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# 8 x 15 (Pick 3)

Topic closed. 52 replies. Last post 2 years ago by adobea78.

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Simi Valley, CA
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July 4, 2014
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 Posted: August 15, 2014, 12:07 am - IP Logged

The key word is PREDICTION, good luck with  algorithm!.  Slicing, subsetting, down-scaling is what every player does because you can't bet 1000 combos.Slicing depends much on a data, what type of data(Historical or extrapolated). Prediction is not a exact science(more intuitive than logic),so keep you line of thought simple, less analytical,  good luck.

I hear what you're saying, but don't necessarily neglect the algorithmic part of the algorithm.

For example: It seems that, generally (key word there: sometimes it's more, sometimes less) the game takes an average of two digits, and "cold plays" them.  Not equally, not all the time, but they tend to be out for long periods and/or rarely played for long periods.

So is it possible the algorithm is programmed to leave two digits (we'll go with for now) more or less cold?  One would have to conclude yes: otherwise, we'd get this too even constant sifting of the same digits, over and over and over... which a purely random distribution (say, balls dropping) wouldn't produce anyway.

So why care about the algorithm, if the same thing occurs either way?  My theory: because the problem with purely random distributions that nevertheless keep all things equal, means the program must produce approved results (i.e., equal distribution over a long period of time); and this often requires the program to go cold on various digits.

And so, what would help one's game, is to figure out (a) what digit's cold, (b) as early as possible in that cold cycle.

Me, I'd say, any digits that drops below the last-five-plays mark - let's say, a digit that has just slipped to the sixth play down (combined) - could be, perhaps? assumed to be going cold.  If so, you simply remove that digit from your plays, until it reappears again (?—better way? do one for both Mid/Evening?).

One cold digit removed from play, shrinks the 120 singles boxes by 36, or 30%; the 45 doubles by 9 boxes, or 20%; and the whole by 46 "units" (this includes a cold digit's triple), or 26%.  Accurately nail two current cold digits, and you can reduce the pool even further.

One more way to sift through those choices....

United States
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September 8, 2011
3941 Posts
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 Posted: August 15, 2014, 12:46 am - IP Logged

I hear what you're saying, but don't necessarily neglect the algorithmic part of the algorithm.

For example: It seems that, generally (key word there: sometimes it's more, sometimes less) the game takes an average of two digits, and "cold plays" them.  Not equally, not all the time, but they tend to be out for long periods and/or rarely played for long periods.

So is it possible the algorithm is programmed to leave two digits (we'll go with for now) more or less cold?  One would have to conclude yes: otherwise, we'd get this too even constant sifting of the same digits, over and over and over... which a purely random distribution (say, balls dropping) wouldn't produce anyway.

So why care about the algorithm, if the same thing occurs either way?  My theory: because the problem with purely random distributions that nevertheless keep all things equal, means the program must produce approved results (i.e., equal distribution over a long period of time); and this often requires the program to go cold on various digits.

And so, what would help one's game, is to figure out (a) what digit's cold, (b) as early as possible in that cold cycle.

Me, I'd say, any digits that drops below the last-five-plays mark - let's say, a digit that has just slipped to the sixth play down (combined) - could be, perhaps? assumed to be going cold.  If so, you simply remove that digit from your plays, until it reappears again (?—better way? do one for both Mid/Evening?).

One cold digit removed from play, shrinks the 120 singles boxes by 36, or 30%; the 45 doubles by 9 boxes, or 20%; and the whole by 46 "units" (this includes a cold digit's triple), or 26%.  Accurately nail two current cold digits, and you can reduce the pool even further.

One more way to sift through those choices....

You might take look at interpretation of  PREDICTION INTERVAL for data inference, it may narrow most of your IF conditions. The're two common way of data inference(Statistics), BAYESIAN and FREQUENTIST. The bayesian relies more or should say need prior data, whereas the frequentist says prior data is not needed for prediction. Bayesian depends more on probability of normal data  distribution (estimate, range), whereas Frequentist  depends more on behavior of chosen parameter which has a discrete value(digits, mean, median..).

United States
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October 16, 2009
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 Posted: August 15, 2014, 1:53 am - IP Logged

I like the charts.

Good Luck.

The Struggle is real!

New Mexico
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January 29, 2010
11161 Posts
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 Posted: August 15, 2014, 2:44 am - IP Logged

You are making Pick 3 too difficult.  The game is Not Rocket ScienceDon't go there.  You don't need to play a large group of numbers to win.

You sound very intelligent.  Has it ever occurred to you to wonder how it is that some people can win Pick 3 or Pick 4, by playing as few numbers as a VTAC string, or even less. Pick 3 VTRAC = 8 numbers. Pick 4 VTRAC = 16 numbers.

I would aim for "EVEN LESS".  Figure that out.

I agree.. Chasing overdue numbers and playing large pools is a very expensive way to play and often with no profit.

Zero in on what sums are due, widths, roots, and bet with you budget. Zoom in on your game. Sit with the last 12 draws in front of you and you will discover many things on how the numbers evolve.

For example

440

189

362

157

The 36 divided by 2 is 18. 1 and 8 is 9.  The 1 and 5 equals 6 and the 3 and 2 equals 5 which gives the 362 draw.

This happens  alot.  The vertical 1 and 3 will gives you 4 which is the mirror of 9.

Winner
South Carolina
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May 31, 2014
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 Posted: August 15, 2014, 2:50 am - IP Logged

I agree.. Chasing overdue numbers and playing large pools is a very expensive way to play and often with no profit.

Zero in on what sums are due, widths, roots, and bet with you budget. Zoom in on your game. Sit with the last 12 draws in front of you and you will discover many things on how the numbers evolve.

For example

440

189

362

157

The 36 divided by 2 is 18. 1 and 8 is 9.  The 1 and 5 equals 6 and the 3 and 2 equals 5 which gives the 362 draw.

This happens  alot.  The vertical 1 and 3 will gives you 4 which is the mirror of 9.

i disagree.

How is his proposed strategy of tracking Wrong Groups looked

at to be Rocket science ...

playing 8 numbers isn't too many..omg..

let him explore.. No strategy is better than the other. If it was.  Show me some winning tickets!!

Winner
South Carolina
United States
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May 31, 2014
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 Posted: August 15, 2014, 2:53 am - IP Logged

Question: Has anyone done this before? Is it possible? And would it even be helpful, or completely useless?

I guess that's three questions, actually….

I've utilized a variation of this myself. This centers on non-doubles/non-triples boxes only, in the Daily 3 game (California, which uses an algorithm method).

1) First, you take all 120 possible boxes, running in lowest-to-highest order. (Here's one site that can easily calculate and lay them out for you, using letters: http://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html )

2) You divide them into 15 sets of 8 ( = 120). However, HOW you divide them, is key—back to this in a minute.

3) You basically track them like you track anything else, seeing what set of boxes is due or not, over a series of plays.

But here's where a math expert's required, not me rummaging around: How would you divide the sets of boxes, PERFECTLY equally? All sets would contain the same number of different digits, or as close as possible; everything would be perfectly even, but not so perfect, or pattern-inducing, as to seem like groupings: all would appear to be a jumble, but would actually be as even a distribution as possible of all similarities in each set.  (Maybe there are mathematical models/formulas/theories/etc. already, for doing this?)

The reason for this, is so that the game can't—or as little as possible—create anomalous "long outs," that cause players to chase endlessly & needlessly after "dues."

The overall purpose would be to, as much as possible, limit the variance between each one firing off, having one of its 8 boxes hit.  It would probably be best in only pursuing "outs": but the pursuit, done right, would be drastically limited (right?).

I figure 8 is the perfect amount of such sets: not too many as to be unwieldy, not too few as to create endlessly long "outs." KEY, most important: totally "mixed up" so as to not allow the algorithm to escape to groupings and trendings and other evasions.

…? Useful, or just stupid?

super great idea...

totally mixed up groups..

no all hhh, lll, ooo,eee,iii,ooo

just 15 groups of everything all mixed up..

New Mexico
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 Posted: August 15, 2014, 3:18 am - IP Logged

I bet a few tickets which allows me to bet on other games such as powerball, hot lotto etc.  Too much emphasis on giant blocks of tickets.

I usually bet 3 to 4 dollars lately on pick 3. A few on roadrunner cash and 4 dollars on megamillion.  That's enough.

I hit 5 of 6 one time in the Texas lotto with one bet.  Its the study and knowing the numbers that counts.

No need to empty the bank.

Winner
South Carolina
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 Posted: August 15, 2014, 8:26 am - IP Logged

I bet a few tickets which allows me to bet on other games such as powerball, hot lotto etc.  Too much emphasis on giant blocks of tickets.

I usually bet 3 to 4 dollars lately on pick 3. A few on roadrunner cash and 4 dollars on megamillion.  That's enough.

I hit 5 of 6 one time in the Texas lotto with one bet.  Its the study and knowing the numbers that counts.

No need to empty the bank.

8 tickets x .50cent =\$4.00

pocketchange.

Your signature says something like. No strategy better than other.

South Carolina
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July 9, 2005
1743 Posts
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 Posted: August 15, 2014, 2:46 pm - IP Logged

So if I believe because of a number of factors 396 will be drawn tonight or in the near future, it looks like you're suggesting I should play 391, 346, 341, 896, 891, 846, 841 too. I really dislike pointing out the obvious, but if I like 3 first digit position, 9 in the second, and 6 in the third, why would I play 841 or waste money by betting against what I believe has a good chance of being drawn?

And I can't figure out how adding 7 more 3 digit numbers plus 6 box combos each is "aiming for even less".

I don't know what you're referring to here.  All I said, was that I personally would NOT play a group of numbers larger than a VTRAC string, which is 8 numbers for Pick 3, or 16 numbers for Pick 4.  I would aim for playing even less, i.e. Filtering.

I play VTRACS STRAIGHT in Pick 3, that's 8 numbers or less.  I would play a Pick 4 VTRAC String Boxed.

If you're certain about only 1 number, then play just 1 number.  That's even better, i.e. aiming for less.

Kentucky
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February 14, 2006
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 Posted: August 15, 2014, 5:39 pm - IP Logged

I don't know what you're referring to here.  All I said, was that I personally would NOT play a group of numbers larger than a VTRAC string, which is 8 numbers for Pick 3, or 16 numbers for Pick 4.  I would aim for playing even less, i.e. Filtering.

I play VTRACS STRAIGHT in Pick 3, that's 8 numbers or less.  I would play a Pick 4 VTRAC String Boxed.

If you're certain about only 1 number, then play just 1 number.  That's even better, i.e. aiming for less.

I guess I wrongly assumed you understood the idea here is to create 15 groups of 8 three digit numbers possibly isolate on a few.

"I play VTRACS STRAIGHT in Pick 3, that's 8 numbers or less."

The question was "How would you divide the sets of boxes, PERFECTLY equally?" and your method of play doesn't come close to answering it.  I'm not sure how to create 8 sets of boxes with perfectly equal three digit numbers, but it can't be VTracs because those results are unequally matched.

"If you're certain about only 1 number, then play just 1 number."

It's more like "if you're certain about a specific 8 line group". It's obvious that when one the three digit numbers hit, the group hits too and it's like 15 number MM or PB wheels, betting 5 of your 15 will be drawn.

Kentucky
United States
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February 14, 2006
7341 Posts
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 Posted: August 15, 2014, 5:43 pm - IP Logged

I agree.. Chasing overdue numbers and playing large pools is a very expensive way to play and often with no profit.

Zero in on what sums are due, widths, roots, and bet with you budget. Zoom in on your game. Sit with the last 12 draws in front of you and you will discover many things on how the numbers evolve.

For example

440

189

362

157

The 36 divided by 2 is 18. 1 and 8 is 9.  The 1 and 5 equals 6 and the 3 and 2 equals 5 which gives the 362 draw.

This happens  alot.  The vertical 1 and 3 will gives you 4 which is the mirror of 9.

"Zero in on what sums are due, widths, roots, and bet with you budget."

You already have five threads started in this Forum so why are are trying to hijack this one?

United States
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September 8, 2011
3941 Posts
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 Posted: August 16, 2014, 10:44 am - IP Logged

I hear what you're saying, but don't necessarily neglect the algorithmic part of the algorithm.

For example: It seems that, generally (key word there: sometimes it's more, sometimes less) the game takes an average of two digits, and "cold plays" them.  Not equally, not all the time, but they tend to be out for long periods and/or rarely played for long periods.

So is it possible the algorithm is programmed to leave two digits (we'll go with for now) more or less cold?  One would have to conclude yes: otherwise, we'd get this too even constant sifting of the same digits, over and over and over... which a purely random distribution (say, balls dropping) wouldn't produce anyway.

So why care about the algorithm, if the same thing occurs either way?  My theory: because the problem with purely random distributions that nevertheless keep all things equal, means the program must produce approved results (i.e., equal distribution over a long period of time); and this often requires the program to go cold on various digits.

And so, what would help one's game, is to figure out (a) what digit's cold, (b) as early as possible in that cold cycle.

Me, I'd say, any digits that drops below the last-five-plays mark - let's say, a digit that has just slipped to the sixth play down (combined) - could be, perhaps? assumed to be going cold.  If so, you simply remove that digit from your plays, until it reappears again (?—better way? do one for both Mid/Evening?).

One cold digit removed from play, shrinks the 120 singles boxes by 36, or 30%; the 45 doubles by 9 boxes, or 20%; and the whole by 46 "units" (this includes a cold digit's triple), or 26%.  Accurately nail two current cold digits, and you can reduce the pool even further.

One more way to sift through those choices....

CAL

Fri, Aug 15, 2014 Thu, Aug 14, 2014 Wed, Aug 13, 2014 Tue, Aug 12, 2014 Drawing Date Pick 3 Pick 4 Midday Evening Midday Evening 1-0-6 5-1-9 7-8-2-7 1-9-4-0 1-5-4 3-4-5 1-2-1-0 3-6-2-3 2-3-1 5-0-6 4-8-0-7 7-3-5-1 4-1-1 1-9-4 3-0-1-9 1-6-3-5 0-0-4 3-1-7 8-0-2-1 9-2-0-2 4-1-3 7-8-6-7 9-8-5 4-8-5 2-3-0-5 3-6-6-3 0-4-2 0-9-9 4-6-6-5 0-4-6-2 9-0-8 8-0-6 8-9-9-4 2-7-7-4 8-8-1 1-2-2 7-6-2-2 4-5-3-8

Concept> recurrent

P3(Day)

881--058> drive 0,5> pair 05,08, 58>hits 908,985

Concept> sequential difference

881>0291-8013> 029183> 81> 0293>drive 0,2> pairs 02,09,03,29,23>hits 908,042,231

P4(Day)

concept> recurrent

7622> 0781> drive 0,7> 078-071-081-781> hits 8021,4807

Concept> sequential difference

7622>1380465702910> 1380456729>filter 762> 1380459>filter 894>1305>drive 1,3>1305-130x-135x-105x-305x> hits 2305, 3019

Concept> recurrent

P3(EVE)

122---819>drive 8,1>pair  81-89-19> hit  194

Concept> sequential difference

122--13800291>138029>12>3809>drive 3,8>pairs 38-30-39-80-89>hit 806

P4(Eve)

concept>recurrent

4538----7480> drive 7,4> 7480-748x-740x-780-480

concept >sequential difference

4538---138024795746>filter 4538>1027976>102796>filter274> 1096> drive 1,0> 1096-109x-106x-196x-096x>hit 1940

In a random setting equal distribution in span of 10000 draws(p4) or 27 years for a straight pick hit does not say much. lets say you opt for  box hits 1234,  with 1 hit in 417 draws which is true for equal distribution(in theory), there is a disconnect between proportionality of 1234 hits , meaning you can hit many 1234 box hits in span of 417 draws.Programming  a 'formula' for random event is uphill task, most formulas are abstract (linear regression, chaos chain etc) due to uncertainty (PREDICTION). Can one predict in a random setting? yes, by finding  a CONCEPT and sticking to a Waging STRATEGY. The above workout is snippet of concepts based on assumptions, the use of the word assumption may brings some critics, but we all do this inherently.

New Mexico
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 Posted: August 16, 2014, 10:49 am - IP Logged

"Zero in on what sums are due, widths, roots, and bet with you budget."

You already have five threads started in this Forum so why are are trying to hijack this one?

Hijack?

This is a open forum  and you have no right to try and dictate what anyone posts.

Kentucky
United States
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February 14, 2006
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 Posted: August 16, 2014, 6:06 pm - IP Logged

Hijack?

This is a open forum  and you have no right to try and dictate what anyone posts.

No thread hijacking: Stick to the topic.If you want to post something unrelated to a particular discussion thread, then start a new topic in the appropriate forum.

You apparently can't even comprehend what "hijacking a thread" means and and/or probably never read the LP posting rules. Do have a suggestion on "How would you divide the sets of boxes, PERFECTLY equally?"

New Mexico
United States
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January 29, 2010
11161 Posts
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 Posted: August 16, 2014, 8:35 pm - IP Logged

No thread hijacking: Stick to the topic.If you want to post something unrelated to a particular discussion thread, then start a new topic in the appropriate forum.

You apparently can't even comprehend what "hijacking a thread" means and and/or probably never read the LP posting rules. Do have a suggestion on "How would you divide the sets of boxes, PERFECTLY equally?"

You need to read the rules.  All you do is complain about how many posts members post etc.  Way off the subject.  Constantly attacking people that are here to help.  You and another poster act as if you run this site.  As far as I remember Todd does. That's food for thought fella.

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