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# Mathematics and the Lottery

646 replies. Last post 23 days ago by SEA-Pick3.

 Page 11 of 44

Can a winning lottery system be created with existing math formulas?

 Yes-It's all in the math books. [ 228 ] [43.02%] No-Anew math for will have to be created. [ 78 ] [14.72%] Math won't beat the lottery regularly. [ 224 ] [42.26%] Total Valid Votes [ 530 ] Discarded Votes [ 54 ]

ORLANDO, FLORIDA
United States
Member #4924
June 3, 2004
5896 Posts
Offline
 Posted: August 20, 2011, 6:23 am - IP Logged

(You mentioned Bayes in your earlier post.)

"I was referring to the term random variable. When you and the other posters write about random, I get the impression there are no variables in random. Am I not reading this right?"

Saying "in random" implies you have a perception of the definition of "random" that is not consistent with that of mathematicians.  "Random" is an adjective in this context.  Nothing can be "in" random.  If you look up definitions of "random" and "variable" and look again at stattrek's definitions of "discrete" and "continuous," I think you will find that, "When the numerical value of a variable is determined by a chance event, that variable is called a random variable." is a pretty good definition!

To get you started:

A "variable" in math is a value that changes (or can change) within a specified range, as opposed to a "constant," which is invariant.

y = 2 * x + 4.7   ( x > 0 )

... defines two variables, x and y, and one constant, 4.7.

x is an independent variable with an infinite range greater than zero.

y is a dependent variable, dependent upon the value of x.

Thanks for the explanation.

Kentucky
United States
Member #32652
February 14, 2006
7302 Posts
Offline
 Posted: August 20, 2011, 5:34 pm - IP Logged

Hi Stack47,

Let me give you a different point of view. There is nothing false about past drawing results. They are what they are and must be delt with accordingly.

Secondly, if someone decides to ignore a 49.52% stat, then, where is the problem. The person with hunches or the math?

Best of Luck

I was looking for a 6/49 game 50/50 stat because coin flips were being discussed, but even if we used math or guessed correctly and eliminated those 6 "cold" numbers, we still have over 7 million combos left. Any 6 six numbers including the numbers from the previous drawing have the same affect. A commonly used combo filter is 0 to 1 of the previous drawing and probably a better way to use 6 "cold" numbers.

Some of system books suggest eliminating many numbers and use the keepers in full or abbreviated wheels and others suggest using all the numbers and filter out low percentage combos IE high/low, even/odd distribution, consecutive, last digit, decades, etc. It depends on how a 6/49 game is viewed because while it's a random drawing and each number has an equal chance of being drawn, it's still a fact 43 numbers won't be drawn.

bgonÃ§alves
Brasil
Member #92564
June 9, 2010
2122 Posts
Offline
 Posted: August 20, 2011, 6:52 pm - IP Logged

Hello, the numbers do not have memory, but often hold positions of certain output   CAD output result, the results are the problem, the solution should be within the same problem! Much of mathematics and other aletoria together! example   Game 49 / 6 result 12,13,24,32,44,47 Mathematics = 12,13,44,47 Random = 24.32   Not necessarily in order of numbers

United States
Member #59354
March 13, 2008
3964 Posts
Offline
 Posted: August 21, 2011, 10:43 am - IP Logged

JKING

Do you consider the information based entirely on the past results or do you look also to the matrix?

My way of thinking is this, Lets say a certain filter has 5 values of which one will hit in the next draw.

Looking at the past history one can see that a certain value has hit in 30% of the past draws and

it is also found in to occure in 30% of the lines in the matrix.  I don't think that the next draw can be

predicted with any certainty but if one chooses what they play based on the probabilities of both the

history and the matrix them I would expect some success.  The 50/50 coin toss is and always will be

50/50 no matter how many flips it it tossed.   Lets say again that heads has come up 10 times in a

row.  Some would bet tails because it is due while others would bet heads because it is hot and the

one who guessed correctly would feel they predicted the outcome even though the next flip is pure

chance unless the flip is biased.  The lottery is quite different because it is taking a randomly selected

sample.   The more random the sample the better the set drawn will follow the matrix in proportion.

All the numbers have a equal chance of being drawn but when looking at the sets one can begin to see

the balance.   A number can go hot or cold without effecting the proportion /distrribution of the matrix.

I have around 100 filters that are used just to gauge a draw and I am suprised many days at how few

repeat from day to day.  Each of these filters can be checked against the matrix and found to fall within

there expected range of values hit.  Take a look at the data below.  There are 10 columbs of digits which

repersent the ten digits 1-2-3-4-5-6-7-8-9-0.  They are broken into three groups F-4 = digits 1-2-3-4.

M-3 = digits 5-6-7 and B-3 = digits 8-9-0.  A zero indicates a no-show and a one equals a hit.  The top

row shows that digits 2-5-7-8 made up the set drawn.   Numbers =  08/20/11   05 07 25 27 28.  Pay

attention to the M-3 and B-3 strings.  A 3 digit binary string has 8 different values from 000 to 111 and

the right columbs show the binary value for each of the three groupings.

##  F-4             M-3        B-3         Binary values

Dig  1  2  3  4    5  6  7    8  9  0

01   0  1  0  0    1  0  1    1  0  0    F=04   M=5   B=4
02   1  0  1  0    1  0  0    1  0  1    F=10   M=4   B=5
03   1  1  1  0    1  0  1    0  1  0    F=14   M=5   B=2
04   0  1  1  1    1  1  0    0  1  0    F=07   M=6   B=2
05   1  1  1  1    1  1  0    1  0  0    F=15   M=6   B=4
06   1  1  0  1    0  1  1    1  0  1    F=13   M=3   B=5
07   1  1  1  1    1  0  0    0  1  1    F=15   M=4   B=3
08   1  0  1  0    1  0  1    0  0  0    F=10   M=5   B=0
09   1  1  0  0    0  1  0    1  0  1    F=12   M=2   B=5
10   1  1  1  0    0  0  1    1  1  1    F=14   M=1   B=7
11   1  1  1  0    1  0  0    0  1  1    F=14   M=4   B=3
12   1  1  1  1    0  0  0    0  0  1    F=15   M=0   B=1
13   1  1  1  0    0  0  0    1  1  0    F=14   M=0   B=6

in the last 13 draws 7 of the 8  binary patterns were drawn for M-3 and all 8 hit in B-3.  F-4 holds

the decade digits which are digits 1-2-3 for a 5-39 and if I separate or remove the decades than

they also follow M-3 and B-3 results.   There seems to be no pattern to the order but what I find

useful is that how they run the gantlet so to speak within a very short span.  One of the guys sent

me a pic this morning of a setup that was done after the fact which I encourage people to do to

learn the most effective setups.  As you know I am a digit man but I think that this shows some of

the logic behind my choices.  Timing the play is maybe the most important factor to system play.

RL

Working on my Ph.D.  "University of hard Knocks"

I will consider the opinion that my winnings are a product of chance if you are willing to consider

they are not.  Many great discoveries come while searching for something else

Trump / 2016 & 2020

United States
Member #59354
March 13, 2008
3964 Posts
Offline
 Posted: August 21, 2011, 10:50 am - IP Logged

JKING

Do you consider the information based entirely on the past results or do you look also to the matrix?

My way of thinking is this, Lets say a certain filter has 5 values of which one will hit in the next draw.

Looking at the past history one can see that a certain value has hit in 30% of the past draws and

it is also found in to occure in 30% of the lines in the matrix.  I don't think that the next draw can be

predicted with any certainty but if one chooses what they play based on the probabilities of both the

history and the matrix them I would expect some success.  The 50/50 coin toss is and always will be

50/50 no matter how many flips it it tossed.   Lets say again that heads has come up 10 times in a

row.  Some would bet tails because it is due while others would bet heads because it is hot and the

one who guessed correctly would feel they predicted the outcome even though the next flip is pure

chance unless the flip is biased.  The lottery is quite different because it is taking a randomly selected

sample.   The more random the sample the better the set drawn will follow the matrix in proportion.

All the numbers have a equal chance of being drawn but when looking at the sets one can begin to see

the balance.   A number can go hot or cold without effecting the proportion /distrribution of the matrix.

I have around 100 filters that are used just to gauge a draw and I am suprised many days at how few

repeat from day to day.  Each of these filters can be checked against the matrix and found to fall within

there expected range of values hit.  Take a look at the data below.  There are 10 columbs of digits which

repersent the ten digits 1-2-3-4-5-6-7-8-9-0.  They are broken into three groups F-4 = digits 1-2-3-4.

M-3 = digits 5-6-7 and B-3 = digits 8-9-0.  A zero indicates a no-show and a one equals a hit.  The top

row shows that digits 2-5-7-8 made up the set drawn.   Numbers =  08/20/11   05 07 25 27 28.  Pay

attention to the M-3 and B-3 strings.  A 3 digit binary string has 8 different values from 000 to 111 and

the right columbs show the binary value for each of the three groupings.

##  F-4             M-3        B-3         Binary values

Dig  1  2  3  4    5  6  7    8  9  0

01   0  1  0  0    1  0  1    1  0  0    F=04   M=5   B=4
02   1  0  1  0    1  0  0    1  0  1    F=10   M=4   B=5
03   1  1  1  0    1  0  1    0  1  0    F=14   M=5   B=2
04   0  1  1  1    1  1  0    0  1  0    F=07   M=6   B=2
05   1  1  1  1    1  1  0    1  0  0    F=15   M=6   B=4
06   1  1  0  1    0  1  1    1  0  1    F=13   M=3   B=5
07   1  1  1  1    1  0  0    0  1  1    F=15   M=4   B=3
08   1  0  1  0    1  0  1    0  0  0    F=10   M=5   B=0
09   1  1  0  0    0  1  0    1  0  1    F=12   M=2   B=5
10   1  1  1  0    0  0  1    1  1  1    F=14   M=1   B=7
11   1  1  1  0    1  0  0    0  1  1    F=14   M=4   B=3
12   1  1  1  1    0  0  0    0  0  1    F=15   M=0   B=1
13   1  1  1  0    0  0  0    1  1  0    F=14   M=0   B=6

in the last 13 draws 7 of the 8  binary patterns were drawn for M-3 and all 8 hit in B-3.  F-4 holds

the decade digits which are digits 1-2-3 for a 5-39 and if I separate or remove the decades than

they also follow M-3 and B-3 results.   There seems to be no pattern to the order but what I find

useful is that how they run the gantlet so to speak within a very short span.  One of the guys sent

me a pic this morning of a setup that was done after the fact which I encourage people to do to

learn the most effective setups.  As you know I am a digit man but I think that this shows some of

the logic behind my choices.  Timing the play is maybe the most important factor to system play.

RL

I forgot to add that the primary run produced 72 sets which are in the temp folder and a second

pass was made using the cover option which reduced to 16 lines.

RL

Working on my Ph.D.  "University of hard Knocks"

I will consider the opinion that my winnings are a product of chance if you are willing to consider

they are not.  Many great discoveries come while searching for something else

Trump / 2016 & 2020

bgonÃ§alves
Brasil
Member #92564
June 9, 2010
2122 Posts
Offline
 Posted: August 21, 2011, 11:39 am - IP Logged

Hello, rl RANDONIC, Fantastico, on the table headings   Rl-randonic, the comparison on the launch of the currency, the crown face   The problem is that the single currency and lanchamento (one currency only) such as the lottery 49 / 6   We have six events (six pairings of two-digit numbers with each number), then   A conbinaçao release (6) in a single event, while the currency has only one, Aexemplo currency only serves to make each digit and also the analysis of the vertical column   Separate the digit number and the termination of frentedo vertically this point we have to understand the difference, many combined events (draw) and currency

United States
Member #5599
July 13, 2004
1184 Posts
Offline
 Posted: August 21, 2011, 7:03 pm - IP Logged

JKING

Do you consider the information based entirely on the past results or do you look also to the matrix?

My way of thinking is this, Lets say a certain filter has 5 values of which one will hit in the next draw.

Looking at the past history one can see that a certain value has hit in 30% of the past draws and

it is also found in to occure in 30% of the lines in the matrix.  I don't think that the next draw can be

predicted with any certainty but if one chooses what they play based on the probabilities of both the

history and the matrix them I would expect some success.  The 50/50 coin toss is and always will be

50/50 no matter how many flips it it tossed.   Lets say again that heads has come up 10 times in a

row.  Some would bet tails because it is due while others would bet heads because it is hot and the

one who guessed correctly would feel they predicted the outcome even though the next flip is pure

chance unless the flip is biased.  The lottery is quite different because it is taking a randomly selected

sample.   The more random the sample the better the set drawn will follow the matrix in proportion.

All the numbers have a equal chance of being drawn but when looking at the sets one can begin to see

the balance.   A number can go hot or cold without effecting the proportion /distrribution of the matrix.

I have around 100 filters that are used just to gauge a draw and I am suprised many days at how few

repeat from day to day.  Each of these filters can be checked against the matrix and found to fall within

there expected range of values hit.  Take a look at the data below.  There are 10 columbs of digits which

repersent the ten digits 1-2-3-4-5-6-7-8-9-0.  They are broken into three groups F-4 = digits 1-2-3-4.

M-3 = digits 5-6-7 and B-3 = digits 8-9-0.  A zero indicates a no-show and a one equals a hit.  The top

row shows that digits 2-5-7-8 made up the set drawn.   Numbers =  08/20/11   05 07 25 27 28.  Pay

attention to the M-3 and B-3 strings.  A 3 digit binary string has 8 different values from 000 to 111 and

the right columbs show the binary value for each of the three groupings.

##  F-4             M-3        B-3         Binary values

Dig  1  2  3  4    5  6  7    8  9  0

01   0  1  0  0    1  0  1    1  0  0    F=04   M=5   B=4
02   1  0  1  0    1  0  0    1  0  1    F=10   M=4   B=5
03   1  1  1  0    1  0  1    0  1  0    F=14   M=5   B=2
04   0  1  1  1    1  1  0    0  1  0    F=07   M=6   B=2
05   1  1  1  1    1  1  0    1  0  0    F=15   M=6   B=4
06   1  1  0  1    0  1  1    1  0  1    F=13   M=3   B=5
07   1  1  1  1    1  0  0    0  1  1    F=15   M=4   B=3
08   1  0  1  0    1  0  1    0  0  0    F=10   M=5   B=0
09   1  1  0  0    0  1  0    1  0  1    F=12   M=2   B=5
10   1  1  1  0    0  0  1    1  1  1    F=14   M=1   B=7
11   1  1  1  0    1  0  0    0  1  1    F=14   M=4   B=3
12   1  1  1  1    0  0  0    0  0  1    F=15   M=0   B=1
13   1  1  1  0    0  0  0    1  1  0    F=14   M=0   B=6

in the last 13 draws 7 of the 8  binary patterns were drawn for M-3 and all 8 hit in B-3.  F-4 holds

the decade digits which are digits 1-2-3 for a 5-39 and if I separate or remove the decades than

they also follow M-3 and B-3 results.   There seems to be no pattern to the order but what I find

useful is that how they run the gantlet so to speak within a very short span.  One of the guys sent

me a pic this morning of a setup that was done after the fact which I encourage people to do to

learn the most effective setups.  As you know I am a digit man but I think that this shows some of

the logic behind my choices.  Timing the play is maybe the most important factor to system play.

RL

Hi RL,

I looked at some of what you where doing from one of your earlier posts, that explained whay you where doing with digits. I fould that combining the first and second digits confusing for me. So I have gone with the last digit distribution, Unique digits, and hi/low digits......

Last Digit Distribution

last digit   (0)  (1)  (2)  (3)  (4)   (5)  (6)  (7)  (8)  (9)
TOTAL = 205 232 227 252 235 229 232 242 243 237

From the standpoint of how I approach it.

I need to have no less than 3 unique digits(UD) and no more than 4 in the hi or low postion.

The boundry conditions postionally:

1.....0-6

2.....0-7

3.....1-8

4.....2-9

5.....4-9

A                 LAST DIGITS (POSITIONAL)                       A

M U               L                                                   0-4    5-9   M

T  D               D   1   2   3   4   5

1  0

2  1               1   163 95   9   0   0

3   65            2   84   125 53   9   0

4   263          3   53   121 91   26   1

5   217          4   23   84   93   64   5

5   9   50   102 77   23

6   9   30   98   87   49

7   0   15   68   126 84

8   0   3   29   115 147

9   0   0   2   42   237

I start with 100% coverage then simply eliminate only the lowest percentages on subsequent passes on all combinations. That is the foundation of what I do. Even when I do have to bit the bullet to eliminate more combinations it is done again by eliminating the next tier of the lowest percentages. The plan is to add more 100% filters and get even more combination reductions.

Yes, I'll set up matricies, but the data and percentages have to be overwhelming for me to use them. No second guessing.

Even though our approach is different, please stay with what you are doing and thanks for posting them. Understading the diversity of different approaches from everbody at the LP is going to be key if anyone of us to find the best system. *S*

You are a slave to the choices you have made.  jk

Even a blind squirrel will occasioanlly find an acorn.

Kentucky
United States
Member #32652
February 14, 2006
7302 Posts
Offline
 Posted: August 22, 2011, 3:22 pm - IP Logged

JKING

Do you consider the information based entirely on the past results or do you look also to the matrix?

My way of thinking is this, Lets say a certain filter has 5 values of which one will hit in the next draw.

Looking at the past history one can see that a certain value has hit in 30% of the past draws and

it is also found in to occure in 30% of the lines in the matrix.  I don't think that the next draw can be

predicted with any certainty but if one chooses what they play based on the probabilities of both the

history and the matrix them I would expect some success.  The 50/50 coin toss is and always will be

50/50 no matter how many flips it it tossed.   Lets say again that heads has come up 10 times in a

row.  Some would bet tails because it is due while others would bet heads because it is hot and the

one who guessed correctly would feel they predicted the outcome even though the next flip is pure

chance unless the flip is biased.  The lottery is quite different because it is taking a randomly selected

sample.   The more random the sample the better the set drawn will follow the matrix in proportion.

All the numbers have a equal chance of being drawn but when looking at the sets one can begin to see

the balance.   A number can go hot or cold without effecting the proportion /distrribution of the matrix.

I have around 100 filters that are used just to gauge a draw and I am suprised many days at how few

repeat from day to day.  Each of these filters can be checked against the matrix and found to fall within

there expected range of values hit.  Take a look at the data below.  There are 10 columbs of digits which

repersent the ten digits 1-2-3-4-5-6-7-8-9-0.  They are broken into three groups F-4 = digits 1-2-3-4.

M-3 = digits 5-6-7 and B-3 = digits 8-9-0.  A zero indicates a no-show and a one equals a hit.  The top

row shows that digits 2-5-7-8 made up the set drawn.   Numbers =  08/20/11   05 07 25 27 28.  Pay

attention to the M-3 and B-3 strings.  A 3 digit binary string has 8 different values from 000 to 111 and

the right columbs show the binary value for each of the three groupings.

##  F-4             M-3        B-3         Binary values

Dig  1  2  3  4    5  6  7    8  9  0

01   0  1  0  0    1  0  1    1  0  0    F=04   M=5   B=4
02   1  0  1  0    1  0  0    1  0  1    F=10   M=4   B=5
03   1  1  1  0    1  0  1    0  1  0    F=14   M=5   B=2
04   0  1  1  1    1  1  0    0  1  0    F=07   M=6   B=2
05   1  1  1  1    1  1  0    1  0  0    F=15   M=6   B=4
06   1  1  0  1    0  1  1    1  0  1    F=13   M=3   B=5
07   1  1  1  1    1  0  0    0  1  1    F=15   M=4   B=3
08   1  0  1  0    1  0  1    0  0  0    F=10   M=5   B=0
09   1  1  0  0    0  1  0    1  0  1    F=12   M=2   B=5
10   1  1  1  0    0  0  1    1  1  1    F=14   M=1   B=7
11   1  1  1  0    1  0  0    0  1  1    F=14   M=4   B=3
12   1  1  1  1    0  0  0    0  0  1    F=15   M=0   B=1
13   1  1  1  0    0  0  0    1  1  0    F=14   M=0   B=6

in the last 13 draws 7 of the 8  binary patterns were drawn for M-3 and all 8 hit in B-3.  F-4 holds

the decade digits which are digits 1-2-3 for a 5-39 and if I separate or remove the decades than

they also follow M-3 and B-3 results.   There seems to be no pattern to the order but what I find

useful is that how they run the gantlet so to speak within a very short span.  One of the guys sent

me a pic this morning of a setup that was done after the fact which I encourage people to do to

learn the most effective setups.  As you know I am a digit man but I think that this shows some of

the logic behind my choices.  Timing the play is maybe the most important factor to system play.

RL

"Lets say again that heads has come up 10 times in a row.  Some would bet tails because it is due while others would bet heads because it is hot and the one who guessed correctly would feel they predicted the outcome even though the next flip is pure chance unless the flip is biased."

A simple formula like A + B = C has no meaning unless we know the value of either A, B, or C. If A = 1 we know if B = 2, than C must = 3 and if C = 4, B must = 3. Before the next three coin flips, we know the outcome will be 3 heads, 0 tails, 2 heads, 1 tails, 1 head, 2 tail, 0 heads, or 3 tails. With no other information we have a 1 in 8 chance of guessing correctly. If the last three results were 3 heads and we know each side has a 50/50 chance of happening, shouldn't we logically assume one of the next three flips will be tails and eliminate 2 of the 8 chances (3 heads and 0 tails) and now have a 1 in 6 chance?

These pro and con coin flip analogies are very problematic because you're saying each side had a 50/50 chance by giving an example showing one side had a zero chance in ten flips. In your example you gave reasons why each player made their choice and both were educated guesses made from known information. With no information they could flip another coin to determine which side to choose but that wouldn't be an educated guess.  We can use math to determine the probably of streaks of three or more consecutive events occurring when the odds are 50/50 or we can look at past results and find a similar number. But either way, we still need the results of at least one outcome before we can make our bet.

Lotteries don't pay even money on 50/50 propositions so that makes the coin flipping analogy even worse. If you give a lottery clerk \$250 betting the first digit in the pick-3 game will be even and win, you'll get your \$250 back and win nothing. Break even or lose is not a sound bet. In pick-4 games there is a 43.2% chance of the winning number having just two matching digits and we can get a 11.1% profit, but the edge is still 8.8%.

"The lottery is quite different because it is taking a randomly selected sample."

The results are randomly selected samples but after more than half the numbers are drawn the odds favor numbers repeating. If you pick 750 numbers and I pick 250 in any pick-3 game, the odds are 3 to 1 in your favor regardless of what the randomly selected sample shows.

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 Posted: August 22, 2011, 5:17 pm - IP Logged

"Lets say again that heads has come up 10 times in a row.  Some would bet tails because it is due while others would bet heads because it is hot and the one who guessed correctly would feel they predicted the outcome even though the next flip is pure chance unless the flip is biased."

A simple formula like A + B = C has no meaning unless we know the value of either A, B, or C. If A = 1 we know if B = 2, than C must = 3 and if C = 4, B must = 3. Before the next three coin flips, we know the outcome will be 3 heads, 0 tails, 2 heads, 1 tails, 1 head, 2 tail, 0 heads, or 3 tails. With no other information we have a 1 in 8 chance of guessing correctly. If the last three results were 3 heads and we know each side has a 50/50 chance of happening, shouldn't we logically assume one of the next three flips will be tails and eliminate 2 of the 8 chances (3 heads and 0 tails) and now have a 1 in 6 chance?

These pro and con coin flip analogies are very problematic because you're saying each side had a 50/50 chance by giving an example showing one side had a zero chance in ten flips. In your example you gave reasons why each player made their choice and both were educated guesses made from known information. With no information they could flip another coin to determine which side to choose but that wouldn't be an educated guess.  We can use math to determine the probably of streaks of three or more consecutive events occurring when the odds are 50/50 or we can look at past results and find a similar number. But either way, we still need the results of at least one outcome before we can make our bet.

Lotteries don't pay even money on 50/50 propositions so that makes the coin flipping analogy even worse. If you give a lottery clerk \$250 betting the first digit in the pick-3 game will be even and win, you'll get your \$250 back and win nothing. Break even or lose is not a sound bet. In pick-4 games there is a 43.2% chance of the winning number having just two matching digits and we can get a 11.1% profit, but the edge is still 8.8%.

"The lottery is quite different because it is taking a randomly selected sample."

The results are randomly selected samples but after more than half the numbers are drawn the odds favor numbers repeating. If you pick 750 numbers and I pick 250 in any pick-3 game, the odds are 3 to 1 in your favor regardless of what the randomly selected sample shows.

Hi Stack47,

I feel my postion and approach is clear enough that it needs no further explination. I feel no need to convince you or convert you to my way of thinking. Whatever mathematical view or approach you take is fine with me and will be respected. I hope it works for you. *S*

Since I don't play the pick 3/4 games, your guess is probably better than mine in that arena.

You are a slave to the choices you have made.  jk

Even a blind squirrel will occasioanlly find an acorn.

bgonÃ§alves
Brasil
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 Posted: August 25, 2011, 3:55 pm - IP Logged
Hello, jking, as this going the theory? Just use this theory. 70% of cold numbers and 30% of hot that more leave, example = you take of a lottery 1/39, separates
The numbers that come out of the 30% 01 39, clear the remaining 70% are cold
Of these 30% separate them 30% of the numbers to mount your bet, the others are at 70% cold, the base or reference is escolher30% of 30% taken from 01 to 39 seems to work! Then separate 13 or 14 numbers that more leave 30% of the numbers 4, 14 = 2 choices of 5 to 14 of the 25 remaining numbers but each choice 4 or 5, and mount a game with 10 numbers and unfold with wheels or filters,
Now to choose 30% of 30% that is the question, filters! And choose 4 to 5 of 70%
Filters!!

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 Posted: August 26, 2011, 12:33 am - IP Logged
Hello, jking, as this going the theory? Just use this theory. 70% of cold numbers and 30% of hot that more leave, example = you take of a lottery 1/39, separates
The numbers that come out of the 30% 01 39, clear the remaining 70% are cold
Of these 30% separate them 30% of the numbers to mount your bet, the others are at 70% cold, the base or reference is escolher30% of 30% taken from 01 to 39 seems to work! Then separate 13 or 14 numbers that more leave 30% of the numbers 4, 14 = 2 choices of 5 to 14 of the 25 remaining numbers but each choice 4 or 5, and mount a game with 10 numbers and unfold with wheels or filters,
Now to choose 30% of 30% that is the question, filters! And choose 4 to 5 of 70%
Filters!!

Hi dr san,

Let me give you a different look of what I think you are saying:

 % of Top Amount of The Amount of Historical Hits Occurring Numbers Last 550 games (8/24/20110) CA Fantasy 5 Payoff % Num. by % that fit % 0 1 2 3 4 5 3 or more Jackpot % 5 2 377 162 11 0 0 0 0.00 0.00 10 3 313 206 29 2 0 0 0.36 0.00 15 4 264 226 53 7 0 0 1.27 0.00 20 10 80 201 196 64 9 0 13.27 0.00 25 21 7 40 137 193 139 34 66.55 6.18 30 28 0 5 44 132 239 130 91.09 23.64 35 34 0 0 3 36 193 318 99.45 57.82 40 36 0 0 0 12 128 410 100.00 74.55 45 37 0 0 0 3 91 456 100.00 82.91 50 39 0 0 0 0 0 550 100.00 100.00 . % of Top To achieve a payoff Occurring 3 or more (approx.) Num. by % hot cold 25 3 2 30 4 1 35 5 0 40 5 0 45 5 0 50 5 0 . % of Top To achieve a jackpot Occurring All Numbers (approx.) Num. by % hot cold 25 0 5 30 1 4 35 3 2 40 3 2 45 4 1 50 5 0

You are a slave to the choices you have made.  jk

Even a blind squirrel will occasioanlly find an acorn.

United States
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July 13, 2004
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 Posted: August 26, 2011, 12:55 am - IP Logged

Hi dr san,

Let me give you a different look of what I think you are saying:

 % of Top Amount of The Amount of Historical Hits Occurring Numbers Last 550 games (8/24/20110) CA Fantasy 5 Payoff % Num. by % that fit % 0 1 2 3 4 5 3 or more Jackpot % 5 2 377 162 11 0 0 0 0.00 0.00 10 3 313 206 29 2 0 0 0.36 0.00 15 4 264 226 53 7 0 0 1.27 0.00 20 10 80 201 196 64 9 0 13.27 0.00 25 21 7 40 137 193 139 34 66.55 6.18 30 28 0 5 44 132 239 130 91.09 23.64 35 34 0 0 3 36 193 318 99.45 57.82 40 36 0 0 0 12 128 410 100.00 74.55 45 37 0 0 0 3 91 456 100.00 82.91 50 39 0 0 0 0 0 550 100.00 100.00 . % of Top To achieve a payoff Occurring 3 or more (approx.) Num. by % hot cold 25 3 2 30 4 1 35 5 0 40 5 0 45 5 0 50 5 0 . % of Top To achieve a jackpot Occurring All Numbers (approx.) Num. by % hot cold 25 0 5 30 1 4 35 3 2 40 3 2 45 4 1 50 5 0

Hi,

As a follow up to the above post...

The amount of hot/cold numbers used should vary by the rating used in column #1. The wonderful thing about using history data is that it lets you know the level of certainty of what your idea is.

Again, thanks dr san. Even though we are on different paths for solving the lottery, your comments make me take a fresh/new/different look at how numbers can be analyzed.

You are a slave to the choices you have made.  jk

Even a blind squirrel will occasioanlly find an acorn.

United States
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July 13, 2004
1184 Posts
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 Posted: August 26, 2011, 2:12 am - IP Logged
 % of Top Amount of The Amount of Historical Hits Occurring Numbers Last 550 games (8/24/2011) CA Fantasy 5 Payoff % Number that fit % 0 1 2 3 4 5 3 or more Jackpot % 5 2 377 162 11 0 0 0 0.00 0.00 10 3 313 206 29 2 0 0 0.36 0.00 15 4 264 226 53 7 0 0 1.27 0.00 20 10 80 201 196 64 9 0 13.27 0.00 25 21 7 40 137 193 139 34 66.55 6.18 30 28 0 5 44 132 239 130 91.09 23.64 35 34 0 0 3 36 193 318 99.45 57.82 40 36 0 0 0 12 128 410 100.00 74.55 45 37 0 0 0 3 91 456 100.00 82.91 50 39 0 0 0 0 0 550 100.00 100.00 . % of Top To achieve a payoff Occurring 3 or more (approx.) Number hot cold 25 3 2 30 4 1 132+239+130=501…(501/550)*100=91%..~ 4 hot 35 5 0 40 5 0 45 5 0 50 5 0 . % of Top To achieve a jackpot Occurring All Numbers (approx.) Number hot cold comb 25 0 5 20k 30 1 4 142k (130/550)*100=23%..~ 1 hot 35 3 2 278k 40 3 2 376k 45 4 1 435k 50 5 0 575k

Hi,

I amended the above table to help with the clarity....

You are a slave to the choices you have made.  jk

Even a blind squirrel will occasioanlly find an acorn.

bgonÃ§alves
Brasil
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June 9, 2010
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 Posted: August 26, 2011, 3:59 am - IP Logged

hello. jking, very good work, this picture paints a thousand words, the great problem of translation from google, mixing the word takes focus images (in my case), this principle   70% 30% changes with the size of the lottery for it interferes with repetitions, a lottery 39 / 5 reps frequently have more than a lottery 49 / 6, it must adapt to each lottery   Utilizing this condition tends to be valued at more than playing in a sweepstakes (About 30 drawings) in a single event (a single draw) is almost impossible!. the numbers   What else should leave when analyzed separately, not in groups now it contributes   (5 in the lottery 39 / 5), seem to repeat in the range of 70% greater numbers of combinations, the law of averages, ie the fewer numbers come out more often in larger numbers of combinations in itself, does not seem a contradiction! !

bgonÃ§alves
Brasil
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 Posted: September 3, 2011, 10:57 am - IP Logged

jking, 20% of numbers are drawn 80% of the time. This theory confirms the need to analyze the frequency of numbers, that is, you should always bet on randomly selected 20% more!

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