How to win the lottery. Grid

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Íà ðóññêîì: http://www.proza.ru/2010/10/31/1175

The principle of this method lies in covering of all the options in a way that will bring a second prize in every draw. This isn’t as simple as it sounds.  For example, using the classic formula:
N! / (K! * (N-K)!)  = 37! / (5! * (37-5)!) =  (37 * 36 * 35 * 34 * 33) / (5 * 4 * 3 * 2 * 1) = 37 * 17 * 11 * 9 * 7 = 435,897
In a draw such as the 6 out of 37, getting 5 out of 6 guesses takes 435,897 guesses (for those who wish to know more, there’s a more detailed description in Wikipedia). Still, many options of 5 repeat themselves, if we see them as a whole system. Personally, I see (although I admit a certain lack of clarity, and generalization, considering the amount of options) changes in the volume, in which each layer is a grid for all the fives, with a range of only 74,993 options (in which there is a promise exactly 5 fives or 1 six-guess). The abovementioned system isn’t a cube shaped one, but closer to dim multy-side, which requires a computerized processing to get certain numbers. And at a first look, it’s hard to believe to how exactly the odds to a second number were multiplied by seven,  and still not a word about it in Wikipedia. However there’s no miracle here: we can build a complex mathematic system (which I’ve built for the SmartClub lottery company) to prove, or just to run head first by the brute force option.
According to this equation, we can grant a win of 4 out of 6 (in a 6 out of 37 draw) instead of through 44,045 tries, by 5,452 attempts alone. As an alternative, the minimal prize (in a 6 out of 37 draw) can be gained instead of by 7,770 tries, through 473 alone, and this isn’t even the single win for each option.
I understand that it’s hard to see a system as a whole, so I’ll offer a simpler example, which each one can test, and through which a training of imager perception in thinking is possible according to the systemaism teaching.
We’ll take a lottery, in which there  are 12 balls, 6 of which are drawn, and you have to guess only 2.
The minimum according to the formula is:
N! / (K! * (N-K)!)  = 12! / (2! * (12-2)!)  = 12! / (2! * 10!) = 12 * 11 / 2 = 66
And still, if we draw the options as a grid with several layers (this can be done in the human mind without the aid of the computer), we’ll get that the minimal layer equals to 6. As an example to such layer, I offer to the inquisitive ones to check, by choosing 6 out of 12 (and each grid will always hold at least 2).
1  2  3  4  5  6 
1  5  6  7  8 12 
1  7  8  9 10 11 
2  3  4  7  8 12 
2  3  4  9 10 11 
5  6  9 10 11 12
I hope that this example is simple enough to understand and supports the promotion of the systemaism issues, by giving a wider range for thinking.


The author is the head of Programming and Statistics department of a lottery company in Israel.

Method "Ranges": http://www.proza.ru/2010/11/14/569
Method "AI": http://www.proza.ru/2010/12/27/1080