ORLANDO, FLORIDA United States Member #4924 June 3, 2004 5896 Posts Offline

Posted: August 20, 2011, 6:23 am - IP Logged

Quote: Originally posted by jimmy4164 on August 20, 2011

(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.

Kentucky United States Member #32652 February 14, 2006 7302 Posts Offline

Posted: August 20, 2011, 5:34 pm - IP Logged

Quote: Originally posted by JKING on August 20, 2011

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 numbersdo not havememory, butoftenholdpositionsof certainoutputCADoutput result, the resultsare the problem, the solution shouldbe within thesame problem!Much ofmathematics andotheraletoriatogether!exampleGame49 / 6result12,13,24,32,44,47Mathematics=12,13,44,47Random=24.32Notnecessarily inorder ofnumbers

bgonÃ§alves Brasil Member #92564 June 9, 2010 2122 Posts Offline

Posted: August 21, 2011, 11:39 am - IP Logged

Hello,rlRANDONIC,Fantastico,on thetableheadingsRl-randonic, the comparisonon thelaunch of thecurrency,thecrownfaceThe problemis that thesinglecurrency andlanchamento(onecurrency only)suchasthe lottery49 / 6We havesixevents(sixpairings oftwo-digitnumbers witheachnumber), thenAconbinaçaorelease(6)in a single event, whilethe currencyhas onlyone,Aexemplocurrencyonly servestomakeeachdigit andalso theanalysisofthe verticalcolumnSeparatethe digitnumberand the terminationoffrentedoverticallythis point we haveto understand the difference,manycombinedevents(draw)and currency

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......

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.

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.

United States Member #5599 July 13, 2004 1184 Posts Offline

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

Quote: Originally posted by Stack47 on August 22, 2011

"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 Member #92564 June 9, 2010 2122 Posts Offline

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%

United States Member #5599 July 13, 2004 1184 Posts Offline

Posted: August 26, 2011, 12:33 am - IP Logged

Quote: Originally posted by dr san on August 25, 2011

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 Member #5599 July 13, 2004 1184 Posts Offline

Posted: August 26, 2011, 12:55 am - IP Logged

Quote: Originally posted by JKING on August 26, 2011

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.

bgonÃ§alves Brasil Member #92564 June 9, 2010 2122 Posts Offline

Posted: August 26, 2011, 3:59 am - IP Logged

hello.jking, very goodwork, thispicture paints athousand words,the great problem oftranslation fromgoogle,mixingthewordtakesfocusimages(in my case),this principle70%30%changes withthe sizeof the lotteryfor itinterferes withrepetitions, a lottery39 / 5repsfrequentlyhave morethan a lottery49 / 6, it must adapttoeach lotteryUtilizingthis conditiontends to bevalued atmore thanplaying inasweepstakes(About 30drawings) in asingle event (a singledraw)isalmost impossible!.the numbersWhat elseshouldleavewhen analyzedseparately, notin groupsnowit contributes(5in the lottery39 / 5),seem to repeatin the rangeof70%greaternumbersofcombinations,the law of averages, ie thefewernumberscome outmore ofteninlarger numbersof combinationsin itself,does not seemacontradiction!!