a wheel of 2

[foreverbob]

Hi guys,

Suppose I play a 6/45 and I have a nice 4 if 3 in 4 wheel.
Now I want to combine that wheel with another wheel next to it, to fill the other 2 numbers next to it.
So I have tickets with 4 numbers (1st wheel). I just need to put in 2 extra number to complete each ticket.
If I would have 2 extra strong keynumbers, I just go and put those numbers on every ticket.
But most of the times I work with repeat and most skip due numbers. This would be a pool of around 6 to 13 numbers max.
A 2 if 2 in 2 wheel is easy to make, but is there another way we can better implement both wheels to have better succesrate in hitting both wheels?

I play my first wheel of adjacent numbers.
In a 6/45 with bonus, most of the time we hit around 2.1 adjacent numbers per draw.
In a 6/45 with bonus, we hit around 1.5 repeats from the previous draw as well.
Sometimes we hit below that average and sometimes above, but there is a very nice balance in this apporach.
I have tracked 2 repeats + 4 adjacents many times. Thats 6 numbers in 6/45 + bonus, so actualy its like a 7/45.
Its all about matching the two wheels.

Best regards,
Bob

 

[foreverbob]

Previous Belgian lotto:
2 6 8 9 26 34 bonus 45
*3 *5 *7 *10 31 43 bonus 19


Hit 4 adjacent numbers this drawing! No repeats.
The trend is your friend!!

Next drawing, I'll be playing the same 4 wheel on all adjacent numbers from previous draw: 2-4-6-8-9-11-30-32-42-44-18-20
with a 3 if 4 guarantee: almost 42 tickets (adding an extra number gives 42 tickets in Covermaster)

The other wheel will contain: all 7 repeats from previous draw together with the most due group: 16-17-18-19-20-21 ( didnt hit in the past 6 games)
This will also result in 42 tickets.

Merging both wheels is the most difficult part, but it still feels like the best numberselection so far in my study of lottonumbers.
If the most due group doesnt hit, it will still be the very best group of numbers to pool in that second wheel together with all numbers of the previous drawing.

If you have a very due skip group, that will do too, but in my game now there is none.

Stick to it, untill it hits.

Its all average math and a good portion of luck. Dont go back too much in studying past results. At the end every chart has something to tell you, but if you put all charts together you will get every single number of your lottery having a good hitpotential.
And we dont want to pool every number now, do we?


Cheers,

Bob

 

[foreverbob]

Ok guys,


3-5-7-10-31-43 (19)
5-26-32-36-37-42 (2)

We hit 1 repeat.
We hit 3 adjacent numbers.
We hit 1 longshot.

My numberselection method does work and I will continue to make it better and better. This pattern that hits is all about average calculations and it is applicable to every past drawing.
You will see that what I explain does happen.

The most dificult part is to match both wheels.

Your opinion is appreciated. Enjoy the weekend!

 

[Icewynd]

Hi Bob,

I'm not exactly sure what you are doing, but it sure sounds interesting.

Just checking history of my 6/49 game, 2 or 3 repeats from the last game come along every 4 games, on average. With adjacent numbers (last game numbers +1 or -1) you can often match 4 or even 5 numbers in this pool. But this would give you a pool of about 20-21 numbers to wheel, no?

So, how are you reducing your pool of numbers? And how do you settle on which 2 key numbers to use?

 

[foreverbob]

Icewynd,

In a 6/49 game the averages are little less. Let me explain in detail:

Adjacent numbers in
6/45 + bonus = 2.1
6/49 + bonus = 2

6/45 without bonus = 1.5
6/49 without bonus = 1.4

Not sure if your game is with bonus, it would be better if it is.
When it comes to adjacent numbers, I always include bonusnumber.
So for the 6/49 with bonus, the average would still be 2 numbers per drawing.

Now I play those adjacent numbers in a pick 4 wheel.
We have max = 14 adjacent numbers. Most of the time (in my 6/45 + bonus game) we have 12 adjacent numbers, because 68% of all drawings have a neighboring pair and that results in less adjacent numbers.
In a 6/49 + bonus it is 60%
In a 6/49 without bonus 53% has no neighboring pair. Just to give you an idea on how much adjacent numbers there are per game.

Ok, so a pick 4 wheel only including adjacent numbers with lets say:
a 3 if 4 guarantee.

Second wheel (pick 2) is about repeats and the other group (=will explain).
I like to include repeats, because they often happen.
My aim is to get one repeat in my game.
So lets say our last drawing was
1-5-10-15-20-25 and bonus 30
then I will have each number combine with each number of the other group.

In my actual game I have around 9 longshots = numbers that have been out >9 games.
1 of these longhots is an adjacent number so I am left with 8 longshots to include in my second wheel.
So every repeat number will combine with every longshot number:
7 * 8 = 56

Longshots do happen regularly, lets say 1 or 2 numbers per drawing are longshots.
If you have a very due skip group, you can work with that group instead of longhots.
If you have a very due ten number group (or narrowed down to a smaller group) use that group.
Its about taking advantage of the average outcome that is to be expected.

My first pick 4 wheel = 28 tickets (12 numbers) or even smaller.
My second pick 2 wheel = 56 tickets.

So now I put my pick 4 wheel next to my pick 2 wheel. I need to do this twice = 28 * 2 = 56.

Conclusion:
If I have some adjacent numbers hitting in my pick 4 wheel
AND I have a repeat + 1 longshot,
I will have 2 in 56 chance that it will be on the same ticket.
If there are 3 adjacent numbers that hit, I win a 5 number prize, which is my goal.

Since this strategy is based upon mathematical averages and not some probability model, I am very convinced of its effectiveness.
The only probability that comes in game is the 2 wheels matching.
2 in 56 pretty ok, IMO. To better implement both wheels we can then rely on the bias trackers to aim for the average again.


Best regards,
Bob

 

[foreverbob]

It is still a work in progress off course.
There are many ways to choose the right wheels.
Also if a longshot number is an adjacent number already, we might even better use it as a longshot number group and use 1 adjacent number less.
If we choose to do so, we have 11 adjacent numbers instead of 12.
Then we can wheel 11 numbers in our pick 4 wheel with an even better 3 if 3 guarantee. This results in 49 tickets.

When having 7 longshots together with our 7 repeats, resulting in 7 * 7 = 49, then we have a 1 in 49 chance to hit both tickets
IF the results match our AVERAGE approach.


Bob

 

[CMF]

A Ride on the Confused Side

I'm quite familiar with the 6/45 Lotto game as it is the most common matrix in Australia. The following discusses your Wheeling knowledge which to say the least is confused and ignores the occultist or numerology content formed from calculations done on random number Lotto results. (You could just as easily and irrelevantly based your calculations on Australian 6/45 draw histories!)

Wheels or Covers are constructed for a block or line size eg Pick 6 or Pick 5. Your post 7/5/2012 proposes to join a Pick 4 Wheel or Cover with a Pick 2 one. As an example consider the Cover C(49,6,3,6,1)=163 which is a merging of the C(22,6,3,3,1)=77 and the C(27.6,3,4,1)=86 BOTH PICK 6 Covers.

There's nothing to stop you mish mashing your numbers however you want to but if you start using established terminology then the meaning should be consistent with current usage to avoid being misleading.

There's a lot more I could say about your posts but I've already covered most topics in detail elsewhere - see http://ColinFairbrother.com

 

[foreverbob]

Hi Colin,

Thanks for your reaction. I will be reading through your website.
However I dont see any problem in using a pick 4 wheel together with a simple pick 2 wheel.
Don't get me wrong, I am just an apprentice trying to study lottery outcomes. Mistakes will be made.
During the last few days I have been using, what I try to explain in my recent posts, with some degree of succes.

Matching 3 numbers in a pick 4 wheel is not always that hard when having 3 adjacent numbers hit in our draw.
Then when that same ticket has a repeat or a longshot as well, at least I could have a 4 number prize.

So far I have been trying all sorts of numberselection methods, and now I feel that I am having a better succes rate.
That is why I wanted to share it on the forum. Maybe I can get some help along.
Sorry if I used wrong terminology.

Best regards,
Bob

 

[CMF]

Using the previous history show me where your calculation has given a winning subset prize two times or more. You may wamt to read http://www.colinfairbrother.com/AnalysisOfLottoDrawHistoryFinalWord.aspx

 

[foreverbob]

8-9-31-33-39-43 30

Hit 3 adjacent numbers (31-33-43)
2 LS numbers (30-39)
No repeats this time.

I was not lucky combining both adjacent and LS numbers on one ticket.
This still remains one dificult task: combining both wheels.

Bob

 

[CMF]

The Light on the Hill ...

A Wheel is just another name for a Cover which has a mathematical meaning. For a Lotto game we have the Pool (v) or numbers (integers) used such as 45, the Pick (k) which is how many integers are put on a line, block or ticket, the Match (t) or Prize which is the subset of a Pick that the lines played are intended to guarantee and the Hits or Combinations of the integers to which the guarantee applies.

In a 6/45 Lotto game a 3if3 guarantee means that for the 14,190 combinations of 3 integers a set of Pick 6 lines will guarantee when played that there is a match with at least 1 of those 14,190 combinations. Each Pick 6 line has 20 distinct combinations of 3 integers. The current record is C(45,6,3,3,1)=852. A 3if3 Cover or Wheel also guarantees a 3if4, 3if5 and 3if6 for the same Pick and Pool.

The least number of lines needed for a 3if6 Cover or Wheel with guarantee where a combination of 3 integers (CombThree) in at least 1 line of the set is guaranteed to match a CombThree in any of 8,145,060 combinations of 6 integers from a Pool of 45 is C(45,6,3,6,1)=131. This Cover or Wheel with Guarantee can be constructed by merging the Steiner C(22,6,3,3,1)=77 with the C(23,6,3,4,1)=54.

One of my contributions to the field of knowledge in this area is to show that using a lesser Pool for a given Lotto game or a merging of two covers as in the example above reduces the Yield or Percentage Return. See http://www.colinfairbrother.com/UsingLesserPoolCoversInLotto.aspx

I have used the term Wheel above without needing to reference the winning draws for a particular Lotto game as they are irrelevant as far as the next draw is concerned. This is so because a Lotto draw is an Independent Event and I guarantee you will win the Noble Prize if you can show a relationship between recent draws for a given Lotto game and the yet to be drawn next winning number that gives consistent results better than probability calculations.

Colin Fairbrother

 

[foreverbob]

Thank you Colin.
How does this Steiner merging works? Why use 2 wheels with both 3if6 guarantee instead of a complete 45number wheel?

Bob

 

[CMF]

Read Uenal Mutlu's article written in 1995 -

http://www.tuco.de/MY168A.TXT

and my article -

http://www.colinfairbrother.com/MergedCoversOrWheelsForPick6Pool49Lotto.aspx

As your knowledge inproves you may want to read -
Search and Enumeration Techniques for Incidence Structures by Paul C. Denny written in 1998.

https://researchspace.auckland.ac.nz/bitstream/handle/2292/3594/085denny.pdf?sequence=1

A lot of the information on Covers was mainly in the realm of academia in the late 1990's but now is freely available on the web.

Colin Fairbrother

 

[foreverbob]

14/07/2012 Belgian draw was
14-24-34-36-38-44 and bonus 3

Hit 3 adjacent numbers
Hit 1 LS
Hit no repeat

 

[CMF]

It's easy after the fact ...

To be meaningful -

Provide the name of the Lotto game, the lines being played before the draw date, the cost per line, the payouts for whatever subsets and a 100 draw recent history.

Show your calculation method and explain your terminology. If adjacent means 1 + or - the integers in the previous draw then this is old hat and easily disproved. Are you playing 64 lines per draw?

Well after the draw date and posting of your "fortune telling" lines post the winning draw number.

You won't be the first to have bragged about consistent wins which disappear when put to a controlled test.

 

[CMF]

Belgium Lotto Draws from 2012/24 or 24/3/2012

Id24 03 04 12 13 18 24 B21
Id25 11 13 29 35 38 42 B08
Id26 05 07 28 29 41 42 B15
Id27 05 13 16 22 25 39 B09
Id28 01 07 10 19 24 43 B20
Id29 03 05 09 15 18 28 B11
Id30 04 11 14 17 32 44 B30
Id31 08 13 19 21 33 36 B34
Id32 03 09 11 24 30 41 B13
Id33 06 09 27 28 40 43 B03
Id34 04 13 20 37 43 45 B27
Id35 14 23 24 29 39 41 B25
Id36 27 30 32 37 42 45 B12
Id37 08 15 16 17 25 45 B06
Id38 15 20 28 34 41 42 B25
Id39 19 30 31 36 38 40 B07
Id40 07 14 20 21 27 36 B39
Id41 11 14 21 29 30 40 B28
Id42 04 11 15 17 40 43 B42
Id43 13 14 17 20 28 35 B24
Id44 07 22 25 41 42 44 B30
Id45 07 15 16 25 33 43 B19
Id46 10 14 17 19 20 23 B27
Id47 01 04 09 34 38 41 B37
Id48 11 23 27 38 40 42 B44
Id48 08 13 15 26 42 43 B18
Id50 05 14 22 31 41 42 B32
Id51 01 04 06 14 23 34 B16
Id52 02 06 08 09 26 34 B45
Id53 03 05 07 10 31 43 B19
Id54 05 26 32 36 37 42 B02
Id55 08 09 31 33 39 43 B30
Id56 14 24 34 36 38 44 B03
Id57 14 15 17 24 36 38 B30

 

[foreverbob]

Hi,

Pitty we can't upload excel files. Hope it comes through a bit.
Respectively for the past results, I've put 4 collumns each representing 4 groups for the starts of id 24.

R = Repeat from previous draw
LS = Long Shot (numbers out more than 9 games)
Ad = Adjacent numbers from previous draw (+1 and -1)
<9 = Numbers that hit in last 9 games.

The "<9" group is the biggest group. A number from this group could be "R" or "Ad" as well, but I separate them to better identify each group.
If "Ad" is a "LS" as well, then it will be noted as "LS" instead...
"R" is 7 numbers, "LS" around 7 numbers, "Ad" is around 7 numbers, all the rest will be "<9" numbers.
The total of all numbers in each collumn is 7 = 6/45 + bonus.

R LS Ad <9
24 0 1 2 4
25 0 1 2 4
26 2 1 2 2
27 1 3 0 3
28 0 2 1 4
29 0 0 3 4
30 0 5 1 1
31 0 1 1 5
32 1 0 1 5
33 2 1 0 4
34 1 2 2 2
35 0 1 1 5
36 0 1 2 4
37 1 0 0 6
38 2 0 0 5
39 0 2 2 3
40 2 0 2 3
41 1 0 3 3
42 2 0 1 4
43 1 1 1 4
44 0 2 1 4
45 2 1 1 3
46 0 2 3 2
47 0 2 1 4
48 0 2 2 3
49 1 0 2 4
50 1 3 2 1
51 1 0 3 3
52 2 0 2 3
53 0 0 4 3
54 1 2 3 1
55 0 2 3 2
56 0 1 3 3
57 4 0 1 2


Observations (no fortune telling):

Choose to play "R" or "LS" as a keynumber. Most of the times when when "R" is 0, then "LS" will be not 0 and vice versa. There are 34 draws here.
11 results contain 0 and 1 (or 1 and 0)
11 results contain 0 and 2 (or 2 and 0)
2 results contain 0 and 0
The rest is discarted in this test.

When you have decided which group to play (R or LS) then we use those numbers as keynumber on each of the tickets produced by our Pick5 wheel to come...

The other 2 groups are "Ad" and "<9".
The total sum of both groups averages 5. There will be less "Ad" numbers than "<9" numbers.
Choose to play 1 or 2 "Ad" numbers together with 4 or 3 "<9" numbers.

If you want to wheel "Ad" and "<9" together, on this last draw, then its 30 numbers. c(30,5,2,5)=15
Guarantee is put to low, because <9 numbers could belong to other groups.
If you would have 6 "LS", combine them with 15 tickets each = 90 tickets.
If you want to play 7 "R", combine them with 15 tickets each = 105 tickets.


Having past results and study them, helps to get a better feeling of whats going on.
I understand that one event has nothing to do with the other.
Therefor, you can put all numbers in 4 separate hats and pick your numbers randomly.

For now, I would choose not to include "R" in my game, so I'll go with the "LS".
I discarted numbers 12 and 42 for some other reason.
My "LS" are then: 20-21-25-28-29-35
My "Adj" are 13-16-18-23-31-37-39
My "<9" are 1-2-3-4-5-6-7-8-9-10-11-19-22-26-27-32-33-34-40-41-43-44-45

Best regards,
Bob

 

[CMF]

Oh, sweet mystery of life at last I've found you ...

Let's consider the facts.

The Belgium Lotto game has 45 identical objects which are marked to distinguish one from the other - the mark can be anything including pictures of animals or Egyptian hieroglyphics. Before an official draw the objects -

are randomly mixed such that one object is not favoured over another -

and then 7 of those objects are -

randomly extracted without replacement such that one object is not favoured over another -

the first six being the main objects and the seventh being the bonus. To participate in the game you nominate 6 of the objects on an entry form which forms a line costing a Euro €.

The odds and rough payouts for a match between the seven objects drawn and the six objects on an entry line are as follows based on probability or proportionality if you like -

1st 6 main objects 1 in 8,145,060 €2,000,000 approx
2nd 5 main objects + Bonus object 1 in 1,357,510 €50,000 approx
3rd 5 main objects 1 in 35,724 €1,800 approx
4th 4 main objects + Bonus object 1 in 14,290 €400 approx
5th 4 main objects 1 in 772 €27 approx
6th 3 main objects + Bonus object 1 in 579 €13 approx
7th 3 main objects 1 in 48 €6 approx
8th 2 main objects + Bonus object 1 in 64 €3 approx

(Why matching 3 objects pays more than matching 2 Main + Bonus where the probability is less I will put down to a Belgium peculiarity.)

Overall chances of winning a prize are 1 in 25.4 - so, playing 25 randomly selected lines per draw will give an average of close to a win per draw.

You need to show that your numerology methods will give better than a 1 in 25.4 chance of winning a prize. I guarantee you will win the Nobel prize if you do. I rate your chances of success at 0 and your progress so far in proving your case as well.

From your last post you are wheeling 36 numbers which can be done with a C(36,6,3,6)=70. However, this diminishes your chances of getting an average win per draw in a 6/45 Lotto game compared to playing just 25 random selection lines. The question also arises who in their right mind wants to throw away €70 per draw?

Colin Fairbrother

 

[CMF]

Playing a 6/45 Lotto game you play a Pick 6 Wheel ...

My first post in this thread was to point out that the Belgium Lotto is a 6/45 game so you need to play a Pick 6 Wheel. Wheels or Covers are built with the objective of guaranteeing a prize and are not constructed on concatenating Pick4 Wheels with Pick2 Wheels etc in some ad hoc manner. Please stop confusing the issue.

Consider a C(15,6,3,6)=4 ie

1 2 3 4 5 6
7 8 9 10 11 12
1 2 3 13 14 15
1 4 5 6 13 14

Now you could use this in a 6/45 Lotto game but instead of using potentially 24 integers you are using only 15 and this is reflected in the Coverage and also the Percentage Return or Yield with a 92% chance of getting nothing compared to close to 90% if you used unique integers in the four lines. Whether you use the integers from 16 to 30 or 31 to 45 or whatever mix doesn't matter - the covers or wheels are isomorphic and
are NOT NUMEROLOGY based.

Saying that the winning integers from the previous draw +1 and -1 can be put in a group of integers which according to you are more likely to be picked is pure numerology and nonsense.

Saying that sometimes you think the previous draw integers can be put in your preferential group and sometimes not without giving a reason is capricious and illogical.

Using terminology such as LS is unnecessary as you are basing it on absence - these are referred to as Cold numbers - use this and everyone knows what you are talking about.

A notable absence in your numerology methods is frequency which generally is used with absence by most aficionados in this area - usually 3 or more occurrences in past 10 draws for Hot, 2 for Warm and 1 for Luke-Warm.

Making a statement such as, "Having past results and study them, helps to get a better feeling of whats going on. I understand that one event has nothing to do with the other." is a contradiction and irrational.

You have given nothing statistically to support what you are advocating.

 

[Icewynd]

Hey Foreverbob,

Have you done any more with this approach. It sounded pretty interesting.