The Repeat "where it happens and how it happens - there is so much undiscovered"

[falkor2k15]

Let's try to take our understanding of the repeat as far as possible and see where we end up through an exhaustive analysis:

Let's use Dozen Cycles for sake of simplicity, excluding zero:



Everything that happens in a Dozen cycle happens in all other pigeonhole universes: EC, Line, Street and other custom cycles, etc. However, once we start to go beyond 12 pigeons then the deadlocks disappear, but a deadlock is just an extreme cycle length such that you can no longer cover all the pigeons and still gain a profit - happens to only cause a problem in Roulette and not PHP/mathematics itself.

Dozens must repeat within 4 spins and are more likely to repeat on spin 3 = CL2. Specifically, the stats are:
CL1 (repeat on second spin): 33%
CL2 (repeat on third spin): 44%
CL3 (repeat on fourth spin): 22%

CL3 represents the aforementioned deadlock situation:
123...

If we bet all 3 dozens then we don't gain any profit, so we no longer have a bet on (only in Roulette; the repeat still has to happen in terms of maths).

However, in Roulette it doesn't actually matter if you bet only CL2 and avoid CL3 bets or even bet just CL1 because we still break even due to the cost of playing each outcome having a proportionate risk/reward. Again, PHP may seem significant in the maths world, but it knows nothing about our break even problem in Roulette and doesn't claim to help us understand that.

When does the repeat happen? Answer: closer to the average cycle length, i.e. CL2, albeit break even, and we don't know when CL2 will happen.

Like CL2, the repeat also happens more on Order 1 and Position 1 based on the starting partition:
1... more chance repeat will be on 1.



Order and Position are both counterparts of each other as demonstrated by the above chart - at least when it comes to Dozen Cycles. Since cycle length 1 only has O1 and P1 available and Cycle Length 2 has only O1-2 and P1-2 available then it follows that overall the repeat will happen on O1 and P1 more often. This is based on the starting partitions already having one appearance of a dozen, so it's easier to get one more to be awarded the repeat compared to the other two dozens that need two appearances in order to overtake the first front runner and be awarded the repeat instead.

Therefore, O1 and P1 are 63% = defined by same.

Like CL2, betting on O1 or P1 doesn't help us escape break even. We don't know when a repeat on CL2, O1 or P1 is likely to happen in the same way we don't know when CL3, Order 3 or Position 3 will happen beyond their maths expectation = break even based on equivalent risk/reward.

Summary
When does the the repeat happen? Answer: closer to average cycle length and order 1/position 1.
When does the average cycle length happen? Answer: Don't know
When does a repeat occur closer to CL2/O1/P1? Answer: Don't know

Dozens are independent. By using the repeat framework to try to predict dozens and gain edge we encounter the following variables that are also independent, resulting in no additional predictability beyond maths expectation and our break even game known as Roulette:
--Cycle
--Cycle Length
--Order
--Position

All independent from the last - just like the dozens that we are trying to predict!

So the only logical thing left to do is to apply similar cycle frameworks to the above variables and see if their averages help us to predict dozen repeats more accurately. If the same thing happens again, i.e. "we don't know" and "independent" then we are simply continuing to describe the unpredictable nature of random without any solution, i.e. nothing left to discover about the repeat.

[Herby]

Everything that happens in a Dozen cycle happens in all other pigeonhole universes: EC, Line, Street and other custom cycles, etc.

I would say :
Everything that happens in a Number cycle happens in all other pigeonhole universes: EC, Line, Street and other custom cycles, etc.
What do you mean different ?

More precise: if you start to record the numbers and a number repeat happens you have a repeat on all other lines, streets, ...

[falkor2k15]

I would say :
Everything that happens in a Number cycle happens in all other pigeonhole universes: EC, Line, Street and other custom cycles, etc.
What do you mean different ?

It doesn't matter which random stream we take when analysing the repeat. It can be a 3 number game, 36 number game or even a custom 72 number game. The PHP framework and variables are still the same, and the result is still break even when excluding the house edge. The only difference is that deadlocks can disappear when there are more pigeons - and instead the repeat in it's current unpredictable state becomes too costly to play.

[falkor2k15]

More precise: if you start to record the numbers and a number repeat happens you have a repeat on all other lines, streets, ...

Sure. When there's a unique dozen then there's also a unique number, and so when there's a repeat on the numbers there will be a repeat on the dozens if both cycles began at the same time. This is simply because the numbers share the same partitions as the dozens. Different streams can overlap differently on the same spin or outcome, but that still makes them independent between spins. By covering only dozens based on a number cycle or numbers based on a dozen cycle - or indeed hedging both together - still results in break even.

In other words, by adding the following streams alongside the existing ones we are no closer to a solution for predicting dozens/dozen repeats:
--Cycle
--Cycle Length
--Order
--Position
--Number
--EC
--Line
--Street
--X Cycle
--X Cycle Length
--X Order
--X position

All are independent to the last regardless of their connection to the Dozen repeat! And all have different stats/averages that results in a break even game with no predictability beyond what is maths expectation.

[ati]

There is a possible solution that Pri used. Play a game within. For example, CL2 dozen cycles have two options on a repeat, position 1 or 2, so play for them to repeat. 50/50 chance, but 2 to 1 payout.
Let's say the last CL2 repeated on pos 1, so the next time when you expect CL2, you bet for pos 1 to repeat. If pos 2 repeats then the next time you bet both (double dozens). One of them must repeat, then you can start over with single dozen bets.
Of course we cannot forget about CL3, which happens just often enough to bring down the profit to break even. So the question is, can you avoid or make CL3's profitable? This is a rhetorical a question, I wouldn't post the solution.

[falkor2k15]

Thanks ati - your idea sounds quite interesting and original!

Let's examine the concepts behind ati's ideas and how they fit into this investigation re: the nature of the repeat - and whether there is much to be discovered or not.

Firstly, when considering custom streams we can often combine the cycle variables to create a number of options. CL has 3 outcomes and position has 3 outcomes - combined they form 9 options. The opposite is to condense Order 1-3 (3 outcomes) to only Same and Different (2 outcomes).

Regardless of how we manipulate the streams and variables connected to the dozen repeat they remain independent and as unpredictable as the dozen/repeats themselves:
--Cycle
--Cycle Length
--Order
--Position
--Defining Element
--Cycle Length+Order
--Cycle Length+Position
--Cycle Length+Defining Element
--Cycle Length+Order+Defining Element
--Number
--EC
--Line
--Street
--X Cycle
--X Cycle Length
--X Order
--X Position
--X Defining Element
--X Cycle Length+Order
--X Cycle Length+Position
--X Cycle Length+Defining Element
--X Cycle Length+Order+Defining Element

The other concept ati discussed almost invalidates the repeat entirely - he suggests playing a random game that is no longer non-Random if I'm not mistaken? ati's idea is not too dissimilar from tracking repeats on dozens 1 and 2 - ignoring dozen 3 (or treating that separately). By splitting the game into tracking two sets of repeats at the same level doesn't really tell us much about the main repeat in the first place, but again appears to invalidate it. With ati's example we are looking at sub-repeats within the main game - even more far-removed from dozen repeats that is the main game.

Previously in this investigation we got as far as this:

Summary
When does the the repeat happen? Answer: closer to average cycle length and order 1/position 1.
When does the average cycle length happen? Answer: Don't know
When does a repeat occur closer to CL2/O1/P1? Answer: Don't know

So the only logical thing left to do is to apply similar cycle frameworks to the above variables and see if their averages help us to predict dozen repeats more accurately.

With ati's example we have already begun to touch on the above (in red) - and I plan to analyse it further - but it's certainly not looking promising based on the continued independence and lack of predictability we encounter with Random continuing to assert itself. It feels like we are very close to a brick wall - and it's these variable repeats (or sub-repeats) that will determine whether Random continues to build an ever complex web of different independent streams - or whether we discover any new information/concepts about the repeat that might lead us closer to predictability instead of more uncertainty.

[MoneyT101]

There is a possible solution that Pri used. Play a game within. For example, CL2 dozen cycles have two options on a repeat, position 1 or 2, so play for them to repeat. 50/50 chance, but 2 to 1 payout.
Let's say the last CL2 repeated on pos 1, so the next time when you expect CL2, you bet for pos 1 to repeat. If pos 2 repeats then the next time you bet both (double dozens). One of them must repeat, then you can start over with single dozen bets.
Of course we cannot forget about CL3, which happens just often enough to bring down the profit to break even. So the question is, can you avoid or make CL3's profitable? This is a rhetorical a question, I wouldn't post the solution.

Falkor isn’t the person for rhetorical questions.  He doesn’t understand....

Please don’t give him anymore direct ideas that were  already shared indirectly.

His mission is to find a winning bet an post it in the forum... keep this mind when giving falkor ideas

[falkor2k15]

reserved

[falkor2k15]

ati's ideas all resulted in break even - based on a number of different variations.

I also created position cycles to see at what stage the dozens repeat - but each stream turned out to be completely independent - resulting in break even based on maths expectation. And dependency on the same spin doesn't help us profit in the long term either.

Since all streams linked to the dozen repeat are similarly independent - this means there is nothing left to discover about the repeat in general.*

I've also tested a lot of stuff surrounding position 1/order 1 based on front runners and betting schemes over 2-4 spins at a time - but it's all break even.

Tell me something new about the basic repeat that I don't already know and haven't already tested extensively?

*Repeats on number-sets above 12 pigeons that seldom result in deadlocks are a special case and may help us in Roulette - but pose a different problem based on table limits - to be explored further in "reverse-engineering the HG" thread.

[Tinsoldiers]

Since all streams linked to the dozen repeat are similarly independent - this means there is nothing left to discover about the repeat in general.

Revelation

Tell me something new about the basic repeat that I don't already know and haven't already tested extensively?

As you asked for it - Stop thinking there is something hiding in repeats. There is nothing there, you might want to use that time to save the earth from another conspiracy.

[falkor2k15]

Unfortunately, when it comes to the bigger conspiracies nobody I've ever met is capable of discussing them, so any time I spend trying to inform others is wasted.

BTW, there was one interesting tidbit I learnt from the latest tests surrounding position 1:



When a cycle length 2 on the dozen repeats coincides with a cycle length 1 on the position repeats then it can only be position 2.

[Tinsoldiers]

Stating the obvious

[falkor2k15]

This one is quite interesting - but not the first time we've encountered outcomes that were locked out:

When a dozen repeat coincides with a position 1 at spin 2 of the position cycle (21 or 31) then it cannot be followed directly by another position 1-based dozen repeat that is PCL1.



In the first example following a dozen repeat on P1 the next double repeat happens to be PCL1 - but it's position 2 instead of position 1.

In the second example following a dozen repeat on P1 the next PCL1 that is position 1 was only possible because there was another double repeat in-between.

Hard to explain that one... doubt there's any exploit though.

[falkor2k15]

Here's a more extreme lockout I found:

In fact, this could be the most extraordinary find in my 5 years of studying Roulette! 

2/5 outcomes are locked out - potentially over many cycles!

This occurs when there's 3 unique positions ending in a position 1 dozen repeat:

[Steve]

Why repeats happen and when is so simple. Why is it so difficult to understand?