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template 32 combos back pairs from width 1 using 945 combinations 5 group

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Heroic City
Indonesia
Member #31689
February 2, 2006
1153 Posts
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Posted: August 26, 2009, 1:13 pm - IP Logged

Greeting Lottery Post Members

only an idea dealing with pairs or we may called template 32 combos back pairs, as we know in pick 3 game we have front pair (12x), split pair(1x3), back pair (x23) and some lotteries don't have @ week (7 draws) but we can use this to track pairs group that hit at least one time per 7 draws

well let get to the idea we use "vtracts" or i prefer 5 groups technique to track the pairs. in 5 groups we clustered number 0123456789 into five for example:

1 = 05
2 = 16
3 = 27
4 = 38
5 = 49

and we have pairs from five groups

different/width 0 = 11  22  33  44  55
different/width 1 = 12  23  34  45  21  32  43  54
different/width 2 = 13  24  35  31  42  53
different/width 3 = 14  25  41  52
different/width 4 = 15  51

different/width mean different between above pairs eg:
0 = |(1-1)=0|
1 = |(1-2)=1|
2 = |(1-3)=2|
3 = |(1-4)=3|
4 = |(1-5)=4|

note: its an absolute value or disregards the minus sign

if we re.convert into numbers 0123456789 we have

pr grp.......pr nbs
-------------------------
11      =   00 05 50 55
22      =   11 16 61 66
33      =   22 27 72 77
44      =   33 38 83 88
55      =   44 49 94 99
-------------------------
12      =   01 06 51 56
23      =   12 17 62 67
34      =   23 28 73 78
45      =   34 39 84 89
21      =   10 15 60 65
32      =   21 26 71 76
43      =   32 37 82 87
54      =   43 48 93 98
-------------------------
13      =   02 07 52 57
24      =   13 18 63 68
35      =   24 29 74 79
31      =   20 25 70 75
42      =   31 36 81 86
53      =   42 47 92 97
-------------------------
14      =   03 08 53 58
25      =   14 19 64 69
41      =   30 35 80 85
52      =   41 46 91 96
-------------------------
15      =   04 09 54 59
51      =   40 45 90 95
-------------------------

so

different/width 0 = 5 x 4 = 20 pairs number
different/width 1 = 8 x 4 = 32 pairs number
different/width 2 = 6 x 4 = 24 pairs number
different/width 3 = 4 x 4 = 16 pairs number
different/width 4 = 2 x 4 =  8 pairs number

and now the hardest challenge we track 32 pairs as groups that hit per 7 draws cycle then find the maximum percentage using 945 possible combinations for 5 groups

here are sample back pairs for Fl Eve using 3500 draws (June 12, 1999 to Jan 09, 2009)

1 => 95.4% = 01 25 34 67 89
2 => 95.7% = 02 15 38 46 79
3 => 95.2% = 03 14 25 69 78
4 => 94.6% = 04 12 35 67 89
5 => 94.7% = 05 12 39 48 67
6 => 94.9% = 06 12 34 57 89
7 => 94.5% = 07 12 39 48 56
8 => 94.8% = 08 15 23 49 67
9 => 94.6% = 09 12 37 45 68

from above stats we may select 2 => 95.7% = 02 15 38 46 79 or the back pair numbers are:

_01   _05
_10   _12   _13   _18
_21   _25
_31   _34   _35   _36
_43   _47   _48   _49
_50   _52   _53   _58
_63   _67   _68   _69
_74   _76
_81   _84   _85   _86
_94   _96

although it have negative or minus pay out (front n back pairs $25) but we know in per 7 draws/week the 32 pair numbers have 95.7% occurrence or to match with front or back pairs pay out we can create =COMBIN(100,25) or  242,519,269,720,337,000,000,000.00 combinations btw is it right calculation why there are xxx,xxx,xxx,xxx,xxx,000,000,000.00 nine zero???

well just a though and thanks for any info

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