The Pick 3 Benford MatrixPrev TopicNext Topic

• WSN1

  

  Anywhere & Everywhere
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  Feb 7, 2007, 12:46 am

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  In response to Blackapple

  Quote: Originally posted by Blackapple on Feb 7, 2007

  Over the same period sums 11-15 hit 354x and sums 16-19 hit 241x

  Jennilynn418.

  Using sums is a very effective way to win in the pick3. The most common group of sums are the 12-15 group. To arrive at a sum just add all three digits together. If you take away doubles you have  only 10 numbers to work in each of these 4 groups. In the month of Jan. for NY there were 16 hits from the 12-15 group. Not bad considering there only 40 possible non-match numbers from these sums. Go back the past couple weeks in your state and look for the hottest sum and most due sum from the 12-15 group, and play the numbers in it. For example in NY the sum 13 is the most due to show, while the 15 has been the hottest this year. Each time a number hits in the 12-15 group mark it down, this way you can see whats hot, and overdue. Here is a list of the 12-15 sums, minus the doubles.

  12-039-048-057-156-147-237-246-345-129-138

  13-049-058-067-139-148-157-238-247-256-346

  14-059-068-149-158-167-239-248-257-347-356

  15-069-078-159-168-249-258-267-348-357-456

   

  *******************

   

  	12	15	6.8%	039 048 057 066 129 138 147 156 228 237 246 255 336 345 444
  	13	15	6.8%	049 058 067 139 148 157 166 229 238 247 256 337 346 355 445
  	14	15	6.8%	059 068 077 149 158 167 239 248 257 266 338 347 356 446 455
  	15	15	6.8%	069 078 159 168 177 249 258 267 339 348 357 366 447 456 555

   

  From

  Input:
  

  000 001 002 003 004 005 006 007 008 009 011 012 013 014 015 016 017 018 019 022 023 024 025 026 027 028 029 033 034 035 036 037 038 039 044 045 046 047 048 049 055 056 057 058 059 066 067 068 069 077 078 079 088 089 099 111 112 113 114 115 116 117 118 119 122 123 124 125 126 127 128 129 133 134 135 136 137 138 139 144 145 146 147 148 149 155 156 157 158 159 166 167 168 169 177 178 179 188 189 199 222 223 224 225 226 227 228 229 233 234 235 236 237 238 239 244 245 246 247 248 249 255 256 257 258 259 266 267 268 269 277 278 279 288 289 299 333 334 335 336 337 338 339 344 345 346 347 348 349 355 356 357 358 359 366 367 368 369 377 378 379 388 389 399 444 445 446 447 448 449 455 456 457 458 459 466 467 468 469 477 478 479 488 489 499 555 556 557 558 559 566 567 568 569 577 578 579 588 589 599 666 667 668 669 677 678 679 688 689 699 777 778 779 788 789 799 888 889 899 999

  Yeah, very good stuff Blackapple...I came to a very similar conclusion, which is why I started a new thread to explore this idea in more detail with a separate thread and keep this topic geared more towards the skip research, Maybe you can add some more insight to the following thread:

  https://www.lotterypost.com/thread/150343/766102?q=WSN1, thanks for your input.....

  I would like to look at applying the sum group with overdue max skip and benford application and see how to best maximize the two concepts together....
• Thoth

  

  Findlay, Ohio
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  Feb 7, 2007, 10:07 pm

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  I thought about tracking the performance of the skips of combos according to root sums and seeing how they hold up to Benfords Law (as I'm sure they will).  This skips for straight numbers that outline Benfords Law can be broken down into 9 groups of 1,111 possibilities each (at least 99.9999% of the time or more anyways). 

  When you break down the Pick 3 combos into root sums you will have 10 groups: 9 of them (sums 1 through 9) will consist of 111 combos each and the 10th (sum 0) will only have one.  The 10th group can be be ignored as it only represents 1 combo out of 1000.  If the law applies, every sum should have around 33+% of its hits come at either one game later (a repeat) or 10 through 19 games later.  I havent taken the time to test this yet but I'd bet money thats how it will be.

  Using the standard sums...0 through 27... makes things more difficult because the different sum groups contain different amounts of numbers which makes certain groups much more probable and frequent than others.  You would have to have different scales to track the different sums on.  

  ~Probability=Odds in Motion~
• WSN1

  

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  Feb 7, 2007, 10:44 pm

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  In response to Thoth

  Quote: Originally posted by Thoth on Feb 7, 2007

  I thought about tracking the performance of the skips of combos according to root sums and seeing how they hold up to Benfords Law (as I'm sure they will).  This skips for straight numbers that outline Benfords Law can be broken down into 9 groups of 1,111 possibilities each (at least 99.9999% of the time or more anyways). 

  When you break down the Pick 3 combos into root sums you will have 10 groups: 9 of them (sums 1 through 9) will consist of 111 combos each and the 10th (sum 0) will only have one.  The 10th group can be be ignored as it only represents 1 combo out of 1000.  If the law applies, every sum should have around 33+% of its hits come at either one game later (a repeat) or 10 through 19 games later.  I havent taken the time to test this yet but I'd bet money thats how it will be.

  Using the standard sums...0 through 27... makes things more difficult because the different sum groups contain different amounts of numbers which makes certain groups much more probable and frequent than others.  You would have to have different scales to track the different sums on.  

  Thoth, very interesting observation. I agree that the sums would be a lil more difficult to track using the current model, due to the variances within the individual sums. Let's took at testing your theory on the root sums, i think it has significance
• WSN1

  

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  Feb 9, 2007, 1:34 am

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  Quote: Originally posted by Thoth on Feb 5, 2007

  I like waiting for one of the "Benford numbers" to hit...then gamble on the 55% chance that another one will hit within the next two games.  Of course other factors have to be included to narrow the picks in that group down

  Thoth, how about we up-the-anty for straight hits with fewer combos by using the Benford skips for each 0-9 digits by position, to get a smaller core groups of most likely digits by position, and use the benford skips with root sums 0-9. You would have only a few numbers to play, with greater success rate
• WSN1

  

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  Feb 9, 2007, 3:08 am

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  I just completed a brief but interesting test I went back to TN eve draw 9/20/06 = 480,
  and located the digits by each position that matched the Benford Skip Model created by Thoth,
  the following explains my breakthrough:

  Positional Benford Digits to wheel:
  POS1= 0,4,5,9   POS2= 7,8,9   POS3= 0,2,5,9   {4x3x4}= 48 st. combos
  If Benford LDR filter is applied:
  Wheel LDR: 2,5,9 final st. combos = 13#s to play.
  9/21/06 draw results=966, we got the POS1 digit correct, however we missed pos2 and pos3 and the LDR

  Day 2, use the same wheel to get all 3 digits by position correct or eliminate #9 from the POS1
  POS1= 0,4,5,9   POS2= 7,8,9   POS3= 0,2,5,9   {4x3x4}= 48#s st. combos/36#s
  Wheel Benford LDR after last draw: 1,5,8,9: final st. combos = 18#s/13#s to play
  9/22/06 draw result=099

  Do you see the winning straight in final selections from these filters and 18#s / 13#s
  or 48#s/36#s with one filter

  072, 092, 470, 489, 492, 495, 579, 582, 972, 975, 982
  080, 090, 099, 585, 595, 979, 990 (18)

  or with less insurance
  072, 092, 470, 489, 492, 495, 579, 582
  080, 090, 099, 585, 595 (13)
• Thoth

  

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  Feb 10, 2007, 11:10 am

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  It's interesting to see someone applying it as a filter to the digits.  I havent really had time to do much testing lately, I'll try to post some more info this coming week. 

  Normally, thier should be three or four digits in each position that meet the criteria.  This would gove you anywhere from 27 to 48 straights to play most the time.   The odds of a "BL digit" hitting in all three positions at the same time to produce a win is going to be somewhere between 27/1000 (1 in 37 reduced odds) to 48/1000 (1 in 21 reduced odds) most of the time.

  Of course if you look at the benford group of straights (there is usually around 320 to 340 at any given time) and you can correctly guess the root sum, that should also cut the pick down to around 37... (330÷9) straights.  Almost every way of looking at it is going to give you between 27 and 48 numbers to play...I havent found a way to reduce the picks below 27 yet.

  ~Probability=Odds in Motion~
• WSN1

  

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  Feb 10, 2007, 2:18 pm

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  In response to Thoth

  Quote: Originally posted by Thoth on Feb 10, 2007

  It's interesting to see someone applying it as a filter to the digits.  I havent really had time to do much testing lately, I'll try to post some more info this coming week. 

  Normally, thier should be three or four digits in each position that meet the criteria.  This would gove you anywhere from 27 to 48 straights to play most the time.   The odds of a "BL digit" hitting in all three positions at the same time to produce a win is going to be somewhere between 27/1000 (1 in 37 reduced odds) to 48/1000 (1 in 21 reduced odds) most of the time.

  Of course if you look at the benford group of straights (there is usually around 320 to 340 at any given time) and you can correctly guess the root sum, that should also cut the pick down to around 37... (330÷9) straights.  Almost every way of looking at it is going to give you between 27 and 48 numbers to play...I havent found a way to reduce the picks below 27 yet.

  Wow, you are correct. I did a quick test to validate. I would be interested to find out what additional information you come up with after additional testing. Based upon the information given to this point, Im thinking a clever way execute the BL#s, after a hit, with about 2-4 Rootsums that fall within the BL skip parameter, this will give you about 75 - 150 #s to play for a straight hit. Im not sure what the reduced odds factor is to hit within 1-2 draws, but I think it pretty significant. Because, at stated before, BL after one hit = 55% chance within 2 draws, then 2-4 in 21 or 37 odds with root sums. If we can isolate the max skip with this concept, play only during the highest probability setup,  and apply the progression where needed, I think this will be a formadible strategy.

  Also, one can apply the BL#s digits positionally or any position for a straight. Cool stuff.

  Thoth, this is giving me another inspiration as I found with the 12-15 sums, I see a hugh odds advantage factor when all odds and evens hit. I will send you a PM, maybe you help me identify the odds factor here.
• WSN1

  

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  Feb 10, 2007, 6:20 pm

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  I also noticed that 25% of the game is ALL HIGH & LOW and of that about 60% of the next draw comes back as a non ALL and 40% comes back as ALL in either direction. Also, about 92% of these ALL HIGH & LOWS will produce non all within 2 draws. That is 750 st. combinations, now if apply BL in of one of two ways we get the following options:

  Play the BL#s which are a mix after ALL high/low from the 1000 poll

  or

  Wait til a BL ALL high or low combo hit, then play the BL# mix group to hit

  I dont how many #s that is but I would assume about 75% of your 330 BL#s =250#s to play for a st. hit. I think that field of numbers is still a little high, however, it statiscally it gives you a great odds advantage. Thoth maybe you or someone can add to this odd stats and or other solid filters, that work.