Roulette Outside the BOX

Morning Frodo,
so summary of it all,

1. track 37 spins.
2. bet all unhits, once they show up.
3. flat up to end of cycle, spin 74.
4. new high, retrack last 37 spins.
5. new high, go back to step 2.
6. no new high, remove all numbers.
7. bet all unhits from cycle 2, once they show
8. plus all 1 hits from cycle 2, once they show
9. proceed to end of cycle, spin 111.
10. continue process until new high.
11. max up to spin 185.

if you add all unhits from cycle 2 and also all 1 hits, you are tracking many numbers, aprox 18+, and they all need to be tracked until they show?
and flat?
thnx

 

isn't ot better to play it this way?

CYCLE 1
bet all unhits once they show
use a possitive progression 1-2-4-8-16
bet up to end of cycle spin 74.
complete cycle, even when you are in profit.
CYCLE 2
add all unhits from spin 37-74 once they show by using above progression until end of cycle spin 111.
proceed until new high, then retrack.
max up to spin 185.

cycle 2, play all unhits from cycle 1 once they show
cycle 2, play all unhits from cycle 1 and 2
using cycle 2 as a fresh cycle.
when you're on spin 74 remove all numbers that have hit more then 2 times from previous cycle. they are on expectation, no need for them to play. so we only proceding with numbers that have not hit or have hit only 1 time, leaving them below expectation.

 

Maybe better at all to forget all your progression and cycles and try to understand why and when numbers hit, why and when they not hit ?

 

Hi, jekh

In the last variation, the main idea is to make the numbers pool refresh.
Not all numbers from the zero hits hit on cycle 2. Those that remain at 1 hit at spin 78 will be included in the pool along the "unawakened" non hits from the first selection.
To give you an example:
Cycle 1(0-37 spins) no bet: pool contains numbers(1,17,18,22,24,26,31,33,35)
spin 78 from all numbers above he had 3 hits on number 17, 2hits on number 33, one hit on numbers 1,35.
The numbers 17 and 33 will be removed from the pool as they gave us profit. The numbers 1,35 will be kept in the pool as they did not produce profit.
Spin 78, BR is still minus
Selection pool:1,35,18,22,24,26,31,35 (non awakened zeros and non performing 1s)
Cycle 2(spins 78-111)
The pool will be refreshed with the above numbers plus all "new" 1 hits
You will bet ONLY after a hit on those numbers.
Lets say #1 hits at spin 81, bet on it until spin 111
#35 awakened again but showed 1 time(again a non performer) keep it in the pool for next cycle.
Spin 111, we are again on minus
#1 hit 2 times on this session and gave us profit, remove it from the pool.
Next cycle check spin 148
Continue with same algorithm until spin 185.
Flat bet.

The pool is refreshed at the end of every cycle.
Numbers that underperformed on cycle 2 can become hot on cycle3.

Hope it helps. I have not tried it with a progression yet although it could be included. The hit rate is very good so far in my tests.

Cheers

 

Frodo, I'll modify the code so that the program plays the way you've explained. So if at the end of any cycle (up to 185 spins) you are in profit do you go back to cycle one, ie re-track last 37 spins and create a pool of unhit numbers?

 

So just to clarify, this means that numbers which hit twice or more in any cycle (above expectation) are not played in the next cycle.

 

Hi, Jerome

You are correct.

After spin 185, the loop finishes. Be it plus or minus from previous 185 spins high.
At Spins 74( end of cycle 2), 111(end of cycle 3), 148 (end of cycle 4) the Bankroll is questioned.
If we are on Plus a the end of any of the cycles, GOTO step 1 ( new 37 numbers /OR last 37 numbers for continuity it doesn't really matter)- the numbers pool is refreshed with the new ZERO HITS numbers, then the algorithm ends after the next 185 spins or when at new high.
That is for the European roulette (37 numbers). For the American wheel, the cycles are of 38 numbers so the checkpoints are at spins 76,114,152,190 STOP.
Flat bet

Let me know if you have any other questions.

 

That is correct. The pool of numbers reduces with every cycle as the non performers and non hits from previous cycles ,hit at least 2 times in our current cycle (1 time to qualify a bet- and at least 1 time to profit)
All profit given numbers (2+ hits during current cycle) will be kept until the end of the current cycle and then removed from the list. The non performers and non hits will remain in the pool for the next cycle.

 

A non performer is a number that is bet on but does not give profit on the current cycle.
Ex: cycle 1(selection cycle) numbers (1,5,11,21,25,27,30,32,33) zero hits go to pool
cycle 2(spins 78-111) #5 hits at spin 93 and we be on it. At spin 111 #5 remains with 1 hit so it becomes a non performer. It will remain in the pool for the next cycle.
Cycle 3(spin 111-148) # 5 hits again at spins #120, #137 , #144 - At spin 148 we will remove it from the pool as in this cycle it hit! 3 times (1 to qualify as a bet and 2 times profit)

 

How many bets have you made so far in your tests and what is the current profit?

 

@Jerome,

If possible, it would be much appreciated if you are able to set this up for the American Wheel of 38 numbers. Many thanks.

 

Hi Jerome,

I tested it manually on RXtreme.

Total spins: 1218
Layouts placed 1532
Profit 631units
flat bet 1 unit

PS: If you are looking to shorten the sessions, once a selected number hits, REMOVE IT from the list. That reduces the pool numbers and the waiting times. Hope this helps.

 

Frodo, I would rather keep to the way you've been playing it, since that is what you've got the results for. The problem with systems is that it's so easy to tweak them, which makes it a nightmare for the programmer. Naturally all tweaks can be coded, but it all takes time and increases the complexity.

I have another question : when you start a new cycle does the waiting time (for the first number in the unhit/single hits pool) count towards the cycle, or do you only count it from when you actually start placing bets? eg. suppose you start cycle 2 with 12 unhit numbers and it takes 10 spins before one of those numbers hits, do you then only bet for 27 (37 - 10) spins or do you ignore the 10 spins waiting time and place 37 bets? (giving a total of 37 + 10 = 47 SPINS).

 

I can do that, but it would make sense to get results for the euro wheel first, since if it doesn't work on the single zero it won't work on the double-zero.

 

Hi Jerome,

All bets are placed inside the current cycle. In your example, you will only bet that number for the remaining spins of the cycle, id est 27 spins.
If the number shows up at spin 36 of the current cycle, you will bet it only for 1 spin.

Cheers

 

@Frodo
Maybe a dumb question, but let's say you have 12 unhit numbers from the first cycle.
7 of them hits only 1 time. then when cycle 3 starts do the remaining unhit become umhits again and need to hit at least twice again in the next cycle to be removed? sorry if you awnswerd it yet.

 

Hi Jekh,

This is a variation of the original system (progression/ play the hit numbers and leave them on the table once they hit first time)
In your example, in cycle 2, if the 7 bet numbers have missed in the 2nd cycle, remove all bets and then in next cycle, once they hit you bet them again.
The idea behind this is to catch the numbers that are close to the expected value and can potentially become hot in the current cycle.
Like i said above, if you want your sessions to be short, remove the numbers from the pool once they hit. I am trying to cut down the time of the attack, and to catch the repeaters in cycles. I need to test the idea on a larger sample of spins to get to a better conclusion.
Flat bet.

Cheers.

 

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Makes sense, Jerome. Thank you.

 

This makes sense as it removes those numbers that peek out only to go back into hiding for extended periods. While they may catch up, the time it takes for that to happen is unplayable in a B&M. Catching those that are ready with 3-4 hits in a cycle is what we’re after. Thank you for spending time of this ...