At what level of information does chaos appear?

Reply #25

Sunglasses — Tue, 20 Jun 2017 22:59:33 GMT

Red, black or zero.

18, 18 or 1 of 37

37 chances

It can be 0 all the time. 0 is not an even number, review your math!

In any case, the payout is better. Cover 18 of 37 chances and double your money.

Reply #24

RJOh — Tue, 20 Jun 2017 22:53:06 GMT

Are you saying flipping a coin can't have a true random result since there are only two possible?

Pay outs for games are designed such that they seldom pay enough to cover all the possible outcomes.

Reply #23

JADELottery — Thu, 25 May 2017 16:25:49 GMT

This is very similar to card counting in Blackjack.

Groups of cards a given a values of -1, 0, +1.

By keeping track of the running total, it's possible to tell when the deck is getting hot for the player or not.

Betting a high when the deck is hot and other strategies is what makes counting cards so useful.

Some of the same can be done here with random selection.

In a random setting, the selections are going to be some order and some chaos.

If one or the other happens more, eve... [ More ]

Reply #22

JADELottery — Thu, 25 May 2017 16:11:03 GMT

If know the running total tends to 0, then based on the last running value of -3, it would goto -2.

To make it goto -2, the bit state would have to be a +1.

Looking at the states available we can see which states would likely fit the possible outcome.

Here we'll transpose the bit states to make it easier to read and highlight the +1's.

0

0

0

0

0

0

0

0

1

1

1

1

1

1

1

1

0

0

0

0

1

1

1

1... [ More ]

Reply #21

JADELottery — Thu, 25 May 2017 15:58:24 GMT

We can add some headers and a running total of the 4 Bit State.

As more samples are added, the running total should tend to gravitate toward 0.

Bit

Stream

4 Bit State

Running

Total

0

1

1

0

-1

-1

1

+1

0

0

-1

-1

0

+1

0

0

+1

+1

1

+1

+2

1

-1

+1

0

-1

0

1

+1

+1

0

Reply #20

JADELottery — Thu, 25 May 2017 15:44:36 GMT

We can transpose the data to make it easier to work with the next process.

0

1

1

0

-1

1

+1

0

-1

0

+1

0

+1

1

+1

1

-1

0

-1

1

+1

0

-1

1

-1

0

-1

1

-1

0

-1

1

-1

1

+1

1

+1

0

+1

0

-1

1

-1

0

+1

Reply #19

JADELottery — Thu, 25 May 2017 15:43:03 GMT

Using the previous example, we can substitute the letters.

0

1

1

0

1

0

0

0

1

1

0

1

0

1

0

1

0

1

1

1

0

0

1

0

1

1

1

0

0

1

1

1

N

Y

N

Y

Y

Y

N

N

Y

N

N

N

N

N

N

Y

Y

Y

N

N

Y

N

Y

Y

Y

N

N

N

Y

This becomes:

0

1

1

0

Reply #18

JADELottery — Thu, 25 May 2017 15:33:33 GMT

Let's reassign the letters with numbers.

Since a order is a lower state and chaos is a higher state, the letter become the following.

Something to notice is order and chaos are equal in frequency and it's symmetrical in form.

0

0

0

0

N

-1

0

0

0

1

Y

+1

0

0

1

0

Y

+1

0

0

1

1

N

-1

0

1

0

0

Y

+1

0

1

0

1

N

-1

0

1... [ More ]

Reply #17

RL-RANDOMLOGIC — Sat, 13 May 2017 10:14:28 GMT

RJ

You left off the word True in your reply. True random does not exist. Any game for which the total

combinations can be calculated, it's possible to buy it out. I said that I doubt playing that way would

make a profit.

RL

Reply #16

RJOh — Fri, 05 May 2017 16:48:01 GMT

I can win any standard pick n of n lottery any time I choose.

Are you saying When you play and don't win anything, it's your choice or You have to play so many combinations to win even though random doesn't exists it might as well ?

If random doesn't exist then winning combinations must exits within a subset of parameters other than those covering all the possible combinations and only picking combinations within them should improve ones odds of winning, it's just matter of defining them... [ More ]

Reply #15

Novan60 — Sun, 23 Apr 2017 08:40:37 GMT

Actually I went as far as to express them as bits, but in the end regardless of how the numbers are expressed or summed the math works out to be the same.

I have finally settled on a method that picks numbers based on choosing the correct conditions from a limited number of options until I have narrowed down the possibilities to one set of numbers. It is combination of multiple methods that I have explored.

Reply #14

Soledad — Sun, 23 Apr 2017 03:59:55 GMT

Interesting reply. Thoughts I've certainly had when playing the lottery. Crazy to think about.

Reply #13

Datapile — Sun, 23 Apr 2017 03:27:59 GMT

Novan, Have you tried expressing all 80 numbers (drawn or not) as a string of hexidecimal numbers

Reply #12

Datapile — Sun, 23 Apr 2017 02:40:38 GMT

Jade, perhaps chaos is simply data without perspective, and converting data into information is the process of giving it that perspective

Reply #11

Novan60 — Tue, 18 Apr 2017 22:42:35 GMT

As of late I have been bumping up against this.

In a Keno pick 10 out of 80 game I have been able to develop a wide variety of strategies that will allow me to accurately predict 4 of 4 specific digits that will appear within the next 7 drawings. My odds of getting the 5th digit right drop down to 56% and continue to decrease exponentially with every additional digit resulting in fraction of a percent for the last digit.

Similarly, if play a strategy where I pick the integer value for all

Reply #10

JADELottery — Fri, 07 Apr 2017 13:18:02 GMT

We point out in the first post that on a very basic level that chaos appears at 5 bits of information.

At that level, chaos over takes the order.

Knowing this, if we look at 4 bits of information, we can see order and chaos are balanced at 8 a piece.

Looking at one of the original streams of {0, 1}'s, 01101000110101010111001011100111, we can examine each at the 4 bit level.

The stream is continuous and starts left and ends right, .

We examine each 4 bit segment at each 4 bits of i... [ More ]

Reply #9

dr san — Sun, 02 Apr 2017 20:32:30 GMT

Hello, where are the variants ?, of the repetition and absence of digits and patterns of the last

Or several draws, followed, then we have repetition absence and delays and cycles as a reference point (the last draw or multiple draws

Reply #8

doodoohead — Sun, 02 Apr 2017 15:16:30 GMT

At what level of information does chaos appear?

Chaos appears at the point information becomes unpredictable or without form.

Reply #7

RL-RANDOMLOGIC — Thu, 30 Mar 2017 20:33:48 GMT

True random does not exist, I can win any standard pick n of n lottery any time I choose. Can I make a profit

playing that way? I doubt it. Any game for which the odds can be calculated for any one outcome is not truly

random. Perception is the art of discovery. Chaos IMHO is perfection in motion, moving so fast it remains outside

our grasp.

RL

Reply #6

Sunglasses — Tue, 28 Feb 2017 23:21:31 GMT

I usually find it funny when people lose money when 111111111111111 . Even more amusing is when people tell you that it isn't random. Spend a night at the casino betting only red or black. Tell me how much you won with it.

Reply #5

Novan60 — Tue, 28 Feb 2017 23:06:16 GMT

if you want to look at short term possibilities you are better off looking at what the potential change is from the baseline. For instance if the set was 00 how many different results are there where one o the new results sets would have only one 1? The answer is 2 and they have the highest probably of appearing thereby narrowing the choice down to two possibilities instead of 4.

Reply #4

JADELottery — Tue, 28 Feb 2017 20:34:19 GMT

You're over thinking this by operating in the continuum, the level we're at for this topic is the instantaneous moments or the instum; as we coined it in one of our previous topics, Chaos-Order Duality.

We know the single bit can change over continuous selections, however, for that instantaneous moment, it does not change.

Many selections result in a string of 1's and 0's; what's the least set of 1's and 0's in that string, that can be reasonably appear to be constant or oscillating in a p... [ More ]

Reply #3

Novan60 — Tue, 28 Feb 2017 20:08:36 GMT

Then I guess I do not get the point you were trying to make because if one only looks at a small subset of data it is impossible to use that as an indicator as 00 could have been the first results. By stating that the first set is 0 and 1 you are no longer talking about random probability.

Reply #2

JADELottery — Tue, 28 Feb 2017 20:02:05 GMT

.

Now, we realize that repeated selection may yield a 1 or a 0, but that's not the point since the beginning group of 1's and 0's could have been selected in sequence or virtually all at once.

.

Reply #1

Novan60 — Tue, 28 Feb 2017 19:52:06 GMT

There is a flaw in your logic, 0 and 1 can't both equal randomness. One has to equate to yes and the other has to be no. In order to prove your theory you would have to flip a coin 5 times at least 1000 times in order to compile a result set that had any meaningful data.

Basically in flipping a coin 5 times there are 32 possible outcomes. The majority of those outcomes will result in 3 of the flips even splitting 1 and 0 in various combinations. The first 5-10 times may possible result in all... [ More ]

At what level of information does chaos appear?

JADELottery — Tue, 28 Feb 2017 18:33:45 GMT

Seems like a simple question.

To keep it simple, we'll use the bit, {0, 1}.

We've probably seen it before.

A string of 1's and 0's.

Like: 01101000110101010111001011100111 or 11011110011000110000110000010111 .

Seems random.

If we start zooming in on a section of bits, we start to see the randomness slowly disappear.

Looking at the first half of each, it appears as: 0110100011010101 or 1101111001100011 .

A little less random.

Now, half that: 01... [ More ]