Hello everyone

Because this post has become so long I want to restate a couple of things for those who are just comming

in. It gets very hard to read the entire post sometimes to find what we are talking about. This system

uses a method of selecting digits to form numbers instead of selecting the numbers themself.

Things to consider.

All numbers are made of one or more digits

All numbers have digits but all digits are not numbers

Our number system uses ten digits "0-1-2-3-4-5-6-7-8-9"

Note! My Grandad use to be fond of saying that "If we had 6 toes and fingers we would have had a much

better numerical system." I am glad we don't, from the lottery perspective anyhow.

This system is based on what I call the (ID) value. "independent digits". This is a very important, often

overlooked, loophole in lottery double digit number games.

example.

In a 5 of 39 lottery, 42,504 sets can be made from the six digits "1-2-3-4-5-6"

However If you select only the sets that have at least 1ea of all 6 digits then

this is reduced to 5,421 sets. Look at the number sets below

01-12-21-31-32 has only digits "1-2-3" = 3 ID

11-13-21-23-32 has only digits "1-2-3" = 3 ID

06-11-21-26-31 has only digits "1-2-3-6" =4 ID

05-13-22-31-35 has only digits "1-2-3-5" =4 ID

All of these sets of numbers are formed with the digits "1-2-3-4-5-6" but none of them have all 6 digits

01-16-24-31-35 does have at least 1 of all 6 digits so ID = 6 for this number.

The digit "0" is not counted as a ID when it is to the left of another digit as in the number "03"

Here is another another example of why this is so important.

Lets say that set "01-16-24-31-35" was drawn

Of the reduced total sets using this method

1371 sets have 2 numbers that matched "pays $1.00 for this game"

408 sets have 3 numbers that matched "pays $10.00 for this game

44 sets have 4 numbers that matched "pays $250.00 for this game"

1 set has 5 numbers that matched pays $50,000.00 + for this game

1,824 sets have at least a $1.00 prize

this means that 1 of every 2.9 tickets will win at least it's money back.

You must first select the correct 6 digits first but which is better select 5 from 39 or 6 from 10

On the other hand what if I miss one of the 6 digits. Lets say that you select 7 instead of 6

this is what would result. The total set reduction is still down to 5421 and the prize paying

tickets are below.

951 sets have 2 numbers that matched "pays $1.00 for this game"

155 sets have 3 numbers that matched "pays $10.00 for this game

4 sets have 4 numbers that matched "pays $250.00 for this game"

0 set has 5 numbers that matched pays $50,000.00 + for this game

1,110 sets have at least a $1.00 prize

this means that 1 of every 4.9 tickets will win at least it's money back. Not to bad.

The ID's can range from 2 to 8 as in the examples below

"01-02-11-12-21" has 2 ID's

"07-18-29-30-34" has 8 ID's

Most sets fall in the ID = 5 to 6 range for most lotterys

For this game lets say ID= 7 then total sets are reduced to 2370

For this game lets say ID= 4 then total sets are reduced to 3216

These values may drop even more depending on the digits selected

a example of removing digit 1 and using "2-3-4-5-6-7" digits gives a

reduction to only 682 sets.

Next, In my 5-39 game digits "1-2-3" are all in a very large percent of the draws

If ID = 6 and I play "1-2-3" then I only have 3 digits to select from 7 most of the

drawings. I know of no other method that can reduce total sets and still retain

this high of prize paying tickets with fewer or easier steps.

There are many good filters that can reduce this to a playable amount.

RL