box Combinations in the Wheeled combinations. The Cost to Win would be $255.00

Making a 3rd Decision and Selecting only the Single Game Combinations would have won the same at a cost of $146.00

I am going to forward test this to catch up to the current game.  I will let you know how I do. 

I am curious about some basic math here.  10x10x10x10 is the odds of picking the right Straight Combination.  If you correctly pick the first 2 numbers there are 100 combinations (of those numbers and 2 more) left. There is a 1:100 Chance of being right. After 2 decisions with this workout there are 255 Combinations left.  I have a 1:20 chance of being right. The odds of winning are 1:5100 at a cost of $255.00.  I am drawing a blank on the General odds of winning when playing 255 tickets.

Waiting for thoughts from the LP Class

    Avatar
    Lincoln, California
    United States
    Member #167130
    June 27, 2015
    256 Posts
    Offline
    Posted: January 5, 2016, 7:30 pm - IP Logged

    The Combinations vary depending on the R Count and the R Hit for each game.  The post is only applicable to a game with an R Count of 7 and hit count 2.  I will follow up with the other  combinations soon.

      Avatar

      United States
      Member #116344
      September 8, 2011
      3928 Posts
      Offline
      Posted: January 5, 2016, 7:56 pm - IP Logged

      I have posted a little bit about this process previously in the Mathematics Forum under the Topic “Predictions from a Sequence”.

      Divide and Conquer is the General idea behind the approach.

      Playing Daily 4 only requires 4 choices, one for each number. 

      Decision 1 (Pick First Number) has 10 possible options as do each succeeding Decision.  I replace the first decision with an option that has only 5 options.  The 5 options are the number of hits expected from 1 of 2 groups of numbers.  Generally the Number Groups could be set in several ways, (0-4 & 5-9; Odd & Even; Alternate Numbers; etc.)  My groups are formed by taking the numbers from the 4 Strings Broken on the previous Draw, placing those in the “R” Group with the remaining numbers placed in the “N” Group.  For each draw there will be from 0 to 4 numbers selected from the “R” Group.    The number of digits in each group varies from 6 to 8 in the “R” Group based on the number of broken strings.  An all Single Draw breaks 15 Strings and places 6 digits in the “R” Group.  A Double Draw breaks 35 Strings and places 7 Digits in the “R” Group.  A Double-Double Draw breaks 70 Strings and places 8 digits in the “R” Group. A Triple breaks 70 Strings and places 8 Digits in the “R” Group.  A Quadruple draw breaks 126 Strings and places 9 digits in the “R” Group (Note his has happened only 3 times in the 2743 games in the data). The following is a summary of the 5 options for Ca. Daily 4 as of Game 2777 (Dec 25, 2015).

       Games in Data   2743      Last Game           2777     

      "R" Hits      Hits     Frequency         Games Out            GO/F

      1                  305            9                         21               234%

      4                  544            5                           7               139%

      3                 1050           2.6                       3                115%

      2                  802            3.4                       1                 29%

      0                  49             56                         2                 4%

      Looking at the options, 0 can be eliminated for a few draws.  I should noted that there are not that many combinations created from this option so it can be played for less than the other 4 options.  Monitoring the Games out Divided by the Frequency shows which options are due or overdue.

      With this decision there is a 1:5 Chance of being right (1:4 if 0 is excluded)

      Decision 2 is what positions will the R and N Groups occupy in the Draw.  The number of options is determined by the first decision as follows: 0=NNNN; 1= NNNR, NNRN, NRNN, RNNN; 2= NNRR, NRNR, NRRN, RNNR, RNRN, RRNN; 3= NRRR, RNRR, RRNR, RRRN; 4= RRRR.   The odds of getting this right (assuming decision 1 is correct) vary from 1:1 for 1:6.

      Looking at Game 2778 there were 7 numbers in the “R” Group.  (Draw 2777 was a Double) The “R” Numbers 0135689; The “N” Numbers 247.  There were 2 numbers from the “R” Group drawn  The position of the group Numbers was RRNN. 

      At this point My Wheel generates all of the Single, Double, Double-Double, Triple and Quadruple Combinations from the First 2 Decisions.  There are 146 Single Combinations, 168 Double Combinations, and 12 Double-Double Combinations. 

      At this point there are 255 Combinations that could be played from the First 2 decisions.   Playing all combinations as a Straight would have produced 1 winning combination.  There was 1 winning Straight ticket that paid $5925 so dividing by 2 Win $2962.50.  Playing all combinations as Boxes would have Won $826.00. (there were 28 Box winning tickets at $236.00 each=$6608.00 total prize. Divided by 32=$206.5 each). There were 4 Box Combinations in the Wheeled combinations. The Cost to Win would be $255.00

      Making a 3rd Decision and Selecting only the Single Game Combinations would have won the same at a cost of $146.00

      I am going to forward test this to catch up to the current game.  I will let you know how I do. 

      I am curious about some basic math here.  10x10x10x10 is the odds of picking the right Straight Combination.  If you correctly pick the first 2 numbers there are 100 combinations (of those numbers and 2 more) left. There is a 1:100 Chance of being right. After 2 decisions with this workout there are 255 Combinations left.  I have a 1:20 chance of being right. The odds of winning are 1:5100 at a cost of $255.00.  I am drawing a blank on the General odds of winning when playing 255 tickets.

      Waiting for thoughts from the LP Class

      Keep things simple, P4 has 10000 combos:

      1.Locate front pairs for 100picks

      2.Locate front triad for 10 picks

      3.Locate  any pair for x picks  (do the calculation)

      4.Locate any triad  x picks( do the calcuation)

      Using the binomial format 10P4  you can filter 3,4.

      Focusing on 1,2,3,4 is all you need, waging is personal choice.

        Avatar
        Lincoln, California
        United States
        Member #167130
        June 27, 2015
        256 Posts
        Offline
        Posted: January 5, 2016, 8:02 pm - IP Logged

        Keep things simple, P4 has 10000 combos:

        1.Locate front pairs for 100picks

        2.Locate front triad for 10 picks

        3.Locate  any pair for x picks  (do the calculation)

        4.Locate any triad  x picks( do the calcuation)

        Using the binomial format 10P4  you can filter 3,4.

        Focusing on 1,2,3,4 is all you need, waging is personal choice.

        So you Have a constant Group 1,2,3,4.  What do you do to pick positions in the Draw

          Avatar

          United States
          Member #116344
          September 8, 2011
          3928 Posts
          Offline
          Posted: January 5, 2016, 8:08 pm - IP Logged

          So you Have a constant Group 1,2,3,4.  What do you do to pick positions in the Draw

          The filter conditions depend on P or C of the binomial format NP or NC .

          P= permutation, meaning repeated digits

          C= Combination, meaning order is not important

          One has to make a choice of sticking to P or C.

            Avatar

            United States
            Member #116344
            September 8, 2011
            3928 Posts
            Offline
            Posted: January 5, 2016, 8:25 pm - IP Logged

            So you Have a constant Group 1,2,3,4.  What do you do to pick positions in the Draw

            1.

            A constant  group 1234 has  16 front pairs considering P

            A constant group 1234 has  6 pairs considering C with no repeat digits

            A constant group 1234 has 10 pairs considering C with a repeat digits

            2.

            A constant group 1234 has 64 front triads considering P

            A constant 1234 has 4 pairs considering C with no repeats

            A constant 1234 has 20 pairs considering C with digit repeats

             

            For 16 front pairs, you have 1600 picks/ draw 

            For 6 any pairs , you have 600 front pairs +any pairs(try to find it out)

            For  10 pairs , you have 1000 front + any pairs

            The choice is reducing the pairs while maintaining the choice of P or C

              Avatar
              Lincoln, California
              United States
              Member #167130
              June 27, 2015
              256 Posts
              Offline
              Posted: January 5, 2016, 8:44 pm - IP Logged

              1.

              A constant  group 1234 has  16 front pairs considering P

              A constant group 1234 has  6 pairs considering C with no repeat digits

              A constant group 1234 has 10 pairs considering C with a repeat digits

              2.

              A constant group 1234 has 64 front triads considering P

              A constant 1234 has 4 pairs considering C with no repeats

              A constant 1234 has 20 pairs considering C with digit repeats

               

              For 16 front pairs, you have 1600 picks/ draw 

              For 6 any pairs , you have 600 front pairs +any pairs(try to find it out)

              For  10 pairs , you have 1000 front + any pairs

              The choice is reducing the pairs while maintaining the choice of P or C

              OK  Thanks for the alternative.  I Looked at what you have.  Did you Look at What I Posted?  Any Ideas related to this process would be appreciated.

                Avatar
                Lincoln, California
                United States
                Member #167130
                June 27, 2015
                256 Posts
                Offline
                Posted: January 5, 2016, 10:06 pm - IP Logged

                  Avatar

                  United States
                  Member #41846
                  June 23, 2006
                  459 Posts
                  Offline
                  Posted: January 6, 2016, 7:18 am - IP Logged

                  AllanB interesting! tell us more. you said you select digits based on broken strings? I don't understand exactly how you are doing this.

                    Tialuvslotto's avatar - Jailin
                    Texas
                    United States
                    Member #150797
                    December 31, 2013
                    815 Posts
                    Offline
                    Posted: January 6, 2016, 8:34 am - IP Logged

                    Going back and reading your post on the math forum made this a lot clearer: https://www.lotterypost.com/thread/296288/2

                    I find this interesting, because I do something similar with skips. 

                    Dividing the skips into Last game (skip 0), second last (skip 1), third last (skip 2), and fourth or more (skip 3+) gives a 'skip signature' for each drawing.  About 75% of the time this will be 2110 or 3100.  1102,1201,2101,2011, 2110, 3001,3100,1003,3010,1300 account for about 2/3 of these or .67*.75=.50.

                    I'm going to try to incorporate the information you have provided here to see if I can't improve my accuracy with this method.

                    "There is no such thing as luck; only adequate or inadequate preparation to cope with a statistical universe."

                    ~Robert A. Heinlein

                      Avatar
                      Lincoln, California
                      United States
                      Member #167130
                      June 27, 2015
                      256 Posts
                      Offline
                      Posted: January 6, 2016, 10:34 am - IP Logged

                      AllanB interesting! tell us more. you said you select digits based on broken strings? I don't understand exactly how you are doing this.

                      The groups are formed by taking the broken Strings from the previous draw.  As an example Game #2782 was a Single Draw (No matches)  7580.  This Combination Broke 15 (4 Strings) as follows:

                      The String Numbers below are found in Winsom's String Excel Sheet (4 Strings)

                      Broken Strings by Number: 85,87,90,92,95,102,107,110,117,127,142,145,152,162and 182. 

                      These 15 Strings contained 6 numbers 123469  These numbers become the "R" Group.  Numbers not included in the 15 broken strings 0578.  These numbers become the "N" Group.

                        Avatar
                        Lincoln, California
                        United States
                        Member #167130
                        June 27, 2015
                        256 Posts
                        Offline
                        Posted: January 6, 2016, 11:15 am - IP Logged

                        The groups are formed by taking the broken Strings from the previous draw.  As an example Game #2782 was a Single Draw (No matches)  7580.  This Combination Broke 15 (4 Strings) as follows:

                        The String Numbers below are found in Winsom's String Excel Sheet (4 Strings)

                        Broken Strings by Number: 85,87,90,92,95,102,107,110,117,127,142,145,152,162and 182. 

                        These 15 Strings contained 6 numbers 123469  These numbers become the "R" Group.  Numbers not included in the 15 broken strings 0578.  These numbers become the "N" Group.

                        Here is a way to determine the broken strings using Excel. 

                        On Winsom's 4 Strings Sheet you will find the complete list of 4 Strings.  There are 210 Strings based on numbers arranged in increasing order from left to right.  In the first 4 columns left of the Broken Strings, in row above the first String enter the numbers from a draw.  Under each number on the row for each string enter a formula that counts the number of times the draw number is present on the String Breakers for the each String number.  If the draw number is included in the string breakers the result will be 1 if not 0.  In a column left of the last drawn digit enter a formula that sums the 4 counts from the String breakers.  A sum of 4 indicates that the String broke on the draw.  Filtering the columns and selecting the sum 4 produces a column of the broken string numbers. 

                        To determine the numbers present on the broken strings create a table using the vlookup function.  List the numbers in rows or columns.  My example uses columns for each String.  To use the vlookup function more easily create a Name for the range containing all 210 4 Strings.  Include at least the first 6 columns ex: "_4_Strings"  Under each String number enter the following formula =vlookup([Cell Containing String Number], _4_Strings, 3) to insert the first number on the string. in the next 3 rows change the last digit in the formula to 4,5,6.  This inserts the remaining 3 numbers on the string.  Do this under all string numbers.  Create a 128 column Table with the formulas.  The unused columns will produce #NA.  Under the used columns the numbers on each string will appear. 

                        Count the number of times each number is present in the table.  Any count greater than 0 indicates a number is included in the table and is in the "R"  Group.  A count of 0 means the number is in the "N" Group.

                        I will post an image of my tables to make this more clear in a few minutes.

                          Avatar
                          Lincoln, California
                          United States
                          Member #167130
                          June 27, 2015
                          256 Posts
                          Offline
                          Posted: January 6, 2016, 11:35 am - IP Logged

                          The first table show the setup to extract the broken strings. 

                          the Second table show the current draw with the previous draws broken strings in the table.  The row containing 0 or 10 is the count for each number in the table.  0 means not present (N), 10 means present (R)

                            Tialuvslotto's avatar - Jailin
                            Texas
                            United States
                            Member #150797
                            December 31, 2013
                            815 Posts
                            Offline
                            Posted: January 6, 2016, 11:40 am - IP Logged

                            Without involving strings, this can easily and quickly be broken down.  The digits that hit in the last draw are the "N" group and the digits that did not  hit in the last draw are the "R" group.

                            The trick is to decide how many of each will appear in the next combination.

                            "There is no such thing as luck; only adequate or inadequate preparation to cope with a statistical universe."

                            ~Robert A. Heinlein

                              Avatar
                              Lincoln, California
                              United States
                              Member #167130
                              June 27, 2015
                              256 Posts
                              Offline
                              Posted: January 6, 2016, 11:47 am - IP Logged

                              Without involving strings, this can easily and quickly be broken down.  The digits that hit in the last draw are the "N" group and the digits that did not  hit in the last draw are the "R" group.

                              The trick is to decide how many of each will appear in the next combination.

                              I new this was simple.  I just could not resist making it complicated with Excel.  Thanks for pointing this out.  It was right there in front of me but I could not see it. 

                              I will be posting how I have tracted positions, and multiples that use these groups.

                              Feeling Stupid but more enlightened, thanks