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P-3 elimination method

Topic closed. 19 replies. Last post 7 years ago by Broulir32.

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Posted: August 10, 2009, 9:01 pm - IP Logged

015 026 037 048 059 060 071 082 093 001           

126 137 148 159 160 171 182 193 104 002         

237 248 259 260 271 282 293 204 215 003

348 359 360 371 382 393 304 315 326 004

459 460 471 482 493 404 415 426 437 005

560 571 582 593 504 515 526 537 548 006

671 682 693 604 615 626 637 648 659 007

782 793 704 715 726 737 748 759 760 008

893 804 815 826 837 848 859 860 871 009

904 915 926 937 948 959 960 971 982 010

 

 

Arizona
 

California5-3-2 
 
Connecticut0-7-0 
Delaware6-2-31-2-1
Florida6-5-91-3-5
Georgia8-4-83-5-3
Idaho
 
Illinois6-0-8 
Indiana4-9-6 
Iowa6-0-8 
Kansas
 
Kentucky3-6-3 
Louisiana
 
Maryland0-6-36-3-9
Massachusetts

Michigan7-1-84-0-9
Minnesota
3-2-5
Missouri2-0-52-1-1
Nebraska
 
New Jersey4-7-73-5-6
New Mexico
 
New York0-2-37-0-4
North Carolina8-3-4 
Ohio0-1-64-4-3
Oklahoma
 
Ontario
 
Oregon (1pm & 7pm)

Oregon (4pm & 10pm)

Pennsylvania6-8-58-8-1
Puerto Rico
 
Quebec
 
Rhode Island

South Carolina8-4-54-4-0
Tennessee9-5-72-5-2
Texas5-7-1 
Tri-State6-2-41-5-4
Virginia7-2-7 
Washington
 
Washington, D.C.0-0-67-8-3
West Virginia
4-8-9
Western Canada
 
Wisconsin
5-3-4

  As you can see, every number in tonight's winning numbers falls within one of those 100 sets above two out of three straight.

What can we do with that info?.....Maybe cross out all the ones which came out. Not all states have shown yet, just using this for example. We'd wait for all states result first.

Maybe somethng else?...C'mon folks...put your thinking caps on...lol

I know if we work on this we can narrow the picks.


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    Posted: August 10, 2009, 9:22 pm - IP Logged

    Be creative. Maybe once we eliminate half of those 100 sets, we'd focus only on sets that have numbers from the last winning number, since most of the time there is a repeating number from before.

    Be creative. Remember the movie "Contact"? They needed to think outside the box to get the 3D image of the "Primer coding" solved.

    Eek

      LAVERNE MALONEY's avatar - smallgirl

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      Posted: August 10, 2009, 11:02 pm - IP Logged

      Hello there Joker17, may I ask did you construct this # setup yourself based upon stats that you collected?

      & are you aware of the # pattern within the rows?

      I'm thinking outloud... 


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        Posted: August 10, 2009, 11:23 pm - IP Logged

        Hello there Joker17, may I ask did you construct this # setup yourself based upon stats that you collected?

        & are you aware of the # pattern within the rows?

        I'm thinking outloud... 

        Hi Laverne.

        I made these numbers up when trying to show Matchmaker that anyone can play 100 sets and get 2 out of 3 everytime.

        But today I was feeling inventive and saw those numbers differently, and maybe it was meant to be. I do see the patterns in the rows Laverne. I feel so alive sometimes, I think I can fly...lol

        I have a few other filters that can be added to this.

        Step one: Eliminate those 100 sets by half.

        Step Two: Only play sets that remain which only have numbers from the last winning draw.

        Step 3: Out of the remaining sets, eliminate all numbers that repeat in the same position from last draw.

        Step 4: ???????

          LAVERNE MALONEY's avatar - smallgirl

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          Posted: August 11, 2009, 12:08 am - IP Logged

          If you are looking at a specific state, then the elimination could be all #s that add up to the previous sum. Now that would be based upon if a state had a pattern of not getting repeat sums back to back. Because I have seen states whose sum's next hit was either up one sum or down one sum from the previous draw.

            LAVERNE MALONEY's avatar - smallgirl

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            Posted: August 11, 2009, 12:39 am - IP Logged

            Joker17, going back to your previous post: Too good to be true P-3 system - Lottery Post Forums 

            Eliminate all pick 3 #s that do not contain the digits in the common front pair, or the common back pair. Only use pick 3s that contain the digits from the common front pair & back pair in any order. For example if the common front pair was 53, then any pick 3 that did not have a 53, a 35 or a 3x5, 5x3 would be eliminated. 

            Are these the types of things for which you are looking?


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              Posted: August 11, 2009, 9:30 am - IP Logged

              Joker17, going back to your previous post: Too good to be true P-3 system - Lottery Post Forums 

              Eliminate all pick 3 #s that do not contain the digits in the common front pair, or the common back pair. Only use pick 3s that contain the digits from the common front pair & back pair in any order. For example if the common front pair was 53, then any pick 3 that did not have a 53, a 35 or a 3x5, 5x3 would be eliminated. 

              Are these the types of things for which you are looking?

              Wow....Right under my nose...lol

              Great Idea Laverne. Gonna look into this later today. Just woke up...lol...

              Need my cup of coffee........Coffee

                Broulir32's avatar - 112

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                Posted: August 11, 2009, 12:38 pm - IP Logged

                are those # alway thesame or they change every day ?? thats cool couse i was just working on a similar thing :)
                at of all 1000 combination(straights) posible: i just did only the one with 0 :and there is 271 posible numbers with zero so if i get it right if u 100% sure that (ex) zero would fold u just should play 271 tickets and win $500 right???? if only u would be able to narrow it down to let say 50 than play one digit return and spend $150 and make 350. any ideas how ??


                by the way cool idea (this thread)Big Smile


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                  Posted: August 11, 2009, 12:53 pm - IP Logged

                  015 026 037 048 059 060 071 082 093 001          

                  126 137 148 159 160 171 182 193 104 002         

                  237 248 259 260 271 282 293 204 215 003

                  348 359 360 371 382 393 304 315 326 004

                  459 460 471 482 493 404 415 426 437 005

                  560 571 532 593 504 515 526 537 548 006

                  671 682 693 604 615 626 637 648 659 007

                  782 793 704 715 726 737 748 759 760 008

                  893 804 815 826 837 848 859 860 871 009

                  904 915 926 937 948 959 960 971 982 010

                   

                   

                  The numbers in red represent the eliminated numbers. They represent all the winning numbers in all states yesterday that were 2 out of 3 in straight form, which fit into the 100 list from above .

                  So the remaining numbers that haven't shown from yesterday are:

                  048   148   160   182   237   248   260   293   371   382   426   526   548   671   759   893   859   860   904   915   926   948   960   971   982   010

                   

                  Next filter is to look at your state's last winning number from the previous eveing draw. In my state Florida it was 135. I want to eliminate any of the 26 sets above that don't have a 1,3,or 5 in them because most of the time, the next winning draw has atleast one of the previous numbers. So the remaining numbers that have a 1,3, or 5 in them are:

                  148   160   182   237   293   371   382   526   548   671   759   893   859   915   971   010

                  That leaves me with 16 sets.

                  Next is to eliminate any set that has a 1, 3 or 5 in the same exact position as 135:

                  Leaving us with: 11 sets

                  293   371   382   526   548   671   759   893   859   971   010 

                   

                  I looked back in my state's history for a clue for the next filter. The one I came up with was overdue numbers. The number 2 in the middle position has been out the longest since 7-14-09, that's 27 days.

                  Still too many numbers that could fit in the above 11 sets.

                  So I just have to make an educated guess. My pick for today's midday Florida is 327.




                   


                    Broulir32's avatar - 112

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                    Posted: August 11, 2009, 12:56 pm - IP Logged

                    015 126 137 148 159 160 171 182 193 104

                    002 093 001 071 059 037 237 259 271 293

                    348 359 360 371 382 393 304 315 326 459

                    471 493 415 437 010 693 615 637 659 759

                    737 715 793 893 815 837859 871915 937

                    959 971 560 571 582 593 504 515 526 537

                    548 215 003 005

                     

                    and those are # for today Midd in Florida
                    best of luck hope u right.


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                      Posted: August 11, 2009, 12:59 pm - IP Logged

                      are those # alway thesame or they change every day ?? thats cool couse i was just working on a similar thing :)
                      at of all 1000 combination(straights) posible: i just did only the one with 0 :and there is 271 posible numbers with zero so if i get it right if u 100% sure that (ex) zero would fold u just should play 271 tickets and win $500 right???? if only u would be able to narrow it down to let say 50 than play one digit return and spend $150 and make 350. any ideas how ??


                      by the way cool idea (this thread)Big Smile

                      Yes, those numbers are a template that stays that way and never changes. Just pick any number from 000-999, and it'll fit in there somewhere.

                      As far as your 271 possible straights with zero....well yes, you could win if you knew for sure zero was due, but like you say, it's safer to narrow them down. Unfortunately with filters, the more you have the more chance you could eliminate the winning set.


                        Broulir32's avatar - 112

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                        Posted: August 11, 2009, 1:11 pm - IP Logged
                        ok  in ur example  there is 64 #'s that include 1-3-5  (one digit return)  64 x .50 cents equal to $192 with chance of winning $250.
                        if i could some how win  $58 a day i would be happy :) but then again not always there is one digit return :(
                        lets wait and see.
                          Broulir32's avatar - 112

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                          Posted: August 11, 2009, 1:46 pm - IP Logged

                           i for got 99 should come up soon in midday

                            Broulir32's avatar - 112

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                            Posted: August 11, 2009, 1:52 pm - IP Logged

                            08.11.09  Midday FL

                            357     2 digit return   :)


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                              Posted: August 11, 2009, 2:24 pm - IP Logged

                              08.11.09  Midday FL

                              357     2 digit return   :)

                              I had 327....so close....lol