P-3 elimination method

Reply #19

Broulir32 — Fri, 14 Aug 2009 18:00:12 GMT

hmmm sorry to say this but ur list is not 100% correct becouse they are number combos which this list does not represent , like 381 for example there is quite few missing .

Reply #18

ZEN — Tue, 11 Aug 2009 19:13:07 GMT

Wow,

Just saw this. Great post!

153 yesterday Illinois Evening!!

Thanks Laverne and Joker!!

-ZEN

Reply #17

LAVERNE MALONEY — Tue, 11 Aug 2009 19:07:48 GMT

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

Reply #16

Broulir32 — Tue, 11 Aug 2009 18:48:15 GMT

next time guys next time

Reply #15

shortlady — Tue, 11 Aug 2009 18:42:53 GMT

I know how you feel I had 359.

Reply #14

joker17 — Tue, 11 Aug 2009 18:24:32 GMT

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

Reply #13

Broulir32 — Tue, 11 Aug 2009 17:52:24 GMT

08.11.09 Midday FL

357 2 digit return

Reply #12

Broulir32 — Tue, 11 Aug 2009 17:46:41 GMT

i for got 99 should come up soon in midday

Reply #11

Broulir32 — Tue, 11 Aug 2009 17:11:44 GMT

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.

Reply #10

joker17 — Tue, 11 Aug 2009 16:59:22 GMT

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.

Reply #9

Broulir32 — Tue, 11 Aug 2009 16:56:07 GMT

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.

Reply #8

joker17 — Tue, 11 Aug 2009 16:53:52 GMT

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 repre... [ More ]

Reply #7

Broulir32 — Tue, 11 Aug 2009 16:38:47 GMT

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 wa... [ More ]

Reply #6

joker17 — Tue, 11 Aug 2009 13:30:51 GMT

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

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

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

Reply #5

LAVERNE MALONEY — Tue, 11 Aug 2009 04:39:51 GMT

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

Reply #4

LAVERNE MALONEY — Tue, 11 Aug 2009 04:08:52 GMT

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.

Reply #3

joker17 — Tue, 11 Aug 2009 03:23:29 GMT

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 t... [ More ]

Reply #2

LAVERNE MALONEY — Tue, 11 Aug 2009 03:02:58 GMT

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...

Reply #1

joker17 — Tue, 11 Aug 2009 01:22:18 GMT

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.

P-3 elimination method

joker17 — Tue, 11 Aug 2009 01:01:39 GMT