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falkor2k15

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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...
"Trotity trot, trotity trot, the noughts became overtly hot! Merily, merily, merily, merily, the 2s went gently down the stream..."¸¸.•*¨*•♫♪:

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ati

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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.

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falkor2k15

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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..?
"Trotity trot, trotity trot, the noughts became overtly hot! Merily, merily, merily, merily, the 2s went gently down the stream..."¸¸.•*¨*•♫♪:

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ati

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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.

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ati

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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.

Person S

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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...

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falkor2k15

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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."
"Trotity trot, trotity trot, the noughts became overtly hot! Merily, merily, merily, merily, the 2s went gently down the stream..."¸¸.•*¨*•♫♪:

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falkor2k15

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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.
"Trotity trot, trotity trot, the noughts became overtly hot! Merily, merily, merily, merily, the 2s went gently down the stream..."¸¸.•*¨*•♫♪:

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falkor2k15

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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.
"Trotity trot, trotity trot, the noughts became overtly hot! Merily, merily, merily, merily, the 2s went gently down the stream..."¸¸.•*¨*•♫♪:

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falkor2k15

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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.
"Trotity trot, trotity trot, the noughts became overtly hot! Merily, merily, merily, merily, the 2s went gently down the stream..."¸¸.•*¨*•♫♪:

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falkor2k15

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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:
CL2-3, CL2-3, CL2o2-CL3, CL3, CL3, CL3...

According to my charts it cannot deadlock and is reminiscent of non-transitive betting.
"Trotity trot, trotity trot, the noughts became overtly hot! Merily, merily, merily, merily, the 2s went gently down the stream..."¸¸.•*¨*•♫♪:

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falkor2k15

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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:



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 2-3 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?
"Trotity trot, trotity trot, the noughts became overtly hot! Merily, merily, merily, merily, the 2s went gently down the stream..."¸¸.•*¨*•♫♪:

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falkor2k15

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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:
"Trotity trot, trotity trot, the noughts became overtly hot! Merily, merily, merily, merily, the 2s went gently down the stream..."¸¸.•*¨*•♫♪:


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