You may be asking how this can help because
you not only have to select the 1 out of the 15 group sets of 6 but you also have to select the correct 4 of
the 6 from the second subset
That is in fact the very question I was asking..........
That, and can you show me an example of a group that hit 5 of 5??
I can't put it in a nutshell so let's say that I use a RNG to generate a string with 6 letters from A to j.
I then run the RNG 15 times so that I have 15 strings all with different arrangments of 6 letters which I
then sort so that they are in order like A-C-E-F-I-J
All 15 strings contain 6 letters of the possible 10 ->ABCDEFGHIJ. If you have downloaded
my software you know that each of these letters repersent groups of 6 numbers each which
is not important for what I am talking about here, the main goal is selecting the correct 3,4 or
5 to put into play for the next draw. Several of the strings will have several letters in common
with other strings. We don't know which groups will hit but using the TG filter, which stands for
"total groups" which is the number of these groups that will provide all the numbers for the
The five winning numbers for the last MM drawing came from groups C-G-H-I with group G providing
two of the numbers. For this example lets say we decide to play 4 groups based on the analysis of
the TG filter. Below is an actual run of the rng. Each run produces different results. I am using a draw
that has already taken place so people can understand the idea.
15 randomly selected groups
A D E F G J =1 G
A C F G H I =4 C-G-H-I
B D E F H J =1 H
A C E F H I =3 C-H-I
B D E F H J =1 H
C F G H I J =4 C-G-H-I
A B C D F H =2 C-H
B C D F H I =3 C-H-I
D E G H I J =2 H-I
A B D F H J =1 H
B C D E G H =3 C-G-H
A B C F H J =2 C-H
B E F G H I =2 G-H-I
E F G H I J =3 G-H-I
B C D F I J =2 C-I
We can see from the above that 2 of the random sets contained all 4 groups that showed in the
last draw and 5 that had 3 of the groups and 4 that had 2 and 4 that had 1 correct. These
results are very close to the expected using the calculations on the previous page match
Match Expected sample +/-
4 1 2 +1
3 4 5 +1
<3 10 8 -2
In the above example we knew which groups to look for but when the draw has not happened yet we have
to find a set of 4 groups that fits as close as possible all the expected values. Before I start to
explain how the program finds the best groups to play I want to remind everyone of something I have
said many times. Every 4of5 needs one wrong number so don't try to set 4 groups to play and block the
other 6, What we need is to put 6 into play and block the remaining 4. This increases the lower prizes
and also gives you room to make a few mistakes and still hit a 5of5. Of the 6 groups we end with here
I suggest setting no more than 3 to play on a TG=5 setup or 2 for a TG=4 setup and let the others 3 or 4
run wild. So for a TG=5 setup you would set 3 to (P) 3 to (W) and 4 to (B).
Now, how the program picks the best four to play.
A D E F G J
A C F G H I
B D E F H J
A C E F H I
B D E F H J
C F G H I J
A B C D F H
B C D F H I
D E G H I J
A B D F H J
B C D E G H
A B C F H J
B E F G H I
E F G H I J
B C D F I J
First we total all the times each groups exist in the random strings. The first step is to take the top 3 or 4 groups
in this case we find they are H-F + B, D, J so we will make 3 sub groups H-F-B, H-F-D, H-F-J. Next find in how many
of the strings these sets of groups appear.
The expected number of strings to have 3 groups is 4. All of these are above the expected number of average hits
so they are not expected to all show in the set to be played. In the last draw C-G-H-I were the correct groups and
none of the above groups of 3 hit and only one of the top hitting values hit in the draw. Almost every time the pool
members sent in there top six choices only 1 of the overall top 3 hit. The lottery officals use probability to calculate
how many tickets will match each of the prizes for any draw, what we are doing here is the same thing only in
reverse. It it works for them then it has to work for us, it has to be one or the other. This is very good stuff and I
hope everyone understands what I am saying. If the draw is not tampered with then this logic should work in a very
It will take too much time to do every tray of groups and sort them by the ones that fit the expected values but the
program does this for trays, pairs and then ones. If at the end of the process the program cannot find a set of four
groups that fits the expected value then it checks for strings that have 4 matching groups and if it finds two or more
that all contain 4 matching groups within another string then it replaces those strings with another randomly generated
Once the program is finished it may find more than one set of four groups to play which it would then try to cover all
of them within one string of six groups. Remember I said I was looking for 6 groups to use in my setup. If the process
finds too many or too few then it starts over by generating 15 new strings and runs the process again. Probability says
that one of the 15 strings should contain a 4of4 so the end set must contain four groups from one of the orgional strings.
If a match of 4 cannot be found then it starts the whole process again.
Working on my Ph.D. "University of hard Knocks"
I will consider the opinion that my winnings are a product of chance if you are willing to consider
they are not.