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Roulettefocused => Main Roulette Board => Topic started by: falkor2k15 on Jan 16, 03:24 PM 2020

We have already touched upon the following in various topics:
*Cycles
*Next spin vs. Next variable spin event, ie. a dozen happens in the next spin  but a dozen repeat happens in the next 24 spins.
*Why Roulette is not about the next spin and how we can make new events, numbersets and payouts over multiple spins, i.e. 72 number Roulette.
*Roulette is a break even game with superadded house edge
No matter whether we play Quads as 3/12 streets per single spin  or 2 ECs stitched over 2 spins  the risk/reward is still 25% with a payout of 1:3.
What do we mean by stitching? If we lose we stop after 1 spin and if we win then we let it ride for the 2nd spin:
Bet H.
H... win +1
Bet L "Let it Ride"!
HL.... win +2
Total: +3
Bet H
H... win +1
Bet H "Let it Ride"!
HL... lose 1
Bet H
L... lose  1
In each case we either won 3 units or lost 1  correlates to 25%/1:3  albeit using multiple spins instead of just a single spin.
So how could stitching bets possibly help us to escape a break even game since the risk/reward is still proportionate as per single spin targets? That's the burning question...
Let's take Dozen cycles, which everyone should be familiar with by now from various topics, such as Random Thoughts. Cycle Length 3 can also generate a payout of just over 1:3  1:3.5 to be precise  with a probability of 22% (instead of 25%).
1... bet 2+3
12... win +1; bet 3 "let it ride"
123... +6
Total: +7
However, there are actually 2 different ways of playing for CL3:
Bet CL3 (A)
1... bet 2+3
12... win +1; bet 3 "let it ride"
123... +6
Total: +7 = 1:3.5 odds
Bet CL3 (B)
1... miss out first spin
12... bet 3
123... win +2
Total +2 = 1:2 odds
If CL3 is 22% then why are the payouts different in each case, and why would we play CL3 one way vs. the other; might that hold the key to predicting Cycle Lengths  hinted to be the prerequisite for exploiting the statistical dependency of front runners in numbers cycles?
You see there's also 35 different ways of betting for Repeat Order 1 (= "Defined by Same"/"Front Runner"; see Random Thoughts, Turbo topic and others) including payout odds of 1:2, 1:3.5 and about 1:25 with a base probability of 62%.

I think it is possible to diversify stitching, for example, in the case of EC  it can be not only the first 2 spins, but 13, 25, etc. This will take the embroidery to a freer area. And the main thread that in my opinion should be present in all this random fabric is displacement.

Yeah you could do that  suggests there might be a way to create additional dependency  but it does get a bit overcomplicated playing several overlapping games at once. We need to try to understand stitching at it's most basic level first to see if it can offer us any advantage with basic cycles and multiple repeats or why anyone would want to use it.

The simplest definition.
1. Connection of several positions ds + ec = street.
2. Connection rates = increase payments.
3. The combination of 1 + 2 definitions.
This is my humble opinion, perhaps there is a deeper level.

1. Using ds + ec to make streets is simply creating a new set of outcomes with the same payout odds as streets. I doubt there's much connection between that and real streets. If there were some parallel game involving when to switch from ds or ec to the stitched combo or even switching payout odds then that would be far too elaborate  so would be better to return to the original example to get a more simplistic understanding first.
2. Connection rates presumably refers to combining the probabilities for the increased payments (always reducing the probabilities in the process)  but again this doesn't stray from the proportionate risk/reward ratio regardless of what we choose to stitch together.
3. I think your suggestions are too deep already, so it would be better to return to the core dozen cycles example and see if we can figure out the basics first  since ati claims he is able to play a basic EC game with cycles and stitching that comes out on top in the graphs.
Remember: stitching doesn't necessarily need the parlaying/let it ride aspect  you can stitch flatbetting  just continue next spin on a win or stop on a loss and wait for the outcome to finish forming before trying again.
The question remains though: what is the difference between the 2 ways of playing CL3  one stitched and the other single spin? Why would you play one or the other? Or why would you play Order 1 differently in different scenarios?
Remember what Priyanka once said:
[19:30][Priyanka]: basically the dozen that defines the previous cycle will define the next cycle 63% of times
[19:42][Priyanka]: It says same will happen 63% of time
[19:42][Priyanka]: different will happen 27%
[19:43][Priyanka]: but there is a catch...
[19:43][Priyanka]: that single dozen could hit either in 1st spin, 2nd or 3rd spin...
[19:43][Priyanka]: So if you keep on doing that blindly, then you will hit the house edge
[19:44][Priyanka]: so we need a breaking point or an entry point
[19:47][Priyanka]: now betting same is a better option
[19:47][Priyanka]: it is not like 10 reds followed by black..
[19:48][Priyanka]: it is slightly different
[19:48][Priyanka]: so you essentially try to find this tipping point
[19:53][Priyanka]: you keep adding the numbers if you are not winning in the same format.. eventually the "same" will catch up.. that is statistics and probability for you
[19:54][Priyanka]: within the same, the same occuring in 2 spins is 88% of that 63%
[19:54][Priyanka]: thats why you play only for two spins
[19:54][Priyanka]: hope it all makes sense now.. thats a simple way to utilise cycles in your play

Whilst waiting for the answer to the above I thought I would have a go at simulating the concept of overlapping games  stitching spin 1 from the first cycle with spin 1 from the second cycle; and spin 2 from the first cycle with spin 2 from the second cycle:
(http://xs://i.postimg.cc/YSjVHCjL/games.png)
From the above template we could check if there's some kind of dependency based on dynamic decisions. Again, this is not really the road I wanted to go down at this stage...

I wrote my assumptions, and they may not coincide with your research.
1. By combining these two, we did not change the payouts, but we changed the statistics. That is, our gaming environment has acquired a new form, and decisions now depend on the past rotation (if we connected 2 spins). We have 2 permanent elements  payout + probability.Playing for 6 lines, the payout does not change, playing in a stitched bet  the payout does not change, but the probability changes. The only person who can give an accurate answer, you know who he is...
2. This is also obvious  in order to confidently use positive progression, it is necessary to increase the likelihood. The only way to use the deviations is with the virtuoso method available only to experienced seekers.
3. You need to do a lot of creative experiments by combining elements.
It is a good idea that by distributing winnings to positions, you can help increase payouts.
I am not a great specialist, and also in searches, do not take this for instructions.
And these messages are from Priyanka, are they from the chat? I study all of her topics, but there are none. If you can share ...

Are you talking about stitching a ds+ec on the same spin as a hedged bet  or betting ds on spin 1 and stitching ec for spin 2? Either way I don't see how stats or payouts becomes distorted  both remain in proportion regardless of what we choose to stitch: probability decreases and payouts increase at the same rate. When you say "past rotation" are you referring to rrbb's dynamically changing positions? If so then that's yet another layer of complexity that could easily distract us from finding the essence.
Sure  positive progressions require at least some edge from flatbetting. I can find no reference to the "virtuoso method"?
I have tried to create lots of things before, including many types of pigeons and holes  often including stitching  but exploiting statistical dependency to escape break even seems more challenging than using functional dependency. That's why it's important to try to understand the fundamental logic behind creating any NonRandom game and why stitching could be needed; it's clearly not used for enhancing edge with a positive progression as the parlaying aspect is seldom used, i.e. most stitching is done flatbetting.
Anyhow, we'll see how my two overlapping games go based on the "displacement"(?) of each set of stitched bets.... will post the results.
For Priyanka's quote see "Random Thoughts a concise reference"; perhaps her description of cycles and front runners (at their most basic level)  not to mention ati's quote about taming variance  might somehow be referring to an application of stitching and how to predict cycle lengths/order with greater accuracy...?

Why is progression an enemy, if it was possible to create a weak but reliable method, it is a good management of funds.
So I do not mean the position of RRBB. We only talk about backs ...
I have a quote from one of the knowledgeable participants.
The only way to win at roulette is if we can come up with a way that the result of the constant bet A depends on the result of the constant bet B. Where B happened earlier in time.
What is a fixed rate? “A fixed bet is a fixed number of chips per fixed set of positions.”
We intentionally use positions instead of numbers, as they can be divided in time.

There's a lot of confusion with the word "position" referring to RRBB's terminology  or just a plain old group or playing position on the carpet layout.
The only way to win at roulette is if we can come up with a way that the result of the constant bet A depends on the result of the constant bet B. Where B happened earlier in time.
Well, we could take rrbb's position 1 of the ds and stitch it 1 spin later to the corresponding EC of the ds  but I doubt the stats would change.

Position  derivative, parallel flow. I had in mind this definition. It is clear that all you do is the same actions, manipulations with cycles. Although you previously read some of your topics, there was something fresh there, but you did not develop it to the end. One idea overlapped another. And in fact, it concerns me the same way. ::) There is always the thought that you need something perfect and reliable, and only then when it will be open can you improve this method.

I've got different projects at different stages  all overlapping with the same concepts. Priyanka claimed she had different ways of winning; and ati recently revealed there are two different types of dependency, hence the need to split things (no pun intended!) ;D

and ati recently revealed there are two different types of dependency, hence the need to split things (no pun intended!) ;D
Did he? Can you point me to that please?
BA

Did he? Can you point me to that please?
BA
It's in the topic called "I think I finally cracked it".
Here's how I am going to test if there's any dependency between game 1 vs. game 2 based on the chart above and stitching 1 with 3 + 2 with 4:
(http://xs://i.postimg.cc/J0HRHRPs/Capture.png)

Nah, I tested it out and the overlapping games are totally independent! :(
Also, it seems impossible to alter statistics but keep the same payout odds. You see, If the addition of 2 probabilities resulted in a skewed result then one of the two source probabilities would need to be distorted itself to contribute to a combined distortion. However, all probabilities and payouts remain proportionately static  individually or stitched.
Therefore, we need to return to CL3 and Order 1 as I think the application for stitching may take a completely different form.

It's in the topic called "I think I finally cracked it".
Cheers

OK let's move on then...
Just like betting front runners, CL2 also has multiple ways of playing. In fact, like CL3, CL2 also offers two different ways of playing the same target: to stitch or not to stitch  that is the question, as the old saying goes.
So now let's bring Martingale back in again so we can get a better idea of extreme situations  strictly for the purpose of testing of course:
Without Stitching
(http://xs://i.postimg.cc/ncZjX6MH/CL2a.png)
With Stitching
(http://xs://i.postimg.cc/qB26MkJb/CL2b.png)
What do you notice...?

Did you call Hamlet?
What we see, we see that we are drowning at a higher speed in the second example.
And this is understandable, you have more bets there. You used CL2, but what did you want from them? To play only in CL2 and become a winner, I think it is naive ... This cycle is part of the other two, it cannot work alone, but you yourself perfectly understand. Such thoughts ...

True: 2nd example has accelerated losses  but this wouldn't be too detectable if we played flatbetting.
That's right: we can win just playing CL2 without changing our bet selection since we are using stitching!
It may seem naive, but what it reveals is the crux of exploiting statistical dependency:
*When to bet or when not to bet
*When to bet normal or when to bet opposite
*When to stitch or when not to stitch
*When to bet one Cycle Length or when to bet another Cycle Length
*When to bet one Cycle Length constant or another
*When to bet a unique or a repeat
*When to bet one dozen or another dozen.
By default all the above are individually break even in the long run.
This cycle is part of the other two, it cannot work alone
Not quite the obvious observation I had in mind, but you're getting there, mate! :thumbsup:
So if we must traverse the losing zone from CL2 to CL2, is the first cycle stitched or not stitched...?

Not quite the obvious observation I had in mind, but you're getting there, mate! :thumbsup:
Falcor sounds like Priyanka ;D
I do not understand the line the first cycle is stitched or not. What do you mean?
What does this design look like?
Cycle 12 is not so difficult to use, everything breaks when it comes 3. So  I suppose the schedule will not smoothly go up, even if there is a method how to play it right.
So, as distortions in the distribution is a very real thing. We should have something finite, relying on a parallel game  which unfortunately has not been disclosed ...

:xd:
When I say the first cycle I mean cycle 1/12 in the following example:
(http://xs://i.postimg.cc/qB26MkJb/CL2b.png)
(http://xs://i.postimg.cc/c4BNWQhH/extreme2.png)
What's the tipping point?

I can not catch ...
If you join (sew) to cycles  this is how to sew positions. CL1 + CL2  1 item
CL2 + CL2  2 items
CL1 + CL1  3 items
There is confusion.
Since your 123 increases difficulty. This only complicates the situation, you can mathematically prove that this approach is correct. If absolutely all probabilities are determined from these data, will it make sense? Your method  how to determine all the probabilities?
We need a foundation  the MATHEMATICAL BASIS, and not Russian dolls, otherwise a deviation from the goal will again arise ...
The turning point is that we risk 54 chips in the HOPE of winning only one)

We are not using probabilities here, we are using dependency, stitching and the tipping point to "go with the flow" based on the variance. We aren't stitching pairs of cycle  but choosing whether to play CL2 with stitching or without stitching.
(http://xs://i.postimg.cc/ncZjX6MH/CL2a.png)
CL1 = 5; CL3 = 5
= 50%/50%
(http://xs://i.postimg.cc/qB26MkJb/CL2b.png)
CL1 = 8; CL3 = 3
= 73%/27%
Here's how I worked out the tipping point:
CL1: 33 = 33%
CL2: 45 = 45%
CL3: 22 = 22%
Total = 100
CL1: 33 = 60%
CL3: 22 = 40%
Total = 55

Ehhh, falkor ...
This tipping point, what is it?
Probability  our absolute everything rests on it, take the birthday paradox, why it works  because with each appearance of a new result, the probability is to move up, so it works ...
What about the tipping point? I do not know such a definition, as I think other people.

Dyslexic, talking about his method. What I wrote  now the CHANCE, make it so high that not one computer in the world is able to calculate them. Odds, not addiction and not the crest of a dispersion wave, but only the odds.

Traps of the mind, artful enough to please them.

Priyanka's quote  remember...?
[19:48][Priyanka]: so you essentially try to find this tipping point
[19:48][Priyanka]: now not every day we will find this tipping point
[19:51][Priyanka]: now coming back to tipping point
[19:52][Priyanka]: you were arranging them in such a manner that it gives you that tipping point of different different diff diff diff
[19:52][Priyanka]: one set reaching 10 is your tipping point
[19:58][Priyanka]: and play only same when tipping point happens
[20:07][Priyanka]: yes.. and the tipping point is on a higher scale there 55% same and 45% different.. so you cant play like what i described above..
[20:08][Priyanka]: and within that find a better tipping point
Search for "tipping point" here too:
xs://x.rouletteforum.cc/index.php?topic=15938.90

Probably these are the features of speech :\
How we can enter the dozen game when that imbalance is in our favour and most likely to result in a positive expectation is the riddle that you need to crack.
The search for this is likely our goal. But pay attention that this is not 100%, but as they say most likely ...
And I do not have this collection of quotes)

Right, so when will you bet and when will you not bet; or when will you stitch and when will you not stitch?

One of the videos has a similar CL2 strategy.
It looks good (but you have to remember about 0). Sometimes I examine the green sheet with dozens, and it looks like chess, only the opponent sometimes goes crazy)

What about positive progression?
But I think we need a method such that when it comes to increasing bankroll, the progression returns to its starting point. But there may be more interesting options.

Exactly. That was only the intro, innit... so now we need to figure out how to improve upon it... not even close to a finished system yet.

hi Guys, I play a lot of mini games stitching and somewhat not . the important thing for me is the following.
the time to do it and not to do is crucial. I want to get a higher return on my winning bets to overcome the long term losses. but at the same time keep things simple enough that i can apply in real games.
so i recognize that the house edge will have to catch up with me but most likely wont if I dont repeat those special bets too often.
I can also play a positive progression that involve from 5 to 8 legs.
example. I play a 1 unit on high, 2,4,8,16. now I am ready to bet a 32 unit bet. i had to wait a while to get to that place.
but now I will not reduce my 32 unit bet but will reduce my real estate to 15 numbers instead of 18 numbers.
I said numbers for the sake of anderstanding . to me a ec bet contain 18 numbers.
but I am willing to reduce my 18 numbers to 15 numbers to get the extra paid off.
I could play in this manner many other ways.
I could play the 32 units on a 3rd column because I am playing a high ec. (maybe it does not matter really which dozen) but the timing is there. if I win one of those high bets it will take the house a long time to catch up.
this is one example. there is no stitching in this example but could easely add one if I wnted to.
the regression of one big bet once you reach the level of a positive progression is a great way to get ahead.
my 2 cents. stitching should happen in rare times to gain its best return in my opinion.
if they get repeated too often the house will catch up. that is why we see those graphs of thousands of spins where the law of large bets and numbers turn everything south.
God bless
Rinad

Hi, and thanks for your contribution to this conversation.
Obviously, the freer you feel, the more confident your bets become. You can see that you have a reasonable method.
You can also share your philosophy of the sewing concept, for example, we use the instructions recommended by experienced players. True, we have them in a conceptual version and are not yet applicable on real battlefields. I also suggest that positive progress is a good way to manage and increase bankroll, respectively, based on a good method.

stitching bets and overlaping bets is a way to extend a strategy for reaching a target.
I am taking a new path in my game for some time now. i have read and study a lot of "Trend followers" playing the markets for decades and i can see that they have done well in building whealth compounding their winnings.
I am certain the same can happen with roulette. but players are looking for a "weekly" paid check and that is a big mistake.
getting 50 unit win per day is almost impossible to achieve and dangerous.
the process is more important then the short term outcome. I am buidting muscles of self control by taking many small losses until the big moves take place and we all know that getting use to the small wins for weeks will make it hard to the average player to take a loss when the day comes. preserving capital is most important.
I think stitching a target may sound useless for some because the randomness of the game tells us that in the long run it all loses, but the fact remains . I want the house to chase me and not the other way around.
most know that playing a neg.prog is a losing game in most instances.
knowing that , why not being on the other side if that is true ?
because for most the outcome is first and the process second. )even know the outcome later on will crash)
mixing a positive progression and overlaping the bets put us in front of the house and we dont have to repeat the same targets over and over once we won.
I am still learning this way of playing and I hope some will see the mistake in playing the traditional way of a system that pays you like a job and not a investor.
anyway I love this topic and hope it wont loses itself over time.
thx
Ronad

Next, we should try to incorporate everything we've learnt from this topic  including Rinad's useful feedback  into combating the variance of EC cycles like a game of Chess:
o1  3:1
CL1  1:1
CL2  1:1
CL2o1  1:1
CL2o2  1:1
CL2o1 stitched  1:3
CL2o2 stitched  1:3
And we should first try to achieve this without using positions or any parallel streams from Dozens, Lines, Streets or Numbers, etc. Also it won't be possible to hedge anything either, so our hands are completely tied to make it more of a challenge.
As Rinad said: the most important thing is when to stitch!

Falkor  let's denote the concept of variance. In a game of cycles, it seems these waves are quieter, in a random stream it can become a tsunami.
Accordingly, the variance has  beginning  middle  end. The duration of the variance is also not predictable, and the value of the chips at these moments is large. There should be entry points somewhere. I am practically not familiar with this monster, and how it behaves in roulette. Surely there are durations or something else.
Peace for everyone!

Well, I think we need to look at 3 things first:
1) Brick design, i.e. stats says they dozens are 33% each and there should be an equal distribution in the long term, i.e. 3 uniques at at time.
2) Law of the Third: we know that repeats will happen quicker than brick design
3) Cycles and the defining element
When it comes to cycles and unequal outcomes, CL1 (50%) is more likely to repeat than CL2o2 (25%) because it has a higher ratio. However, if CL2o2 defined the cycle then it only requires one more appearance to be awarded the repeat, so we get 25% CL2o2 vs. CL1 + CL1 (1/2 * 1/2) = 25%
So if both are 25% derived, which one is likely to define the next cycle?
Order 1 is 75% so that would beat CL2o2  even if the latter defined the cycle.
So once we've studied what's meant to happen normally  only then can we move onto the variance, i.e. what happens when things go pearshaped.
However, we have special tools to help  namely stitched versions of CL2o1 and CL2o2 that are akin to Super Mario Bros. powerups!
We also have more than one way of betting for a unique or a repeat thanks to multispin cycles.
I think we should begin this challenge by initially looking at what happens when the 7 individual bets encounter a losing streak from hell like what we did with Dozens CL2 vs. stitched CL2.

I've uploaded simulations of the 7 bets here in case anyone is interested in analyzing what happens with other parts of the EC cycles when they lose:
Download 7bets.zip... (http://x://surviving2023.com/7bets.zip)

hello Guys, I think of stitching as a way to prolongue a losing dozen, losing for 2 spins or 1 cycle , by also transforming the losing dozen into 2 double streets within that same dozen, and therefore adding 2 extra spins for the play.
stitching dont always need to be a exterior play. so by adding the 2 double streets into play after losing the dozen I am creating a way to save the early proposition. stitching a losing dozen to 2 ds.
in the same manner i could stitch 2 double streets into 4 single streets, ect...
at the end I could lose only if a single number dont show within a cycle. or 36 units loss playing flat bet.
just a thought. it could be viewed as a parachute but not in the traditional way it is played.
Rinad

Hey Rinad, parachuting to dozens or lines is cheating! ;) There is already a lot happening in just EC cycles alone:
If aiming for a repeat or unique you can choose a combo to bet on behalf of or an individual target:
WINNING STREAK
o1  extreme o1, more CL1, more CL2o1
CL1  extreme CL1, extreme o1
CL2  extreme CL2, more CL2o1, more CL2o2
CL2o1  extreme CL2o1, extreme o1, more CL1 (missed out)
CL2o2  extreme CL2o2, more CL1 (missed out), more o1
CL2o1 stitched  extreme CL2o1, extreme CL2, extreme o1
CL2o2 stitched  extreme CL2o2, extreme CL2
EXTREME o1 (3:1)
o1 (3:1)
CL1 (1:1)
CL2o1 (1:1)
CL2o1 stitched (1:3)
MORE o1 (3:1)
CL2o2 (1:1)
EXTREME CL1 (1:1)
CL1 (1:1)
MORE CL1 (1:1)
o1 (3:1)
CL2o1 (CL1 missed out) (1:1)
CL2o2 (CL1 missed out) (1:1)
EXTREME CL2 (1:1)
CL2 (1:1)
CL2o1 stitched (1:3)
CL2o2 stitched (1:3)
MORE CL2 (1:1)

EXTREME CL2o1 (1:1)
CL2o1 (1:1)
CL2o1 stitched (1:3)
MORE CL2o1 (1:1)
o1 (3:1)
CL2 (1:1)
EXTREME CL2o2 (1:1)
CL2o2 (1:1)
CL2o2 stitched (1:3)
MORE CL2o2 (1:1)
CL2 (1:1)
If you lose the repeat or unique then you can guage what is likely to happen on a loss:
LOSING STREAK
o1  extreme CL2, extreme CL2o2
CL1  extreme CL2, more CL2o1, more CL2o2
CL2  extreme CL1, extreme o1
CL2o1  extreme CL2o2, more CL1 (missed out), more o1
CL2o2  extreme CL1o1, extreme o1, more CL1 (missed out)
CL2o1 stitched  extreme CL1, extreme o1
CL2o2 stitched  extreme CL1, extreme o1
Given that different targets have different options at different payouts, what kind of strategy could we devise..?

It is difficult to understand what is written here, why not switch to a more accessible way of presentation?
You claimed that CL1o1 + CL2o1 = 75%. But if we track cycles and only after they appear make a bet, then the statistics will change. For example  CL1 / CL2o1 / CL2o1 / CL2o2 /  here CL1 appeared, but there was no bet, (we waited for cl2) a bias in the statistics. And what about CL3?

I agree that the presentation could be simpler . I appreciate the hard work you have done with charting, meaning you are very passionate about finding a real method.
however sometime the mind has to be free from too much infos in order for the imagination to see a clear pathway to a solution.
the greater benefit to stitching I have learned is this; it enables the player to not get eaten by the law of large numbers.
second; it enables the player to exit a play when he brakes even caused by the stitched play.
third; it increases the odds of winning period since we are creating more then one out, like a poker hand where your cards can get you a flush,straight, a full house, from the same first 3 cards.
so it can always lose but that is part of any method. exept that creating more "outs" can turn a losing proposition into a brake even one most of the time, which should be the first target after a loss.
the goal is to find a serie of plays that can be formed into a Method. since there are so many alternatives we need to reduce them to a few plays. maybe ec ,doz. dz ,streets.....???
we know that what should work for one should work for others. that would be the easy work.
thanks Guys for sharing your ideas and interests in this underrated Topic which hold a real key to winning I know.
Rinad

Theoretically, at the initial stage, the goal of sewing is to increase the number of chips. The combination of events is not because the probability of coincidence is higher, but rather the speed and increase in the payout ratio.

It is difficult to understand what is written here, why not switch to a more accessible way of presentation?
You claimed that CL1o1 + CL2o1 = 75%. But if we track cycles and only after they appear make a bet, then the statistics will change. For example  CL1 / CL2o1 / CL2o1 / CL2o2 /  here CL1 appeared, but there was no bet, (we waited for cl2) a bias in the statistics. And what about CL3?
There's no CL3 with EC cycles. That's why it's the most simple cycle formation I know of  made up only of 2 equallylikely pigeons. However, I will be using the outcomes of that to create new Outer Cycles of 3 unequal pigeons that can reach up to CL3.
Yes, we are betting after an appearance that defined the previous cycle and forms the starting partition of the next cycle/outer cycle. That's why I brought up the example of 50% vs. 25% really being 25% vs. 25%. The reason we do this is to create some kind of butterfly effect and keep within the finite cycles framework  to avoid curve fitting.
It seems complicated, but it's not really; just 2 pigeons creating outer cycles of 3 pigeons, albeit with 7 different bets  many of which can cover 2 pigeons instead of 1 and influence a winning or losing stream in the long run.
We have 3 different deadlock situations:
CL1, CL2o1, CL2o2 = a proper CL3 deadlock.
(CL2o1, CL2o2) or (CL1, CL2) could also act as deadlocks or early CL2 repeats (example o1 repeat: CL2o1, CL2o2, CL1) where we miss out CL3  or even "keyframes" for considering whether to risk encountering a proper CL3 deadlock by continuing the mini game, i.e. current cycle.
And a stitched CL2o1 immediately following the appearance of CL2o1 would give us an outer CL1 repeat @ increased payout odds, so perhaps that's when it should be used  depends how our fixed betting template matures and evolves over time visàvis analysis of the stats vs. variance.
Given all the situations that can arise with EC outer cycles and 7 bets we should potentially be able to find an optimal way to change our fixed template and betting plan with a finetuned edge in mind.

I played 500 spins doing just the following.
1 chip on the first dozen.
if losing, place another chip on the first dozen and 1 chip on the low ec.
repeating this exact play I was able to net = + 20 chips.
drawdown was never more then 23 chips. so all flat betting. i encountered a lot of 0/00 during this sessions which was a good sign.
my wins also over power my losses.
I am risking 3 chip if no wins happen in those 2 spins.
I win always 2 chips when I win on the first bet, win 3 chips when it happends within the dozen because of the addon of the low ec.
and brake even on the second bet if I miss the second dozen attempt but win the low ec bet with the 13/16 double street.
I like the payout and dont mind losing 3 chips at most.
Cheers,
Rinad

Here is a quote from a red dwarf.
3. A “game of anticipation of victory” should be avoided at all costs.
It turns out that playing in cycles, we also expect victory. And so the charts are flying down, and the flat rate is not working. The only way is not to wait or wait, but to rely on a variable that will tell you when to bet.

I'm not sure if "waiting" is actually a problem. There are possibly various non random ways to win.
Yeah, reddwarf said that waiting must be avoided, and we should not bet on events. But remember, he also said that his method is different from Priyanka's. And long time ago he also said that inside bets must be used at one point. I'm sure that statement is a bit outdated.
Pri said that we should bet for events that will definitely happen, even in the videos and excels he played for repeats. Redd said we shouldn't bet on events, but on a process of events, so it's obvious that their way of play is different.
While studying the posts and clues, I try not to mix these different approaches. It can cause a lot of confusion.

Well, maybe redd meant a random game for example on triggers. And waiting when we are inside the cycle is part of a nonrandom game. But in my opinion this is also an expectation. Although there is a statistical advantage in favor of some events over others, the sequence of events is still not uniform. For example CL1 / CL2  I need to cover these two and I expect someone to be the winner, but CL3 appears  and I lose, but here you can start using variations with defining positions and make the necessary corrections, but I'm still in standby mode your intended event. I agree that Prie pointed to another way, but he also answered someone that the game of waiting leads to the bottom.

the sequence of events is still not uniform
That's true. But I think some of these events can help us adapt our bets to the variance of random. Like cycle lengths. For example a dozen sequence like this 1212133231122 is a random sequence of individual dozens, but all cycles are length 2, so we can bet for that to repeat and ride on the run of CL2's as long as they last.
Note: this is just an idea, not a hint. It's definitely not enough to win.
Btw, in my desperation, I started to question the basics, and now I think that maybe we are trying to use cycles in their most basic form and it will never get us anywhere. Pri gave a clue long time ago that we can "define" cycles, meaning that we can create our own cycles. What if we say that a cycle ends only after 3 repeats of a dozen? Or only ends if two dozens have repeated once? After all we can be, and we must be creative. We can always extend the already infinite possible ways to play, and it gives me a headache. :D

You can expand, for example, to a length of 4, and it will look like this  1221/1112 / well, and not a subordinate of 1231. What will it lead to? My opinion is that this is just an increase in combinations, so the laws of large numbers will be weaker, but I need a strategy to learn how to win. And we don’t possess this, and all of Falkor’s futile attempts to break through the wall have failed, I believe that he is the best specialist in cycles, in the technical part. (Not flattery))
The reasons given to us are quite stable, but they are not predictable. This is a form for study and possible development in this area. But it becomes like a philosophy and a magic stone, which was sought after by great efforts, and for many it was not found.
Sophistry is the clever use of arguments that seem true, but are actually a lie.
You know, I want to believe the opposite🔒+🔑=🤗 but for now 🍏+🍒=🤬🤬🤬

I'm not sure if "waiting" is actually a problem. There are possibly various non random ways to win.
Yeah, reddwarf said that waiting must be avoided, and we should not bet on events. But remember, he also said that his method is different from Priyanka's. And long time ago he also said that inside bets must be used at one point. I'm sure that statement is a bit outdated.
Pri said that we should bet for events that will definitely happen, even in the videos and excels he played for repeats. Redd said we shouldn't bet on events, but on a process of events, so it's obvious that their way of play is different.
While studying the posts and clues, I try not to mix these different approaches. It can cause a lot of confusion.
Some authorities also talk about taming variance  but Priyanka said she can win within a finite number of spins and that variance doesn't come into it.
Since each cycle is independent it's hard to understand how to control variance  unless we are measuring it based on the repeat of an outer cycle as a keyframe otherwise cycles will lose the keyframes over time and become random. And I doubt the fact they are not equallylikely and multispin helps us with the variance of random outcomes as a byproduct of closing the nonrandom cycles.
The other thing about cycles is the defining element. Perhaps the variance could be measured based on the defining element, so it would have to be something like same1, diff2, same2, diff1, diff3, etc... might retain some kind of structure, but same, diff, same, diff is defo independent.
It seems Priyanka was hinting at multiple repeats as the next stage of cycles. Normally we need special tracking for multiple repeats proper  and at the same time the MLE increases:
1... 62% chance repeat will be on 1, i.e.:
11
121
1231
131, etc.
121... 71% chance 2nd repeat will be on 1, i.e.:
1 2 1 3 1
1 2 1 1
1 2 1 1
1 2 1 1
1 2 1 2 3 1
1 2 1 1
1 2 1 3 2 3 1
1... 44% chance repeat will be on Cycle Length 2, i.e.:
121
122
CL2... 76% chance repeating cycle length will be CL2, i.e.:
CL2 CL2
CL2 CL1 CL2
CL2 CL1 CL3 CL2
With cycles we could take the defining element of EC or Dozen cycles  or it's Cycle Length  and extend the game to several repeats thereby keeping within a NonRandom framework. Unfortunately, cycles is not accurate at tracking multiple repeats in terms of the pigeons that are trailing behind  but the stats for the front runners participating in the race may yield similar stats to normal.

Just to expand on one point above:
High or Low is independent of previous outcomes:
HHLHLHLLLHHLH
Same and Diff  referring to the defining element of cycles  are also independent:
Same, Diff, Diff, Same, Same, Same
d1 cannot follow s1 so there exists dependency:
s1, d2, d3, s3, s3, d2
CL1d cannot follow CL1s there exists dependency:
CL1s, CL2s, CL3s, CL1d, CL2d
Just because the outcomes are dependent though does that make them finite...?
Or could the most important thing be: dependency has to be created from finite outcomes?
If we play the following based on the first to reach 3 repeats or the first to reach 4 repeats then the game is definitely both finite and dependent between sets of cycles:
s1, d2, d3, s3, s3, d2

We get a little creativity.
3
25
35
14
12
23
34
Another event by adding up two results.

A cycle of length 5, and here it is possible to change permutations.
Falcor, you can see in your software how many times in a row cycle 6 appeared on the DS?
I believe that he could not appear 6 times in a row ...

We get a little creativity.
3
25
35
14
12
23
34
Another event by adding up two results.
Good idea  but perhaps it's just generating another random stream like positions  not directly applicable. I wonder if each stream might be dependent on each other without being finite or nonrandom?

A cycle of length 5, and here it is possible to change permutations.
Falcor, you can see in your software how many times in a row cycle 6 appeared on the DS?
I believe that he could not appear 6 times in a row ...
Again, I don't really see an application. It's like saying that a number will usually repeat within 8 spins or a single number is less likely to keep repeating  doesn't really help.

Statistics for DS it became interesting to me how many times in a row CL6 will appear  if its probability is 11.5%. Within all cycles, a rather rare event. And this is just a request ...
The flow obtained from the addition of two dozen  for example, the sum of 2  1 combination and the sum of 6  1 combination of dozens. 345  can be composed in 2 ways, and this means that events are uneven. Just an idea ...

Yeah, we know there's rare events, but it doesn't help us against risk vs. reward, i.e. the break even game and seems out of context.
The last sentence is too vague. I don't know how you went from adding dozens using 2 columns to 345 and two ways... lost me there.

Sorry, I made a mistake, exclude some combinations.
Since our parallel stream consists of the sum of the addition of two dozen. The combinations will look like this 
dozen 1+ dozen 1 = 2, 1 + 2 = 3
2 + 3 = 5
3 + 3 = 6, we can do this in only one way.
But 2 + 2 and 3 + 1 = 4, two ways.
Rather, it's just an idea ...

Ummm... Yeah, a parallel stream is essentially what it is, albeit with added dependency...
So we could have dozens + halves:
Stream 1: 3,1,2
Stream 2: H,L,?
Dozen 2 could be H or L in the same way that Person S' 2nd stream 4 could be formed from 1+3 or 2+2.
Both streams are dependent on the same spin  but Person S' has that additional dependency from the previous spin. However, if you started with 2 expecting to make 4 then the only way that can happen is with another 2; or if we play 2 spins in advance then we could stitch 2+2 or 1+3?
The same concept appears to apply more closely to one of my previous examples:
Stream 1: 1, 2, 3, 3, 3, 2
Stream 2: s, d, d, s, s, d
Different depends on the previous dozen (1 = same) and could be a choice of either of the two unhit dozens (2 or 3 = diff), so if we wanted to stitch SD in two different ways then we could bet 1,2 or 1,3. Again, that's the same as stitching 1,3 or 2,2 in Person S' example based on the same kind of dependency.
Both examples happen to be random play since there's no repeat framework or vdw framework  though since my example uses the defining element then the streams are at least generated from nonrandom cycles (finite only when open)  the only significance being that they are not equallylikely contrary to Person S' example.
Again, when the games start becoming complex like this then we lose the nonrandom/finite aspect since rrbb said we are meant to "win or break even within a cycle" or could the actual purpose of cycles be to generate dependency and not equallylikely outcomes that could then be exploited with stitching...?
There's still the old riddle... you have to know when your play is random or nonrandom  comprising some steps of each.

rrbb said we are meant to "win or break even within a cycle"
Within the straight cycle, not any cycle. It would be impossible to win every dozen cycle. Pri said sometimes it takes him over 40 spins to take a profit.
Who knows if it's even possible to win every straight cycle?
First it was Dyk who claimed his system is 100% guaranteed to win in 38 spins. But his clues are very confusing. He said that you cannot wait for a future winning event, you have to bet on a process, yet he said he's system wins as soon as a number repeats... And how could a bankroll requirement for a flat bet 38 spin system be nearly 3000 units?? He must have used very complex money management, and definitely some kind of progression. He talked about "held bet" and wrote that a held bet is not necessarily a flat bet.
Anyway, I'm not looking to win every cycle, but I want to see guaranteed profit in say, 100 spins.

xs://x.rouletteforum.cc/index.php?topic=20411.msg212019#msg212019
xs://x.rouletteforum.cc/index.php?topic=19290.msg180531#msg180531
Well, there is more evidence of victory.
And I don’t know to believe it, but maybe it's a carrot on a stick :question:

My opinion, neither of those charts are real.
I mean absolutely no disrespect, but I highly doubt Blueprint has a winning system, and I think MoneyT made a mistake somewhere. That was in 2017, and the way he described his play 2 years later, would definitely not result in a chart like that. He may have the HG, but I'm not convinced. Only he knows, and it doesn't matter what I think.

Within the straight cycle, not any cycle. It would be impossible to win every dozen cycle. Pri said sometimes it takes him over 40 spins to take a profit.
Who knows if it's even possible to win every straight cycle?
First it was Dyk who claimed his system is 100% guaranteed to win in 38 spins. But his clues are very confusing. He said that you cannot wait for a future winning event, you have to bet on a process, yet he said he's system wins as soon as a number repeats... And how could a bankroll requirement for a flat bet 38 spin system be nearly 3000 units?? He must have used very complex money management, and definitely some kind of progression. He talked about "held bet" and wrote that a held bet is not necessarily a flat bet.
Anyway, I'm not looking to win every cycle, but I want to see guaranteed profit in say, 100 spins.
Right... well said!
Incidentally, we can keep dozens finite if we play first to 10 repeats of the defining element or try to win within 5 outer cycles.
Since we cannot guarantee a win on an individual dozen cycle  but could potentially win within a straight cycle  then it should be possible to test whether parachuting to EC or Lines within each dozen cycle is a better proposition than continuing with the same game. However, exactly how a full system would work is anybody's guess... if parachuting to any parallel stream can offer any edge whatsoever then I would guess that is the winning mechanism regardless of whether you complete the straight cycle or not. And as we've established: there are already several parallel streams possible with just EC cycles alone  excluding neighboring groups from the carpet.
The above might also support the idea that if we keep our outcomes dependent, parallel, stitched and possibly unequal too then we could fight variance without maintaining a nonrandom framework throughout the entire duration of the game. A fractal has a couple of these properties  but does it have any finite framework?

Sorry, I made a mistake, exclude some combinations.
Since our parallel stream consists of the sum of the addition of two dozen. The combinations will look like this 
dozen 1+ dozen 1 = 2, 1 + 2 = 3
2 + 3 = 5
3 + 3 = 6, we can do this in only one way.
But 2 + 2 and 3 + 1 = 4, two ways.
Rather, it's just an idea ...
Haven't really noticed anything with this yet other than some combinations are less likely to happen.
When applied to the defining element 44 is most common  comprised mostly of cycles ending in 2, i.e not many 3+1s or 1+3s.
Single hit
5 or 3 are less likely than 2,4,6
2hit Combo
35 or 53 is the rarest
3hit Combo
335
453
355
533
354
435
534
553
I guess if we encountered the above combos then we've effectively strayed off path and were more likely to return to something like:
352 or 536
And I doubt the risk/reward is impacted much when choosing an odd/even path?

Hi Folkner,it will b best you put your logic in a program and do a bulk data testing,, if that's working and results are good then u can take it to next level...

Dyksexlic was a troll, who constantly contradicted himself because he had no clue. He's gone now, probably under a different username. And only inexperienced fools remain his followers.

It was also noted that the system should have a low drawdown. It works like a safety spring, but it may look like a ladder down. For example, a loss of 20 units, then a gain of 10, then again a loss of 15, then a gain of 7, etc. It becomes clear that W should prevail over L.
Let W> L by 10%, this means that the edge will not catch up. So you need to look for a line of the game where this parameter exceeds L.
And if W> L by 2%, I would be inclined to progression.
Peace for everyone!

Hmm, a hedge bet. That's probably the part of the game I have analyzed the least. I have never seen why and how a hedge bet could be useful. If we bet also on the opposite side, it reduces the loss, but also reduces the wins, so there is no point in doing that. By minimizing the negative permutations we are minimizing the positive permutations.
However, if our hedge bet is not just a simple opposite side bet, but some kind of a parallel non random game, it could be useful.
the method must be such that all permutations are covered OR the impact of permatuations that will cause a loss are minimized. Because roulette is a Dutch bet, we are not able to cover all permutations. So we must minimize the impact of "negative"permutations.
This minimization must follow the rules of a winning system: we must be 100% that the loss due to a negative permutation is less than the loss without covering for the negative permutation

Good, but there is a script.
A bet is made until all permutations appear, otherwise it is a game in anticipation of a certain trigger with a certain permutation.
For instance:
1 spin  1 permutation.
2 spin  2 permutation.
etc...
If the rules say that every spin is placed, we will be in a situation where we will be accompanied by a losing streak. And with each loss, the amount of chips will go down, in this scenario, you need to raise bets after each lost permutation, and this is the way to progression.
In short.
We do not know on which back a certain permutation will play. It can be spin 2, or maybe spin 15.

Passion, well, tell us your thought. So far, your words are a free retelling of what you read.

I'm trying to understand how NonRandom is meant to help overcome variance. So far I cannot spot any difference compared with random variance.
We could bet 11 out of 12 streets @ 92% or we could cover all outcomes of a dozen cycle except CL3 order 3 @ 93%.
Both require same risk to get the same reward = break even in the long run.
By rights we should win 11/12 times and lose 1/12.
If there is dispersion then we will lose 2/12 times or more = negative variance. Chasing losses is no different in each case.
If there is concentration then we will win 12/12 or more = positive variance. Stopping after X wins in a row is no different in each.
I guess if the Law of Large Numbers kicks in then it means we've been measuring the outcomes for too long then our betting decisions start to have lesser and lesser effect on the percentages, you could say.
So why might NonRandom be better? Only a few possible things come to mind:
1) Dozen Cycle outcomes occur across 3 spins and produce a variable result as opposed to a constant result  only when you take the results of several winning cycles or several losing cycles can you begin to measure the variance; for example, if 11 outcomes were all winners but only on spin 1 then it's not really accurate because the 93% is based on average winnings across all 3 spins.
2) Most cycle outcomes can be stitched or not stitched. By choosing whether to stitch a CL2 or something you are inadvertently betting for/against something else. Another thing to explore here is Person S' example about dozen pairs that total 4 being made up of 1+3 or 2+2.
3) The average cycle length for a repeat can be increased when certain outcomes are locked out. And if something takes longer to repeat then it means that stitching becomes more effective.
Without bringing in parallel streams I really cannot think of anything else that could possibly help fight variance in a NonRandom game compared to a random game?

It’s time to go to bed, so I will briefly express my opinion.
Tens have short cycles, which means they are more dispersive.
In the first example Prnky, on quadrangles it was 9%, provided that n = 4 + 1, and on dozens n = 3 + 1 ...
Accordingly, dozens are subject to more severe deviations.

It’s time to go to bed, so I will briefly express my opinion.
Tens have short cycles, which means they are more dispersive.
In the first example Prnky, on quadrangles it was 9%, provided that n = 4 + 1, and on dozens n = 3 + 1 ...
Accordingly, dozens are subject to more severe deviations.
Throughout my extensive testing, all individual (constant) cycle outcomes are subject to the same deviations as any official carpet selection with a similar probability ratio  notwithstanding my point about a variable outcomes, i.e. when we take several constant outcomes and combine them to produce a higher ratio, collectively  albeit based on a variable number of spins as opposed to finding out the result on the same cycle spin each time.
Incidentally, the variable outcome of NonRandom cycles could be compared to the following random betting plan:
Spin 1: bet dozen 1
Spin 2: bet dozen 1+2
Spin 3: bet dozen 1+2
If we always play up to 3 spins and stop on a win then we've effectively created the same type of variable outcome of 93%; or what difference is there? I guess without measuring the front runners we cannot play:
4) Multiple repeats of the defining element

I just realised something quite important actually: we've never been able to study variance because of random and curve fitting, i.e. different results across different data sets. However, cycles can provide us a framework to study it where the results will be the same across all data sets.
The cycles will need to stay finite in order to offer us keyframes that can be used as checkpoints:
First to 1 repeat; 2 repeats; 3 repeats (..) of either the Cycle Length or the Defining Element.
First to 1 repeat of inner cycle; first to 1 repeat of outer cycle; first to 2 repeats of outer cycle...
Returning to some earlier examples, I couldn't actually tell you right now if each of the outcomes represent a finite series or not  but I am guessing they are random as there's no defined limit:
s1, d2, d3, s3, s3, d2
CL1s, CL2s, CL3s, CL1d, CL2d
Anyway, with variance once a series of losses/dispersion kicks in the first rule is to wait them out because we don't know when it will end. Once the wins start to come back in, and under what frequency, we could start to measure the variance all the while keeping tabs on the law of large numbers. Whatever the stats are for variance it ought to hold true across any finite series of outcomes.

Falcor, according to RRBB, Priyanka ...
At first, they used a tracker, then, having learned the rules, they played without it.
How are you going to follow the dispersion and the law of large numbers, and even take into account the cycles? You create something cumbersome and complex, which you can vryat can use when coming to the casino. Well, add another bell, SSD, parallel wheel table  and you’ll get a monster at all. There are 5 theories in the theory of quantum mechanics, and all these 5 describe one  the M theory is called, but only by different methods and formulas. I suppose that the method, as we were told, should be easier.

The file has a difficult session, try playing dozens, you can use a weak progression and say the result.

Falcor, according to RRBB, Priyanka ...
At first, they used a tracker, then, having learned the rules, they played without it.
How are you going to follow the dispersion and the law of large numbers, and even take into account the cycles? You create something cumbersome and complex, which you can vryat can use when coming to the casino. Well, add another bell, SSD, parallel wheel table  and you’ll get a monster at all. There are 5 theories in the theory of quantum mechanics, and all these 5 describe one  the M theory is called, but only by different methods and formulas. I suppose that the method, as we were told, should be easier.
Just as they did  start out complex and then see if there's an underlying concept or simple pattern to overcoming dispersion. And to return to your earlier assumption: so far there's nothing to support the claims that dozens vs. quads or even random vs. nonrandom results in different variance behaviours or provides additional tools to tame it. All cycles are break even regardless of the number of pigeons, and subject to the same extreme variance as per random; nobody has been able to demonstrate otherwise. A steady weak progression doesn't help random so why would it help nonRandom?

If so, then my conclusion is incorrect. I meant that the wider the field of events, the more the probability changes.
Example
RR / BB / RB / BR  there is a 25% probability for each event. But if you do RRR / BBB / RRB / RBR / RBB / BBR / BRB / BRR  hmm and here 25% for RRR / BBB. So there is no advantage here, but I had a thought that in the second example CL8  it would seem only in 1%.

Ok, here are my results. The game was conducted with a progression of + 1 / 1.
Not cycles, but some principles from there.

Ok, here are my results. The game was conducted with a progression of + 1 / 1.
Not cycles, but some principles from there.
Looks good 👍🏻

If so, then my conclusion is incorrect. I meant that the wider the field of events, the more the probability changes.
Example
RR / BB / RB / BR  there is a 25% probability for each event. But if you do RRR / BBB / RRB / RBR / RBB / BBR / BRB / BRR  hmm and here 25% for RRR / BBB. So there is no advantage here, but I had a thought that in the second example CL8  it would seem only in 1%.
That's about right for 8 pigeons. Likewise, it's rare that 12 unique streets will show, but that doesn't help us with escaping break even or controlling variance. Since each of your pigeons are comprised of 2 or 3 stitched outcomes then that would match the first of the 4 concepts I posted previously  still to be tested  otherwise there's nothing else in it to support the variance claims:
"1) Dozen Cycle outcomes occur across 3 spins and produce a variable result as opposed to a constant result  only when you take the results of several winning cycles or several losing cycles can you begin to measure the variance; for example, if 11 outcomes were all winners but only on spin 1 then it's not really accurate because the 93% is based on average winnings across all 3 spins."

In Person S' previous example we couldn't cover more than a few of pigeons anyway because there would be too many deadlock situations:
RBR
BRB
Straight away that's a deadlock  even betting for uniques!

That's about right for 8 pigeons. Likewise, it's rare that 12 unique streets will show, but that doesn't help us with escaping break even or controlling variance. Since each of your pigeons are comprised of 2 or 3 stitched outcomes then that would match the first of the 4 concepts I posted previously  still to be tested  otherwise there's nothing else in it to support the variance claims:
"1) Dozen Cycle outcomes occur across 3 spins and produce a variable result as opposed to a constant result  only when you take the results of several winning cycles or several losing cycles can you begin to measure the variance; for example, if 11 outcomes were all winners but only on spin 1 then it's not really accurate because the 93% is based on average winnings across all 3 spins."
Who wrote that? I have never seen that quote.
Btw, in my opinion the game isn't always break even with uncontrollable variance. Maybe your approach is wrong.
Try this if you still have your vdw codes. Split up the game into 9 spins, and play for vdw only on a color that started the 9 spin cycle and run it for a 50100K spins. Maybe the result will be different from random, but I'm not sure.

That was my quote from this topic in a previous reply, as we've pretty much exhausted all approaches in trying to ascertain if there's something to the claims about variance in a NonRandom context  though a few of those remaining ideas are still pending in terms of testing (previously numbered 14).
@ati, If VDW can perhaps help us understand statistical dependency, NonRandom variance, and exploiting PHP cycles then wouldn't it make more sense to retrack on the first AP and then carry over the defining EC to the next cycle?
(http://xs://i.postimg.cc/7h2bSmZd/vdwcycles.png)
If playing all 9 spins then that is comparable to playing all EC PHP cycles to 3 spins regardless of any early CL1 repeats  happens to not be the usual way of playing cycles  not with PHP at least.
(http://xs://i.postimg.cc/QVZCm1Bt/vdwcycles2.png)
So which approach is meant to be the right one if we are discover something new about variance  "all 9 spins" or "up to 9 spins"? Or are both approaches equally valid?

wouldn't it make more sense to retrack on the first AP and then carry over the defining EC to the next cycle?
I don't know, it probably would make more sense. Waiting for the 9 spin cycle to end is a waste of time and betting opportunity. I did it that way because it was more simple.
I haven't tried other ways, and honestly I'm not a big fan of vdw as it is too difficult to track and code. But my test result of 80K no zero spins showed a very slight edge and no crazy variances. Biggest DD was 70 units at around the middle, but that's over 2000 spins. So I thought that if you decide to test it and have similar results, than maybe you would also come to the conclusion that not every result is random and there is still hope. But it's nothing playable of course, it was just a test I did and will probably not return to this.
(http://xs://pichost.net/i/2020/02/08/Capture6fb09a466ace790b.png)

The stats are quite interesting:
(http://xs://i.postimg.cc/63scqTbt/vdwstats.png)

VDW  if you use it even for events with an edge, for example 75/25, after a while it also loses. I tried this approach while playing EC loops.
Example RR / RBR / RBB
VDW  looks like this: RRB  L.
When statistics say that the EC will be the same as the previous one at 75%. Inserting this event into VDW, you need to make two bets since we don’t know what rotation R will come in combination number 3 ( RBB).

Who wrote that? I have never seen that quote.
Btw, in my opinion the game isn't always break even with uncontrollable variance. Maybe your approach is wrong.
Try this if you still have your vdw codes. Split up the game into 9 spins, and play for vdw only on a color that started the 9 spin cycle and run it for a 50100K spins. Maybe the result will be different from random, but I'm not sure.
I tested playing vdw only on a color that started the 9 spin cycle, but the result is still break even without any negative or positive edge as per betting front runners in standard cycles.

VDW  if you use it even for events with an edge, for example 75/25, after a while it also loses. I tried this approach while playing EC loops.
Example RR / RBR / RBB
VDW  looks like this: RRB  L.
When statistics say that the EC will be the same as the previous one at 75%. Inserting this event into VDW, you need to make two bets since we don’t know what rotation R will come in combination number 3 ( RBB).
Could you elaborate? I under EC cycles is 75% defined by same. What's that got to do with vdw or encountering a loss or two bets? You lost me again...

I tested playing vdw only on a color that started the 9 spin cycle, but the result is still break even without any negative or positive edge as per betting front runners in standard cycles.
That's interesting, I don't know why my results were different. I'm not sure how many spins you tested, I recommended at least 50K, because the break even period that you see in the middle of my chart is actually around 32,000 spins.
But it's quite likely that I was just lucky, because I don't see the dependencies in that game. In the past few months I have really started to understand more and more the dependencies rrbb was talking about, and how every bet must be dependent on a previous outcome. I even tend to agree with him that to find out if a system is a winner or not, requires zero spins of testing. Just by the description you could tell if the system utilizes the dependencies between events or not.
But unfortunately I have still not been able to create a system that has a steady win rate and won't collapse if an unlucky sequence of numbers come out.

That's interesting, I don't know why my results were different. I'm not sure how many spins you tested, I recommended at least 50K, because the break even period that you see in the middle of my chart is actually around 32,000 spins.
But it's quite likely that I was just lucky, because I don't see the dependencies in that game. In the past few months I have really started to understand more and more the dependencies rrbb was talking about, and how every bet must be dependent on a previous outcome. I even tend to agree with him that to find out if a system is a winner or not, requires zero spins of testing. Just by the description you could tell if the system utilizes the dependencies between events or not.
But unfortunately I have still not been able to create a system that has a steady win rate and won't collapse if an unlucky sequence of numbers come out.
But isn't the statistical dependency we are exploring here meant to be based on the starting color within each cycle so how can you say that you don't see the dependency? Or was he tallying Uniques vs. Repeats or something each and every spin across multiple cycles..?

Could you elaborate? I under EC cycles is 75% defined by same. What's that got to do with vdw or encountering a loss or two bets? You lost me again...
My guess is that he tried to utilize the statistical imbalance of an EC event in a vdw sequence. Unfortunately it cannot be done.
We cannot have equal risk and reward ratio for two different cycle lengths. And we have to bet every spin. The 75% only applies to the set of outcomes, never to the next spin. We will always (?) have 50/50 chance of winning on the next spin. There are certain dependencies that makes question this, but for now it remains a fact that the next spin on any EC is always 50/50.

so how can you say that you don't see the dependency?
Think about this, the color that started the cycle, has over 50% chance to end the cycle. Therefore during the cycle, there is less than 50% chance to see that color. So in theory the other color should have more chance to win. That's why I said that I don't see the dependency. The result was the opposite of what I expected.
The issue here is that above mentioned over 50% fact applies to number cycles, and not to a vdw cycle. There can be a repeat within the 9 spin, which would change what the outcomes are dependent on.

Could you elaborate? I under EC cycles is 75% defined by same. What's that got to do with vdw or encountering a loss or two bets? You lost me again...
As we know EC it is 50/50. But the cycles are already 75/25 in favor of the determining one. Using VDW, I tried to use cycles, since this is not an equally possible event, but an advantage event. But as I wrote this did not bring success, dispersion waves carried the win into the abyss. Remember the recipe  you need to combine 2 variables. One independent is events, the second is dependent. Which we are all looking for...

My guess is that he tried to utilize the statistical imbalance of an EC event in a vdw sequence. Unfortunately it cannot be done.
We cannot have equal risk and reward ratio for two different cycle lengths. And we have to bet every spin. The 75% only applies to the set of outcomes, never to the next spin. We will always (?) have 50/50 chance of winning on the next spin. There are certain dependencies that makes question this, but for now it remains a fact that the next spin on any EC is always 50/50.
That is reminiscent of my previous quote about a concept still to be tested:
"1) Dozen Cycle outcomes occur across 3 spins and produce a variable result as opposed to a constant result  only when you take the results of several winning cycles or several losing cycles can you begin to measure the variance; for example, if 11 outcomes were all winners but only on spin 1 then it's not really accurate because the 93% is based on average winnings across all 3 spins."

Think about this, the color that started the cycle, has over 50% chance to end the cycle. Therefore during the cycle, there is less than 50% chance to see that color. So in theory the other color should have more chance to win. That's why I said that I don't see the dependency. The result was the opposite of what I expected.
The issue here is that above mentioned over 50% fact applies to number cycles, and not to a vdw cycle. There can be a repeat within the 9 spin, which would change what the outcomes are dependent on.
That's the mirror opposite of betting for repeats  both break even  and rrbb said betting either uniques or repeats alone is a losing proposition. I doubt one approach or the other would assist us with battling variance or gaining edge. One thing that is interesting is that when there's a 75% defined by same on EC cycles there's more chance of a defined by same on the VDW cycle  but again it's still break even regardless of the 2 similar dependencies involved in the Russian doll.

As we know EC it is 50/50. But the cycles are already 75/25 in favor of the determining one. Using VDW, I tried to use cycles, since this is not an equally possible event, but an advantage event. But as I wrote this did not bring success, dispersion waves carried the win into the abyss. Remember the recipe  you need to combine 2 variables. One independent is events, the second is dependent. Which we are all looking for...
It's already been shown that uniques and repeats depend on the starting partition  resulting in "statistical dependency" as opposed to "functional dependency". It follows then that if we've already identified the dependency involved then it's a different concept entirely that we are looking for to complete the recipe?
Of course the defined by same dependency continues throughout each spin of the cycle:
1... more chance repeat will be on 1
12... more chance repeat will be on 1 or 2, etc.
Likewise, ati describes dependency on each spin, and hasn't yet denied the above as the dependency involved in the recipe  but it just seems like he ignores it and perhaps chooses to end some cycles prematurely regardless of the repeat? Speculating further: such a method would be the opposite of choosing whether to bet a CL2 by missing out CL1 vs. stitching a CL2 across both spins when concentrating primarily on repeats instead of uniques.

I was studying variance today... it certainly can be measured between cycles  but there doesn't appear to be anyway of overcoming it  even if you wait out the dispersion. Each "wave" is independent of the last.

I've got a good idea... we could always try betting for a sleeper based on all the previous cycles that have accumulated only single appearances or repeats. Such a game would also progress from covering many options to fewer options:
Example cycle sequence:
CL1, CL1, CL2o1, CL2o2, CL2o1, CL1....
We would bet:
CL23, CL23, CL2o2CL3, CL3, CL3, CL3...
According to my charts it cannot deadlock and is reminiscent of nontransitive betting.

The previous example as well as the variance test showing independent waves got me thinking...
To return to the variance problem as well as whether we should stitch an outcome or not and other concepts discussed herein, I am starting to think there might be a riddle to solve here:
(http://xs://i.postimg.cc/N0C0NSCx/SameDiff.png)
1) The outcome of the dozen cycle is not officially set in stone until the cycle has closed.
2) The cycle outcome can occur across 3 different spins and we have various ways of playing Same and Different based on stitching or missing out spins.
3) The repeat depends on the uniques; Same/Diff also depends on the spin number. For example, dozen 1 has more chance of repeating if it's in the starting partition; Diff has more chance of resulting if the cycle is on spin 2 as opposed to spin 1 or spin 3; Same has more chance on spin 1 followed by spin 2, etc.
4) Spin 1/CL1 can only be missed out and not stitched, so spin 23 are probably our main weapons.
5) Each cycle outcome is independent of the previous cycle besides the carrying over of the defining element; Same/Diff probability/maths expectation depends on several cycles as opposed to a single cycle.
6) Same is not necessarily better than Different because each bet is a break even bet in the long run (notwithstanding MLE)  but the dependencies differ in different situations.
7) Different cannot happen unless the cycle remains open beyond spin 1; Same on spin 3 cannot happen unless the cycle remains open beyond spin 2, etc.
8 ) We don't yet know why we should stitch, not stitch, play a spin or miss out a spin during an open cycle for the purpose of exploiting the variance  only becomes official when the cycle is closed.
Since finding out that waves of variance between cycles are completely independent of previous waves, and the wheel has no memory, it then follows that solving this riddle must go something like this...
We need to find a way to bet for multiple things to happen over the next X cycles  more specifically so that Same ends up totaling more than Different via each individual spin result (also tallying up to maths expectation, respectively). If we try to bet Same only then we reach the house limits when using progression or we get eaten by dispersion; if we try to bet Same on first spin only the same thing also happens, so we cannot force just a single part of the system. Perhaps we need to find some way to alternate a "betting wave" and make it dynamic somehow.
To test this we really need to think outside the box: I'm guessing we need to measure spin 1 vs. spin 2 vs. spin 3 tally when, say, Same is 5 and Different is 3; something like that might provide us the right betting template?

OK I've got some interesting findings...
1... 62% defined by dozen 1 (same)
same... 86% defined by same
same (CL1)... about 50% defined by CL1 for defined by same
same (CL2o1,CL2o2 or CL3 option)... less than 50% defined by same option for defined by same
ati said: we should be able to describe why a HG should win without actually playing it.
Well, we know that 62% requires a risk of up to 3 dozens with a max return of +2.
I can reveal that 86% requires a risk of up to 9 dozens with a max return of +4.
However, if we start an outer cycle with different then we have 60% chance of finishing on different  but we risk only 2 dozens through stitching with a max return of at least +100!
And depending on the diff option for the starting partition we could refine that edge further. So that's the riddle solved.... :thumbsup:

I use a dozen loop methods. I recently made over $ 1000 from this. So I can safely advise everyone to use this strategy!