Help Needed: Observable Pattern

Reply #26

pumpi76 — Fri, 21 Sep 2007 05:29:55 GMT

Apparently we will not be able to write the wheel with the same code multiple times and get multiple 5 out of 6's...But i think we might be able to gather the group of 5's grouping which i am thinking it will be less than 8, 5's grouping and write (actually i am thinking it will be less than 4) a regular wheel out of the group of 30...This way we will not have to worry about getting the exact 6 numbers 5's groupings...For example, if i get 2-3-4-5-6-7 and what plays is 2-2-3-4-5-6, i will still... [ More ]

Reply #25

JADELottery — Thu, 20 Sep 2007 18:05:19 GMT

Hope this helps with your pattern quest.

Reply #24

JADELottery — Thu, 20 Sep 2007 13:20:25 GMT

...

6

A B C D E F

1 2 3 4 5 6

15625

1 2 3 4 5 7

15625

1 2 3 4 5 8

15625

1 2 3 4 5 9

6250

1 2 3 4 6 7

15625

1 2 3 4 6 8

15625

1 2 3 4 6 9

6250

1 2 3 4 7 8

15625

1 2 3 4 7 9

6250

1 2 3 4 8 9

6250

1 2 3 5 6 7

15625

1 2 3 5 6 8

15625

1 2 3 5 6 9

6250

1 2 3 5 7 8

15625

1 2 3 5 7 9

6250

1 2 3 5 8 9

6250

1 2 3 6 7 8

1562... [ More ]

Reply #23

JADELottery — Thu, 20 Sep 2007 13:17:54 GMT

...

5

A A B C D E

A B B C D E

A B C C D E

A B C D D E

A B C D E E

Total

1 2 3 4 5

6250

6250

6250

6250

6250

31250

1 2 3 4 6

6250

6250

6250

6250

6250

31250

1 2 3 4 7

6250

6250

6250

6250

6250

31250

1 2 3 4 8

6250

6250

6250

6250

6250

31250

1 2 3 4 9

2500

2500

2500

2500

625

10625

1 2 3 5 6

6250

6250

6250

6250

6250

31250

Reply #22

JADELottery — Thu, 20 Sep 2007 13:15:33 GMT

...

4

A A A B C D

A A B B C D

A B B B C D

A A B C C D

A B B C C D

A B C C C D

A A B C D D

A B B C D D

A B C C D D

A B C D D D

Total

1 2 3 4

1250

2500

1250

2500

2500

1250

2500

2500

2500

1250

20000

1 2 3 5

1250

2500

1250

2500

2500

1250

2500

2500

2500

1250

20000

1 2 3 6

1250

2500

1250

2500

2500

1250

2500

2500

2500

1250

20000

1 2

Reply #21

JADELottery — Thu, 20 Sep 2007 13:13:00 GMT

...

3

A A A A B C

A A A B B C

A A B B B C

A B B B B C

A A A B C C

A A B B C C

A B B B C C

A A B C C C

A B B C C C

A B C C C C

Total

1 2 3

125

500

500

125

500

1000

500

500

500

125

4375

1 2 4

125

500

500

125

500

1000

500

500

500

125

4375

1 2 5

125

500

500

125

500

1000

500

500

500

125

4375

1 2 6

125

500

500

125

500... [ More ]

Reply #20

JADELottery — Thu, 20 Sep 2007 13:10:18 GMT

Ok, next are the break down of distributions by Basic Permutational Form and the associated Combinatorial Form. I'll have to post one at a time, so look for additional posts. Keep in mind these are only for a Pick 6 of 42 lottery broke down into your groups of 1-5, 6-10, 11-15, 16-20, 21-25, 26-30, 31-35, 36-40, and 41-42 only. A different lottery with a different pick size and/or number pool size and/or group pattern is going to have a different set of distributions. Anyway, here are the distri... [ More ]

Reply #19

JADELottery — Wed, 19 Sep 2007 20:33:52 GMT

Once again... work calls... have get back to this later... have fun.

Reply #18

JADELottery — Wed, 19 Sep 2007 20:28:04 GMT

Previously I showed you the distribution of the Basic Permutational Forms, in addition, there is a Combinatorial Form Distribution that shows how the groups are relative to each combination. They are as follows.

Combinatorial Form Distribution

123456

101 22101 2 343751 2 3 4200001 2 3 4 5312501 2 3 4 5 615625

201 32101 2 443751 2 3 5200001 2 3 4 6312501 2 3 4 5 715625

301 42101 2 543751 2 3 6200001 2 3 4 7312501 2 3 4 5 815625

401 52101 2 643751 2 3 7200001 2 3 4 8312501 2 3 4

Reply #17

JADELottery — Wed, 19 Sep 2007 20:08:12 GMT

Pumpi76,

You can work backward to find the original group of numbers using the Basic Permutational Forms and the Combinatorial Forms. As an example, let's say we knew the Basic Permutational Form is A B C C D E and it's associated Combinatorial Form is 2 3 4 6 8, we can work backward by assigning each letter a number in order of ascending numbers and alphabetic letters: A = 2, B = 3, C = 4, D = 6 and E = 8. This leads to the original group of 2 3 4 4 6 8.

Continues...

Reply #16

JADELottery — Wed, 19 Sep 2007 19:22:36 GMT

I'll get back to this a little later... family and work calls...

Reply #15

JADELottery — Wed, 19 Sep 2007 19:19:46 GMT

The Basic Permutational Forms are sub-divided into 6 parts, these parts are then associated with the different Combinatorial Forms. Below are the Basic Permutational Forms in the separate associated Combinatorial Forms.

Separate Basic Permutational Forms

11AAAAAA

22AAAAAB

3AAAABB

4AAABBB

5AABBBB

6ABBBBB

37AAAABC

8AAABBC

9AABBBC

10ABBBBC

11AAABCC

12AABBCC

13ABBBCC

14AABCCC

15ABBCCC

16ABCCCC

417AAABCD

18AABBCD

19ABBBCD

20AA... [ More ]

Reply #14

JADELottery — Wed, 19 Sep 2007 18:26:08 GMT

...

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... [ More ]

Reply #13

JADELottery — Wed, 19 Sep 2007 18:08:23 GMT

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

Reply #12

pumpi76 — Tue, 18 Sep 2007 23:58:29 GMT

thanks Thoth...

Reply #11

Thoth — Tue, 18 Sep 2007 06:15:11 GMT

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 tha... [ More ]

Reply #10

JADELottery — Mon, 17 Sep 2007 20:54:05 GMT

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.

Reply #9

JADELottery — Thu, 13 Sep 2007 03:43:30 GMT

pumpi76,

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

Reply #8

pumpi76 — Wed, 12 Sep 2007 21:50:05 GMT

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 gu... [ More ]

Reply #7

pumpi76 — Wed, 12 Sep 2007 20:59:59 GMT

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... [ More ]

Reply #6

JADELottery — Tue, 11 Sep 2007 17:08:47 GMT

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 ad... [ More ]

Reply #5

JADELottery — Tue, 11 Sep 2007 15:45:02 GMT

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... [ More ]

Reply #4

pumpi76 — Tue, 11 Sep 2007 15:03:38 GMT

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... [ More ]

Reply #3

pumpi76 — Tue, 11 Sep 2007 14:48:19 GMT

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... [ More ]

Reply #2

time*treat — Tue, 11 Sep 2007 05:29:41 GMT

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.

Reply #1

pumpi76 — Tue, 11 Sep 2007 00:10:35 GMT

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....

Help Needed: Observable Pattern

pumpi76 — Mon, 10 Sep 2007 23:57:41 GMT

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 exa... [ More ]