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quiz pick-4 daily game odds

Topic closed. 48 replies. Last post 10 months ago by RL-RANDOMLOGIC.

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RL-RANDOMLOGIC's avatar - usafce

United States
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March 13, 2008
3971 Posts
Offline
Posted: January 23, 2016, 9:04 pm - IP Logged

SergeM

I seem to doing that quite a bit since I started messing around with the daily games.

With the number games, I know what I am doing but the daily games require both

combinations and permutations.  I need to think about things more before jumping

in.  I am starting to get the feel but still got a ways to go. 

RL

Working on my Ph.D.  "University of hard Knocks"

I will consider the opinion that my winnings are a product of chance if you are willing to consider

they are not.  Many great discoveries come while searching for something else

USAF https://en.wikipedia.org/wiki/Prime_Base_Engineer_Emergency_Force

  US Flag Trump / 2016 & 2020  

    Avatar

    United States
    Member #116344
    September 8, 2011
    3927 Posts
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    Posted: January 23, 2016, 9:42 pm - IP Logged

    P.S.

    Playing large is a option as my 2 positional digit hits are better than 1 to 1 per play overall.  To cover

    a 6-way p-4 pair bet require 486 lines.  MO pays $6K for a $1.00 P-4 str8, the only downside is filling

    out 486 boards every day or twice a day if mid and evening games are played.

    Here are a couple pics of my positional wheel and the prize matches.  winning line = 7664

     

    P.S. What I mean by the above hit rate is that I play anywhere from 1 to 5 lines per game and I

    often manage more than one line with 2 positional digit hits and in a few games every line had

    two positional digits.

     

    RL

    RL, you're on right track. You need to make a choice between the options ' front pair, front triads and any pair, any triad', each has different lines. Good Luck

    Front pair> 100lines

    Any pair> filter target pair contained in 10000picks for a straight hit

    front triad> 10 lines

    any triad>filter target triad contained in 10000 picks for a straight hit.

      Avatar
      Lincoln, California
      United States
      Member #167130
      June 27, 2015
      256 Posts
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      Posted: January 23, 2016, 10:56 pm - IP Logged

      The Chart Below show the Number of possible Straight Combinations that are Possible for 0,1,2,3 or 4 Numbers Repeating from the Previous Draw.  In the Left Column the number left of the decimal point is the number of Repeat.  The number Right of the decimal point is the number of digits that can repeat. 

      From the chart it is possible to see the number of combinations resulting from Your answer the First Question.  How many numbers from the previous game Repeat?

      Next you can see the number of combinations resulting from your answer to the Second Question.  Will the Draw be a No Match, Double, Double-double, Triple or Quadruple Draw.  The Value in the Chart under each Type is for 1 Set of Number Positions.  The number in the Right Column is the number of position combinations for the Answer to the First Question.  For Example, 2 repeating Digits has 6 position Combinations.

      I am not yet sure What the Odds are Yet.  My gut tells me that there is a 1:5 Chance of Answering Question 1 so the number of combinations in the Chart would be multiplied by 5 to give the True Odds. 

      I hope this Chart is helpful and Correct.

        RL-RANDOMLOGIC's avatar - usafce

        United States
        Member #59354
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        Posted: January 24, 2016, 8:11 am - IP Logged

        Guys

        Thanks for the info.  What I am doing now is this.  Using my steps program and my S/C chart for analysis

        I pick one digit for each of the four positions.   I never know which digits are correct until after the drawing

        but I have a very high hit rate for at least two of my selections showing in the correct spot.   There are 6

        positional pairs in a 4 digit line.  AB, AC, AD, BC, BD and CD.  It takes 486 lines to cover my 4 digit selections

        to cover all six possible pairings.  If any two of my digits selected show in the correct position then a straight

        will be in the lines generated.

         

        I know the 6-way pair will pay off big as long as future hit rates remain consistent with my test.  My test show

        a 77% win rate for the 486 line 6-way pair bet.  $.50 bets pays $3K so cost = $8,748.00 for 18 plays, winning

        $41,580 for $32,796 profit.  My current estimate for playing the 4.1 line average using percents based on the

        77%, 486 line 6-way pair hits is 1 in 631 tickets / 4.1 = 1 in every 154 games.  Playing the evening draw I could

        expect to hit a straight around once every 5 months.

         

        So far I have a very good success rate hitting 2 positional digits in my first line.  Based on my test data and using

        the 6-way pair option I would have made a very nice profit over the 18 game.

         

        The Problem.  I don't want to play 486 lines and hitting once in around every 154 attempts does not sound

        that great.

         

        Using the S/C chart and the steps program mentioned on page 2 of this topic I had 3 games out of 18 where I hit

        11 of the 12 steps.  This leads me to believe that a 12 of 12 play would not take 154 attempts. 

         

        I hope someone can shed some light on my estimates to help me decide on a betting strategy. 

         

        Thanks in advance.

        RL

        Working on my Ph.D.  "University of hard Knocks"

        I will consider the opinion that my winnings are a product of chance if you are willing to consider

        they are not.  Many great discoveries come while searching for something else

        USAF https://en.wikipedia.org/wiki/Prime_Base_Engineer_Emergency_Force

          US Flag Trump / 2016 & 2020  

          JADELottery's avatar - MeAtWork 03.PNG
          The Quantum Master
          West Concord, MN
          United States
          Member #21
          December 7, 2001
          3675 Posts
          Offline
          Posted: January 24, 2016, 11:37 am - IP Logged

          Step-1 breaks the digits into 3 groups,  1=(012), 2=(3456) and step-3=(789)

          Step-2 is odd or even 1=odd 2=even

          Step-3 is 1 = low and 2=high.

          What numbers are associated with each group: even, odd, low and high?

          Presented 'AS IS' and for Entertainment Purposes Only.
          Any gain or loss is your responsibility.
          Use at your own risk.

          Order is a Subset of Chaos
          Knowledge is Beyond Belief
          Wisdom is Not Censored
          Douglas Paul Smallish
          Jehocifer

            RL-RANDOMLOGIC's avatar - usafce

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            Posted: January 24, 2016, 9:29 pm - IP Logged

            Jade

            In step-1 we have 3 choices.  Group-1 = (0,1,2)  Group-2 = (3,4,5,6) and Group-3 = (7,8,9)

            If step-1 is thought to be 2  the the pool is reduce to digits (3,4,5,6)

            Lets say that I then set step-2 = 2 the pool is reduce to odd digits (3,5)

            Lets say that I then set step-3 to 1 the pool is reduced to the lowest of the two remaining

            digits.   Step code = 2-2-1 = digit 3

             

            Note! Codes 1-1-2 and 3-2-2 are not used so there are 10 possible outputs, 0 to 9. 

             

            The overall odds are the same but analyzing the steps data seems to perform

            better than a standard digit analysis.   I also use what I call a S/C chart where

            I analyze the break points for a step value to either stay the same or change to

            another value.  Steps 2 and 3 only have two choices and for my P-4 most of the

            time I hit them all.  Most of my misses are with step-1 which has a range of 1 to 3.

             

            Below is a sample of the S/C chart.  I only use around 45 games to make my choice

            to play the same value or change it.   If step-1 is believed to change then I use the

            steps tool to help decide which of the remaining two choices to play.  Steps 2 and 3

            are always 50/50 as they have 2 possible choices.  Sometimes depending on the

            step-2 and step-3 values set, Step-1 is limited to only one choice when a change is

            expected.  This happens when steps 2 and 3  are set to the values in the unused

            codes 1-1-2 and 3-2-2 mentioned above.

             

            The S/C chart is a hand workout where I try to hit the break-points for a value to

            change from the last game.  I also have a S/C wheel where I can try to predict the

            number of steps that will change overall then set a few that look certain and wheel

            to cover.  I analyze the data vertically, top line is most current game.     

            P-1      P-2      P-3      P-4

            --------------------------------
            2 1 2    3 2 1    1 2 1    2 2 1     <-  step values that hit in last game, set 5804.
            --------------------------------

            s s s    s c c    c c c    c c c         c = 8   s = 4   <- Same / change totals
            s s s    s c c    c s s    c c s         c = 5   s = 7
            c s c    c s c    c c s    c s c         c = 8   s = 4
            c s s    c c s    c s c    s c s         c = 6   s = 6
            c s s    c c s    s s c    c c c         c = 7   s = 5
            c c s    s s s    c s c    c c c         c = 7   s = 5
            c s c    c s c    s c c    s s s         c = 6   s = 6
            s s c    s c s    c c s    c c c         c = 7   s = 5
            s c s    c s c    c c c    s s s         c = 6   s = 6
            c c s    c c s    c s c    c c c         c = 9   s = 3
            c s c    c c c    c c c    s c s         c = 9   s = 3
            c c s    c c s    c s c    c s c         c = 8   s = 4
            s c c    s c c    c s s    s c c         c = 7   s = 5
            s s s    s s s    s s c    s s s         c = 1   s =11
            c s c    c c s    s s c    c s s         c = 6   s = 6
            c c s    c s s    c s c    c c s         c = 7   s = 5
            s s c    c c c    c s c    c s c         c = 8   s = 4
            c s s    s c s    c c c    c c c         c = 8   s = 4
            c s s    c s s    s c c    c c c         c = 7   s = 5
            c c c    s s c    c c s    c s c         c = 8   s = 4
            c c s    s s c    c c s    c s s         c = 6   s = 6
            c c s    c s s    c c c    s c s         c = 7   s = 5
            s c s    c s c    c s s    s s c         c = 5   s = 7
            s c c    s s s    s s s    c c c         c = 5   s = 7
            c s s    c s s    c s s    c s s         c = 4   s = 8
            c s c    c s c    s s s    c s s         c = 5   s = 7
            c s c    c s s    c c s    s c s         c = 6   s = 6
            c c c    c s s    c c c    c s c         c = 9   s = 3
            s s c    s s s    c c s    c c c         c = 6   s = 6
            c s c    c c s    c s c    s s c         c = 7   s = 5
            c s s    c c c    c c c    c c c         c =10   s = 2
            c c c    c c c    s c s    c c c         c =10   s = 2
            s s c    c c c    c c s    c c c         c = 9   s = 3
            c c c    s c c    c s c    c c c         c =10   s = 2
            s s s    c c c    c c s    s c s         c = 6   s = 6
            c c s    c c c    c c s    c s c         c = 9   s = 3
            c c s    c c c    c c s    c s s         c = 8   s = 4
            c s c    s s c    s s s    c s s         c = 4   s = 8
            c s c    c s s    c c c    c s c         c = 8   s = 4
            c c c    s c c    c c s    s c c         c = 9   s = 3
            c c s    c c c    s s s    c c s         c = 7   s = 5
            c s c    s s s    s c c    s s s         c = 4   s = 8
            c c s    c s s    c s s    c c s         c = 6   s = 6
            c s s    c s s    s s s    s s s         c = 2   s =10

            RL

            Working on my Ph.D.  "University of hard Knocks"

            I will consider the opinion that my winnings are a product of chance if you are willing to consider

            they are not.  Many great discoveries come while searching for something else

            USAF https://en.wikipedia.org/wiki/Prime_Base_Engineer_Emergency_Force

              US Flag Trump / 2016 & 2020  

              RL-RANDOMLOGIC's avatar - usafce

              United States
              Member #59354
              March 13, 2008
              3971 Posts
              Offline
              Posted: January 25, 2016, 1:52 am - IP Logged

              I need to correct a error

              Lets say that I then set step-2 = 2 the pool is reduce to odd digits (3,5)

              should read

              Lets say that I then set step-2 = 1 the pool is reduce to odd digits (3,5)

              Step code = 2-1-1 = digit 3

              RL

              Working on my Ph.D.  "University of hard Knocks"

              I will consider the opinion that my winnings are a product of chance if you are willing to consider

              they are not.  Many great discoveries come while searching for something else

              USAF https://en.wikipedia.org/wiki/Prime_Base_Engineer_Emergency_Force

                US Flag Trump / 2016 & 2020  

                JADELottery's avatar - MeAtWork 03.PNG
                The Quantum Master
                West Concord, MN
                United States
                Member #21
                December 7, 2001
                3675 Posts
                Offline
                Posted: January 25, 2016, 9:41 am - IP Logged

                Jade

                In step-1 we have 3 choices.  Group-1 = (0,1,2)  Group-2 = (3,4,5,6) and Group-3 = (7,8,9)

                If step-1 is thought to be 2  the the pool is reduce to digits (3,4,5,6)

                Lets say that I then set step-2 = 2 the pool is reduce to odd digits (3,5)

                Lets say that I then set step-3 to 1 the pool is reduced to the lowest of the two remaining

                digits.   Step code = 2-2-1 = digit 3

                 

                Note! Codes 1-1-2 and 3-2-2 are not used so there are 10 possible outputs, 0 to 9. 

                 

                The overall odds are the same but analyzing the steps data seems to perform

                better than a standard digit analysis.   I also use what I call a S/C chart where

                I analyze the break points for a step value to either stay the same or change to

                another value.  Steps 2 and 3 only have two choices and for my P-4 most of the

                time I hit them all.  Most of my misses are with step-1 which has a range of 1 to 3.

                 

                Below is a sample of the S/C chart.  I only use around 45 games to make my choice

                to play the same value or change it.   If step-1 is believed to change then I use the

                steps tool to help decide which of the remaining two choices to play.  Steps 2 and 3

                are always 50/50 as they have 2 possible choices.  Sometimes depending on the

                step-2 and step-3 values set, Step-1 is limited to only one choice when a change is

                expected.  This happens when steps 2 and 3  are set to the values in the unused

                codes 1-1-2 and 3-2-2 mentioned above.

                 

                The S/C chart is a hand workout where I try to hit the break-points for a value to

                change from the last game.  I also have a S/C wheel where I can try to predict the

                number of steps that will change overall then set a few that look certain and wheel

                to cover.  I analyze the data vertically, top line is most current game.     

                P-1      P-2      P-3      P-4

                --------------------------------
                2 1 2    3 2 1    1 2 1    2 2 1     <-  step values that hit in last game, set 5804.
                --------------------------------

                s s s    s c c    c c c    c c c         c = 8   s = 4   <- Same / change totals
                s s s    s c c    c s s    c c s         c = 5   s = 7
                c s c    c s c    c c s    c s c         c = 8   s = 4
                c s s    c c s    c s c    s c s         c = 6   s = 6
                c s s    c c s    s s c    c c c         c = 7   s = 5
                c c s    s s s    c s c    c c c         c = 7   s = 5
                c s c    c s c    s c c    s s s         c = 6   s = 6
                s s c    s c s    c c s    c c c         c = 7   s = 5
                s c s    c s c    c c c    s s s         c = 6   s = 6
                c c s    c c s    c s c    c c c         c = 9   s = 3
                c s c    c c c    c c c    s c s         c = 9   s = 3
                c c s    c c s    c s c    c s c         c = 8   s = 4
                s c c    s c c    c s s    s c c         c = 7   s = 5
                s s s    s s s    s s c    s s s         c = 1   s =11
                c s c    c c s    s s c    c s s         c = 6   s = 6
                c c s    c s s    c s c    c c s         c = 7   s = 5
                s s c    c c c    c s c    c s c         c = 8   s = 4
                c s s    s c s    c c c    c c c         c = 8   s = 4
                c s s    c s s    s c c    c c c         c = 7   s = 5
                c c c    s s c    c c s    c s c         c = 8   s = 4
                c c s    s s c    c c s    c s s         c = 6   s = 6
                c c s    c s s    c c c    s c s         c = 7   s = 5
                s c s    c s c    c s s    s s c         c = 5   s = 7
                s c c    s s s    s s s    c c c         c = 5   s = 7
                c s s    c s s    c s s    c s s         c = 4   s = 8
                c s c    c s c    s s s    c s s         c = 5   s = 7
                c s c    c s s    c c s    s c s         c = 6   s = 6
                c c c    c s s    c c c    c s c         c = 9   s = 3
                s s c    s s s    c c s    c c c         c = 6   s = 6
                c s c    c c s    c s c    s s c         c = 7   s = 5
                c s s    c c c    c c c    c c c         c =10   s = 2
                c c c    c c c    s c s    c c c         c =10   s = 2
                s s c    c c c    c c s    c c c         c = 9   s = 3
                c c c    s c c    c s c    c c c         c =10   s = 2
                s s s    c c c    c c s    s c s         c = 6   s = 6
                c c s    c c c    c c s    c s c         c = 9   s = 3
                c c s    c c c    c c s    c s s         c = 8   s = 4
                c s c    s s c    s s s    c s s         c = 4   s = 8
                c s c    c s s    c c c    c s c         c = 8   s = 4
                c c c    s c c    c c s    s c c         c = 9   s = 3
                c c s    c c c    s s s    c c s         c = 7   s = 5
                c s c    s s s    s c c    s s s         c = 4   s = 8
                c c s    c s s    c s s    c c s         c = 6   s = 6
                c s s    c s s    s s s    s s s         c = 2   s =10

                RL

                Well, we deciphered the steps.

                We just wanted you to confirm.

                 

                Step 1: Group 1 {0, 1, 2}, Group 2 {3, 4, 5, 6}, Group 3 {7, 8, 9}

                Step 2: Group 1 {1, 3, 5, 7, 9}, Group 2 {0, 2, 4, 6, 8}

                Step 3: Group 1 {0, 1, 3, 4, 7, 8}, Group 2 {2, 5, 6, 9}

                 

                Here's the step chart:

                Presented 'AS IS' and for Entertainment Purposes Only.
                Any gain or loss is your responsibility.
                Use at your own risk.

                Order is a Subset of Chaos
                Knowledge is Beyond Belief
                Wisdom is Not Censored
                Douglas Paul Smallish
                Jehocifer

                  JADELottery's avatar - MeAtWork 03.PNG
                  The Quantum Master
                  West Concord, MN
                  United States
                  Member #21
                  December 7, 2001
                  3675 Posts
                  Offline
                  Posted: January 25, 2016, 10:17 am - IP Logged

                  The percent probability of one group remaining the same is as follows:

                  Step 1: Group 1 30%, Group 2 40% Group 3 30%

                  Step 2: Group 1 50%, Group 2 50%

                  Step 3: Group 1 60%, Group 2 40%

                   

                  The percent probability of one group changing to any other is as follows:

                  Step 1: Group 1 70%, Group 2 60%, Group 3 70%

                  Step 2: Group 1 50%, Group 2 50%

                  Step 3: Group 1 40%, Group 2 60%

                  Presented 'AS IS' and for Entertainment Purposes Only.
                  Any gain or loss is your responsibility.
                  Use at your own risk.

                  Order is a Subset of Chaos
                  Knowledge is Beyond Belief
                  Wisdom is Not Censored
                  Douglas Paul Smallish
                  Jehocifer

                    SergeM's avatar - slow icon.png
                    Economy class
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                    Member #123700
                    February 27, 2012
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                    Offline
                    Posted: January 25, 2016, 1:12 pm - IP Logged

                    Why do you guys stick to groups?

                      SergeM's avatar - slow icon.png
                      Economy class
                      Belgium
                      Member #123700
                      February 27, 2012
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                      Offline
                      Posted: January 25, 2016, 1:15 pm - IP Logged

                      Well, we deciphered the steps.

                      We just wanted you to confirm.

                       

                      Step 1: Group 1 {0, 1, 2}, Group 2 {3, 4, 5, 6}, Group 3 {7, 8, 9}

                      Step 2: Group 1 {1, 3, 5, 7, 9}, Group 2 {0, 2, 4, 6, 8}

                      Step 3: Group 1 {0, 1, 3, 4, 7, 8}, Group 2 {2, 5, 6, 9}

                       

                      Here's the step chart:

                      Step 1: Group 1 {0, 1, 2}, Group 2 {3, 4, 5, 6}, Group 3 {7, 8, 9}

                       

                      How about Group 1 {0, 9, 2}, Group 2 {3, 4, 5, 6}, Group 3 {7, 8, 1} ?

                        RL-RANDOMLOGIC's avatar - usafce

                        United States
                        Member #59354
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                        3971 Posts
                        Offline
                        Posted: January 25, 2016, 10:01 pm - IP Logged

                        SergeM

                        The way the digits are distributed in the groups has no bearing other than that's the way the codes

                        are developed.

                        Step-1 has 3 choices.

                        If set to 1 then we are limited to the digits in group 1  or digits (0-1-2)

                        If set to 2 then we are limited to the digits in group 2  or digits (3-4-5-6)

                        If set to 3 then we are limited to the digits in group 3  or digits (7-8-9)

                         

                        Step-2 has 2 choices 1 = play odd and 2=play even

                         

                        Step-3 has 2 choices 1=lowest remaining digit or 2 = highest remaining digit.

                         

                        Lets say that I set step-1 to 2, I am then working with the digits in group #2, (3-4-5-6)

                        Lets say that I set step-2 to 2, I have then reduced the digits in group #2 to  (4 and 6)

                        Lets say that I set step-3 to 1, I now select the lowest remaining digit from step-2, (4)

                        The step code I chose in the 3 steps is 2-2-1 which decoded = (4) so 4 is the digit I would play.

                         

                        It's my findings that breaking the digit selection into a 3-step process does better than making

                        one choice from a pool of 0 to 9.   This is nothing new as daily game players have been trying

                        to select hI/low, odd/even ect..  almost since the game first came into existence.  The steps tool

                        just organizes everything and provides a simple means to analyze the data by using a 3 digit

                        code. 

                         

                        There are 2 step codes that require  two steps settings to reduce to a single value.  These are

                        codes 1-1-1 and 3-2-1.  Anytime step-1 and step-2 are both set to 1 then step 3 must be a (1).

                        If step-1 and step-2 are set to 3-2 then step 3 must be a (1)

                         

                        I am trying to formulate a betting strategy based on my 18 game test.   What I want to do is estimate

                        the number of lines I need to play over so many games to hit a P-4 straight.  I don't want to use the

                        published odds but create a new set of odds based on my current hit rates as they seem to be consistent.

                        Using the steps and S/C chart my 3-positional hits are 11 x better than the odds would suggest.  I think

                        I am even doing better than 11 x as on average I am hitting 10 or 11 out of the 12 steps. 

                         

                        I am having trouble with my calculations because of the number of variables.   If I select group-2 in step 1

                        my odds are 1 in 4 for hitting the next two steps but if I select group 1 or 3 in step-1 then the odds are 1 in

                        4 for hitting steps 2 and 3 except when step-1 is set to 1 and step-2 is set to 1 then the odds are 1 on 2.

                        It's the same if step-1 is set to 3 and step-2 is set to 2. 

                         

                        I need to reverse engineer a set of odds for winning a P-4 straight based on my hit rates for the 18 game

                        test. 

                         

                        Lets say we start with a base line for the number of expected step hits is 8 of 12.  We have to offset the odds

                        because the odds for hitting steps 2 and 3  are 50/50 except when the codes for step1 and 2 are 1-1 or 3-2.

                        If we set the base line at 8 can we throw away those 8 and then set the odds for hitting 4 of 4 based on the

                        4 remaining steps and their prospective odds.   What if we set the base line up to 9?   We manage 9 or more

                        step values over 95% of our plays.  Using this information it should be easy to build a new set of odds for

                        reaching the 12 of 12 hit. 

                         

                        The bottom line of my question is this.  If I were working with only 4 steps then what are the average odds 

                        based on some of these 4 coming from a mixture of steps 1, 2 or 3.  It seems logical to me that if my average

                        number of hits, game to game is 9 then I should be able to calculate my odds for hitting the remaining 3 for a

                        straight hit.   

                         

                        It may be that I am way off in my assumptions but since I have hit 11 of 12 step values 3 times in 18 attempts it

                        seems logical that a straight hit should not take 10K lines.   I could play until I hit but I would much rather have

                        a number to aim for.  If I am wrong I might end up playing 10K lines without a win.

                         

                        RL

                        Working on my Ph.D.  "University of hard Knocks"

                        I will consider the opinion that my winnings are a product of chance if you are willing to consider

                        they are not.  Many great discoveries come while searching for something else

                        USAF https://en.wikipedia.org/wiki/Prime_Base_Engineer_Emergency_Force

                          US Flag Trump / 2016 & 2020  

                          RL-RANDOMLOGIC's avatar - usafce

                          United States
                          Member #59354
                          March 13, 2008
                          3971 Posts
                          Offline
                          Posted: January 25, 2016, 11:02 pm - IP Logged

                          Jade

                          Based on my test play where the average hit rate is 9 of 12 step values, is it possible that

                          I can calculate my odds for hitting all 12 by calculating the odds for hitting 3 of 3 steps.  In

                          my test play I managed 3 attempts with 11 out of 12.  The other games ranged 8 to 10 

                          the overall average was 9 of 12.   

                          Most of my misses are step-1 values which are always 1 in 3. It seems logical that I could

                          get a good estimate by calculating the odds for hitting 3 or 4 step-1 values.

                          3 * step-1 ( .333 * .333 * .333) = 0.0369. 

                          4 * step-1 ( .333 * .333 * .333 * .333) = 0.0123

                          In my test play of 18 games and 74 total lines the average was 4.1 lines per play.  In those

                          74 lines we can see that the expected for hitting 3 step-1's would be around 3.7 in 100 lines

                          which is on target with my play.  Since my play falls very close to the expected then is it logical

                          that I could expect to hit around 1 straight for every 100 lines played based on the average

                          hit rates of the test play? 

                          In the ticket posted earlier I hit all 4 step-1 values and missed position 4 step-3 so I consider

                          this getting very close and also that was the only play if I remember correctly where I hit all

                          4 step-1's. 

                          This estimate is far better than the one based on the 77% I calculated previously.  1 in 631

                          vs 1 in 102.   I could expect a win around 1 in every 25 plays.

                          I may be way off here but if my test hit rates remain consistent then this seems logical and will

                          return a nice profit. 

                          RL

                          Working on my Ph.D.  "University of hard Knocks"

                          I will consider the opinion that my winnings are a product of chance if you are willing to consider

                          they are not.  Many great discoveries come while searching for something else

                          USAF https://en.wikipedia.org/wiki/Prime_Base_Engineer_Emergency_Force

                            US Flag Trump / 2016 & 2020  

                            JADELottery's avatar - MeAtWork 03.PNG
                            The Quantum Master
                            West Concord, MN
                            United States
                            Member #21
                            December 7, 2001
                            3675 Posts
                            Offline
                            Posted: January 26, 2016, 1:18 pm - IP Logged

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                            Any gain or loss is your responsibility.
                            Use at your own risk.

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                            Jehocifer

                              RL-RANDOMLOGIC's avatar - usafce

                              United States
                              Member #59354
                              March 13, 2008
                              3971 Posts
                              Offline
                              Posted: January 26, 2016, 10:24 pm - IP Logged

                              Jade

                              I can calculate the odds for each S/C value.  I feel that I am not adequately explaining what I am trying to do.

                              I am trying to offset the odds using my average hit rate as a starting point to calculate an estimate for being

                              able to hit all the values.  This is kind of quasi-math guess at best but since my baseline hits are far above the

                              expected, I would feel comfortable as long as I know I am not missing something in my calculations.  I am not 

                              a mathematician so I can stray outside the box without feeling I am sinning against God so to say. 

                              RL

                              Working on my Ph.D.  "University of hard Knocks"

                              I will consider the opinion that my winnings are a product of chance if you are willing to consider

                              they are not.  Many great discoveries come while searching for something else

                              USAF https://en.wikipedia.org/wiki/Prime_Base_Engineer_Emergency_Force

                                US Flag Trump / 2016 & 2020