Hello, rl randonic, you could see also see this filter, an example of a game 50/5 Filter Ace Terminations of Sums ranging from (0-45).
Ex: a termination: 1-11-21-31-41 sum = 5 terminator 2: 2-12-22-32-42 = sum 10 terminating 3: sum 3-13-23-33-43 = 15 terminating 4: = sum 4-14-24-34-44 20 termination 5: 5-15-25-35-45 sum = 25 6 Termination: 6-16-26-36-46 = sum 30 terminating 7: = sum 7-17-27-37-47 35 termination 8: 8-18-28-38-48 sum = 40 termination 9: = sum 9-19-29-39-49 45 terminating 0: 10-20-30-40-50 = sum Zero (0)
We can then conclude that the sum of the endings of any key draw, will be included between 0 and 45.
For testing I have analyzed, there are many key outputs with the same sum, and that most are between the sums 15 and 30.
Taking this small added value, we can reduce some bets and be more drawn into the possible key, not implying that this week we can also fail, of course, but walk above the minimum and maximum output.
Ex key will last win of the year 2011:
Key drawn :16-36-43-44-50
Sum of endings: 6 +6 +3 +4 +0 = 19 And here's the data:
Probabilities of each output Sum of terminations S.T. Q.Ch. % S.T. Q.Ch. % S.T. Q.Ch. % --------------------------- | ---------------------- ------- | --------------------------- 0.00005 1 0 | 67 851 3.20239 15 | 2.63008 31 55 725 1 25 0.00118 | 3.79939 16 80 500 | 44 400 32 2.09557 2125 0.00590 | 4.38936 17 93 000 | 34 375 33 1.62241 3375 0.01770 | 4.95337 18 104 950 | 25 625 34 1.20943 4900 0.04248 | 5.44422 19 115 350 | 18 501 35 0.87320 5 1776 0.08382 | 5.84073 20 123 751 | 12 775 36 0.60295 3250 0.15339 6 | 6.11679 21 129 600 | 37 8525 0.40236 7 5400 0.25487 | 6.25838 22 132 600 | 38 5400 0.25487 8 0.40236 8525 | 132600 6.25838 23 | 0.15339 39 3250 9 12775 0.60295 | 6.11679 24 129 600 | 40 1776 0.08382 18 501 0.87320 10 | 5.84073 25 123 751 | 41 900 0.04248 25 625 1.20943 11 | 5.44422 26 115 350 | 42 375 0.01770 34 375 1.62241 12 | 4.95337 27 104 950 | 43 125 0.00590 2.09557 13 44 400 | 93 000 4.38936 28 | 0.00118 44 25 2.63008 14 55 725 | 80 500 3.79939 29 | 0.00005 45 1 | 30 3.20239 67 851 |
In which: - S.T. = Sum of terminations; - Q.Ch. Number of keys = Sum of each of the terminations in 2,118,760 possible keys; -% = Percentage of the number of keys each Sum of terminations in 2,118,760 possible keys.
In fact, as you say, the sums of the terminations between 15 and 30, are most likely to leave, totaling 80.01% of probability.
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