• Welcome to #1 Roulette Forum & Message Board | www.RouletteForum.cc.

## News:

Odds and payouts are different things. If either the odds or payouts don't change, then the result is the same - eventual loss.

## Outside the box: a different view on roulette numbers

Started by rrbb, May 30, 08:46 AM 2016

0 Members and 3 Guests are viewing this topic.

one more!

cold? looks cool

#### Person S

Do you know what the red numbers mean?
The guidebook is not very clear whether this is good or bad :вопрос:

#### alexlaf

Net win and net you mean!

YES

#### alexlaf

If wins are coming where is the problem with those nr!
The last test was advance on betting usually I start from 1 unit if needed go up on a win.

#### Person S

The analogy with a bell-shaped curve - naturally the most bread will be in the middle, but it all depends on the baker. ;Д

#### Person S

Hey people know there are cycles 1,2,3.
For example, cycle 1 is common to all dozens and is equal to 33%. If you look at the cycle for each dozen - it will not be 33%, but 11%.
I don't know yet how it can be useful, just out loud...

#### winkel

Quote from: Person S on Sep 27, 01:46 PM 2022Hey people know there are cycles 1,2,3.
For example, cycle 1 is common to all dozens and is equal to 33%. If you look at the cycle for each dozen - it will not be 33%, but 11%.
I don't know yet how it can be useful, just out loud...

Can you give an example, please
There is always a game

#### Person S

We know that cycle 1 can be on any of the 3 dozen. For example - 11,22,33.
Probability of this cycle = 33%.
But this probability covers all 3 combinations.
Correctly? My idea is that if we take a lot of cycles - for example 100 and look not at all 3 dozen together, but SEPARATELY - then for each there will be only 11%.
100-200 spins can be generated and see how many times the combination 11 is repeated, how many times -22, and -33.

#### Person S

Here, for example, cycle 1 on a dozen 2 - appeared 6 times.
On a dozen 3-3 times. A dozen 1 did not work at all yet.

#### winkel

perhaps this stats can help

`Form count % 111 443 3,18112 499 3,58113 501 3,59121 460 3,30122 532 3,82123 513 3,68131 508 3,64132 558 4,00133 494 3,54211 474 3,40212 495 3,55213 541 3,88221 513 3,68222 523 3,75223 549 3,94231 527 3,78232 560 4,02233 542 3,89311 526 3,77312 509 3,65313 520 3,73321 535 3,84322 531 3,81323 561 4,02331 527 3,78332 518 3,72333 483 3,46 tot 13942 `
There is always a game

#### Person S

No, it won't help.
A cycle of length 1 is a combination of exactly 2 dozen.
Here is an example
121 -CL2
11 -CL1
1323 -CL3
33 -CL1
322 -CL2
22 -CL1

#### alexlaf

on the first three spins, we have (1 2 1) What is the probability that the fourth is ( 3 )?

#### Person S

If you count by probabilities, the probability will be 33% - on the 4th spin or on the 14th, this probability will be unchanged.
But I'm talking about cycles - let's take a cycle of length 2, its probability is 44%. With a rough approximation of 100 cycles, we will have 44 cycles of length 2. In these 44 cycles, there will be all 3 dozen. So we can divide 44/3 = 14.6. This means that each of the defining dozens will appear an average of 15 times. Maybe we can keep count. It will be like a regression.

-