Non-Random vs. the Brick-Layer

[falkor2k15]

Our best weapon to beat the game comes from knowing that a repeat will happen before all uniques have been laid out...


37 numbers that have yet to appear each have equal probability for their initial appearance, but we know that one MUST repeat before all 37 uniques have had their fair share of single appearances. And if we can wait long enough for 24 unique numbers to appear before a repeat then we reach a point of almost certainty that the next spin will be that repeat - if not the spin afterwards.

This goes against brick design! The human brick-layer will logically complete all 37 bricks for the bottom layer before moving onto layer two and proceeding to add a second brick on top of each of the 37 single foundation bricks.


Random, on the other hand, would choose to lay out only around 7 or 8 bricks before unexpectedly expectedly moving onto the second layer (the bottom course probably won't become complete till way into the 5th layer+).


Even with a wall only 3 bricks wide - comparable to Dozens - Random still has more chance of repeating before completing the bottom course: 121, 133, etc. Unlike the human brick-layer, with his logical design working one wall at a time, Random refuses to rest with it's repetitive behaviour: evolving more and more brick walls of different types, being built in parallel, albeit to the same blueprint...

EC
Dozen
Quad
Line
..

     .
       .
   . ;.
    .;
     ;;.
   ;.;;
   ;;;;.
   ;;;;;
   ;;;;;
   ;;;;;
   ;;;;;
   ;;;;;
..;;;;;..
  ':::::'
    ':`

Cycle Length (a constant with distance/length properties for a repeat)
Order (a constant based on a repeat's relationship to the single appearers)
Defining Element / Reverse Order (the main repeat outcome value; RO includes its relationship to past repeats)
Halves (the sleeper attributes of a cycle)

     .
       .
   . ;.
    .;
     ;;.
   ;.;;
   ;;;;.
   ;;;;;
   ;;;;;
   ;;;;;
   ;;;;;
   ;;;;;
..;;;;;..
  ':::::'
    ':`

Win/Lose Streaks (a constant with cluster properties)
Gaps (the sleeper attribute between cycles)

[RouletteGhost]

We know this

But waiting for 24 unique may take a lifetime

[falkor2k15]

We know this

But waiting for 24 unique may take a lifetime

For every person who waits a lifetime for 24 uniques there will be another person who also waits a lifetime for 24 uniques... most likely ahead of 2 to 3 people who will one day all be waiting parallel lifetimes for 24 uniques... do you understand? 

[FreeRoulette]

In business 90% is best usually. Getting that last 10% to perfection will cost more than getting to the first 90%.

In roulette, it is better to optimize. You don't need a guarenteed win, you just need that optimal spot where it tips in your favor.

I would guess most repeats happen between 8 and 12 unique numbers in a row.

[thelaw]

Falkor, I preferred this method when you called it Pattern Breaker.............uh..........I mean, when "John Legend" called it Pattern Breaker. 

Whoops!

[Turner]

This goes against brick design! The human brick-layer will logically complete all 37 bricks for the bottom layer before moving onto layer two and proceeding to add a second brick on top of each of the 37 single foundation bricks.

Theres Mortar this than meets the eye

[falkor2k15]

In business 90% is best usually. Getting that last 10% to perfection will cost more than getting to the first 90%.

In roulette, it is better to optimize. You don't need a guarenteed win, you just need that optimal spot where it tips in your favor.

I would guess most repeats happen between 8 and 12 unique numbers in a row.

It's around 7-8, but let's say you only target that spin for a repeat - notwithstanding zero - will you do better than break even? The problem is: you are only targeting 1 repeat, but if you were also expecting a repeat on a street or split at the same time as the number might we gain advantage?

[falkor2k15]

Let's say the next 5 repeats happen on these spins:

spin 8
spin 10
spin 8
spin 12
spin 14

If you bet spin 8 you would have won on 2 sequences but lost on the other 3.

1, 12, 32, 24, 7, 19, 10...

We can target that one of those uniques will repeat "soon" (mostly @ spin 8 = average) because of how random behaves contrary to the brick-layer, but how can we also predict when the repeat is likely fall on spin 10 or spin 9 instead of just hoping for the average and ignoring the rest of the sequences that will compensate for our winnings? In other words, we know the repeat "will" happen but we just don't know "when" (nor which number, etc.) - barring some rough guidance coming by way of the average stats. But it wasn't any stats per se that told us the repeat will happen, so instead of stats what mechanism(?) will give us the additional information we seek in order to gain greater predictability?

[Priyanka]

In other words, we know the repeat "will" happen but we just don't know "when" (nor which number, etc.) - barring some rough guidance coming by way of the average stats. But it wasn't any stats per se that told us the repeat will happen, so instead of stats what mechanism(?) will give us the additional information we seek in order to gain greater predictability?

Falkor - these questions are very thought provoking. Please do continue as it seems very interesting.

[maestro]

i guess what you will try to say next....and would be that you will try to catch repeat in actual spin value...just guess and if so then you will have to wait average four or five spin values till repeat

[Herby]

Calculated Probabilities of the first repeat of any shown number (in %)

{{1,2.7027},{2,5.25931},{3,7.46254},{4,9.14329},
{5,10.1935},{6,10.5792},{7,10.341},{8,9.58236},
{9,8.44931},{10,7.10453},{11,5.70282},{12,4.3717},
{13,3.2},{14,2.23535},{15,1.48879},{16,0.944243},
{17,0.569417},{18,0.325898},{19,0.176651},{20,0.0904612},
{21,0.0436414},{22,0.0197706},{23,0.00837944},{24,0.00330845},
{25,0.00121086},{26,0.000408421},{27,0.000126093},{28,0.0000353412},{29,8.90354×10-6},{30,1.99147×10-6},{31,3.89324×10-7},
{32,6.51701×10-8},{33,9.08199×10-9},{34,1.01159×10-9},
{35,8.44331×10-11},{36,4.69435×10-12}}

[Herby]

The diagramm to the calculation above:

link:://:.rouletteforum.cc/index.php?topic=17018.msg156596#msg156596