Quick Links You last visited December 9, 2016, 2:19 pm All times shown are Eastern Time (GMT5:00)  Help Needed: Observable PatternHonduras Member #20982 August 29, 2005 4715 Posts Offline  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, 15 = 1; 610= 2; 1115= 3; 1620 = 4; 2125 = 5; 2630= 6; 3135= 7; 3640 = 8 and 4142 = 9....So that the combinations for example 061924252730 equal in 5's= 245566 051420374042 equal in 5's = 134889 030509212741 equal in 5's = 112569 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).. 82124303538 which equals to: 255678 Jan 3rd, 2007 081011151618 which equals to: 223344 040916192541 which equals to: 124459 081216232532 which equals to: 234557 010818252638 which equals to: 124568 081315222631 which equals to: 233567 102425303841 which equals to: 255689 020507153740 which equals to: 112388 072023273741 which equals to: 245689 040814182225 which equals to: 123455 010713282941 which equals to: 123669 010208133641 which equals to: 112389 041328294042 which equals to: 136689 050916182636 which equals to: 124468 021626323536 which equals to: 146778 040511152232 which equals to: 113357 061821243239 which equals to: 245578 050812162329 which equals to: 123456 071420213033 which equals to: 234567 080913162931 which equals to: 223467 March10, 2007 050609263741 which equals to: 122679 102126283038 which equals to: 256668 061518192033 which equals to: 234447 010308121638 which equals to: 112348 040508213137 which equals to: 112578 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 112348 plays at least 2 numbers will play on the following draw and this is what plays on the following draw: 112578....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....   
Honduras Member #20982 August 29, 2005 4715 Posts Offline  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....   
United States Member #13130 March 30, 2005 2171 Posts Offline  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.   
Honduras Member #20982 August 29, 2005 4715 Posts Offline  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: 233567 and 223567 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...   
Honduras Member #20982 August 29, 2005 4715 Posts Offline  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...   
The Quantum Master West Concord, MN United States Member #21 December 7, 2001 3675 Posts Offline  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  1  2  3  4  5  6  1  749398       2  658008  91390      3  575757  164502  9139     4  501942  221445  25308  703    5  435897  264180  46620  2664  37   6  376992  294525  71400  6300  180  1  7  324632  314160  98175  11900  525  6  8  278256  324632  125664  19635  1190  21  9  237336  327360  152768  29568  2310  56  10  201376  323640  178560  41664  4032  126  11  169911  314650  202275  55800  6510  252  12  142506  301455  223300  71775  9900  462  13  118755  285012  241164  89320  14355  792  14  98280  266175  255528  108108  20020  1287  15  80730  245700  266175  127764  27027  2002  16  65780  224250  273000  147875  35490  3003  17  53130  202400  276000  168000  45500  4368  18  42504  180642  275264  187680  57120  6188  19  33649  159390  270963  206448  70380  8568  20  26334  138985  263340  223839  85272  11628  21  20349  119700  252700  239400  101745  15504  22  15504  101745  239400  252700  119700  20349  23  11628  85272  223839  263340  138985  26334  24  8568  70380  206448  270963  159390  33649  25  6188  57120  187680  275264  180642  42504  26  4368  45500  168000  276000  202400  53130  27  3003  35490  147875  273000  224250  65780  28  2002  27027  127764  266175  245700  80730  29  1287  20020  108108  255528  266175  98280  30  792  14355  89320  241164  285012  118755  31  462  9900  71775  223300  301455  142506  32  252  6510  55800  202275  314650  169911  33  126  4032  41664  178560  323640  201376  34  56  2310  29568  152768  327360  237336  35  21  1190  19635  125664  324632  278256  36  6  525  11900  98175  314160  324632  37  1  180  6300  71400  294525  376992  38   37  2664  46620  264180  435897  39    703  25308  221445  501942  40     9139  164502  575757  41      91390  658008  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  1  2  3  4  5  6  1 {1,2,3,4,5}  2921002  741517  81067  3367  37   2 {6,7,8,9,10}  1418592  1584317  626567  109067  8237  210  3 {11,12,13,14,15}  610182  1412992  1188442  452767  77812  4795  4 {16,17,18,19,20}  221397  905667  1358567  933842  293762  33755  5 {21,22,23,24,25}  62237  434217  1110067  1301667  700462  138340  6 {26,27,28,29,30}  11452  142392  641067  1311867  1223537  416675  7 {31,32,33,34,35}  917  23942  218442  882567  1591737  1029385  8 {36,37,38,39,40}  7  742  21567  250642  1258812  2215220  9 {41,42}      91390  1407406 
I'll add more in a bit, busy. 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   
The Quantum Master West Concord, MN United States Member #21 December 7, 2001 3675 Posts Offline  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  1  2  3  4  5  6  1  14  4      2  7  8  3  1    3  3  7  6  2    4  1  4  6  4  1   5   2  5  6  3  1  6   1  3  6  6  2  7    1  4  8  5  8     1  6  11  9       7 
Here is your actual Distribution. Group Distribution for 25 Draws in Pick 6 of 42  Groups  Column  1  2  3  4  5  6  1  14  5      2  11  9  5     3   5  7  5    4   3  6  6  3  1  5   3  4  6  5  1  6    3  7  7  1  7     1  5  6  8      5  9  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  1  2  3  4  5  6  1  0  1      2  4  1  2  1    3  3  2  1  3  0   4  1  1  0  2  2  1  5   1  1  0  2  0  6   1  0  1  1  1  7    1  3  3  1  8     1  1  2  9       0 
... 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   
Honduras Member #20982 August 29, 2005 4715 Posts Offline  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: 234567, 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... 223456 (but is missing the 7) 233456 (but is missing the 7) 234456 (but is missing the 7) 234556 (but is missing the 7) 234566 (but is missing the 7) we also have 345677 (but is missing the 2) 234677 (but is missing the 5) 235677 (but is missing the 4) 245677 (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...   
Honduras Member #20982 August 29, 2005 4715 Posts Offline  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..   
The Quantum Master West Concord, MN United States Member #21 December 7, 2001 3675 Posts Offline  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. 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   
The Quantum Master West Concord, MN United States Member #21 December 7, 2001 3675 Posts Offline  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's60'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. 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   
Findlay, Ohio United States Member #4855 May 28, 2004 400 Posts Offline  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!/(354)!/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 "NoMatch," "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. TYPE  EXAMPLE  COUNT  CHANCE  NoMatch  1234  21,875  41.78%  Double  1123  26,250  50.13%  Double/Double  1122  2,100  4.01%  Trips  1111  2,100  4.01%  Quads  1111  35  0.067% 
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~   
Honduras Member #20982 August 29, 2005 4715 Posts Offline  Posted: September 18, 2007, 7:58 pm  IP Logged  
  
The Quantum Master West Concord, MN United States Member #21 December 7, 2001 3675 Posts Offline  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, 245566, 134889 and 112569. 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... 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   
The Quantum Master West Concord, MN United States Member #21 December 7, 2001 3675 Posts Offline  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  1  A  A  A  A  A  A  2  A  A  A  A  A  B  3  A  A  A  A  B  B  4  A  A  A  B  B  B  5  A  A  B  B  B  B  6  A  B  B  B  B  B  7  A  A  A  A  B  C  8  A  A  A  B  B  C  9  A  A  B  B  B  C  10  A  B  B  B  B  C  11  A  A  A  B  C  C  12  A  A  B  B  C  C  13  A  B  B  B  C  C  14  A  A  B  C  C  C  15  A  B  B  C  C  C  16  A  B  C  C  C  C  17  A  A  A  B  C  D  18  A  A  B  B  C  D  19  A  B  B  B  C  D  20  A  A  B  C  C  D  21  A  B  B  C  C  D  22  A  B  C  C  C  D  23  A  A  B  C  D  D  24  A  B  B  C  D  D  25  A  B  C  C  D  D  26  A  B  C  D  D  D  27  A  A  B  C  D  E  28  A  B  B  C  D  E  29  A  B  C  C  D  E  30  A  B  C  D  D  E  31  A  B  C  D  E  E  32  A  B  C  D  E  F 
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 Distribution  Form Count  1  A  A  A  A  A  A  0  2  A  A  A  A  A  B  156  3  A  A  A  A  B  B  1440  4  A  A  A  B  B  B  2800  5  A  A  B  B  B  B  1400  6  A  B  B  B  B  B  140  7  A  A  A  A  B  C  8400  8  A  A  A  B  B  C  33600  9  A  A  B  B  B  C  33600  10  A  B  B  B  B  C  8400  11  A  A  A  B  C  C  29400  12  A  A  B  B  C  C  58800  13  A  B  B  B  C  C  29400  14  A  A  B  C  C  C  28000  15  A  B  B  C  C  C  28000  16  A  B  C  C  C  C  7000  17  A  A  A  B  C  D  115500  18  A  A  B  B  C  D  231000  19  A  B  B  B  C  D  115500  20  A  A  B  C  C  D  231000  21  A  B  B  C  C  D  231000  22  A  B  C  C  C  D  115500  23  A  A  B  C  D  D  189000  24  A  B  B  C  D  D  189000  25  A  B  C  C  D  D  189000  26  A  B  C  D  D  D  87500  27  A  A  B  C  D  E  525000  28  A  B  B  C  D  E  525000  29  A  B  C  C  D  E  525000  30  A  B  C  D  D  E  525000  31  A  B  C  D  E  E  393750  32  A  B  C  D  E  F  787500  Total  5245786 
Continues... 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   
