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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
United States
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July 4, 2014
671 Posts
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 Posted: August 12, 2014, 1:34 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?

bgonÃ§alves
Brasil
Member #92564
June 9, 2010
2122 Posts
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 Posted: August 12, 2014, 12:05 pm - IP Logged

Peer hello, pick the 3 you have to look for the pairs posicioniais
Example = 1st 2nd 3rd
1 or 2 5 8 3 6
Well = 1st 2nd 1ª3ª 2ª3ª
Well each pick 3 has 3 positions of pairs with pairs of digits = even / odd
Example = 851 = odd pair at position 51 = 2nd 3rd
Or example = 263 pair, 1ª2ª 2 = 6
Then the apres can be odd and even transiting under the three positions (1.2 1.3 2.3)
odd
13
15
17
19
35
37
39
57
59
79
pairs
24
26
28
20
46
48
40
68
60
80

bgonÃ§alves
Brasil
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June 9, 2010
2122 Posts
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 Posted: August 12, 2014, 12:07 pm - IP Logged

ex= 24x, 2x4,  x24=    3º digit odd

South Carolina
United States
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July 9, 2005
1704 Posts
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 Posted: August 12, 2014, 12:08 pm - 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?

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.

Stone Mountain*Georgia
United States
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November 2, 2002
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 Posted: August 12, 2014, 1:51 pm - IP Logged

Before VTRAC's were thought up....I just called them "Mirrors." Wasted a lot of time on VTRAC's  then dropped them for just that reason. Redundant.

You can only use just so many things at this point of our state of the art play.  Still don't get pairs either. I have tried but don't get pairs. I use them for confirmations ...after everything else is done.....but just can't use them as "front line targets".   They just produce too many numbers as primary filters. Just never got it.

I broke down pairs into a pretty good grouping chart back when. Now those had and still have some promise. Ghost Pairs... Mirror Pairs... etc Chart. Placed them on my Blog years ago but haven't posted them in a long time.

The only real failure .....is the failure to try.

Luck is a very rare thing....... Odds not so much.

Odds never change .....but probability does.

Win d

Wyncote,Pa
United States
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January 3, 2004
60721 Posts
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 Posted: August 12, 2014, 2:39 pm - IP Logged
 0 1 2 3 4 5 6 7 8 9 037 029 039 049 013 014 015 016 017 018 127 038 057 058 059 023 079 025 035 027 136 047 129 067 149 069 169 034 125 036 145 056 138 148 158 078 178 124 134 045 235 128 147 238 167 168 259 269 189 126 389 146 156 247 239 249 349 278 279 289 479 236 237 256 257 258 358 368 369 469 569 245 589 346 347 267 367 458 378 478 578 489 679 689 356 348 457 467 459 568 019 678 345 012 789 456 123 089 567 234 028 137 048 139 068 159 024 179 026 135 046 579 246 157 248 357 268 359 468 379 0 1 2 3 4 5 6 7 8 9

This centers on non-doubles/non-triples boxes only, in the Daily 3 game

1) First, you take all 120 possible boxes

WinD has done this before.. sectioned them in quarters

Wyncote,Pa
United States
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January 3, 2004
60721 Posts
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 Posted: August 12, 2014, 2:44 pm - IP Logged

Published: October 3, 2004, 2:15 pm

 A B 0 1 2 3 4 5 6 7 8 9 1 037 029 039 049 013 014 015 016 017 018 1> 2 127 038 057 058 059 023 079 025 035 027 3 136 047 129 067 149 069 169 034 125 036 4 145 056 138 148 158 078 178 124 134 045 2> 5 235 128 147 238 167 168 259 269 189 126 6 389 146 156 247 239 249 349 278 279 289 C D 7 479 236 237 256 257 258 358 368 369 469 3> 8 569 245 589 346 347 267 367 458 378 478 9 578 489 679 689 356 348 457 467 459 568 10 019 678 345 012 789 456 123 089 567 234 4> 11 028 137 048 139 068 159 024 179 026 135 12 046 579 246 157 248 357 268 359 468 379 0 1 2 3 4 5 6 7 8 9
backwoods ga
United States
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May 31, 2014
1887 Posts
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 Posted: August 12, 2014, 5:15 pm - 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?

Great idea.

my name Lil Darryl   you got some Milk

Simi Valley, CA
United States
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July 4, 2014
671 Posts
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 Posted: August 12, 2014, 7:52 pm - IP Logged

Thanks, all, for the great suggestions and additions.

Dr. San - I'm sorry, I'm just almost completely lost by what you're explaining - I'm still new here, and don't quite get the jargon or gist of things all the time. It looks like you have a well-worked out theory, though. Could you explain it step by step?

DestinyCreations - thank you, that's very kind, but see my response to Dr. San above.   I'm not really familiar with Vtracs (nor mirrors, WinD), but I'm meaning to get to them, since so many here - just like you say - have shown amazing skills in utilizing them. So there must be something to that.

BlackApple - That's great, you took the 120 and found a way to make 10s, but unlike how I was doing it - just slicing it up as it comes - it seems like this indeed does evenly distribute them.

The interesting thing about slicing them up into 10s starting from lowest to highest digit - 012, 013, 014, etc., all the way to 789 (the terminus point) - is that - as you study the way they pay off or not - you realize some sets will fire off quickly, while some go on for long, even long, stretches.  So I wonder if this itself isn't a clue to the secret algorithm: maybe (like someone once told me concerning Apple's own super-secret iPod shuffling algorithm [true or not? unknown]) the game doesn't randomize from the whole; but rather, concentrates on certain areas (or, avoids others), sort of "dancing around" in those before moving on to another section. (e.g., the D3 will hover for a time around the 3 digits, but avoid the 7 digits.) Maybe it even varies the time it spends with, away, or long away, from certain sections, for total randomization of the game considered as a whole (i.e., not blip by blip, play by play, as we encounter it).

The standard slicing, seems to "reveal" this compelling data; but if one could (and hopefully all your examples here are indicative of this) mess up that algorithmic pattern, one wouldn't get caught in long-outs, or miss all the short-outs.

Of course, just becoming savvy with the way the game DOES randomize… maybe that's good enough as it is?

Lidaryyl: Thanks!

Simi Valley, CA
United States
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July 4, 2014
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 Posted: August 12, 2014, 8:09 pm - IP Logged

Correction - I messed up there - those last two are WinD's charts!  Sorry about that, my friend.

But thanks, Blackapple, for re-posting them.

nj
United States
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August 10, 2013
974 Posts
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 Posted: August 12, 2014, 9:15 pm - 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.

THE V'\$-even less ,odd more to b played one can win more and loose less.

Kentucky
United States
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February 14, 2006
7308 Posts
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 Posted: August 14, 2014, 6:55 pm - IP Logged

"How would you divide the sets of boxes, PERFECTLY equally?"

You can't do it by dividing the 120 box combos by 8, but what about by 12 giving you 12 groups with 10 lines? 830, 941, 052, 163, 274, 385, 496, 507, 618, 729 is an example of one of the unique 12 groups. To get the 12 groups, start with the lowest sum with 3 different digits, 012, add 1 to each digit, 012, 123, 234, 345, 456,, 567, 678, 789, 890, 901 and that line has 10 lines with consecutive digits. Next 013, 124, 235, 346, 457, 568, 679, 780, 891, 902.

That's three of the lines and you can get the other nine if you're interested. Each each group has a 1 in 22 probability. The groups starting with 024 and 026 each have 5 all even and 5 all odd lines.

Kentucky
United States
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 Posted: August 14, 2014, 7:40 pm - 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.

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".

United States
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September 8, 2011
3926 Posts
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 Posted: August 14, 2014, 8:20 pm - IP Logged

Thanks, all, for the great suggestions and additions.

Dr. San - I'm sorry, I'm just almost completely lost by what you're explaining - I'm still new here, and don't quite get the jargon or gist of things all the time. It looks like you have a well-worked out theory, though. Could you explain it step by step?

DestinyCreations - thank you, that's very kind, but see my response to Dr. San above.   I'm not really familiar with Vtracs (nor mirrors, WinD), but I'm meaning to get to them, since so many here - just like you say - have shown amazing skills in utilizing them. So there must be something to that.

BlackApple - That's great, you took the 120 and found a way to make 10s, but unlike how I was doing it - just slicing it up as it comes - it seems like this indeed does evenly distribute them.

The interesting thing about slicing them up into 10s starting from lowest to highest digit - 012, 013, 014, etc., all the way to 789 (the terminus point) - is that - as you study the way they pay off or not - you realize some sets will fire off quickly, while some go on for long, even long, stretches.  So I wonder if this itself isn't a clue to the secret algorithm: maybe (like someone once told me concerning Apple's own super-secret iPod shuffling algorithm [true or not? unknown]) the game doesn't randomize from the whole; but rather, concentrates on certain areas (or, avoids others), sort of "dancing around" in those before moving on to another section. (e.g., the D3 will hover for a time around the 3 digits, but avoid the 7 digits.) Maybe it even varies the time it spends with, away, or long away, from certain sections, for total randomization of the game considered as a whole (i.e., not blip by blip, play by play, as we encounter it).

The standard slicing, seems to "reveal" this compelling data; but if one could (and hopefully all your examples here are indicative of this) mess up that algorithmic pattern, one wouldn't get caught in long-outs, or miss all the short-outs.

Of course, just becoming savvy with the way the game DOES randomize… maybe that's good enough as it is?

Lidaryyl: Thanks!

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.

Simi Valley, CA
United States
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July 4, 2014
671 Posts
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 Posted: August 14, 2014, 11:51 pm - IP Logged

"How would you divide the sets of boxes, PERFECTLY equally?"

You can't do it by dividing the 120 box combos by 8, but what about by 12 giving you 12 groups with 10 lines? 830, 941, 052, 163, 274, 385, 496, 507, 618, 729 is an example of one of the unique 12 groups. To get the 12 groups, start with the lowest sum with 3 different digits, 012, add 1 to each digit, 012, 123, 234, 345, 456,, 567, 678, 789, 890, 901 and that line has 10 lines with consecutive digits. Next 013, 124, 235, 346, 457, 568, 679, 780, 891, 902.

That's three of the lines and you can get the other nine if you're interested. Each each group has a 1 in 22 probability. The groups starting with 024 and 026 each have 5 all even and 5 all odd lines.

This is a very good separating, Stack47!  The test would be to backtrack and see how it fares; hopefully the lags between groups would be as close as possible.  I will try it out when I get some time, thanks.

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