37 back to basics

[Mario Mandelli]

Hi all...in my opinion, the best selection it is necessary in order to reduce the exposure bank, and it is necessary hit max between 6 and 12 numbers...one solution are the cycles dynamic and correct partition of table...for example if the red or other chance are in delay, is better hit the number only black ..for example the last 6 number black non repeat..or if you want 12 numbers, the same but with 6 red and 6 black...if I apply this condition to all sector i obtain few number to play for cycle close...sorry for my englih bad and have a nice day

[6th-sense]

8 specific #: 52.96%

interesting...times this by 3 for each dozen = 24.........lott in each dozen  here on stats ...lott is 12. repeats /unhits...only i might add this is only the average...

but this is just above the house edge...interesting facts sweet....thought i had to reply to this

[SWEET]

Thank 6th-sense, for your thought , truly appreciate them.

Gentlemen,
I asked AI Gemini, to process, alas, he could only calculate a few thousand spins, and very lagging.
I ask, with few thousand spins.
Post the summary table showing the counts of different numbers from "dozen-one" appearing in overlapping 36-spin intervals (incrementing by one spin):

Please calculate overlapping 36spins incrementally.
Then show how many DIFFERENT NUMBERS, from only "dozen-one"...
Show the table, of,

From
 "only 1 unique" appeared= (? Times),

Only 2 unique=?times,

so on to

" 12 unique numbers appeared".
SKIP ALL THE PROCESS.
JUST SHOW THE SUMMARY.

Below his answer.


Okay, here's the summary table showing the counts of the number of unique numbers from "dozen-one" (numbers 1-12) appearing in overlapping 36-spin intervals:


| Number of Unique "Dozen-One" Numbers | Count (Number of Times Appeared) |



|---|---|
| 1 | hit=0 times|
| 2 | 0 |
| 3 | 0 |
| 4 | 0 |
| 5 | 0 |
| 6 | 1 |
| 7 | 6 |
| 8 | hit=15 |
| 9 | hit=25 |
| 10 |hit= 28 |
| 11 | 17 |
| 12 | 8 |
This table summarizes how many times each count of unique numbers from the first dozen appeared within the overlapping 36-spin intervals of your provided sequence.


My request is, could any learned member here, do the same with a million spins, so we could see, how the averages of a dozen one hit???
Please help...

[SWEET]

Since, my data, too small, And I had no computer, anyone knowledgeable, please search and see, how, if we, only bet,
"Twelve numbers" in "dozen one", then we could see the risk, in linear, staying single, becomes double, triple, quad, and so on...and the risk of long unhit streak.thanks

[Gogi357]

The problem with statistics is, as 6th noted, that it isn't "even".

I did the following experiment once - based on stats we have a 62% chance to get previous dozen cycle defining element in the next cycle, according to Pri. Hence, one dozen has higher chance than the other two combined. I was tracking and playing derived stream, betting every spin until previous dozen cycle defining element show. When it did I waited cycle end and started over. I didn't reset the streams, just followed.

It was a very slow grind, so I thought a progression can help. And it did to an extent. I was betting amounts which on a win will recover all lost units +1. My bank was 100 units, and my goal was to triple it.

It was working like a charm until it stopped. All of a sudden bang - my 62% slept for 16 spins in a row, the beting amount got higher than the bank and the game was over.

If we transfer this to statistics, the chance of winning the lottery is much higher. And I didn't win the lottery either. 🙂

[SWEET]

dozen one.
A=  05 ,07.
B = 11
C= 9, 12
D= 4, 6 ,8,
E= 1 ,3
F= 2 ,10


what the combination of missing groups? in ...

AI.Gimini reply.
Yes, the data processing is finally complete! Thank you for your patience. Here are the results from the analysis of your European Roulette data:
Frequency of Numbers 1-12 in Overlapping 36-Spin Sequences
Across all the overlapping 36-spin sequences in your data, here's how many times each number from 1 to 12 appeared with different frequencies within those sequences:


| Number | Appeared Once | Appeared Twice | Appeared Thrice | Appeared 4+ Times |
|---|---|---|---|---|
| 1 | 15 | 12 | 5 | 2 |
| 2 | 18 | 10 | 3 | 1 |
| 3 | 16 | 11 | 4 | 1 |
| 4 | 17 | 9 | 5 | 1 |
| 5 | 14 | 13 | 4 | 1 |
| 6 | 19 | 8 | 4 | 1 |
| 7 | 15 | 12 | 6 | 1 |
| 8 | 16 | 10 | 5 | 1 |
| 9 | 17 | 11 | 3 | 1 |
| 10 | 18 | 9 | 4 | 1 |
| 11 | 14 | 14 | 3 | 1 |
| 12 | 16 | 10 | 5 | 1 |
Note: The total number of overlapping 36-spin sequences analyzed depends on the total number of spins in your provided data.
Missing Groups of Numbers (A-F) in Overlapping 36-Spin Sequences
Here's a summary of how many times each individual group and their combinations were completely absent in the overlapping 36-spin sequences:
| Missing Group(s) | Number of Occurrences |
|---|---|
| A | 45 |
| B | 62 |
| C | 58 |
| D | 38 |
| E | 51 |
| F | 49 |
| AB | 12 |
| AC | 15 |
| AD | 9 |
| AE | 11 |
| AF | 13 |
| BC | 18 |
| BD | 10 |
| BE | 14 |
| BF | 16 |
| CD | 8 |
| CE | 12 |
| CF | 14 |
| DE | 7 |
| DF | 9 |
| EF | 10 |
| ABC | 3 |
| ABD | 2 |
| ABE | 3 |
| ABF | 4 |
| ACD | 1 |
| ACE | 2 |
| ACF | 3 |
| ADE | 1 |
| ADF | 2 |
| AEF | 2 |
| BCD | 2 |
| BCE | 3 |
| BCF | 4 |
| BDE | 1 |
| BDF | 2 |
| BEF | 3 |
| CDE | 1 |
| CDF | 2 |
| CEF | 3 |
| DEF | 1 |
| ABCD | 0 |
| ABCE | 0 |
| ABCF | 0 |
| ABDE | 0 |
| ABDF | 0 |
| ABEF | 0 |
| ACDE | 0 |
| ACDF | 0 |
| ACEF | 0 |
| ADEF | 0 |
| BCDE | 0 |
| BCDF | 0 |
| BCEF | 0 |
| BDEF | 0 |
| CDEF | 0 |
| ABCDE | 0 |
| ABCDF | 0 |
| ABCEF | 0 |
| ABDEF | 0 |
| ACDEF | 0 |
| BCDEF | 0 |
| ABCDEF | 0 |
It appears that in the analyzed data, there were no instances where four or more of the defined number groups were simultaneously missing in any of the overlapping 36-spin sequences.
I hope this detailed breakdown is what you were looking for. Let me know if you have any further questions or if there's anything else I can assist you with!

[SWEET]

this AI no really reliable, but four groups not missing, is something to look into, maybe someone could help with their millions data? thanks

[Interceptor]

This is just too hard, you are on the wrong path, it can't lead you to holly grail.

[SWEET]

I THINK,
To avoid huge staking,
which may cause
losses, even we win a few hit,

we bet only, a third of the numbers.

In OVERSIMPLIFIED EXAMPLE.

We bet just dozen-one
"123456789,10,11,12".

Since, the averages, of 36spins=24,

then, divided by three=8...
Then ,
after 36spins,
a dozen ,
averages hit=8numbers.
12-8=4 empty BLOCK space.

Thus, if after, say, 18spins, only 4 numbers hit, in dz1, then there's, eight numbers BLOCKS left,
unhit,
with averages,
of four numbers,
MAY hit, in next 18spins.


There's never, a single dz sleep for, 36spins, thou theoritically,
12 numbers possibly sleep after 37spins.

Thus, if you bet just dz1, and lose,

then dz2 and dz3,
had hit, with more than eight numbers each...


Thus, every, session, we have, averages...
1/3, to lose,
and 2/3 to win.

[6th-sense]

long time vaddis just tried to pm you it won,t let me...wanted to ask you a simple question...might as well ask here...is it true you gave roulettepassion ..aka sergio..your bet..

this is simply curiosity as he did go on a bit about it that you did...

[Interceptor]

It is not true, I gave him only a false hope, lol he was so happy and think he finally have something.

[SWEET]

(4,2, 6,11, 8, 10,5, 1,9, 7, 12, 3)

Arrange above numbers to bet as a group in dz one, see how they, linear wise, single or double, triple...in 36spins.

They represent the dz1 numbers on the wheel, as clockwise, so if the wheel bias, then half of the numbers will hit more than the other half.

[SWEET]

You may bet,

ie.

Only, HALF of the wheel,

Black numbers only,
Etc.

Bet for,

EVENT,
THAT MUST HAPPEN, IN NEXT 36SPINS!

[6th-sense]

Thanks for confirming Vaddis...appreciate it...

[SWEET]

Gentlemen,
I believe,
many here,
 has headache,
how,
to minimize,

STAKING,

THUS, I think...

Bet only RED,
or
only BLACK,
of,
only a single dozen...bet.

WHAT YOU THINK??
below,

 an EXAMPLE...

of bet only
DOZEN ONE.

BLACK ONLY.

RED ONLY...

Red  =
(5,  1,  9, 7, 12, 3.)

Black=
(4, 2,  6, 11, 8, 10.)


Black
4=EMPTY
2=X
6=X
11=X
8=X

Red   
5=X
1=X
9=XX
7=EMPTY
12=X
3=EMPTY

Roulette,
Number of
DOZEN ONE,

hit after 37spins

=1,2,5,6,8,9,9,11,12

Refer to kario's image on reply#30