HELLO, garyo and more = this is what this
no link, translated by google, is not very important for the dynamic filter concept =
Following the advice of several of the forum, I started reading the most varied post, making my notes, tests and drawing my conclusions (wrong or right). The fact is that at first glance I come to the conclusion that:
1) We can expect luck to hit us (playing games for a long time and hoping that one day it will be drawn) or
2) We can try to hit the luck (trying to "predict" which will be the dozens drawn in each contest)
Well, these phrases are not mine, I've read in some of the hundreds of posts I've read in the last few days, but I have to agree with them. And at that moment I decided to opt for the 2 option.
To try to hit the next draw I studied the historical series of results and from there I observed the following facts:
1) In more than 85% of the contests we have 4 to 7 dozens that came out in the last 3 draws (the current + the previous 2)
2) We have 3 more dozen that only came out in the last draw (totaling 7 to 10 along with those of point 1)
3) We have between 2 to 5 tens that came out only in the penultimate draw
4) We have on average 3 dozen who left neither last nor second to last. Usually here are the late dozens. And that a dozen takes on average 3-4 draws to go out. So if a dozen are more than 3 games without leaving is a good option.
With that I divided my number selection into 4 sequencing groups of 18 dozens of which:
5 to 7 Dozens come from group 1
3 to 5 Dozens of group 2
4 to 6 Dozens of group 3
3 to 4 tens of group 4 (here we always have 1 fixed, the latest)
the number of ten of each group varies according to the amount of ten in each of them
Let's take the 1090 draw to pick the numbers for the 1091:
Group 1) 3 8 13 14 15 21 22 24 (left at 1090 and 1089) of these would make groups of 5 (C (5.8) = 56)
Group 2) 5 9 10 11 17 18 20 (only left in 1090) of these would make groups of 5 (C (7.5) = 21)
Group 3) 2 4 6 12 16 19 23 (only left in 1089) This would also make groups of 5 (C (7,5) = 21)
Group 4) 1 7 25 (left neither 1090 nor 1089) would make 1 group of 3.
By exchanging the groups we have 56x21x21x1 = 24696 distinct groups of 18 dozen and there is the problem. I am usually getting between 8mil to 24mil groups of 18 numbers, which is particularly a lot for me. And at this point I ask your help to decrease the number of combinations by 18?
1) Currently I am fixing only the last dozen, can you fix other tens without losing quality?
2) I am not comfortable with excluding dozens, but do you have any technique to do this with an 85% chance of success.
3) I've read in other posts that has combinations of 18 numbers that guarantee 15 points much lower than my 24 thousand. Does anyone have for me to evaluate and / or apply these filters above.
4) Any additional filters that I can do that I'm not doing?
