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Modified; If you can pick one digit

Topic closed. 5 replies. Last post 7 years ago by LANTERN.

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United States
Member #63692
August 5, 2008
83 Posts
Posted: July 2, 2009, 10:27 am - IP Logged

Hello friends. I don't know if this has been discussed before but I have been doing some work with this new technique and on paper have hit five out of seven times. This is for pick 3. I call it the one will one won't method. Basically you pick one digit to hit, and one digit not to hit. That knocks your combo's down to 28. You know one digit from the night before WON'T hit otherwise you would have an exact repeat. That narrows your choices of no hit numbers down to 3. Very good odds, especially after doubles or a digit has had a long string of hits. Now for some that may still seem like too many numbers but I feel anything under 40 numbers is okay. The thought being if your a straight and box player, and can keep hitting box till you nail a straight then it's like playing for free. And minimizing losses, even if it means spending a little more up front may be worth it. Naturally, getting the straight is key but here also, if you can pick where one digit will fall, it only leaves two ways to come in. As usual, doubles will spoil the party. Any thoughts?

    ZEN's avatar - DiscoBallGlowing
    United States
    Member #40436
    June 1, 2006
    180 Posts
    Posted: July 2, 2009, 4:57 pm - IP Logged


    Please gave a working example?




      Member #20982
      August 29, 2005
      4715 Posts
      Posted: July 2, 2009, 5:04 pm - IP Logged

      you gave me an idea but i don't know if this is what you are talking about...

      Let me see...You say that 1 number repeats from the previous numbers, the the other 2 or 3 depending whether is pick3 or pick4 if is pick3 then 2 numbers will not repeat so that takes away 2 numbers leaving you with 8...I really don't understand...Which is a good idea...

      what exactly do you mean?

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        United States
        Member #63692
        August 5, 2008
        83 Posts
        Posted: July 2, 2009, 11:00 pm - IP Logged

        Okay. Let me clarify. This is a take-off of the, if you can pick one number (Digit) system. You pick one digit. However you do it is fine.  For me, a number that is due, has a pattern established, or looks like it will come. Okay, now we have chosen one digit we think will hit. (If your any kind of pick 3 player you really should be able to pick one digit). That leaves us with 36 combo's. For me that's a little too many. So... how to narrow it down? Pick one digit that WON'T hit.  There's a 99.9 percent chance that ONE digit from the night before WON't hit. That's my way. Sure there's 7 other digits that won't hit either, and you may choose one of them if you feel strongly about it. Such as if one digit has hit 5 days in a row or something of that nature. But your gaurenteed one digit from the night before WON'T hit. Get it? Because if at least one digit from the night before does not hit,  you would have an exact match of the previous draw. And how often does that happen. Twice a year! Hope this helps and good luck.

          United States
          Member #63692
          August 5, 2008
          83 Posts
          Posted: July 2, 2009, 11:48 pm - IP Logged

           Try to remember that picking a winning combo isn't just about picking digits that will hit. Eliminating digits that WON'T hit is crucial as well. That's what this is geared towards okay. One of the three digits that came in last night won't hit tonight. Really, how often do two digits repeat? In my state, on average every 20 days or so. A three digit repeat, like I said earlier, two or three times a year? So what I'm saying is in your pool of digits that your trying to eliminate, make sure to include one of the digits that hit in the previous draw. Hope this helps.

            LANTERN's avatar - kilroy 28_173_reasonably_small.jpg
            United States
            Member #4570
            May 4, 2004
            5180 Posts
            Posted: July 3, 2009, 2:37 am - IP Logged

            Digits repeat by straight position from last to next draw 1/10th of the time average, so on 100 draws there will be 10 repeats by each straight position and about 30 repeats by all 3 straight positions.

            This is because there are 10 possible digits from 0 to 9 by position and each digit has the same 1/10 % chance of being next to come out by straight position.


            By boxed position this should multiply I guess by 3 as there are 3 positions.

            If a 0 has a 1/10 % chance to come out next on position 1, it also has the exact same chance to come out on position 2 and also the same on position 3.

            So 1/10 + 1/10 + 1/10 = 3/10 % to come out on any of the 3 positions.

            So then any digit of the 10 from 0 to 9 has or should have on the long run an average of a 3/10 chance to come out on any one of the 3 positions.

            As the digits and or the balls have no memory this might also apply to digits from last to next draw.

            LotSoft gives this info also:

            # digitsTimesCur. skipAv. skipMed. skipMax. skip
            Pos 1226510.26660
            Pos 2246119.43574
            Pos 3238149.74656

            Take a look at the average skip.

            This info is for Pa Midday Pick 3.


            Pa Midday 1 Month's data there on that picture just above.

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