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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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August 29, 2005
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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....

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....

United States
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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.

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...

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...

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 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.

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Jehocifer

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 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

 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

...

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Order is a Subset of Chaos
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Jehocifer

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...

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..

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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Order is a Subset of Chaos
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Jehocifer

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

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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.

 TYPE EXAMPLE COUNT CHANCE No-Match 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
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 Posted: September 18, 2007, 7:58 pm - IP Logged

thanks Thoth...

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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Order is a Subset of Chaos
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Jehocifer

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 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...

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