If you look at what they are doing there is actually figuring out the span between each number leaving the first number as a starting point of reference. If you knew what the lowest number was going to be in a drawing and you knew what the span was going to be between the first number and the second number in ascending order was, you would know the first two numbers in ascending order, the same goes for the other ascending numbers drawn.

The problem is that you don't know what the smallest number drawn is going to be and you don't know what the span between each number in ascending order will be. This would be like taking 15 #'s and wheeling them into combinations of 5 if 5.

A 5 if 5 using 15 numbers is 3003 combinations in ascending order now I don't know this one but if you take each 5# combination and arranged them into every sequence possible and multiply that by 3003, well you get the picture.

Below is an example using 7 recent draws from the Oh 6/49.

7 recent Oh 6/49 draws

5/4 26 29 30 34 37 49 (14)

5/8 07 16 17 29 30 38 (42)

5/11 05 20 36 40 41 45 (37)

5/15 01 07 18 22 30 35 (05)

5/18 04 17 28 41 42 47 (18)

5/22 09 18 21 22 24 31 (23)

5/25 19 22 29 30 33 45 (42)

In the next set of numbers the 1st # is the first # of each of the 7 draws above the lowest number drawn.

The #'s after that are the span #'s or "Deltas" for the 7 draws above. If you take the first or lowest number and add the span number to it you will have the 2nd number of the drawing - 26+03=29, 29+01=30, 30+04=34 and so on.

1st# #Span

26: 03 01 04 03 12

07: 09 01 12 01 08

05: 15 16 04 01 04

01: 06 11 04 08 05

04: 13 11 13 01 05

09: 09 03 01 02 07

19: 03 07 01 03 12

The next chart shows the spans or deltas values from 1-15. The 1st # being the span value and the next 5#'s being the amount of times occurring in each position.

Position meaning p1-p2, p2-p3, p3-p4, p4-p5, p5-p6.

The last number is the total span value or delta occurrences.

This is based on the last 40 drawings of the Oh 6/49

Span Occurrence Total value

Value per position occurrence

1 02 08 08 08 03 29

2 03 03 05 03 05 19

3 06 03 02 05 04 20

4 04 03 08 04 03 22

5 02 02 02 04 05 15

6 05 05 02 03 03 18

7 01 02 02 02 05 12

8 02 04 03 01 03 13

9 03 01 01 01 02 08

10 03 01 01 01 01 07

11 01 03 00 01 00 05

12 01 01 02 02 02 08

13 04 01 02 01 01 09

14 00 00 01 00 00 01

15 01 01 00 01 00 03

As the span values get higher the occurrence gets lower.

In a 6/49 the avg. number by position is 7,14,21,28,35,42 but to play this set of numbers together all the time we would never win a jackpot. You would want to play a mix of the most common occurring spans and that still most likely wouldn't capture a jackpot as you would have to guess what the lowest number drawn was going to be and what the spans were going to be and if you did know the first two you still would have to know what order the spans would appear.

An interesting point is that if you add all of the span numbers together the sum will be the spread between the lowest and highest numbers in the drawing plus, the sum will not be any higher then the highest number in the field of numbers in the lotto game you're playing.

Man, I think I would be better off playing the auto-pick.

Good Luck to all and may all of your deltas be in rivers.

Scott Rudy

To which Nick Koutras added . . .

The probability of the smallest delta that can be found in a 6/49 game can be calculated by the following equation:

n=delta

(49-5n)-choose-6 minus (49-5(n+1))-choose-6

-------------------------------------------

49-choose-6

This gives us the following table.

delta p(min delta) p(as dec) 1/p

--------------------------------------------

0 6924764 /13983816 0.495198449 2.02

1 3796429 /13983816 0.271487339 3.68

2 1917719 /13983816 0.137138461 7.29

3 869884 /13983816 0.062206482 16.08

4 340424 /13983816 0.024344142 41.08

5 107464 /13983816 0.007684884 130.13

6 24129 /13983816 0.001725495 579.54

7 2919 /13983816 0.000208741 4790.62

Please note that n=0 for the discussion I've seen here means delta of 1

> Nice work Nick, It looks like we will have to make

> sure that future entries contain at least 1 delta

> of no greater than 5 ;-) Sean B

Yes!

But imagine if we "assume" that the smallest delta will

be >3 We have eliminated 90% of the combinations!

>

> A nice exercise for someone would be to confirm

> the above, by checking that their history contains

> 1.725% of draws with a min. delta of 6

>

> Sean B

>

> >

> >

> >Robert Perkis wrote:

> >

> >> I was out on the web reviewing some lottery site

> >> links and came across http://www.use4.com/lotto.html

> >> which talks about converting the 49 numbers to 15

> >> delta numbers dependent on sequence, predicting from

> >> within the 15 and than restoring them to combinations.

> >>

> >> Does this make sense and does anyone see potential in

> >> this method?

> >>

> >> Thanks. Robert Perkis / http://www.lotto-logix.com/

--

Nick Koutras

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