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
No. of endings |
Distribution |
Example |
2 |
3-3 |
13, 23, 43, 49, 9, 29 |
2 |
4-2 |
13, 23, 43, 49, 9, 23 |
2 |
5-1 |
13, 23, 43, 49, 3, 23 |
3 |
2-2-2 |
13, 26, 36, 39, 3, 19 |
3 |
3-2-1 |
13, 23, 43, 22, 42, 8 |
3 |
4-1-1 |
13, 23, 43, 33, 22, 8 |
4 |
2-2-1-1 |
13, 23, 44, 4, 6, 2 |
4 |
3-1-1-1 |
13, 23, 43, 17, 25, 19 |
5 |
2-1-1-1-1 |
13, 23, 22, 48, 5, 17 |
6 |
1-1-1-1-1-1 |
13, 24, 22, 48, 5, 17 |
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