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Outside the box: a different view on roulette numbers

Started by rrbb, May 30, 08:46 AM 2016

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0 Members and 6 Guests are viewing this topic.

Firefox

So what is the next bet, 1 or 2? They are both 50% (- the house edge if 0 (s) is/are present)

The next spin is all you can bet on. The previous even chances have absolutely no bearing on it. You get no cash from the casino for forming an arithmetic progression of even chances or whatever other nonsense is being used as a basis.

One can ask all the greatest mathematicians and scientists on the planet, and they will give you exactly the same answer.

No wonder the gurus hide behind a pile of secrecy and mumbo jumbo - because they have no advantage whatsoever!

Person S

Quote from: Firefox on Mar 24, 10:28 AM 2019
So what is the next bet, 1 or 2? They are both 50%

The next number is all you can bet on. The previous even chances have absolutely no bearing on it.

One can ask all the greatest matheticians and scientists on the planet, and they will give you exactly the same answer.

No wonder the gurus hide behind a pile of secrecy and mumbo jumbo - because they have no advantage whatsoever!

Chances are constant, this is absolute. But if I try to look at things in combinations, I do not interpret them as chances for the next spin.
I begin to think that they are not so important. Of course, this is my opinion, if Math-guys confirm this, I will only be glad

Blueprint

As soon as you try to predict, you fall into the trap and the law of large numbers will always catch up. 

I’m not predicting. 

Madi

Quote from: Blueprint on Mar 24, 04:09 PM 2019
As soon as you try to predict, you fall into the trap and the law of large numbers will always catch up. 

I’m not predicting.

I have seen u mention this in different place. What that? Only one case its possible if u play 37 number.

Firefox

If one can't predict, one can't play roulette.

Unfortunately you have to place a bet before the outcome of a spin. Any bet you place is your prediction.

You can call it anything you want, and use any amount of prevarication or double speak you want, but if you don't bet,  you don't play. You may as well just sit watching the numbers till paint dries and you grow old.

Blueprint


MoneyT101

Quote from: Firefox on Mar 24, 04:58 PM 2019
If one can't predict, one can't play roulette.

Unfortunately you have to place a bet before the outcome of a spin. Any bet you place is your prediction.

You can call it anything you want, and use any amount of prevarication or double speak you want, but if you don't bet,  you don't play. You may as well just sit watching the numbers till paint dries and you grow old.

Predicting isn’t the only way to play...

look at blueprint example...

67% no predicting!  Everything is tied to these numbers.

Dependence creates a bias

Multiple dependent events going on you just have to pick which ones will give you a profit overall
Simple once you get it!  Chased all the pigeons away and they were already in their hole

The General

Quote from: MoneyT101 on Mar 24, 05:28 PM 2019
Predicting isn’t the only way to play...

look at blueprint example...

67% no predicting!  Everything is tied to these numbers.

Dependence creates a bias

Multiple dependent events going on you just have to pick which ones will give you a profit overall

If each spin of the wheel is an independent trial, then how is that possible?  ::)

Can you provide an example on how it gives the player an edge?
Basic probability and The General are your friend.
(Now hiring minions, apply within.)

The General

QuoteDependence creates a bias

By the way, there is no dependence between roulette spins.  Each spin is independent of the previous spins.
Basic probability and The General are your friend.
(Now hiring minions, apply within.)

Blueprint

Quote from: The General on Mar 24, 09:30 PM 2019
By the way, there is no dependence between roulette spins.  Each spin is independent of the previous spins.

I’m not here to convince you.  Good luck to you.

Firefox

Quote from: MoneyT101 on Mar 24, 05:28 PM 2019
Predicting isn’t the only way to play...

look at blueprint example...

67% no predicting!  Everything is tied to these numbers.

Dependence creates a bias

Multiple dependent events going on you just have to pick which ones will give you a profit overall

11
121
122

I do know what he is doing, he is just listing  cycles in your terminology when 1 comes first. Cycles where a repeat is formed,  I didn't ask for cycles though, but as usual he answers another question. Yes 2/3 of cycles have both 1 and 2.  Cycle length 2.... order 1 and order 2. 67%.  Not too useful.

But to play along with your terminology for a bit. If you take the first number as 1 and then play for the end of the cycle and bet on 1:

First line 1 wins (W). EOC. C1 01 ...... Win 1 (50%)
Second line 2 wins (L) then 1 wins (W). EOC. C2 02 ....Break even (25%)
Third line  2 wins (L) then 2 wins (L). EOC. C2 01.....Lose 2 (25%)

E(x) = Sum xPr(x) = 1 x 0.5  + 0 x 0.25 - 2 x 0.25 = 0.5 - 0.5 = 0

So we played a cycle and the expectation is 0 units. If we'd played on 2 to the end, we'd have won 2 units 25% and lost 1 unit 50%, and broken even 25%. Same result, expectation 0 units.

As according to the prevailing conditions, as green zero did not come up and take a rake.

I asked for a situation where non-random would give an edge and he failed to demonstrate it.

So, still waiting....




MoneyT101

Quote from: Firefox on Mar 25, 12:05 AM 2019
11
121
122

I do know what he is doing, he is just listing  cycles in your terminology when 1 comes first. Cycles where a repeat is formed,  I didn't ask for cycles though, but as usual he answers another question. Yes 2/3 of cycles have both 1 and 2.  Cycle length 2.... order 1 and order 2. 67%.  Not too useful.

But to play along with your terminology for a bit. If you take the first number as 1 and then play for the end of the cycle and bet on 1:

First line 1 wins (W). EOC. C1 01 ...... Win 1 (50%)
Second line 2 wins (L) then 1 wins (W). EOC. C2 02 ....Break even (25%)
Third line  2 wins (L) then 2 wins (L). EOC. C2 01.....Lose 2 (25%)

E(x) = Sum xPr(x) = 1 x 0.5  + 0 x 0.25 - 2 x 0.25 = 0.5 - 0.5 = 0

So we played a cycle and the expectation is 0 units. If we'd played on 2 to the end, we'd have won 2 units 25% and lost 1 unit 50%, and broken even 25%. Same result, expectation 0 units.

As according to the prevailing conditions, as green zero did not come up and take a rake.

I asked for a situation where non-random would give an edge and he failed to demonstrate it.

So, still waiting....

You finally did one step into the correct direction.

But no one here is going to give things to you on a platter.  But take my last response to you where I told you what I would play as a gift and work that out using this idea. 

See what you can come up with...

Quote from: MoneyT101 on Mar 24, 06:43 PM 2019
To be honest with you depends how I’m playing.

But I would have won way before 33 uniques and multiple wins atleast a minimum of 2 wins

33 uniques equals

Atleast 10 ec cycles
Atleast 8 dozen cycles
Atleast 4 double streets cycles
Atleast 2 street cycles

When I say atleast it means that’s the minimum. Worst case scenario!  So normally it’s more.

Also my current game play is streets mixed with unique numbers.  So 33 uniques is profitable for me!

The way I would play is to remove a unit off any number that showed up within the street I am playing.

The system profits flatbet over 60 spins.  Then playing uniques compliments and gives it the edge to boost it. 

My street bet wins flatbet the more spins I bet and the unique play gives me the real profit

Two systems at play all based on info I’ve talked about  :thumbsup:
Simple once you get it!  Chased all the pigeons away and they were already in their hole

Firefox

Well I just showed playing an EC  cycle has no advantage. You've said yourself here and elsewhere that playing DS cycles alone has no advantage. Priyanka said the same things about streets and dozens.

So why should playing two things at once, both with a disadvantage, factoring in zero, make something with an advantage? The maths does not stack up.

How have you confirmed this advantage

(a) Calculation
(b)  Computer testing over many 100,000s of spins
(c) Perception and records from manual testing over a small number of spins.
(d) Advantage not quantitavely confirmed, only qualitatively.

MoneyT101

Quote from: Firefox on Mar 25, 10:18 AM 2019
Well I just showed playing an EC  cycle has no advantage. You've said yourself here and elsewhere that playing DS cycles alone has no advantage. Priyanka said the same things about streets and dozens.

So why should playing two things at once, both with a disadvantage, factoring in zero, make something with an advantage? The maths does not stack up.

How have you confirmed this advantage

(a) Calculation
(b)  Computer testing over many 100,000s of spins
(c) Perception and records from manual testing over a small number of spins.
(d) Advantage not quantitavely confirmed, only qualitatively.

The advantage is math based and holds.  It’s hard to spot it because of how big and how stacked the odds are against you in this game.

But if you use paper and pencil and work out information it makes sense and you can see it easier.  Then you can apply more spins and a lot of data to confirm it.

Yes The advantage is very small and can be seen with ec.  But requires minimum of 60 spins to see it start taking effect.

I know every spin in roulette is independent!

But there are dependent events happening.  If you find these dependent events then they just might help you gain an advantage.

For instance I shared 2 different dependent events. 

1. Dozen that defines the cycle repeats 60%

2. 93% of Line repeats in position 123 of rrbb chart

These events show dependency!  Can be used to you’re advantage!

I believe I’ve shared Enough info for everyone to come up with a way to win.   

Figure out how these events show dependency and figure out what you can do to try to win and avoid some loses. 

And you have your self a method that wins!
Simple once you get it!  Chased all the pigeons away and they were already in their hole

Firefox

Well the odds are not stacked against you very much. Only 2.7%. Which is why I can win under the right conditions using VB. I only need to consistently eliminate a small area where the ball falls less. Just a handful of pockets is enough.

Assuming for the moment, a small advantage hypothetically exists on a single zero wheel, would the same method be enough to overcome the double zero wheel?

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