The Non-Random riddle solved?

[falkor2k15]

Over the past 2 weeks I've been re-reading some of the Non-Random corpus of information from reddwarf and priyanka before re-running a few tests. Something then occurred to me that could solve the riddle.

First let's cover the problems I have been unable to resolve:
1) Turn deadlocks (all uniques showing without a repeat) into winning game? No idea... not even an inkling.
2) Numbers + Positions exploit. Have analysed both streams ala "outside the box" for countless days without seeing much connection let alone an exploit - seems like any 2 official streams that cannot be exploited either.

I was testing out a few more wacky "play your own game" ideas without trying to predict, but haven't been able to tilt the edge either way. I played all kinds of custom streams made up of same/different/open/closed/standard/position cycles and constant combinations.... resulted in NOTHING!

We cannot predict the next spin, so prediction is meant to be a special case of gathering more wins than losses from grouping characteristics during a cycle. Well, if each spin is independent then is it even possible to win, say, 4/6 bets over 6 spins played non-randomly vs. 1/2 spins played randomly? I can't think of any example setup of how that might be achievable? Though that seems to be the claim...

Priyanka confused us with different stats and more chance of red balls vs. green balls, when in actuality each bet selection has a break even outcome in the long term - doesn't matter which is more likely or least likely - and this is more in line with reddwarf's "fruit universes" where stats are abandoned in favour of fruits/pigeons and distances/positions.

So how does Non-Random actually help us then since this is meant to be the pre-requisite for finding the exploit - based on playing for something that has to happen?

Incidentally, some steps in Priyanka's games were betting for uniques instead of repeats. Now, this appeared to be legal providing we end each cycle on a repeat, resulting in a framework with consistent stats; and since we always end up breaking even, playing for uniques or repeats seemed no different to each other within that Non-Random framework. Red even said that uniques or repeats on their own are a losing proposition. However, the repeat has to happen in a defined limit - yet playing for uniques has no limit - seems to contradict Non-Random. In their defense, they did leave us to ponder the question of "when it is ok to play some steps random"?

Anyway, going back to basics. Each spin is independent. Non-Random says that stats are unimportant - only the number of unique objects/pigeons in a numberset/universe and their relationship. Erdos - governs laws surrounding unique numbers before any repeat - implies that some permutations/sequences are more likely than others. Is that true? When it comes to repeats, although everything is independent, it's implied that some magical relationship must exist between those independent observations.

It's worth pointing out here that Non-Random - describing patterns relating to any series of numbers - does not take into account any of the following:
1) How the numbers are generated, i.e. random or deliberately selected with bias
2) Any statistics/probabilities attached to those numbers/results
3) Roulette or any other games of probabilities.

Pigeonhole principle simply says a repeat has to happen - and this one small fact means that no random stream of numbers are completely random! Again, it doesn't know anything about or that we are playing Roulette. If we have equally-likely dozens we can reach deadlocks over and over where all uniques show without any playable repeats:
123123123123123123...

We could create our pigeons with different stats and keep playing for pigeon 1 to "catch up" or become the leader:
1 - 63%
2 - 29%
3 - 8%

But there's no guarantees pigeon 1 will win - nor any uniques will win - only that a repeat will occur.

Since there isn't really much that Non-Random can offer us there can't be many ways to use it to gain advantage. It's a separate branch of maths that has no knowledge of roulette - we are simply trying to apply a basic repeats principle from it.

Approaching a solution: equally-likely pigeons, such as official numbers/groups, have been shown to deadlock repeatedly; yet these deadlocks are reduced with pigeons that are not equally-likely. However, even with numbersets that are 6,9,12,18 or 24, non-equal pigeons can still reach deadlock or break the bank before the repeat occurs. Of course playing for a repeat on 37 numbers can definitely reach table limits before the repeat!

Something else then occurred to me: VdW is very much like the 2nd repeat of PHP! Let's say we have an AP of 123, then this is the 2nd repeat. It could be 246 and it's still like a 2nd repeat in the PHP world - VdW is meant to be an application of PHP after all!  Repeats must happen regardless of how the distribution strays from their maths expectation probabilities for each pigeons - 2 or 3 appearances of the same pigeon "cancel out" so to speak proving the Non-Random theorems to be true. So the idea that comes to mind is not to bet uniques or start switching to the pigeon that is most likely over the next pigeon - but simply increase the span for the number of repeats and keep within a closed system.

If we look at Six Dozen Options as an example, a repeat must happen within 6 cycles. We can still encounter a deadlock of 6 cycles on the first repeat, but the 2nd repeat has yet to result in a 2nd deadlock based on my tests and observations. The maximum number of cycles is therefore 11, and because of the staggered nature of the betting scheme, we don't always bet every spin nor require double dozens every spin. If we did play more double dozens over single then perhaps playing opposite would be better. And this is exactly how Priyanka suggested we play a "game within a game" - using opposite bets and multiple repeats - not what some people have claimed a "game within a game" to mean. It's been noted that a standard dozen and it's position could count as 2 pigeons but require only a 1 dozen bet on some occasions, thereby losing "apples and pears", so that could potentially help further. Anyhow, here's how an 11 cycle game looks:


The bet selections needed to win are as follows:
1 dz
1 dz
1 dz
1 dz
1 dz
1 dz
1 dz
2 dz
2 dz
1 dz
2 dz
1 dz
1 dz
1 dz
1 dz
1 dz
1 dz

Translating the above into a negative stepped progression, would it break the bank or stay within the table limits? I haven't calculated it. However, simply looking at the BR from flat-betting it looks suspiciously like negative edge, so perhaps flat-betting opposite would result in positive edge.

[Bigbroben]

Something you said that makes sense ...
it seems that you are seeing something ...

Hast Du was verstanden überhaupt?

[falkor2k15]

Some interesting relationship patterns here:

All 231s, 213s, 312s, 321s (800 each every 100K spins) NEVER deadlock!


123/132 deadlocked only ONCE in 100K spins.


Most 2nd repeats had at least two CL3s in the first 3 cycles and more x3s than x1s: