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
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.
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.
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.
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:
(49-5n)-choose-6 minus (49-5(n+1))-choose-6
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
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/