Repost: 649 Patterns.

[magusprovernum]

I origionally posted this is the 649 discussion forum mistakingly.

---

Hello everyone,

This will be my first post on the forum. As far as I can tell from looking around on the net nobody has tried this.

What I did was split each draw up into two sets of numbers; tens and ones.

eg, 01 23 26 28 33 40 (The last 649 draw numbers)

becomes, 0 2 2 2 3 4 (tens, or "decades")
and, 1 3 6 8 3 0 (ones)

I then looked to see how often numbers repeated in each set. I found that in the decades group the pattern A A B B C D accounts for 5,400 of the total 15,625 possible combinations or 34.6% So should be drawn once in roughly three draws.

In the ones group the pattern A A B C D E accounts for 453,600 combinations of the total 1,000,000 or 45.4% and should be drawn roughly once in every two draws.

Both patterns should be drawn togeather about once in every six draws. With the total number of sets with both patterns being 2,196,634. Did I calculate this correctly? My math skills are not that great so if I made a mistake please let me know.

My theory has been plauged by human error since day 1. Because of my poor math skills and dissorganized way of keeping notes I mistakingly played the wrong pattern in the decades group last saturday on the Ontario49 (which I use to test my theories because it's cheaper to play). Instead of A A B B C D, I played A A A B B C. Luckily that pattern happened to hit along with the pattern for the ones. I played random numbers in the patterns and filtered them by sums so I ended up with 16 sets. I ended up with 29 unique numbers in all the sets I played. I scored 4/6 numbers in the draw. Unfortunately because I made an error in the program I wrote (it never picked 0,) to pick the numbers plus I chose the wrong sum, I only got a maximum of two draw numbers in any set.

I really believe I've hit on something unique and believe that with a little fine tuning my system could be the key to winning. I was going to keep this to myself.. but because I obviously need help with this I decided to post it here.

All feedback and suggestions welcome.

 

[FANTASY2REALITY]

I'm slightly confused,okay I'm lying.Actually I'm totally confused but if your consistent in your attempt to evolve this,you probably will have a nice win one day.I'm not going to offer you some advice,sadly because I have none lol,but if I played o49,I would do this.I have a similar strategy for keno,except there you can see a pattern if you do your homework,and if I played 2 weeks longer,would of won 125gs but hey,better luck next time.So,you pick 3 numbers and see on overage how often they arrive through all the years the game has been in play.Than,add multiple and multiple combos of 3 to the original 3 you selected.The key is tryin to catch those numbers,so that when those numbers come,your halfway home baby.Tell me what you think.

 

[Springbok]

"I then looked to see how often numbers repeated in each set. I found that in the decades group the pattern A A B B C D accounts for 5,400 of the total 15,625 possible combinations or 34.6% So should be drawn once in roughly three draws."

What do you mean by A,B,C,D? Does A = (first decade- 1 to 10)? Where is E for the 5th decade.

 

[magusprovernum]

Placeholders.

For example, the set: 03 09 11 15 23 34

Would be broken down to:

0 0 1 1 2 3 (tens)
3 9 1 5 3 4 (ones)

or

A A B B C D
A B C D A E (A A B C D E)

The letters are just placeholders for the numbers and have no value.

A A B C D E could be;

0 0 1 2 3 4
0 0 1 2 4 5
...
...
...
...

...
9 9 5 6 7 8

When your talking about the tens group (0-4) and you have a set such as:

02 11 23 35 46 48 then the pattern would look like this: A B C D E E

It just looks like A represents the first decade because the tens are always listed in numerical order. But if you look at the same set randomized (3 4 0 1 2 4) Then the pattern would be A B C D E B.

A B B C D E is the same thing as A A B C D E. Look at it this way, instead of saying the pattern is A A B C D E, I'll say it's 2 1 1 1 1. That just means that one number is drawn twice with four other unique numbers.

Is that clearer?

 

[magusprovernum]

Simplified.

The basics of my system is this:

Nearly 45% of all draws will have two numbers that end in the same digit and four that don't. If you play this pattern then every other draw your odds of winning the jackpot will become 1 in 7.7 Million.

35% of all draws will have two numbers from one decade, two numbers from a second decade and two numbers from two others. With one decade left out.
If you play this pattern your odds of winning the jackpot become 1 in 9.1 Million every third draw.

Once in every six draws these two patterns will show up togeather. When this happens your odds of getting 6 out of 6 numbers is reduced from 1 in 13.9 Million to 1 in 2.19 Million.

This can be applied as a filter. Say that you want to play 18 numbers. There are 18,564 combinations of 18 numbers. The odds of winning the jackpot are 1 in 753.28. If you apply the first pattern you can eliminate 10,210 sets and your odds will remain unchanged every second draw.

If you apply both patterns you can eliminate another 6,637 sets and your odds of winning will remain 1 in 753.28 in six draws. Out of 18,564 you will only need to play 1,717 sets.

Now, from a money management standpoint you *DO* end up paying 2% more for your tickets if it takes six draws to match the patterns for the same odds.

For example; Instead of playing 100 tickets in one draw you play 17 filtered tickets in six draws. 17*6=102.

However, if it only takes three draws to match then you are spending only 51% of what it would cost to play all 100 sets in one draw for the same odds of winning the jackpot.