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Outside the box: a different view on roulette numbers

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

Previous topic - Next topic

WeWillWin, thocxo2207 and 1 Guest are viewing this topic.

alexlaf


Person S


Person S

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

alexlaf


Person S


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 3,18
112 499 3,58
113 501 3,59
121 460 3,30
122 532 3,82
123 513 3,68
131 508 3,64
132 558 4,00
133 494 3,54
211 474 3,40
212 495 3,55
213 541 3,88
221 513 3,68
222 523 3,75
223 549 3,94
231 527 3,78
232 560 4,02
233 542 3,89
311 526 3,77
312 509 3,65
313 520 3,73
321 535 3,84
322 531 3,81
323 561 4,02
331 527 3,78
332 518 3,72
333 483 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.

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