SPUTNIK, I think it is better that I bring new ideas to better structure your approach and it will be useful also for those who play progressions, because the goal is to find "quickly" gaps, and then play with more security. The combinatorics between ECs must be used. To make it very simple it is necessary to take values red = 1; black = 2; odd = 3; pair = 4 ...

We must now establish a maximum of combinations. The first 1 2 3 4 5 6; 0 1 2 3 4 5 6; 0 0 1 2 3 4 5 6; 0 0 0 1 2 3 4 5 6; 0 0 0 0 1 2 3 4 5 6; 0 0 0 0 0 1 2 3 4 5 6. This is the first block. These sequences must be vertically aligned with the spins notation. Each zero indicated corresponds to the offset of a line of the combination. For example with 0 0 1 2 3 4 5 6 this combination starts its notation at spins number 3 and as it is number 1, we will tell if red has come. Since we are only shifting at the beginning, this combination will be 0 0 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 ... Another block of combinations now consists of taking only 5 EC: 1 2 3 4 5; 0 1 2 3 4 5; 0 0 1 2 3 4 5 ... Then you have to create another block with 5 other numbers, example 1 2 3 4 6.

There are so many possible blocks. But each new block always needs at least one different number or fewer numbers. This is important because the former authors who have worked on combinatorics have apparently not understood that if we make all the combinatorial possibilities with the 6 EC, there is a "circular effect". Which means that we would make combinations that would oppose each other in different ways, at different times. It's an amazing thing. The solution I found is to have at least one different number in each block. Making these combinations on Excel is better.

For the moment we must understand the principle. The combinatory + a good way to note the spins on the combinations, makes it possible to find interesting gaps very quickly. This is a first tool to try with paper and pen at home before thinking of writing formulas on Excel to get incredibly fast results ... to check his theories. Of course, we can do the same thing with D and C.