GENERAL INFORMATION

Acces is forbidden to persons under 18 years old.

This site is addressed to people with IQ above average and in particular persons with mathematical skills.

This program is a mathematical (statistical) system to reduce variants.

To create a reduction system it is not enough to understand mathematics and the rules of the 6/49 lottery game system; besides these, one needs creativity, vision and the dedication of the lottery players.

The statistics presented lead to a system of reduction efficient as ratio between: played numbers / resulted combinations (how much money you bet) / winning combinations (how much money is won).

On the whole, the number of winning combinations as against standardized reductions is at least triple and the costs are at a quarter. Multiply…

This reduction system does not tell you which numbers will be extracted BUT guides you not to throw away your money on combinations of very low probability.

This type of reduction allows you to play as many numbers as you want but, following the reduction one does not have a mathematical impediment with a FIX number of winning combinations.

In order to fill in the playing tickets, a printing program specific for each country is available (the playing ticket differs with each country).

The reduction system of the number of combinations is based on the history of the draws. One can notice that, as the number of draws analyzed is bigger, the covering graph is very close to the theoretical graph of probability of number occurrence. So, in our opinion, we consider that ALL LOTTO PROGRAMS ARE INACCURATE because they give to the players instructions based on the history of thousands of draws. The principle we rely on in this paper is:

This is valid up to a certain limit and probability. And it is possible when one relies on the history of 490 draws in parallel with the recent history of the draws. We will detail each of the statistical chapters.

IMMINENT QUESTION: If you think you’re so smart, why don’t you play and win by yourself?

ANSWER TO THE IMMINENT QUESTION: If we applied a 90% reduction to the 13,983,816 combinations, we would still have to play about 1,400,000 combinations. If we applied a 99% reduction, we would still have to play 140,000 combinations.

Now multiply those with the price for a combination and you will get the ANSWER.

If there are 1,000 players for 140 combinations each, then everything seems credible, doesn’t it?

By applying the same conditions of reduction, this system will generate different results because the database is modified with each draw; we update this database after each draw.

If a single LOTTO draw was made weekly and each possible combination was extracted once only, then it would take us 268,919.53 years to extract all possible combinations!!!!

The TABLE below displays the amount of resulted combinations dependent on how many numbers were played.

The combination of the 49 numbers will give, according to the formula C⁶₄₉=A⁶₄₉/P₆ , a number of 13,983,816 possible combinations. For "n" numbers played we have C⁶_n=A⁶_n/P₆.

The next table displays the values calculated for all the "n" numbers combinations played:

FREQUENCY

It shows how many times each number appeared in the last 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 50, 500, 1000 draws.

If we look at the 1000 draws graph, we will see that, according to the BIG NUMBERS THEORY, the graph tends to balance the frequencies of all numbers, therefore the indication of the numbers that could appear is not only very weak, almost inexistent, but even baffling, not to say FALSE; this is what all LOTTO sites do. If we look at the graphs of the last 5, 10, 25 draws the signals for the numbers occurrence are evident. From the draw made today we have maximum one or two and very seldom three numbers out of the numbers extracted LAST TIME.

For example, if in the last 5, 10 or 25 draws we find a number which was not extracted, then we know that it is a high probability of its occurrence; it is possible that, looking at the 1000 draws graph, the frequency of this number to be higher than that of other numbers and the indication of occurrence based on all draws to be incorrect.

The chapter on frequency will help selecting the numbers played. Avoid the mistake of picking only the numbers played in the range 1-31 (birthdays, dates of events etc.). We still have 18 numbers from 32 up to 49 to play.

We think it odd whenever these numbers appear, but their probability of occurrence is absolutely equal to that of the other numbers.

It is also the reason why we had more winners playing at random than playing with selected numbers. When playing randomly there is a chance for numbers from 32 to 49 to be drawn.

Everybody, absolutely everybody, relies on good luck. One cannot say that luck does not exist. The idea is that LUCK can be SUBSTANTIALLY assisted.

All those rejecting this idea on the grounds that if you are to win you will, if not, you won’t, have a very low IQ.

The graph displays on the abscissa (horizontal) all the numbers from 1 to 49 and on the vertical a column for each number from 1 to 49. The height of the column corresponds to the number of occurrences of that number (at the top of the column we have the number of occurrences).

You can check the frequency of occurrences analyzing the graph for the last 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 50, 100, 250, 500, or 1000 draws.

PARITY

Numbers can be divided into even numbers (2, 4, 6, 8, 10, 12, . . . 46, 48) and odd numbers (1, 3, 5, 7, 9, 11, 13, . . . 47, 49).

So, there are 24 even numbers and 25 odd numbers. Considering this criterion, the 6 numbers extracted may appear in the following sequence in which the first number is the even number and the second number is the odd number.

0-6 (no even numbers and six odd numbers)

1-5 (1 even number, 5 odd numbers)

2-4 (2 even numbers, 4 odd numbers)

3-3 (3 even numbers, 3 odd numbers)

4-2 (4 even numbers, 2 odd numbers)

5-1 (5 even numbers, 1odd number)

6-0 (6 even numbers, no odd numbers)

Look at the graph:

Why playing combinations with low extraction probability?

You are right, all combinations have a chance of occurrence in the draw but the 0-6 or 6-0 combinations should be left to others.

Personally I play 2-4, 3-3, 4-2.

On the horizontal there are alternatives of combination of even numbers with odd numbers.

Beside each even-odd combination there is a column indicating the number of occurrences of that combination (it is displayed on top of the column).

ENDINGS

We have 10 groups of endings. Those ending in

0 ( 10 , 20 , 30 , 40 ) 4 numbers

1 ( 1 ,11 , 21 , 31 , 41 ) 5 numbers

2 ( 2 , 12 , 22 , 32 , 42 ) 5 numbers

3 ( 3 , 13 , 23 , 33 , 43 ) 5 numbers

4 ( 4 , 14 , 24 , 34 , 44 ) 5 numbers

5 ( 5 , 15 , 25 , 35 , 45 ) 5 numbers

6 ( 6 , 16 , 26 , 36 , 46 ) 5 numbers

7 ( 7 , 17 , 27 , 37 , 47 ) 5 numbers

8 ( 8 , 18 , 28 , 38 , 48 ) 5 numbers

9 ( 9 , 19 , 29 , 39 , 49 ) 5 numbers

When looking at the numbers extracted we notice the endings. One can see that in more than half of the combinations, the numbers have a doubled ending.

E.g. 1:     2 , 42 , 19 , 38 , 31 , 20

Here we have 5 endings 0, 1, 2, 8, 9. The ending 2 is doubled (for numbers 2 and 42); the other endings: 9 (at 19), 8 (at 38), 1 (at 31) and 0 (at 20) occur only once.

We have the formula for the number of endings: 2 - 1 - 1 - 1 - 1

E.g. 2:     48, 16, 33, 18, 28, 5

Here we have 4 endings: 3, 5, 6, 8

The ending 8 is tripled (48, 18, 28)

So we have 3-1-1-1

E.g. 3:     13, 19, 46, 33, 49, 2

In this case we have 4 endings: 2, 3, 6, 9

The endings 3 and 9 are doubled (13, 33; 19, 49)

So, we have 2-2-1-1

The chapter on endings must be very well studied because it is an outstanding source of reduction of the combinations number.

The graph displays on the abscissa (horizontal) the groups of possible endings

Beside each group of endings there is a column on top of which it is written the number of occurrences of this combination of endings.

Before playing we have to study the occurrence graphs and corroborated with the theoretical probability we can select the playing combination.

If, in accordance with the statistics presented, one deliberately eliminates the maximum probability area, then one practically rejects getting to the combination of 6 winning numbers but the reduction is very big and the chances of having 5 winning numbers are also very high (theoretically even in this case there are chances to get 6 numbers but very weak ones).

Recommendation: 2-2-1-1

DECADES

We divide all the numbers from 1 to 49 into 5 groups of 10 numbers each called decades.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10

11, 12, 13, 14, 15, 16, 17, 18, 19, 20

21, 22, 23, 24, 25, 26, 27, 28, 29, 30

31, 32, 33, 34, 35, 36, 37, 38, 39, 40

41, 42, 43, 44, 45, 46, 47, 48, 49

Each group has 10 numbers except for the last group 41-49 that has only 9 numbers. We will still call it a decade.

The numbers extracted are distributed in 1, 2, 3, 4, 5 decades. Generally, in 4 decades, but whoever wants a higher reduction can go to 3 decades.

The graph displays on the abscissa the combinations with the number of decades 1,2,3,4,5, and beside each combination, a column shows how many times the respective combination was extracted (the number is displayed above).

PENTADS

We divide all the numbers from 1 to 49 in 10 groups of 5 numbers each called pentads.

1 , 2, 3, 4, 5;

6, 7, 8, 9, 10;

11, 12, 13, 14, 15;

16, 17, 18, 19, 20;

21, 22, 23, 24, 25;

26, 27, 28, 29, 30;

31, 32, 33, 34, 35;

36, 37, 38, 39, 40;

41, 42, 43, 44, 45;

46, 47, 48, 49;

Each group has 5 numbers except for the last group 46-49 that has 4 numbers but which we will still call a pentad. The numbers extracted are distributed in 2,3,4,5, or 6 pentads out of the total of 10 pentads. In most of the cases they are distributed in 5 pentads meaning that in a single pentad we have 2 numbers of each and in the others we have one number of each. The reduction is big!

And yet, whoever rejects the idea of having 6 winning numbers and thinks of having 5 winning numbers will distribute the numbers in 4 pentads. It’s only a hint...

The graph displays on the abscissa the combinations with pentad numbers 2, 3, 4, 5, 6, and beside each combination there is a column indicating how many times the respective combination was extracted (the number in displayed above).

TRIADS

We divide all the numbers from 1 to 49 into 16 groups of 3 numbers each called triads:

1, 2, 3;

4, 5, 6;

7, 8, 9;

10, 11, 12;

13, 14, 15;

16, 17, 18;

19, 20, 21;

22, 23, 24;

25, 26, 27;

28, 29, 30;

31, 32, 33;

34, 35, 36;

37, 38, 39;

40, 41, 42;

43, 44, 45;

46, 47, 48, 49.

Each group has 3 numbers except for the last group that has 4 numbers. We will still call it a triad.

The numbers extracted are distributed in 2, 3, 4, 5 or 6 triads.

If you look at the statistics you will notice that it is a very high probability for the numbers to be distributed in 5 or 6 triads.

We would select 5 triads to increase our chances of getting 5 numbers.

The graph displays on the abscissa the combinations with the number of triads 2, 3, 4, 5, or 6 and beside each combination there is a column showing how many times the respective combination was extracted (the number is displayed above the column).

LINES and COLUMNS

We distribute the 49 numbers in a 7x7 square.

We have 7 lines (horizontal).

As presented in the graph, the distribution of the extracted numbers enlists in the matrix L-C 7x7 on the abscissa in the following combinations (the first figure – no. of Lines, the second figure – no. of Columns).

From the huge number of possible combinations, look at the graph of 1000 draws and you will see the most suitable combination.

We would play L: min 4 – max 5; C: min 4 – max 5

Beside each L-C combination we have a column on top of which there is the number of occurrences of this combination.

VALUES OF LINES-COLUMNS REDUCTION

In the 6 lines and 6 columns table one finds the results of the Lines-Columns reduction for each pair of Line-Column values (from 1 to 6 lines, respectively from 1 to 6 columns).

The table below displays the number of combinations left out of the total of 13,983,816 combinations after the Lines-Columns reduction.

The next table displays the probability of occurrence of each Line-Column combination.

DISTANCE

Suppose the numbers extracted are: 24, 33, 48, 2, 12, 25.

What is DISTANCE?

To answer this question, one has to rewrite the combination with the numbers in ASCENDING ORDER : 2, 12, 24, 25, 33, 48

D6 (for D6 there is only one value).

Distance "6" – is the value of the difference between the first number (the smallest – in our case, number 2) and the sixth number (the biggest – in our case, number 48).

So, D6 = 48 – 2; D6 = 46;

D5 (for D5 there are two values)

D5 the first value:

Distance "5" – is the value of the difference between the first number (the smallest – in our case, number 2) and the fifth number in ascending order (in our case, number 33).

So, D5 = 33 – 2; D5 = 31 (the first value)

D5 the second value:

Distance "5" – is the value of the difference between the second number in ascending order (in our case, number 12) and the sixth number in ascending order (in our case, number 48).

So, D5 = 48 – 12; D5 = 36 (the second value)

Hence, for D5 there are two values: D5=31 (the first value) and D5=36 (the second value)

D4 (for D4 there are three values)

D4 the first value:

Distance "4" – is the value of the difference between the first number (the smallest – in our case, number 2) and the fourth number in ascending order (in our case, number 25).

So, D4 = 25 – 2; D4 = 23 (the first value)

D4 the second value:

Distance "4" – is the value of the difference between the second number in ascending order (in our case, number 12) and the fifth number in ascending order (in our case, number 33).

So, D4 = 33 – 12; D4 = 21 (the second value)

D4 the third value:

Distance "4" – is the value of the difference between the third number in ascending order (in our case, number 24) and the sixth number in ascending order (in our case, number 48).

So, D4 = 48 – 24; D4 = 24 (the third value)

Hence, for D4 there are 3 values: D4 = 23 (the first value), D4 = 21 (the second value), D4 = 24 (the third value).

The graph displays on the abscissa the values of the distances from 5 to 48.

For each distance graph for 6 numbers, 5 numbers or 4 numbers there is a column with the number of occurrences on top corresponding to each distance. Consequently, this criterion can be decisive for the selection made which means that you play so as to get the combination of 6 numbers or the combinations of 5 numbers or that of 4 numbers. It is possible to play 4 numbers and get 6 numbers but the probability is low. The intermediate solution would be to play so as to get 5 winning numbers in the 20 – 35 area of distance where the reduction is optimal and if this time we do not get 5, we will certainly get 4.

We have to add that, for this type of reduction you can play as many numbers as you want and, following the reduction, there will not be a mathematical impediment with a fix number of winning combinations.

When applying the "DISTANCE 6" criterion, the next table displays the values for the number of combinations left out of the total of 13,983,816 combinations corresponding to each distance; the last column displays the probability of success (in percentage) for the selected distance.

FIGURES

The numbers extracted may be made up of one or two figures. One can have in a draw from 6 to 12 figures.

E.g.1: 8, 2, 5, 7, 1, 9 (6 figures)

E.g.2: 19 , 23 , 33 , 48 , 41 , 12 (12 figures)

The figures 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 0 will be divided into two categories.

Category A : 1 , 2 , 3 , 4

Category B : 5 , 6 , 7 , 8 , 9 , 0

During the draws, a ratio between the number of figures from the first category and the number of figures from the second category will be kept.

In the previous examples we find:

Draw of six figures: 8 , 2 , 5 , 7 , 1 , 9 has 2 figures from category A (on 1 and on 2) and 4 figures from category B (on 5,on 7,on 8, and on 9).

Draw of 12 figures: 19 , 23 , 36 , 48 , 41 , 12 has 9 figures from category A (on 1, on 2, on 3, on 3, on 4, on 4, on 1, on 1 and on 2) and 3 figures from category B (on 9, on 6 and on 8).

When analyzing this criterion on the graph one finds the groups of 6, 7, 8, 9, 10, 11 and 12 figures. Each combination variant of numbers from the first category (A) with those from the second category (B) is represented by a column; the number above each column shows how many times that figures combination was extracted. It results from the graph that the most probable intervals (in our opinion) are: for Group A (1,2,3,4) minimum 6, maximum 9 and for Group B (5,6,7,8,9,0) minimum 3, maximum 5.

TABLE 6 X 7

This table has 42 little boxes, each of them containing two values: a minimum and a maximum one.

What are these values?

The study of all the conditions imposed so far is concentrated in this table with 42 little boxes. The checking program will allow you to verify the values corresponding to each draw made until now, except for the first 420, respectively 490 draws from the beginning of the 6/49 lottery game.

This table finds its place among other elements of singularity of this system and that is why we are not going to disclose the sources but only the manner of action, of play.

The success rate of processing the table is over 90% in terms of a reduction over 30%.

ROWS OF NUMBERS

Assume one can consider "n" elements from a numbers row of "N" elements, with "n" chosen between a minimum "a" and a maximum "b" values.

E.g.:

Let N be 16 and let our numbers row consist of the elements 2, 6, 9, 13, 15, 18, 22, 23, 26, 29, 31, 32, 37, 42, 44 and 49. If we set "a" and "b" to 2 and respectively 4, "n" would take a value between a=2 and b=4, which means that we would consider at least 2 and at most 4 numbers from the described numbers row.

By setting these conditions, a significantly reduction will be made.

STEP BY STEP REDUCTION

Variants reduction is done by selecting the variants from the .csv file with the chosen ”step” ”n”.

E.g. step n=2

Out of the variants 1,2,3,4,5,6,7,8,9,10,11,12, ... from the variants file you have selected, there will remain the variants 1,3,5,7,9,11, ... .

E.g. step n=3

Out of the variants 1,2,3,4,5,6,7,8,9,10,11,12, ... from the variants file you have selected, there will remain the variants 1,4,7,10, ... .

ENDINGS GRIDS

The chapter ENDINGS deals with the ”number” of endings from a draw; in this case the player ”nominates” the combinations of endings.

E.g. We select 4 sets of endings:

1, 3, 4, 8 ( 4 endings )

2, 3, 7, 8, 9 ( 5 endings )

3, 4, 5, 8, 9 ( 5 endings )

1, 4, 6, 7, 8, 9 ( 6 endings )

The variants from results.csv which do not comply with these conditions are eliminated.