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Vaddis Holy Grail

Started by RFMAXX, Aug 20, 03:35 AM 2015

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0 Members and 51 Guests are viewing this topic.

nottophammer

Good to see ya posting IG
Now those 40 spins.
in spins 1-10 average says 1 repeat, the spins 11-40, so 30 spins average shows repeats are 3,5,7; so 15 repeats and 15 singles.
But BBB says it's only average. Then go to spin 60 and on average it's 30 repeats and 30 non-hit. So peeps should look to 3,5,7
How do you win at roulette, simple, make the right decision

ignatus

Quote from: nottophammer on Mar 06, 07:54 PM 2019
Good to see ya posting IG
Now those 40 spins.
in spins 1-10 average says 1 repeat, the spins 11-40, so 30 spins average shows repeats are 3,5,7; so 15 repeats and 15 singles.
But BBB says it's only average. Then go to spin 60 and on average it's 30 repeats and 30 non-hit. So peeps should look to 3,5,7

Alright,  :) thx notto!  O0
If you like to donate link::[url="//paypal.me/ignatus1"]//paypal.me/ignatus1[/url]

"Focus on predicting wheel sectors where the ball is expected to land" ~Steve

RiseAgainst

Hi Ignatius thanks for that rxcoded repeater system!
I tried to RX a similar system by BBB posted on page 149. He played 4 numbers plus the - and + neighbors on the table.. And the table neighbors where which I got stuck.. Does RX has a way to do this.
I only get across wheel neighbors which are defined by span function.
Thanks

Herby

Quote from: PassionRuleta on Mar 05, 05:25 PM 2019Does not it make you think to start with 4 numbers?
If 2 comes out, he says he plays 2 and 3.
How to predict future results after 4 turns?

If you have 4 different numbers the probability for any repeater in the next 4 numbers is 40.7 %.
The probability for any repeater after that is 10.17 %.

"If 2 comes out, he says he plays 2 and 3."   ???
There is no natural connection between the numbers ( in roulette) , so you could make any artificial connection you like.


Frodo

What IF?

What if we don't need Vaddi. What if we don't need his/her clues?
What if we look at the table and see the rounded up integers:
37/4=9.25 4 spins spaces of max 10
37/8=4.62 8 spaces of max 5
37/12=3.08 12 spaces of max 4
37:24= 1.54 24 spaces of max 2
37:37=1 37 surprise surprise!

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

First 4 numbers
9
18
27
36

How convenient!
What if all we need is to look at the spaces between the numbers that hit.

What IF you bet the numbers above expectation(2+hits) and the spaces around them a/p the above matrix?

Where will you be at the end of the 37 spin cycle?:)


Herby

Quote from: Frodo on Mar 07, 05:08 AM 2019What if we look at the table and see the rounded up integers:
Good idea, but the topic here is ...

Frodo

Quote from: Herby on Mar 07, 05:14 AM 2019
Good idea, but the topic here is ...

To persevere. To be relentless. To think outside the box. To never give up..

Herby

Quote from: Frodo on Mar 07, 05:08 AM 2019What IF you bet the numbers above expectation(2+hits) and the spaces around them a/p the above matrix?

I dont't get what you mean with "numbers above expectation". Can you give an example please.
I already programmed your idea ...


Frodo

Quote from: Herby on Mar 07, 05:22 AM 2019
I dont't get what you mean with "numbers above expectation". Can you give an example please.
I already programmed your idea ...
In a 37 spin cycle, 1 hit= expected value for the individual numbers. Anything above that 2, 3,, 4.... In the same cycle. 2+.

Frodo

Here's a sample of 37 spins cycle from the mighty random.org

link:s://:.random.org/integers/?num=37&min=0&max=36&col=1&base=10&format=html&rnd=new

20
23
11
24
3
4
26
27
31
22
6
14
10
8
7
16
6
20
23
11
1
11
30
12
26
24
23
4
30
7
22
1
13
27
0
17
4

All numbers in this cycle are at least at the expected value(1hit).
Those above it(repeaters):#20(2 hits) , #23(3 hits) etc are ABOVE expectation.

Herby

Thanks Frodo,
I got it.
You had 2 different dimensions (I think) in your post above. This was 1 dimension too much for me. :xd:

Frodo

No worries
Imagine a hypercube.37 at the power of 37 :)
Fact: you will never ever ever ever see a 37 number cycles without a repeat. Never. Ever!
Fact: you will never ever ever ever see a 37 cycle with 13 continuous spaces between 2 hit numbers. Never ever ever ever. Never ever!
Fact: you will nevrer ever ever see a number repeat 37 times in a 37 spin cycle. Never ever!
Facts
-------------------------------
1 million cycles of 37 spins
2 hit 100%
3 hit 98.405% times
4 hit 50.990% times
5 hit 10.939% times
6 hit 1.599% times
7 hit 0.192% times

-------------------------------
1 million cycles of 74 spins
3 hit 100%
4 hit 99.982% times
5 hit 90.1054% times
6 hit 45.47% times
7 hit 13.78% times
8 hit 3.22% times
9 hit 0.637% times

--------------------------------
1 million cycles of 111 spins
4 hit 100%
5 hit 99.9998% times
6 hit 98.277% times
7 hit 73.8% times
8 hit 34.35% times
9 hit 11.587% times
10 hit 3.301% times
11 hit 0.812% times

---------------------------------
1 million cycles of 148 spins
5 hit 100%
6 hit 100%
7 hit 99.693% times
8 hit 88.718% times
9 hit 55.022% times
10 hit 24.392% times
11 hit 8.699% times
12 hit 2.753% times
13 hit 0.7899% times

----------------------------------
1 million cycles of 185 spins

7 hit 100%
8 hit 99.938% times
9 hit 95.324% times
10 hit 71.416% times
11 hit 38.747% times
12 hit 16.700% times
13 hit 6.194% times
14 hit 2.090% times
15 hit 0.637% times


nottophammer

I don't know about millions of cycles. But this one cycle performs perfect.
Generals bucket of KFC ( KTF ) +55.
Vaddi spins 1-4 no repeat. So bet last 4 singles 3,24,11,23 and their mat pairing; win
How do you win at roulette, simple, make the right decision

Herby

Quote from: Frodo on Mar 07, 05:52 AM 2019Imagine a hypercube.37 at the power of 37 :)

37^37 = 10555134955777783414078330085995832946127396083370199442517    :sad2:

= 1.05551*10^58

Calculation can be done.
Imagination ... I'm working on it :wink:


Possibilities for 37 different outcomes in 37 spins:
37 ! = 13763753091226345046315979581580902400000000 = 1.37638*10^43   O0

Herby

Quote from: Herby on Mar 07, 06:10 AM 2019
37^37 = 10555134955777783414078330085995832946127396083370199442517    :sad2:

= 1.05551*10^58

Calculation can be done.
Imagination ... I'm working on it :wink:


Possibilities for 37 different outcomes in 37 spins:
37 ! = 13763753091226345046315979581580902400000000 = 1.37638*10^43   O0

1.37638*10^43 / 1.05551*10^58 = size(nucleus) / size(atom)

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