0 Members and 2 Guests are viewing this topic.

*

ati

  • 500+ posts Member
  • *****
  • 533
  • Roulette Forum .cc | Lurker
  • Rated: +47
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.

*

falkor2k15

  • 2500+ posts member!!
  • *****
  • 2502
  • Roulette Forum .cc | Member
  • Rated: +111
The stats are quite interesting:
"Trotity trot, trotity trot, the noughts became overtly hot! Merily, merily, merily, merily, the 2s went gently down the stream..."¸¸.•*¨*•♫♪:

Person S

  • 100+ posts Member
  • ***
  • 196
  • Member
  • Rated: +16
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).

*

falkor2k15

  • 2500+ posts member!!
  • *****
  • 2502
  • Roulette Forum .cc | Member
  • Rated: +111
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 50-100K 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.
"Trotity trot, trotity trot, the noughts became overtly hot! Merily, merily, merily, merily, the 2s went gently down the stream..."¸¸.•*¨*•♫♪:

*

falkor2k15

  • 2500+ posts member!!
  • *****
  • 2502
  • Roulette Forum .cc | Member
  • Rated: +111
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..."¸¸.•*¨*•♫♪:

*

ati

  • 500+ posts Member
  • *****
  • 533
  • Roulette Forum .cc | Lurker
  • Rated: +47
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.

*

falkor2k15

  • 2500+ posts member!!
  • *****
  • 2502
  • Roulette Forum .cc | Member
  • Rated: +111
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..."¸¸.•*¨*•♫♪:

*

ati

  • 500+ posts Member
  • *****
  • 533
  • Roulette Forum .cc | Lurker
  • Rated: +47
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.

*

ati

  • 500+ posts Member
  • *****
  • 533
  • Roulette Forum .cc | Lurker
  • Rated: +47
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

  • 100+ posts Member
  • ***
  • 196
  • Member
  • Rated: +16
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...

*

falkor2k15

  • 2500+ posts member!!
  • *****
  • 2502
  • Roulette Forum .cc | Member
  • Rated: +111
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..."¸¸.•*¨*•♫♪:

*

falkor2k15

  • 2500+ posts member!!
  • *****
  • 2502
  • Roulette Forum .cc | Member
  • Rated: +111
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..."¸¸.•*¨*•♫♪:

*

falkor2k15

  • 2500+ posts member!!
  • *****
  • 2502
  • Roulette Forum .cc | Member
  • Rated: +111
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..."¸¸.•*¨*•♫♪:

*

falkor2k15

  • 2500+ posts member!!
  • *****
  • 2502
  • Roulette Forum .cc | Member
  • Rated: +111
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..."¸¸.•*¨*•♫♪:

*

falkor2k15

  • 2500+ posts member!!
  • *****
  • 2502
  • Roulette Forum .cc | Member
  • Rated: +111
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..."¸¸.•*¨*•♫♪:

 

Popular pages: