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Sequencial Groups Predicting

Topic closed. 7 replies. Last post 11 years ago by paint1.

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JKING's avatar - Kaleidoscope 3.gif

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Posted: December 8, 2005, 5:03 pm - IP Logged

For 5/39 take a game spread of 23. Then get occurrences, by lotto number, in the segments 1-22, 23-46, 47-68, 69-91.



At that point I calculated the overall average occurrence by lotto number using the four segments.



To predict, I took the occurrences for each number over the last 20 games and found the highest differences from the average.



The first test got 3 out 5....we'll see how it goes.







note: there were a couple additional filters, but this was the core of the program.

Any comments?

You are a slave to the choices you have made.  jk

Even a blind squirrel will occasioanlly find an acorn.

    RJOh's avatar - chipmunk
    mid-Ohio
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    Posted: December 8, 2005, 5:49 pm - IP Logged

    How do those figures compare to an actual 5/39 game like Ohio Rolling Cash5?

     * you don't need to buy more tickets, just buy a winning ticket * 
       
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      JKING's avatar - Kaleidoscope 3.gif

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      Posted: December 8, 2005, 6:14 pm - IP Logged

      Rjoh,

          I don't know, I have barely enough time for CA. Theoretically, it should work for all 5/39 games.

      You are a slave to the choices you have made.  jk

      Even a blind squirrel will occasioanlly find an acorn.

        RJOh's avatar - chipmunk
        mid-Ohio
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        Posted: December 8, 2005, 7:36 pm - IP Logged

        In that case, how well has it worked with California Fantasy5?  I didn't understand what you meant by a game spread of 23 for occurrences of numbers in segments 1-22,23-46,47-68,69-91 for a 5/39 game.

         * you don't need to buy more tickets, just buy a winning ticket * 
           
                     Evil Looking       

          JKING's avatar - Kaleidoscope 3.gif

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          Posted: December 8, 2005, 8:43 pm - IP Logged

          I've just started using it and had success. I had an earlier post where I said that I would detail the strategy more completely. I hope the following will anser you questions:

          How many times did each number occur the last 23 games(1-22), how many times for the 23 games before that(23-45), how many times for the 23 games before that(46-68), and finally how many times did each number occur for the 23 games before that (69-91). So you are measuring occurrances of each lottery number 4 times in equal game ranges. Once you have the 4 occurrance values (over equal pick spans) you take the average of the 4 numbers. Now that you know how often each number is occuring over that range, on the average it time to apply the information. We need something to compare our new found average against. So, I found out the amount of occurrances for each lotto number from the last 20 games. Being that 23 is more than 20 the amount of occurrances that happen over 20 games should try to approach the amount of occurrance of 23 games. So the average occurrance values of 23 games minus the occurrances values of 20 games gives an indication of which numbers are on track to appraoching your caclulated 4 game average. At this point you simple select the numbers with the greatest positive difference.

          Sorry if I've gotten a little wordy. It really is fairly simple if your programming. A bit more complicated if your using excel (use the countif function).

          My latest efforts are to try to best determine when a number might occur. The above method accommodates cold as well as hot numbers simultaneously. If you have the means, try it.

          As always, thanks for your interest Rjoh.

          You are a slave to the choices you have made.  jk

          Even a blind squirrel will occasioanlly find an acorn.

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            Posted: December 8, 2005, 8:50 pm - IP Logged

            Hi Jking, for a 6/49 game, would you still go with 23 games as a game spread?

              JKING's avatar - Kaleidoscope 3.gif

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              Posted: December 9, 2005, 12:19 am - IP Logged

              dynodyno,

              I'm not familiar with 6/49. If it's a 5 ball +1 bonus ball then:

              49/5=9.8

              9.8*3=29.4

              I would probably go for a 29 game spread. With an average hit of 3 per number, practically every number should have a hit value of at least 1+.

              dynodyno, this system is new and in progress. I'm trying to make a break from the norm here and change the mindset. It will become more refined. I feel the core concept isn't very far off the mark. 

               

              You are a slave to the choices you have made.  jk

              Even a blind squirrel will occasioanlly find an acorn.

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                Manitoba
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                Posted: December 10, 2005, 4:32 pm - IP Logged

                I used to do something very similar to this for the 6/49 game. It often gave 3 or 4 numbers never more.After we finish Lantern's filtering lessons who knows. Might have to come back to this.