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Help Needed: Observable Pattern

Topic closed. 26 replies. Last post 9 years ago by pumpi76.

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Honduras
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Posted: September 10, 2007, 7:57 pm - IP Logged

 I need help with a math problem...Any help will be appreciated....I have discovered a pattern in Ga's Win for Life (Pick6/42), which you all probably already knew or didn't....But first you have to convert the numbers into Six sections of 5's....

First you must divide and convert the entire range of Ga's Win For Life (Pick6/42 / Lotto6/42) into 5's....For instance, 1-5 = 1; 6-10= 2; 11-15= 3; 16-20 = 4; 21-25 = 5; 26-30= 6; 31-35= 7; 36-40 = 8 and 41-42 = 9....So that the combinations for example 

06-19-24-25-27-30  equal in 5's= 2-4-5-5-6-6

05-14-20-37-40-42 equal in 5's = 1-3-4-8-8-9

03-05-09-21-27-41 equal in 5's = 1-1-2-5-6-9

The numbers on your left are actual draws from Jan.3rd, 2007 'til March 28, 2007 (in descending order) and the numbers on your right are those draws converted into 5's numbers (Fives).. 

8-21-24-30-35-38 which equals to: 2-5-5-6-7-8         Jan 3rd, 2007

08-10-11-15-16-18 which equals to: 2-2-3-3-4-4

04-09-16-19-25-41 which equals to: 1-2-4-4-5-9

08-12-16-23-25-32  which equals to: 2-3-4-5-5-7

01-08-18-25-26-38 which equals to: 1-2-4-5-6-8

08-13-15-22-26-31 which equals to: 2-3-3-5-6-7

10-24-25-30-38-41 which equals to: 2-5-5-6-8-9

02-05-07-15-37-40 which equals to: 1-1-2-3-8-8

07-20-23-27-37-41 which equals to: 2-4-5-6-8-9

04-08-14-18-22-25 which equals to: 1-2-3-4-5-5

01-07-13-28-29-41 which equals to: 1-2-3-6-6-9 

01-02-08-13-36-41 which equals to: 1-1-2-3-8-9

04-13-28-29-40-42  which equals to: 1-3-6-6-8-9

05-09-16-18-26-36  which equals to: 1-2-4-4-6-8

02-16-26-32-35-36  which equals to: 1-4-6-7-7-8

04-05-11-15-22-32 which equals to: 1-1-3-3-5-7

06-18-21-24-32-39 which equals to: 2-4-5-5-7-8

05-08-12-16-23-29 which equals to: 1-2-3-4-5-6

07-14-20-21-30-33 which equals to: 2-3-4-5-6-7

08-09-13-16-29-31 which equals to: 2-2-3-4-6-7      March10,  2007

05-06-09-26-37-41 which equals to: 1-2-2-6-7-9

10-21-26-28-30-38  which equals to: 2-5-6-6-6-8

06-15-18-19-20-33 which equals to: 2-3-4-4-4-7

01-03-08-12-16-38 which equals to: 1-1-2-3-4-8

04-05-08-21-31-37 which equals to: 1-1-2-5-7-8       March 28th, 2007

And this is the pattern you can see..From each draw to the next at least 2 to 3 numbers repeat on the following draw...For example, if 1-1-2-3-4-8 plays at least 2 numbers will play on the following draw and this is what plays on the following draw: 1-1-2-5-7-8....If you notice on this one 3 numbers played on the following draws and those were 1, 2 and 8....I've color coded them in yellow...From January 3rd 2007 'til March 28th, 2007 there has been 25 Win For Life draws, out of those 25 draws, 22 draws repeat at least 2 to 3 numbers on their consecutive draws.....

Here is the MATH PROBLEM... If you notice the 5's numbers are from 1 to 9 is similar to pick4 but you could say that this is pick6 from 1 to 9 taken 6 at a time instead of 4...I need to know the following...If 6 numbers play out of 9, what are the next numbers that would play if i know that out of those 6, 2 or 3 numbers will repeat, what i am saying is how many sets will i have taken 6 at a time out of 9 and 2 to 3 numbers will repeat...I hope you understand what i am saying...If you don't or do understand what i am saying feel free to say it....

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    Honduras
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    Posted: September 10, 2007, 8:10 pm - IP Logged

    did i mentioned that i have a theory that i want to post on the forum and on my blog on how to beat Win For Life (pick6/42 & Pick6/44), but i need to know the answer to that problem first....

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      Posted: September 11, 2007, 1:29 am - IP Logged

      The number of sets will depend on how much repeating you allow among your "5s" and how many groups of 2 or 3 of the same number block you allow in the "playable set". 

      If you are doing this by hand, you would need to calculate for each of those possibilities and then add them together. Write carefully.

      If you are coding this, you would just let your code generate each type of string for you (that would also let you drop any "ugly" sets) and have it count the total.

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        Honduras
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        Posted: September 11, 2007, 10:48 am - IP Logged

        I think this may be the question too.....A wheel size 9 for pick6, how many sets would it have for it to have a 2out of 6 and a 3 out of 6....Once you know the answer you can start writing the combinations with those specific codes/backbones in one direction...For example if you had as an answer: 2-3-3-5-6-7 and 2-2-3-5-6-7 you will beging to write the wheel with those codes/backbones...you will beging to write you wheel with only those 2, Six 5's numbers....Remember a wheel has many 5's numbers (perhaps hundreds or thousands) but you are writing your wheel with 2 only...

        but before i go further i need to know if you have a wheel for pick6, size 9, how many sets would it have for it to have 2 out of 6 and 3 out of 6...I don't know but i think this is the question....

        YOu basically will be writing a wheel in a different code and the wheel when done will be very LETHAL....If you are successful in writing the wheel and you are correct the consequences are that just imagine the number of 5of 6 that you will get...To be modest i say you get at least like 30 5of 6.....And just imagine that...I have all this written down on my theory the only thing i don't have is the answer to the problem...

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          Honduras
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          Posted: September 11, 2007, 11:03 am - IP Logged

          i was looking at the wheel store and i notice that a wheel size 12 for pick6 with 2, 4out of 6 win if 6 in 12 will be 14 sets...so i know a wheel size 9 with 3 of 6 and 2 of 6 should be less than 14 sets...

          I also saw a wheel  size 11 with 5, 4of 6 will give you 11 sets, so i know a wheel size 9 with ONLY 1 3 of 6 or 2 of 6 should be less than 11 sets...At least this is great news....

          But this is not counting though the repetitions because i notice that the 5's numbers do repeat at least 1 number of the Six 5's numbers...HOw to deal with that problem i still don't know...

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            The Quantum Master
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            Posted: September 11, 2007, 11:45 am - IP Logged

            pumpi76,

            The method of analysis you are referring to is called 'Grouping'. It's when you break down the full set of playable numbers into smaller subsets; typically of nearly equal items in each set. Your subsets or groups are 8 with 5 items and 1 with 2 items for the full set of 42 items. Also, the groups are in a structured order, the items in each group are ascending and from group to group it's ascending. This is good, because it helps to define a distribution for each column the numbered groups are in. Below is a table that shows the total distribution of all the numbers in a Pick 6 of 42 lottery. As you'll see, when you group each subset, it will also produce a distribution by adding each quantity in each column for each group. From a post I made before, Combinatorial Distribution, we can find the total distribution of a Pick 6 of 42 lottery as follows with each group highlighted.

            Distribution for Pick 6 of 42

             

            Numbers

            Column
            123456
            1749398     
            265800891390    
            35757571645029139   
            450194222144525308703  
            543589726418046620266437 
            63769922945257140063001801
            732463231416098175119005256
            827825632463212566419635119021
            923733632736015276829568231056
            10201376323640178560416644032126
            11169911314650202275558006510252
            12142506301455223300717759900462
            131187552850122411648932014355792
            1498280266175255528108108200201287
            1580730245700266175127764270272002
            1665780224250273000147875354903003
            1753130202400276000168000455004368
            1842504180642275264187680571206188
            1933649159390270963206448703808568
            20263341389852633402238398527211628
            212034911970025270023940010174515504
            221550410174523940025270011970020349
            23116288527222383926334013898526334
            2485687038020644827096315939033649
            2561885712018768027526418064242504
            2643684550016800027600020240053130
            2730033549014787527300022425065780
            2820022702712776426617524570080730
            2912872002010810825552826617598280
            307921435589320241164285012118755
            31462990071775223300301455142506
            32252651055800202275314650169911
            33126403241664178560323640201376
            3456231029568152768327360237336
            3521119019635125664324632278256
            3665251190098175314160324632
            371180630071400294525376992
            38 37266446620264180435897
            39  70325308221445501942
            40   9139164502575757
            41    91390658008
            42     749398

             

            When we add up each column in each group, the distribution changes to as follows.

             

            Group Distribution for Pick 6 of 42

             

            Groups

            Column

            123456
            1
            {1,2,3,4,5}
            292100274151781067336737 
            2
            {6,7,8,9,10}
            141859215843176265671090678237210
            3
            {11,12,13,14,15}
            61018214129921188442452767778124795
            4
            {16,17,18,19,20}
            221397905667135856793384229376233755
            5
            {21,22,23,24,25}
            6223743421711100671301667700462138340
            6
            {26,27,28,29,30}
            1145214239264106713118671223537416675
            7
            {31,32,33,34,35}
            9172394221844288256715917371029385
            8
            {36,37,38,39,40}
            77422156725064212588122215220
            9
            {41,42}
                913901407406

             

            I'll add more in a bit, busy.

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              JADELottery's avatar - MeAtWork 03.PNG
              The Quantum Master
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              Posted: September 11, 2007, 1:08 pm - IP Logged

              From your first post, you have a total of 25 draws. Even before those draws were made, you could have estimated what the distribution of each group would be like based on the total group distribution. We can solve this by dividing each groups distribution in each column by the total number of group counts in each column. That just happens to be the total number of combinations for Pick 6 of 42 or C(42,6) = 5245786. Next, we multiply by the number of draws we want to estimate for, 25. See also additional information in my post, Relative Combinatorial Distribution. Below is a table of the Relative Group Distribution for 25 Draws.

               

              Relative Group Distribution for 25 Draws in Pick 6 of 42

              Groups

              Column

              123456
              1144    
              27831  
              33762  
              414641 
              5 25631
              6 13662
              7  1485
              8   1611
              9     7

              Here is your actual Distribution.

              Group Distribution for 25 Draws in Pick 6 of 42

              Groups

              Column

              123456
              1145    
              21195   
              3 575  
              4 36631
              5 34651
              6  3771
              7   156
              8    59
              9     7

               

              Fairly close to the actual.

              The next table shows how far away the estimate is from the actual.

              Distribution (Estimate - Actual)

              Group

              Column

              123456
              10-1    
              2-4-1-21  
              332-1-30 
              4110-2-2-1
              5 -110-20
              6 10-1-11
              7  133-1
              8   112
              9     0

              ...

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                Honduras
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                Posted: September 12, 2007, 4:59 pm - IP Logged

                Thanks you very much jadelottery but i realized i needed another question answer which was: suppose you have the 5's numbers: 2-3-4-5-6-7, i'll like to know how many kinds like that i will have that will 1 number of the 5's repeat..with pen and paper i can deduce the followings...

                2-2-3-4-5-6 (but is missing the 7)

                2-3-3-4-5-6 (but is missing the 7)

                2-3-4-4-5-6 (but is missing the 7)

                2-3-4-5-5-6 (but is missing the 7)

                2-3-4-5-6-6 (but is missing the 7)

                we also have 3-4-5-6-7-7 (but is missing the 2)

                2-3-4-6-7-7 (but is missing the 5)

                2-3-5-6-7-7 (but is missing the 4)

                2-4-5-6-7-7 (but is missing the 3)

                (i know there is more numbers to optimally to cover) (what i am saying is to cover all possible combinations of 2 numbers repeating withing the 5's group....)

                notice that in each case 1 number repeats (there is a double) but 1 number is left out...I would need to find out how may sets will such way have as to cover all positions available optimally, so as to cover all numbers)....I'll like to know only with 2 repetitions not with 3 repetitions)....If you create a wheel with only those 5's numbers as your backbone, i am deducing that you should be like 6, 5/6 from each 5's numbers that you make into a wheel...That will be like 6 X 8= 48....YOu would be getting 48 (at least i am saying) 5/6 and in half of them you would get the second prize of Win For Life which pays 1,000 dollars a week for 1 year, that times 24....And that's only in 1 drawing...The only problem is that is going to cost you a little more to make the wheel...

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                  Honduras
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                  Posted: September 12, 2007, 5:50 pm - IP Logged

                  But i was talking more of a full wheel with those 5's group...I might have mentioned a conditional abbreviated or full wheel but is more like a conditional abbreviated or full wheel with parameters/constraints.....And when i mentioned the backbone i refer to the frame actually...I was talking if you make a full or close to full wheel with each of those 9, 5's group from the group that will have at least 1 set with 2 or 3 out of 6 with a wheel sized 9......Each of those combinations which i am guessing will be like 6 or 7 will have like 15 (though i wrote 9 that i know) and use them as a frame and start creating a full or close to full wheel with them using them as the frame of the wheel....It will be a very LETHAL, potent wheel...I said a wheel that will give you 48, 5/6 but i think is more than that (and that's per drawing)....You could say that it will be close to guaranteed, you don't have to worry if you match the numbers or not...It will be sort of a guaranteed wheel...Guaranteed to win....But the problem is the cost...How much such wheel will run you about..., I am hoping is not more than 5,000 sets..

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                    The Quantum Master
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                    Posted: September 12, 2007, 11:43 pm - IP Logged

                    pumpi76,

                    I'll be busy with family related activities and will get back to this problem in a few days.

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                    Jehocifer

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                      The Quantum Master
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                      Posted: September 17, 2007, 4:54 pm - IP Logged

                      Pumpi76,

                      I'm back from my camping trip with the family in Lac du Flambeau, WI. We had a great time tent camping in freezing night time conditions. Overall, it was good weather during the day, 50's-60's. However, that's to be expected in northern Wisconsin.

                      I'll be a little busy for just a few more days while I catch up with events going on here at home. Till then good luck and skill.

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                        Thoth's avatar - binary
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                        Posted: September 18, 2007, 2:15 am - IP Logged

                        Pumpi...

                        Here is how all the possible combinations (using 5 number groupings) break down for the Texes Two Step that we talked about.  Like a regular Pick 4 game, there are 5 combination types.  The table below outlines only the four white ball possibilities and does not take into consideration the "bonus ball."  When picking four balls (or numbers) from a pool of 35, there are 35!/(35-4)!/4! combinations....this equals 52,360 total combinations.  As you can see - there are more outcomes that contain one "repeating" group number than all the other types combined: The "Doubles" account for 26,250 of the total combinations, while the "No-Match," "Double/Doubles," "Triples," and "Quads" account for a combined total of 26,110 combinations!

                        So, it is slightly more likely to see exactly two digits drawn from a single group (a repeat) and the other two digits each drawn from a different group than it is to see four different groups and/or more than one repeat.

                         

                         

                        TYPEEXAMPLECOUNTCHANCE
                        No-Match123421,87541.78%
                        Double112326,25050.13%
                        Double/Double11222,1004.01%
                        Trips11112,1004.01%
                        Quads1111350.067%

                         

                         

                        Also, suppose you are tracking repeating pairs within a combination...like 1334.  Each of the seven pairs (within Double types of combinations only) can be found within 3,750 different real combinations.  Again -this does not including the bonus ball.  This means that the chance of seeing a Double type of combo drawn with say, two repeating digits from group 1 is about 7.162%.

                        The repeat median of such a combo is 9.327 games...lets just round that up to 10 even.  In a nutshell, suppose the last combo drawn (as it appears in grouped format) was 2256.  There is a little better than a 50% chance than another Double type of combo containing two 2's will be drawn within the next 10 games.  The same probability applies to all the Double combos that contain any of the pairs (11, 22, 33, etc.... through 77).

                        Hope this helps a little

                        ~Probability=Odds in Motion~

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                          Honduras
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                          Posted: September 18, 2007, 7:58 pm - IP Logged

                          thanks Thoth...

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                            The Quantum Master
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                            Posted: September 19, 2007, 2:08 pm - IP Logged

                            pumpi76,

                            I have some interesting data to show you, but before I can do that, we need to look at your data in a different way. The groups you have are too large to analyze and get a good picture of the repeating patterns. In order to help see those patterns, we need to break down your groups into two forms that help show these patterns a little more easily. These forms are the Basic Permutational Form and the Combinatorial From. The Basic Permutational Form uses Letters to represent the first basic permutation of all the possible permutations. The Combinatorial Form uses Numbers that are substituted into the Letters to get back to the original set of numbers or group of numbers.

                            Here's an example using your first three examples, 2-4-5-5-6-6, 1-3-4-8-8-9 and 1-1-2-5-6-9. We'll transform these into two different forms of Basic Permutational and Combinatorial.

                            Basic Permutational Form

                            1. Start with the leftmost number and replace it with an 'A'.

                            2. Next, move to the number to the right; if the number is not the same, change the letter to the next alphabetic letter, else if it is the same, keep that letter as the substitute.

                            2 4 5 5 6 6 becomes A B C C D D
                            1 3 4 8 8 9 becomes A B C D D E
                            1 1 2 5 6 9 becomes A A B C D E

                            Combinatorial Form

                            1. Remove any duplicate numbers, however, keep in order of ascending to the right.

                            2 4 5 5 6 6 becomes 2 4 5 6
                            1 3 4 8 8 9 becomes 1 3 4 8 9
                            1 1 2 5 6 9 becomes 1 2 5 6 9

                            These two forms are only related to the set of numbers (groups) your derived it from. A B C C D D is associated with 2 4 5 6 by example. Also, the number of different letters must match the quantity of numbers in the combinatorial form. Example, A A B C D E has 5 different letters and 1 2 5 6 9 has 5 different numbers.

                            With that explained we can continue a little further with the total possible Basic Permutational Forms.

                            Continues...

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                              The Quantum Master
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                              Posted: September 19, 2007, 2:26 pm - IP Logged

                              ...

                              For your case, you are dealing with a Pick 6; the total possible Basic Permutational Forms are as follows:

                              Pick 6 Basic Permutational Forms
                              1AAAAAA
                              2AAAAAB
                              3AAAABB
                              4AAABBB
                              5AABBBB
                              6ABBBBB
                              7AAAABC
                              8AAABBC
                              9AABBBC
                              10ABBBBC
                              11AAABCC
                              12AABBCC
                              13ABBBCC
                              14AABCCC
                              15ABBCCC
                              16ABCCCC
                              17AAABCD
                              18AABBCD
                              19ABBBCD
                              20AABCCD
                              21ABBCCD
                              22ABCCCD
                              23AABCDD
                              24ABBCDD
                              25ABCCDD
                              26ABCDDD
                              27AABCDE
                              28ABBCDE
                              29ABCCDE
                              30ABCDDE
                              31ABCDEE
                              32ABCDEF

                              I wrote a program to search the entire Pick 6 of 42 combinations, a total of 5245768 combinations, and find all the Basic Permutational Forms; analyze them to get the distribution of those forms. Below is the total distribution of Basic Permutational Forms for your case.

                              Pick 6 Basic Permutational Forms DistributionForm Count
                              1AAAAAA0
                              2AAAAAB156
                              3AAAABB1440
                              4AAABBB2800
                              5AABBBB1400
                              6ABBBBB140
                              7AAAABC8400
                              8AAABBC33600
                              9AABBBC33600
                              10ABBBBC8400
                              11AAABCC29400
                              12AABBCC58800
                              13ABBBCC29400
                              14AABCCC28000
                              15ABBCCC28000
                              16ABCCCC7000
                              17AAABCD115500
                              18AABBCD231000
                              19ABBBCD115500
                              20AABCCD231000
                              21ABBCCD231000
                              22ABCCCD115500
                              23AABCDD189000
                              24ABBCDD189000
                              25ABCCDD189000
                              26ABCDDD87500
                              27AABCDE525000
                              28ABBCDE525000
                              29ABCCDE525000
                              30ABCDDE525000
                              31ABCDEE393750
                              32ABCDEF787500
                              Total5245786

                              Continues...

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