GENERAL INFORMATION

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

This program is a REDUCTION SYSTEM for the variants, based on statistics.

To create a reduction system it is not enough to understand mathematics and the rules of the 6/49 lottery game system, but to also be creative, have a vision and the dedication of lottery players.

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

On the whole, the number of winning combinations as opposed to standardized reductions is at least triple and the costs are at a quarter. Do the calculations…

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

This type of reduction allows you to play as many numbers as you want but, and following the reduction, you do not have a mathematical impediment with a FIXED 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 chart is very close to the theoretical chart of probability of number occurrence. So, in our opinion, we consider that ALL LOTTO PROGRAMS ARE INACCURATE because they give instructions to the players, 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, yourely 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 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 Lottery draw was made weekly and each possible combination was extracted onlyonce, 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 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’ chart, we will see that, according to the BIG NUMBERS’ THEORY, the chart 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 confusing, not to mention FALSE; this is what all Lottery websites do. If we look at the charts of the last 5, 10, 25 draws, the signals for the numbers occurrences are evident. From the draw made today, we only have maximum one or two and very seldom three numbers out of the numbers drawn LAST TIME.

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

The chapter on frequency will help select the numbers played. Avoid the mistake of only picking the numbers played in the 1-31 range (birthdays, event dates etc.). There are still 18 numbers from 32 up to 49 to be played.

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 chart 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 chart 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 chart:

Why play combinations with low probability to be drawn?

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 has 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 chart displays on the abscissa (horizontal) the groups of possible endings

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

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

If, in accordance with the statistics presented, you deliberately eliminate the maximum probability area, then you practically reject 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, which only has 9 numbers. We will still call it a decade.

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

The chart displays the combinations with the number of decades 1, 2, 3, 4, 5 on the abscissa, and a column shows how many times the respective combination was extracted, next to each combination (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 we will still call it a pentad. The numbers extracted are distributed into 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 chart 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 into 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 chart displays the combinations with the number of triads 2, 3, 4, 5, or 6 on the abscissa and there is a column showing how many times the respective combination was extracted next to each combination (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 chart, the distribution of the extracted numbers enlists L-C 7x7 on the abscissa in the following combinations (the first figure – no. of Lines, the second figure – no. of Columns) in the matrix.

From the large number of possible combinations, look at the chart 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.

REDUCTION VALUES OF LINES-COLUMNS

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 for each Line-Column combination.

DISTANCE

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

What is DISTANCE?

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

D6 (there is only one value for D6).

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

For each distance chart for 6, 5 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 middle solution would be to play so that we 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:

The 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).

The 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 chart 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 chart 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

Assuming that you one can consider "n" elements from a row of numbers 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 ”n step”.

E.g. step n=2

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

E.g. step n=3

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

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.