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Guaranteeing an infinite loss on BV Roulette?

Started by falkor2k15, Dec 23, 09:41 PM 2018

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falkor2k15

I was thinking about how to guarantee continuous losses on BV, which - when playing randomly - should break even in the long run.

After pondering it for some time I reckon I've found a solution!

How might you stop yourself winning, resulting in negative edge and escaping break even?

If we were to find a losing BV method then might there be a way to reverse it and turn it into positive edge?
"Trotity trot, trotity trot, the noughts became overtly hot! Merily, merily, merily, merily, the 2s went gently down the stream..."¸¸.•*¨*•♫♪:

donik7777

Sure Falkor!

Perfect loser we can convert to perfect winner.

falkor2k15

OK, so here's the perfect loser strategy:

1) Keep betting on the sleepers of a number cycle, starting with 35 numbers then 34 then 33...
2) When you win you must parlay the winnings plus 1 unit onto the next sleeper bet, letting it ride so to speak.
3) When the repeat comes you lose all your winnings minus 1 unit, so the cycle ends at least -1

But how to tackle to house edge when you try to stop yourself winning 100K?  >:D

Note: negative/positive progression doesn't work, so what is the opposite of parlaying? Hint: outer cycles?  :question:
"Trotity trot, trotity trot, the noughts became overtly hot! Merily, merily, merily, merily, the 2s went gently down the stream..."¸¸.•*¨*•♫♪:

Roulettebeater

What’s the point from your post ?

What are you trying to say ?
A dollar won is twice as sweet as as a dollar earned

nottophammer

Quote from: falkor2k15 on Dec 23, 10:08 PM 20181) Keep betting on the sleepers of a number cycle, starting with 35 numbers then 34 then 33...
Nimo he's catching on  :smile:
How do you win at roulette, simple, make the right decision

Shogun

Hope it’s ok to ask this question here.
Let’s say you get the perfect losing system. Let’s say you manage to reverse it into a system with a small positive edge for you. Let’s also say it is flat betting on lines   :wink:

How would you go about altering this system if you wanted to play at a casino with zero ?
I don’t think uk citizens can play at BV anymore ?

Thanks.

falkor2k15

The zero would be of later concern, but even without a zero it seems impossible to gain any kind of edge.

Unfortunately, this system reached the table limits at spin 17, which I believe is about 120 units for inside bets?

2669 - 2280 still leaves some profit
Edit: I might be 1,000 out here, as I was meant to reset to session BR to zero.  :-[

After spin 17 we can no longer guarantee a loss?

How about if we played 72 number roulette? Would we then be able to guarantee a loss every cycle?
"Trotity trot, trotity trot, the noughts became overtly hot! Merily, merily, merily, merily, the 2s went gently down the stream..."¸¸.•*¨*•♫♪:

6th-sense

Won’t work tried it a long time ago nicksmi used to reverse my systems ..
Think of it as an oscillating wave

You can either hop on
Or hop off
Only thing in this thinking is when it’s time to jump on or off the bus

falkor2k15

What consistent wave have you encountered? Since all bets are independent there doesn't appear to be a right entry or exit moment.

The only thing I haven't tested is jumping from different universes in the event of a repeat or deadlock. For this I need to construct many streams that are dependent on each other, but I've been too lazy to do it.
"Trotity trot, trotity trot, the noughts became overtly hot! Merily, merily, merily, merily, the 2s went gently down the stream..."¸¸.•*¨*•♫♪:

Steve

Only way to guarantee loss on bv is keep transacting. Even when you win, the math is against you when you withdrawal.
"The only way to beat roulette is by increasing the accuracy of predictions"
Roulettephysics.com ← Professional roulette tips
Roulette-computers.com ← Hidden electronics that predicts the winning number
Roulettephysics.com/roulette-strategy ← Why most systems lose

falkor2k15

I reckon I've figured out the secret...

Every number cycle beyond, say, 7 spins, has a cycle length 1/order 1/position 1 on a lesser stream:
High-Low
Dozens
Columns
Quads
Lines
Streets
+ other custom streams (may not even be needed).

Many of those streams can be hedged for an immediate repeat before the number cycle repeats - to be tested.
"Trotity trot, trotity trot, the noughts became overtly hot! Merily, merily, merily, merily, the 2s went gently down the stream..."¸¸.•*¨*•♫♪:

donik7777

Quote from: falkor2k15 on Dec 30, 09:01 PM 2018
I reckon I've figured out

Many of those streams can be hedged for an immediate repeat before the number cycle repeats - to be tested.

My opinion it right path. Only with 1 stream impossible beat random.
I think parachuting is similar.

falkor2k15

I'm testing these ideas using a positive progression instead of negative, as it's more easier to play.

With zero included we can only hedge 1 EC and 1 Dozen or Column at a time.

When a Dozen or Column CL1 is not available then we can hedge the Quads instead

When a Dozen, Column and Quad CL1 bet is not available then we can only bet zero + EC.

For this strategy to be consistent, therefore, we are required to have enough of the right streams that are compatible for hedging CL1 bets - there has to be enough CL1 bets available to fill the bet selection to capacity = a brand new concept.

Also, when the Quads are not available then the Lines and Streets are seldom available either, so Non-Random is possibly giving us direction in terms of selecting and parachuting streams!

HAPPY NEW YEAR!  O0

"Trotity trot, trotity trot, the noughts became overtly hot! Merily, merily, merily, merily, the 2s went gently down the stream..."¸¸.•*¨*•♫♪:

falkor2k15

I've had some interesting "random thoughts" today in the form of another brand new concept, which is leading on from the previous new concept about selecting the right streams.

When a number repeats within 4 spins then the dozens are also likely to repeat. Likewise, when a number repeats within 6 spins then the lines are also more likely to repeat. This has been referred to as dependency (on the same spin). So if we were hedging many streams together then when one loses the rest tend to follow suit - so why would anyone want to bet those together? Wouldn't it make more sense if we chose streams that are unrelated and likely to express their most likely outcomes without being dependent on another stream? That way we could hedge several independent streams and most of them should obey MLE.

Dependency allows us to bet on behalf of numbers using, say, dozens or lines. However, once those repeats have happened beyond spins 4-6 and the number cycle still remains open then we can no longer "bet on behalf" so to speak... or can we? I wonder if the number repeat past spins 4-6 would carry some kind of dependency with the 2nd or 3rd repeats of the aforementioned streams - or better still - the "shadow" of those streams. And you know what I am referring to by "shadow".... of course it's the positions streams. That whole idea is one concept I have not ever thought about or analysed till now. We do have the streets - but that can only guarantee dependency up to spin 12. We need some custom streams or new views of old streams that are going to take us into the latter part of a number cycle.
"Trotity trot, trotity trot, the noughts became overtly hot! Merily, merily, merily, merily, the 2s went gently down the stream..."¸¸.•*¨*•♫♪:

Kattila

No need for more than one stream, just create the perfect loser stream (only hit few certain positions)
and the  W will hit most of the times. The other important  positions will not late ....
123412341234 here only hit positions  4 and 8 ( then maybe pos 12 )
Will pos 1 or 2 or 3  hit in the next few spins ?
Use  the classic splits or *created * splits /groups for that pattern 1234......

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