As suggested, I've going through previous posts, checking out associated links, reading articles...etc. And one comparison proprosed by several people go something like:
A) when picking numbers, skip the second game and choose numbers from the third game.
B) Pick numbers from the first 5 games.
C) And all variety of other combinations based on the last time individual number are picked,
I'm having problems with how any of these assumptions are valid. The attached summary sheet for the California Fantasy 5 shows a very evenly spread distribution. And although not shown, I'm getting equivalent result for the 6/49 game:
|
LST |
|
GAME |
|
PCK |
|
|
|
|
75G |
|
|
|
|
L1 |
L2 |
L3 |
L4 |
L5 |
TOT |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
0 |
134 |
33 |
1 |
0 |
0 |
168 |
1 |
65 |
59 |
17 |
2 |
0 |
143 |
2 |
27 |
53 |
34 |
8 |
0 |
122 |
3 |
17 |
41 |
44 |
19 |
2 |
123 |
4 |
8 |
29 |
28 |
18 |
6 |
89 |
5 |
3 |
16 |
30 |
18 |
5 |
72 |
6 |
5 |
10 |
29 |
26 |
10 |
80 |
7 |
2 |
9 |
20 |
24 |
14 |
69 |
8 |
0 |
3 |
19 |
16 |
11 |
49 |
9 |
0 |
2 |
15 |
20 |
11 |
48 |
10 |
0 |
2 |
8 |
22 |
15 |
47 |
11 |
0 |
2 |
5 |
20 |
17 |
44 |
12 |
0 |
2 |
8 |
12 |
6 |
28 |
13 |
0 |
0 |
0 |
13 |
18 |
31 |
14 |
0 |
0 |
0 |
13 |
12 |
25 |
15 |
0 |
0 |
0 |
4 |
10 |
14 |
16 |
0 |
0 |
1 |
7 |
10 |
18 |
17 |
0 |
0 |
0 |
8 |
13 |
21 |
18 |
0 |
0 |
1 |
3 |
10 |
14 |
19 |
0 |
0 |
1 |
4 |
10 |
15 |
20 |
0 |
0 |
0 |
2 |
9 |
11 |
21 |
0 |
0 |
0 |
1 |
7 |
8 |
22 |
0 |
0 |
0 |
0 |
11 |
11 |
23 |
0 |
0 |
0 |
0 |
6 |
6 |
24 |
0 |
0 |
0 |
0 |
6 |
6 |
25 |
0 |
0 |
0 |
0 |
4 |
4 |
26 |
0 |
0 |
0 |
1 |
4 |
5 |
27 |
0 |
0 |
0 |
0 |
3 |
3 |
28 |
0 |
0 |
0 |
0 |
6 |
6 |
29 |
0 |
0 |
0 |
0 |
2 |
2 |
30 |
0 |
0 |
0 |
0 |
1 |
1 |
31 |
0 |
0 |
0 |
0 |
1 |
1 |
32 |
0 |
0 |
0 |
0 |
0 |
0 |
33 |
0 |
0 |
0 |
0 |
3 |
3 |
34 |
0 |
0 |
0 |
0 |
2 |
2 |
35 |
0 |
0 |
0 |
0 |
3 |
3 |
36 |
0 |
0 |
0 |
0 |
2 |
2 |
37 |
0 |
0 |
0 |
0 |
1 |
1 |
38 |
0 |
0 |
0 |
0 |
5 |
5 |
39 |
0 |
0 |
0 |
0 |
1 |
1 |
40 |
0 |
0 |
0 |
0 |
1 |
1 |
41 |
0 |
0 |
0 |
0 |
1 |
1 |
42 |
0 |
0 |
0 |
0 |
1 |
1 |
43 |
0 |
0 |
0 |
0 |
0 |
0 |
44 |
0 |
0 |
0 |
0 |
0 |
0 |
45 |
0 |
0 |
0 |
0 |
0 |
0 |
46 |
0 |
0 |
0 |
0 |
0 |
0 |
47 |
0 |
0 |
0 |
0 |
1 |
1 |
48 |
0 |
0 |
0 |
0 |
0 |
0 |
49 |
0 |
0 |
0 |
0 |
0 |
0 |
50 |
0 |
0 |
0 |
0 |
0 |
0 |
All I seem to find is the basic hot/cold mix analysis that you would use occurences. What am I missing?
What I have looked at is:
|
LST |
|
SRT |
|
|
|
|
|
|
|
|
75G |
|
|
|
|
AMT |
CLC |
|
L1 |
L2 |
L3 |
L4 |
L5 |
|
|
PCK |
AMT |
|
|
|
|
|
|
|
TOT |
>9 |
>9 |
COL |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
2 |
4 |
4 |
7 |
10 |
|
27 |
1 |
11 |
4 |
0 |
3 |
4 |
5 |
7 |
|
19 |
0 |
8 |
5 |
0 |
2 |
4 |
11 |
47 |
|
64 |
2 |
9 |
5 |
1 |
2 |
4 |
4 |
11 |
|
22 |
1 |
9 |
4 |
4 |
7 |
8 |
11 |
14 |
|
44 |
2 |
9 |
5 |
0 |
1 |
5 |
12 |
14 |
|
32 |
2 |
9 |
5 |
1 |
2 |
6 |
7 |
20 |
|
36 |
1 |
7 |
5 |
0 |
1 |
2 |
9 |
10 |
|
22 |
1 |
8 |
5 |
0 |
1 |
3 |
6 |
10 |
|
20 |
1 |
7 |
5 |
1 |
2 |
3 |
7 |
20 |
|
33 |
1 |
5 |
5 |
1 |
2 |
3 |
6 |
15 |
|
27 |
1 |
5 |
5 |
6 |
6 |
7 |
7 |
19 |
|
45 |
1 |
6 |
3 |
7 |
9 |
10 |
12 |
37 |
|
75 |
3 |
7 |
5 |
0 |
4 |
7 |
10 |
41 |
|
62 |
2 |
8 |
5 |
0 |
0 |
1 |
2 |
13 |
|
16 |
1 |
8 |
4 |
1 |
5 |
6 |
9 |
19 |
|
40 |
1 |
8 |
5 |
Is sums(TOT)-To wide a range to establish any effective limiting.
The amount of numbers that have not been picked in ten games or better (AMT PCK). And the the associated amounts of current lotto numbers with values of 10 games or more(AMT CLC). And lastly, to track how many different values occur during any one pick.To conclude:
|
COL |
>9 |
|
CNT |
AMT |
|
|
|
|
|
|
0 |
|
59 |
1 |
0 |
92 |
2 |
3 |
86 |
3 |
21 |
18 |
4 |
98 |
6 |
5 |
139 |
0 |
TOT |
261 |
261 |
You can expect each pick to have 4 or 5 different times when they were last picked. And that the amount of picks during any one draw will only have a top value under ten games (the 59) only about %22 percent of the time.
Given this kind of data, how can anybody find a skip game approach to selecting then next set of lotto numbers?