Quote from: Person S on Aug 27, 09:29 AM 2022He probably meant the primitive systems that wait for 9 reds to start betting on black.

Maybe. But that would imply that a winning method is not fundamentally different from primitive systems, just more "complex".

let me tell you the ultimate truth with roulette:

in 37 spins and with 37 possible results there are 37^37 possible outcomes or results.

every trigger you find, triggers also the loss.

the only thing you can do is to watch the permutation and follow the possible next step.

for example:

we know: that the first repeat might happen at spin 6 7 or 8.
we know: always betting the hit numbers at that points will make you lose
we know: the chance to win at these spins is about 16% (for each bet)
we know: repeats can also happen at the 2nd, 3rd, 4th spin
we know: all depends on the starting with a first spin.

what can we do:

36
25
15       
21
8       
4       

we could watch at these spins a different way:
we could have started with the 2nd spin
we could have started with the third spin and so on.

36   
25  25 
15  15   15    
21  21  21  21
8    8    8   8   8
4    4   4   4   4

for the first column we expect a repeater. but the repeater can also be a result of the other 4 columns.
why should we bet all 6 numbers of the first column it mustn´t be the number 36 or 25 to become the repeater. and then bet 7 numbers when there was no repeat and then 8 and then 9 and so on.

So why don´t we just bet the last 4 numbers (2/3rd by accident?)
We know we can also be wrong, but we won´t risk 6 7 8 9 10 (40 units) but only 4 4 4 4 4 (20 units) and still be in profit in a win.

the spins came up with
6        lose 4 (15 21 8 4)
36 oh dear we were wrong lose 8 (21 8 4 6)
28        lose 12 (8 4 6 36)
28    lose 16 win 36 (4 6 36 28)

because we know within 13 spins there will be 2 reapeaters.
because we know after 6 spins there might appear a repeater ( 6+6 =12 -> two repeaters)

You also can bet any other count of numbers:

you can bet last 5 number for 7 times
last 4 numbers for 9 times
last 3 numbers for 11 times


this kind of betting you can transfer to every other chance except EC

Is this flat winkel , i got a gap to hit lets say 20 round
20 x 4 or after 4 x 9 do i reset!

@alexlaf

I only design strategies with flatbet.

What most people don´t realize:
If I give you a strategie, which winns 80 times in 100 bets, that also means you lose 20 bets.
It is not said that the 20 losses appear every 5th spin.
They also can happen 20 times in a row.
Nobody can tell if the losses are at the beginning or in the middle or at the end of the 100 spins.
It can happen they are at the end of 100 and in the next cluster at the beginning. so you can lose 40 times in a row.

Keep that always in mind.

Quote from: Herbyx on Aug 27, 02:12 AM 2022Let's make a Gedankenexperiment (thought experiment):

You have a result at the end of every
37 spins = Count(Unhits) + Count(1 Hits) + Count(Repeats without the first hit)

Important: Count(Repeats without the first hit): the first hit of the number is not counted

Now you take one of the repeats and put it to an unhit position, and again, until all unhit positions are filled with one ball.  => no repeats are left over, only single 37 single hit

Important: you start with 37 - you end with 37, the number stays constant

The other way round: the repeats can only come from the number of unhits - always !

Herbyx should the first stage look like this or no?
0   0   0   
1   0   0   
2   2   1   
3   3   2   r
4   0   0   
5   0   0   
6   3   2   r
7   0   0   
8   1   1   
9   0   0   
10   1   1   
11   1   1   
12   0   0   
13   1   1   
14   1   1   
15   0   0   
16   3   2   r
17   0   0   
18   0   0   
19   0   0   
20   4   3   r
21   1   1   
22   1   1   
23   1   1   
24   0   0   
25   0   0   
26   0   0   
27   1   1   
28   3   2   r
29   1   1   
30   0   0   
31   2   1   
32   2   1   
33   2   1   
34   2   1   
35   1   1   
36   0   0   

Quote from: saihtaM on Aug 27, 09:42 AM 2022

Quote from: Person S on Aug 27, 09:29 AM 2022He probably meant the primitive systems that wait for 9 reds to start betting on black.

Maybe. But that would imply that a winning method is not fundamentally different from primitive systems, just more "complex".

Correction, I should have said:

"Maybe. But that would imply that the winning method he was talking about is not fundamentally different from primitive systems, just more 'complex'."

Quote from: Herbyx on Aug 27, 02:12 AM 2022Now you take one of the repeats and put it to an unhit position, and again, until all unhit positions are filled with one ball.  => no repeats are left over, only single 37 single hit

I also don't get this part:-\

Hi alex,

look at Person's  pic:

13  14  10  13

"rouletteforum.cc/index.php?action=dlattach;attach=46702"

With Ayk's tracker, 37 spins the first and last number (I colored red) have to be the same. (not really important, just a fact, I think - as I know less and always lesser )

I think it's better to go back to 6-th statements and posts. 

Ok, I got this, but how do I decide whether UNQ or REP to get that?

Quote from: Herbyx on Aug 28, 06:58 AM 2022Person's  numbers are more general:

13  14  10  13

You see:   13 14 10        :    Sum = 37
                     14 10  13  :    Sum = 37

Without knowing something now I would compare with 6th's recently shown diagrams, if I only would remember where they are.

This is the end of my klingon language, or was it latin ?   

Quote from: Herbyx on Aug 28, 10:51 AM 20226th's recently shown diagrams

Can someone repost the diagrams from 6th!

I meant the second pic

rouletteforum.cc/index.php?msg=254615

After 37 or until 37?

I may be veering off topic, but an interesting experiment can be conducted. It's about how long we have to wait for all 37 numbers playing one number in a cycle of 37 spins.
So spin 1 rolls 27 we bet for 37 spins on the number 27.
If we win or lose it does not matter.
If the cycle ends on the number 27 we do nothing, and continue to bet on this number.
But for example the cycle ended on number 36 we bet on 27 and 36 for 37 spins.
If the next cycle does not fall within these two numbers, we add a new number to the bet and continue to spin 37 spins.
I wonder how many cycles it will take to cover all 37 numbers.
Although I did not conduct this test because I do not know how to code, but I am sure that these 37 will come out not within 37 cycles, but much further. It could be 100 cycles or more.