This thread is based on a simple ideal> Predict just one digit, preferable position 1 for subsequent draw/s for probably straight hit. The picks are not for the risk averse, but for those who consider waging as residual income, so am not going to indulge in discussion about cost and what not(Every member here is of legal age).

I am open to discuss concepts behind the method of pick selection, give and gather suggestions to enhance this method, I will post picks for validation sake, not on request, is best to learn the ideal and do your own workout.

Method> Predict next position digit + Np3, N is your data size for certain cycles, p is your Permutation for triads, basically you're grouping all your triads out of N. The size of N will determine the number of picks to wage.

Hint> 10P4= 5040 , 10P3=720, predict a number on 10P3 for a straight hit (that's ideal, be mindful that N is not static!)

Quote: Originally posted by adobea78 on November 24, 2014

This thread is based on a simple ideal> Predict just one digit, preferable position 1 for subsequent draw/s for probably straight hit. The picks are not for the risk averse, but for those who consider waging as residual income, so am not going to indulge in discussion about cost and what not(Every member here is of legal age).

I am open to discuss concepts behind the method of pick selection, give and gather suggestions to enhance this method, I will post picks for validation sake, not on request, is best to learn the ideal and do your own workout.

Method> Predict next position digit + Np3, N is your data size for certain cycles, p is your Permutation for triads, basically you're grouping all your triads out of N. The size of N will determine the number of picks to wage.

Hint> 10P4= 5040 , 10P3=720, predict a number on 10P3 for a straight hit (that's ideal, be mindful that N is not static!)

Quote: Originally posted by mrlotto-nc on November 24, 2014

Adobea,

Not too familiar with Pick4 so have a question.

This leading digit chart contains all the permutations of the 3 remaining digits behing the leading digit?

If so, do you have the charts for the other key digits 0-9?

I get your drive, but am not going to give that chart out. I'll prefer to share the ideal , which is clearly stated than a finish 'product'. I had a situation some months ago of someone peddling my file on the net, at this juncture, is best to discuss concepts.

NB> you can deduced the triads from the picks just posted, but N is not static.

Quote: Originally posted by adobea78 on November 24, 2014

I get your drive, but am not going to give that chart out. I'll prefer to share the ideal , which is clearly stated than a finish 'product'. I had a situation some months ago of someone peddling my file on the net, at this juncture, is best to discuss concepts.

NB> you can deduced the triads from the picks just posted, but N is not static.

Ok cool. Yeah this idea isnt bad, but not for the faint at heart. But not a bad idea at all buddy.

Quote: Originally posted by mrlotto-nc on November 24, 2014

Ok cool. Yeah this idea isnt bad, but not for the faint at heart. But not a bad idea at all buddy.

The reason I said not for the 'risk averse', it pays off big time if consider it as return on investment. Look at the P3 thread that was locked yesterday, most picks were straight hits within 2 draws for huge returns, is all about choice-my strategy is straight bet or nothing, win or lose, there no middle ground.

Quote: Originally posted by adobea78 on November 24, 2014

This thread is based on a simple ideal> Predict just one digit, preferable position 1 for subsequent draw/s for probably straight hit. The picks are not for the risk averse, but for those who consider waging as residual income, so am not going to indulge in discussion about cost and what not(Every member here is of legal age).

I am open to discuss concepts behind the method of pick selection, give and gather suggestions to enhance this method, I will post picks for validation sake, not on request, is best to learn the ideal and do your own workout.

Method> Predict next position digit + Np3, N is your data size for certain cycles, p is your Permutation for triads, basically you're grouping all your triads out of N. The size of N will determine the number of picks to wage.

Hint> 10P4= 5040 , 10P3=720, predict a number on 10P3 for a straight hit (that's ideal, be mindful that N is not static!)

Method> Pos 1 +Np3

Hint> Keep the size of N to 5 or 6 string length, ie 01254, 724013 and reduce triads to simple permutation without repeat. 6P3>120triads, 6P3>216 (with repeat digit)

Method> pos 1 +120triads or Position 1 +216 triads for straight hit within a short span

Task> Generate your string for N, members of N are not fixed.

Quote: Originally posted by adobea78 on November 25, 2014

Method> Pos 1 +Np3

Hint> Keep the size of N to 5 or 6 string length, ie 01254, 724013 and reduce triads to simple permutation without repeat. 6P3>120triads, 6P3>216 (with repeat digit)

Method> pos 1 +120triads or Position 1 +216 triads for straight hit within a short span

Task> Generate your string for N, members of N are not fixed.

You are truly good with these number systems.I have been thinking about this concept for a couple of days, and it has merit.

I don't know if I can explain what I am thinking, but here goes....

N = first digit (we need a method/workout to determine this)

I have been looking at BlackApple's thread regarding the first digit. He shows how one can use Vtrac short sums to narrow the field of numbers from 10 (0-9) to 4 (1-2-5-7, 3-8-4-9, etc) based on last 4 draws.

Vtracs 1(5-0), 2 (1-6), 3 (2-7), 4 (3-8), 5 (4-9)

From these remaining digits, one could reduce digits further to 2 digits by determining which of the 4 digits are the longest out. Now we have only 2 digits to choose from, giving a 50/50 chance of success.

This was one method I thought about.

The next idea was using Dr Miracle's number followers:

3-5-8

1-4-6-0

2-7-9

These numbers tend to follow each other. I have seen it work as a single digit or sum.

Btw, I am still a little confused on the string concept. Based on the number list published, I do understand the sequence. I just don't understand how to represent it as a string. I am thinking that the string is based on the digits not drawn in the last x draws?.

N =first digit

N012.....N019 (horizontal flow - first 3 digits constant, last digit changes through 9)

N023.....N029 (vertical flow - first & second digit constant, third and fourth digit = +1 of previous digit, continue through 9)

Quote: Originally posted by Pheonix57 on November 26, 2014

You are truly good with these number systems.I have been thinking about this concept for a couple of days, and it has merit.

I don't know if I can explain what I am thinking, but here goes....

N = first digit (we need a method/workout to determine this)

I have been looking at BlackApple's thread regarding the first digit. He shows how one can use Vtrac short sums to narrow the field of numbers from 10 (0-9) to 4 (1-2-5-7, 3-8-4-9, etc) based on last 4 draws.

Vtracs 1(5-0), 2 (1-6), 3 (2-7), 4 (3-8), 5 (4-9)

From these remaining digits, one could reduce digits further to 2 digits by determining which of the 4 digits are the longest out. Now we have only 2 digits to choose from, giving a 50/50 chance of success.

This was one method I thought about.

The next idea was using Dr Miracle's number followers:

3-5-8

1-4-6-0

2-7-9

These numbers tend to follow each other. I have seen it work as a single digit or sum.

Btw, I am still a little confused on the string concept. Based on the number list published, I do understand the sequence. I just don't understand how to represent it as a string. I am thinking that the string is based on the digits not drawn in the last x draws?.

N =first digit

N012.....N019 (horizontal flow - first 3 digits constant, last digit changes through 9)

N023.....N029 (vertical flow - first & second digit constant, third and fourth digit = +1 of previous digit, continue through 9)

Please share your thoughts.

P57

Is just a simple proposition that come with cost, most people want to win, but how badly do you need this win? If you can device a predictive method with 50% degree of certainty, then COST becomes investment.

P4 has 10,000 combinations for straight hit for net lost of $5000

Waging 5040 straight combo is still a lost

Single box combinations of 210 picks is also lost by default.

Now take a look at All triads of P4 without a digit repeating>10!/7! (720 picks), so all it takes to predict 1 or 2 digits for next five draws for a straight hit.

Take any State draws, eg Tx

Drawing Date

Pick 3

Pick 4

Midday

Evening

Midday

Evening

Wed, Nov 26, 2014

0-7-9

0-9-4-5

Tue, Nov 25, 2014

8-1-5 0-2-0

7-5-6 4-1-4

3-3-1-3 6-1-2-8

7-8-9-9 1-4-7-4

Mon, Nov 24, 2014

2-5-9 2-4-3

8-0-2 8-3-3

1-6-5-2 3-2-6-5

3-6-7-6 4-6-2-6

Sat, Nov 22, 2014

2-9-1 4-7-7

9-4-0 0-3-0

5-5-4-3 2-2-0-1

1-7-6-1 1-7-0-2

Fri, Nov 21, 2014

1-2-0 9-7-7

3-1-2 6-6-8

6-1-0-7 0-2-6-6

8-2-3-9 6-4-1-3

Thu, Nov 20, 2014

3-7-4 5-1-8

4-9-0 1-1-7

8-6-5-6 8-6-4-6

6-6-1-6 4-0-2-8

Wed, Nov 19, 2014

4-2-0 5-2-9

5-7-2 8-6-7

7-0-6-9 6-0-0-6

8-3-4-7 6-8-2-9

Tue, Nov 18, 2014

4-8-5 8-1-6

1-6-6 3-1-6

4-8-9-2 3-0-7-3

3-0-4-7 2-0-2-4

Mon, Nov 17, 2014

7-9-4 7-3-9

0-9-3 6-7-1

8-9-2-6 1-4-2-5

3-6-0-5 4-9-1-0

Sat, Nov 15, 2014

7-3-0 9-3-2

8-1-9 8-2-7

3-7-0-9 8-8-7-2

3-4-0-1 7-9-6-3

draw 7963 will give me two key digits 2,4 for position + 720 triads at cost of 1440(do not be scared if certainty is over 50%). You had str hits 2024, 4028 , is it cost effective? depend on waging strategy(waging combined or a type draw). Draw 8892> key 0,5 for str hits ,5543(Betting is Morn/day draws)

Draw>3401> key 8,1 for str hit 8347, 8239,1702,1761

Inference> The propensity of hits confirms the degree of certainty, is it profitable? you decide.

Now the 720 triads is too much or you feel uncomfortable , then step down with simple 10C3> 120 triads

120 triads

draw key hits

8872 0,5 box 3709,7069>>> check cost for box hits

7963 2,4 str 2024, 4028>> check cost for str hits

3401 8,1 str 8347, 8239, Box 1702,1761> check cost for str hit within such time frame.

Why not find a middle ground between 720/120 triads by considering 10P3 with a digit repeat for 220 triads, predict a key digit for potential straight.

Reducing the size of N based on current draw:

This is locating a group (string) based on current draw and setting up Np3, it can be done by setting parameters,eg, sums for prediction points, the number points becomes the size of N. So for a draw 8872,

my points will be 0.2,1,3,5,7 for N=6, my triads will be 6P3(120 triads) or 56 triads with one repeat digit.

NB> I suggest, you leave the string part and focus on predicting your key digits with 220 triads. You can check the net for permutation table, LP has it.

Quote: Originally posted by adobea78 on November 24, 2014

This thread is based on a simple ideal> Predict just one digit, preferable position 1 for subsequent draw/s for probably straight hit. The picks are not for the risk averse, but for those who consider waging as residual income, so am not going to indulge in discussion about cost and what not(Every member here is of legal age).

I am open to discuss concepts behind the method of pick selection, give and gather suggestions to enhance this method, I will post picks for validation sake, not on request, is best to learn the ideal and do your own workout.

Method> Predict next position digit + Np3, N is your data size for certain cycles, p is your Permutation for triads, basically you're grouping all your triads out of N. The size of N will determine the number of picks to wage.

Hint> 10P4= 5040 , 10P3=720, predict a number on 10P3 for a straight hit (that's ideal, be mindful that N is not static!)

Strategy> 10P3 with one repeat for 2 key digits, strictly straight for 5 draws

Assumption> At LEAST a hit, recover cost or lost , based on 50% degree of certainty

Quote: Originally posted by adobea78 on November 24, 2014

This thread is based on a simple ideal> Predict just one digit, preferable position 1 for subsequent draw/s for probably straight hit. The picks are not for the risk averse, but for those who consider waging as residual income, so am not going to indulge in discussion about cost and what not(Every member here is of legal age).

I am open to discuss concepts behind the method of pick selection, give and gather suggestions to enhance this method, I will post picks for validation sake, not on request, is best to learn the ideal and do your own workout.

Method> Predict next position digit + Np3, N is your data size for certain cycles, p is your Permutation for triads, basically you're grouping all your triads out of N. The size of N will determine the number of picks to wage.

Hint> 10P4= 5040 , 10P3=720, predict a number on 10P3 for a straight hit (that's ideal, be mindful that N is not static!)

Quote: Originally posted by adobea78 on November 24, 2014

This thread is based on a simple ideal> Predict just one digit, preferable position 1 for subsequent draw/s for probably straight hit. The picks are not for the risk averse, but for those who consider waging as residual income, so am not going to indulge in discussion about cost and what not(Every member here is of legal age).

I am open to discuss concepts behind the method of pick selection, give and gather suggestions to enhance this method, I will post picks for validation sake, not on request, is best to learn the ideal and do your own workout.

Method> Predict next position digit + Np3, N is your data size for certain cycles, p is your Permutation for triads, basically you're grouping all your triads out of N. The size of N will determine the number of picks to wage.

Hint> 10P4= 5040 , 10P3=720, predict a number on 10P3 for a straight hit (that's ideal, be mindful that N is not static!)

For folks with limited budget, use the concept of recurrence or sequential difference to establish the size of N, hence NP3 for your triads, this will reduced your picks to just 24 picks. The two concepts basically gives a range of prediction points(prediction interval). Say, my draw is 1234, recurring points will be 8197 for 4P3 triads, the points 8,1,9,7 has size 4 (4 elements) for 24 triads >>>>>>>>>>>>>>>>>>>>>>>>>>

Quote: Originally posted by adobea78 on November 27, 2014

For folks with limited budget, use the concept of recurrence or sequential difference to establish the size of N, hence NP3 for your triads, this will reduced your picks to just 24 picks. The two concepts basically gives a range of prediction points(prediction interval). Say, my draw is 1234, recurring points will be 8197 for 4P3 triads, the points 8,1,9,7 has size 4 (4 elements) for 24 triads >>>>>>>>>>>>>>>>>>>>>>>>>>

Predict next position 1 digit/s and add it to your 24 triads, you can also consider 4P3 with one repeat digit for 64 triads

NB> The two concepts has been exhausted in most of my threads, even with charts, so I will not repeat again.

Adobea78,

I have been reading this thread over and over and over, and reading some of your previous threads that are my favorites, and I think I have it now. I will refer to a previous example that you gave. I believe this is what you are referring to, but in this thread, you are going at a differently, but with the same end result.

Take any State draws, eg Tx

Drawing Date

Pick 3

Pick 4

Midday

Evening

Midday

Evening

Wed, Nov 26, 2014

0-7-9

0-9-4-5

Tue, Nov 25, 2014

8-1-5 0-2-0

7-5-6 4-1-4

3-3-1-3 6-1-2-8

7-8-9-9 1-4-7-4

Mon, Nov 24, 2014

2-5-9 2-4-3

8-0-2 8-3-3

1-6-5-2 3-2-6-5

3-6-7-6 4-6-2-6

Sat, Nov 22, 2014

2-9-1 4-7-7

9-4-0 0-3-0

5-5-4-3 2-2-0-1

1-7-6-1 1-7-0-2

Fri, Nov 21, 2014

1-2-0 9-7-7

3-1-2 6-6-8

6-1-0-7 0-2-6-6

8-2-3-9 6-4-1-3

Thu, Nov 20, 2014

3-7-4 5-1-8

4-9-0 1-1-7

8-6-5-6 8-6-4-6

6-6-1-6 4-0-2-8

Wed, Nov 19, 2014

4-2-0 5-2-9

5-7-2 8-6-7

7-0-6-9 6-0-0-6

8-3-4-7 6-8-2-9

Tue, Nov 18, 2014

4-8-5 8-1-6

1-6-6 3-1-6

4-8-9-2 3-0-7-3

3-0-4-7 2-0-2-4

Mon, Nov 17, 2014

7-9-4 7-3-9

0-9-3 6-7-1

8-9-2-6 1-4-2-5

3-6-0-5 4-9-1-0

Sat, Nov 15, 2014

7-3-0 9-3-2

8-1-9 8-2-7

3-7-0-9 8-8-7-2

3-4-0-1 7-9-6-3

Let's take draw 8872 (sum=25,rt=7).

Based on one of your previous charts, this draw (8872) brings sets 2489:9670. Since the draw contains pair 28 from set 2489, our concentration should be on set 9670. From set 9670, we get triads 967x,960x,970x, and 670x. For x, we could choose digits not within the range 0-9 that have not been drawn. Range: (0,1,2,3,4,5,6,7,8,9)

So our 20 Picks would be 9670,9671,9673,9675,9676,9600,9601,9603,9606,9605,9700,9701,9703,9706,9705,6700,6701,6703,6706,

6705. Now these picks are for box plays only. For straight plays, these would need to be played 12 (6)/24 (14) ways for 408 picks.

Now these picks could be reduced by focusing on the first triad 967x; 5 picks boxed, 108 picks straight. Straight hit with draw 7963.

Quote: Originally posted by Pheonix57 on November 27, 2014

Adobea78,

I have been reading this thread over and over and over, and reading some of your previous threads that are my favorites, and I think I have it now. I will refer to a previous example that you gave. I believe this is what you are referring to, but in this thread, you are going at a differently, but with the same end result.

Take any State draws, eg Tx

Drawing Date

Pick 3

Pick 4

Midday

Evening

Midday

Evening

Wed, Nov 26, 2014

0-7-9

0-9-4-5

Tue, Nov 25, 2014

8-1-5 0-2-0

7-5-6 4-1-4

3-3-1-3 6-1-2-8

7-8-9-9 1-4-7-4

Mon, Nov 24, 2014

2-5-9 2-4-3

8-0-2 8-3-3

1-6-5-2 3-2-6-5

3-6-7-6 4-6-2-6

Sat, Nov 22, 2014

2-9-1 4-7-7

9-4-0 0-3-0

5-5-4-3 2-2-0-1

1-7-6-1 1-7-0-2

Fri, Nov 21, 2014

1-2-0 9-7-7

3-1-2 6-6-8

6-1-0-7 0-2-6-6

8-2-3-9 6-4-1-3

Thu, Nov 20, 2014

3-7-4 5-1-8

4-9-0 1-1-7

8-6-5-6 8-6-4-6

6-6-1-6 4-0-2-8

Wed, Nov 19, 2014

4-2-0 5-2-9

5-7-2 8-6-7

7-0-6-9 6-0-0-6

8-3-4-7 6-8-2-9

Tue, Nov 18, 2014

4-8-5 8-1-6

1-6-6 3-1-6

4-8-9-2 3-0-7-3

3-0-4-7 2-0-2-4

Mon, Nov 17, 2014

7-9-4 7-3-9

0-9-3 6-7-1

8-9-2-6 1-4-2-5

3-6-0-5 4-9-1-0

Sat, Nov 15, 2014

7-3-0 9-3-2

8-1-9 8-2-7

3-7-0-9 8-8-7-2

3-4-0-1 7-9-6-3

Let's take draw 8872 (sum=25,rt=7).

Based on one of your previous charts, this draw (8872) brings sets 2489:9670. Since the draw contains pair 28 from set 2489, our concentration should be on set 9670. From set 9670, we get triads 967x,960x,970x, and 670x. For x, we could choose digits not within the range 0-9 that have not been drawn. Range: (0,1,2,3,4,5,6,7,8,9)

So our 20 Picks would be 9670,9671,9673,9675,9676,9600,9601,9603,9606,9605,9700,9701,9703,9706,9705,6700,6701,6703,6706,

6705. Now these picks are for box plays only. For straight plays, these would need to be played 12 (6)/24 (14) ways for 408 picks.

Now these picks could be reduced by focusing on the first triad 967x; 5 picks boxed, 108 picks straight. Straight hit with draw 7963.

Please advise if this is correct.

I am still confused about the key digits though.

P57

Use the sum parameters for your workout, while you study this permutation method, this method does not come cheap, though the reward is huge, you have to have confidence and knowledge of the method.

See the draw 8827 with keys 2489, 9670, since the digits of the key sets are positional, start filtering/deleting digits 8827 from key sets but maintain the structure> x4x9 and 96x0>>94x9, 96x0,96x9, now fill in x starting with the draw digits 8,2,7 etc