Chaos Theory and Roulette - Raindrops and Sleepers.

[I have cookies]

Every trail is independent as the wheel has 37 degree of freedom.
But all numbers wont have one show each - it can happen - but I assume not during my life time.

Like raindrops - falling at the same spot witch occurs before the whole region or ground of pockets get wet,
Well I just find this amusing that every wheel is a victim due towards fluctuation and bias.

It does not matter how we twist things - not for me - but I don't know about you.
One thing that I find amusing is how nothing is due to happen.

The last and most previous feels more attractive then chasing sleepers.
I never understood does who believe that the raindrops has to hit does empty pockets.

/.\

[I have cookies]

When I think of it - its imbalance that is more attractive.

My mind strike what Wizard of odds mention - if I remember it correct - same as Caleb also known as Snowman.
That if some one would play numbers they would be better of use the last five.

It boils down to if we quote Caleb that it does to certain degree reduce the house edge - but it still remain negative.
What is the conclusion.

First it does feels appealing and like common sense that the most previous matters and that past result to certain degree have a certain effect on future outcomes.
Then it boils down to that it should be better to follow certain events then play against them - feel free to correct me if I am wrong.

To illustrate this using even money bets.

I read that Frank Barstow mention the 5 count.
Play black three to four times after 5 reds or the other way around using 5 black.
Also know during observation that during 9 trails there is a very high probability for a change or that a series of 9 appears once in 500 trails - still nothing is due to happen.

But due to what I mention above it should be better to take this example 5 previous red numbers - with no repeat -and play red.
Just to take the even money bets that is a 50/50 situation - is there a confilict here?

See attach file.

Note.
It looks like the frequencies of attack would be beast using three or four attempts if some one would use the example above and that follow the last 5 would be a much better alternative then playing against them.

[I have cookies]

As I see it one could divide and use 18 numbers using the wheel layout or the even money bets position as one extra parameter - I can also think of other parameters - but I assume I get back regarding that.

Just run for fun some short sampels.
13 attacks where 8 include one repeat and all won.

1
5
4
7
8 Previos 5 Low

3 Won
22
35

- - -

20
29
33
31
26 Previos 5 High

7  Loss
32 Won
26 Repeat

- - -

28
19
34
24
25 Previos 5 High

32 Won
1
24 Repeat

- - -

26
28
31
27
23 Previos 5 High

20 Won
25
26 Repeat

- - -

35
20
29
32
23 Previos 5 High

3  Loss
27 Won
17

- - -

28
33
20
35
29 Previos 5 High

30 Won
1
15

- - -

9
11
1
7
18 Previos 5 Low

11 Won Repeat
9  Repeat
9  Repeat

- - -

30
24
26
21
27 Previos 5 High

28 Won
25
32

- - -

22
29
30
19
25 Previos 5 High

11 Loss
14 Loss
21 Won

- - -

24
28
21
22
31 Previos 5 High

0  Loss
36 Won
32

- - -

22
27
20
25
29 Previos 5 High

27 Won Repeat
11
23

- - -

24
23
33
22
31 Previos 5 High

29 Won
36
35

- - -

10
17
12
5
18 Previos 5 Low

17 Won Repeat
20
31


/.\

[I have cookies]

I assume with out testing that some one could use previous 5 using 5 x 3 numbers = 15 numbers or use previous 6 and get 18 numbers.
That way the repeat or the previous neighbors would have a hit - using the wheel layout - that is also one parameter among others.

[toby]

Systems on hot/punter numbers have been proven to sink.

You don´t know when to stop playing the punters and you lose when it becomes a sleeper.

[I have cookies]

Systems on hot/punter numbers have been proven to sink.

You don't know when to stop playing the punters and you lose when it becomes a sleeper.

Thanks for making that clear - this is not a post about roulette systems or hot numbers - even if it might sound like it - but you are correct that its about last and most previous numbers being equal to hot numbers if that is the word you prefer.

The main topic here is that there exist extra parameters in many variations that is in conflict with average common probability - is like why would some one ride one horse when he can use two or three parameters.

I believe this increase the overall hit ratio and also delay certan events to become due.
I believe as i state above with my example that is better to follow then play against.

So the question is - does it change probability to use two or three parameters if you chose to follow or play against.

[I have cookies]

 
/.\

[toby]

Suppose you take the law of the thirds and play according of what it says on strait bet, dozens and EC. In addition to that, you play a sleeping sector or layout group that is -3 or -4 standard deviations.

Chances are in your favour(it looks like) but you will not win long term.

Trying to beat the wheel mathematically is a waste of time.

Nobody has never proved that randomness can be read(yet).

To start you have to predict hits, for instance, that go over +4 SDb as Laurance said last year at GG.

Most of us have been in roulette for years(25) and tested most of the known systems and procedures to try to achieve what millions state is imposible.

Some people have achieve the goal.

br

[I have cookies]

Yes true - but i will still continue to play around with different parameters - have a nice day Toby.

[I have cookies]

What would be the option to use three parameters - it feels like its above me - but it does not prevent me from giving it a try.

This is when we use the wheels layout.
Experimenting with the 5 count.

Parameter 1. Bi-modal effect 18 pockets.
Parameter 2. Neighbors using sectors 0 to 11.
Parameter 3. 5 clear numbers/sectors with no repeat.

Note.
First is not 100% floating using each individual number as we will deal with sectors.

Sectors of three using the wheel layout - where zero comes with first sector - so when you observe and play even money bets you will at some times play 19 numbers and not 18 numbers.

0. 26 0 32 15
1. 19 4 21
2. 2 25 17
3. 34 6 27
4. 13 36 11
5. 30 8 23
6. 10 5 24
7. 16 33 1
8. 20 14 31
9. 9 22 18
10. 29 7 28
11. 12 35 3

Now you use and divide the sectors into four areas using the bi-modal effect - the opposite sector and its neighbors.

The crossovers or the bi-modal sectors are as follows.

0-6
1-7
2-8
3-9
4-10
5-11

So if you have sector 1 hitting then sectors 0 1 2 and 6 7 8 is your 19 numbers.
The other 18 numbers is the opposite bi-modal areas.

Illustration.

[albertojonas]

I can also think of other parameters - but I assume I get back regarding that.

may you evolve on this?

[albertojonas]

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