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pick3 even-odd numbers

Topic closed. 3 replies. Last post 5 years ago by dr san.

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New Member
NewYork
United States
Member #68392
December 20, 2008
21 Posts
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Posted: November 19, 2011, 12:27 am - IP Logged

Hi guys,

I need some help in developing this method. out of 1000 numbers

125 - all odd( ex: 135 113 357 975 997 .....)

125 - all even (ex: 246 268 880 ....)

375 - two even and one odd (ex: 467 986 883 029 ....)

375 - two odd and one even (ex: 974 556 396 ....)

My concentration is on only 375 numbers which are of odd even (both) . More than 85% of the time either    two even one odd  or  two odd one even  comes.

I need help in picking only one 375 numbers  on exact day which comes depending on previous draw or day or date or anything..

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    Member #0
    January 1, 2000
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    Posted: November 19, 2011, 8:29 am - IP Logged

    Check history and see how many draws the longest before they switch. If you see long draws between switches, then go back in history again and see how often it happens. If it happens on regular basis, wait a few draws then switch. but still very risky.

    ex:

    In 3 month history, 12 times a long wait happens before a switch. let's say 4 or 5 draws before switch. this means it happens on regular basis, which is about once a week. still too risky unless you plan to narrow down the 375 and play online.

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      bgonçalves
      Brasil
      Member #92564
      June 9, 2010
      2126 Posts
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      Posted: November 19, 2011, 2:22 pm - IP Logged
      Hello, ahtnana, you can also see it =
      Low = 1,2,3,4,5
      High = 6,7,8,9,0
      If the third digit of 1,2,3,4,5 pick3 are all low, we have 10 conbinçoes
      Also 10 pairs combinations
        Or pick three high 6,7,8,9,0
        We have 10 conbinaçoes.
      10 high pairs conbinaçoes
      Ahtnana note that the groups have two low pairs = 2.4
      And in the high group 3 pairs = 6,8,9
      You can also pair formation with low single digits and another high example = 28
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        bgonçalves
        Brasil
        Member #92564
        June 9, 2010
        2126 Posts
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        Posted: November 19, 2011, 2:41 pm - IP Logged
        Hello, this use has 25 pairs, and 90% of the results,
        Course within the three positions, the third digit is random
        These 25 pairs are formed with a group down and typed a digit in the high group ok
        Here = 25 pairs =
        1 6
        1 7
        1 8
        1 9
        1 0
        2 6
        2 7
        2 8
        2 9
        2 0
        3 6
        3 7
        3 8
        3 9
        3 0
        4 6
        4 7
        4 8
        4 9
        4 0
        5 6
        5 7
        5 8
        5 9
        5 0