Outside the box: a different view on roulette numbers

[MoneyT101]

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?

Stacked against you because house edge is 2.7

But if you try predicting, due to variance and law of large numbers it’s very hard to stay close to even and only lose to the house edge using a mechanical method.  Things get worse if you add a progression. 

VB is great but you can only win under the right conditions. 

Just playing ec alone you can get 60% win rate.  That overcomes bothe wheels.

Imagine it like this for example playing straights  1/36 but winning at 1/12 odds when choosing

You still lose some spins but your choosing at better odds then you should be. 

[Firefox]

Just playing ec alone you can get 60% win rate.  That overcomes bothe wheels.

There seems to be some inconsistency there.

I showed above just playing a cycle of even chances did not lead to an edge. And you confirmed that playing DS alone even with your 93% statistic did not of itself lead to an advantage.

I thought the assertion was that two or more losing dependent cycles had to be played in parallel to achieve a small advantage. Yet you say just EC alone gives a 60% win rate.

[Person S]

 technically a cycle either starts on, and ends before a repeat, or it starts after a repeat and ends on a repeat. That kind of depends on the view.

Can anyone from experienced participants give an example of these cycles?

[Person S]

short test

[ati]

The forum is full of examples. Even the message that you copied from rrbb is a reply to examples of cycles.

What he means is that in a cycle there are only unique numbers. So on a repeat the cycle has ended and the new cycle has already started.

It would look like this for dozen cycles. There are cycles of length 1,2 and 3.

132/321/3/3/31/3/321...

[MoneyT101]

There seems to be some inconsistency there.

I showed above just playing a cycle of even chances did not lead to an edge. And you confirmed that playing DS alone even with your 93% statistic did not of itself lead to an advantage.

I thought the assertion was that two or more losing dependent cycles had to be played in parallel to achieve a small advantage. Yet you say just EC alone gives a 60% win rate.

You didn’t play correctly!  You just had an idea and I’m telling you you’re heading in the right direction.

You can achieve a win with just 1.  But with more you can improve the win

No inconsistency, still saying the same thing.

I told you what to look for.  You don’t need to run around in circles.  I pointed at it for you. 

Now you want me to show you how step by step?

[MoneyT101]

The forum is full of examples. Even the message that you copied from rrbb is a reply to examples of cycles.

What he means is that in a cycle there are only unique numbers. So on a repeat the cycle has ended and the new cycle has already started.

It would look like this for dozen cycles. There are cycles of length 1,2 and 3.

132/321/3/3/31/3/321...

For someone who understand better they would read it like how you showed

But for someone starting out it’s easier like this

1323... carrry over the 3 to start new cycle

3213.....carry over the 3 to start new cycle

33.....carry over the 3 to start new cycle

[Person S]

Thank you boyzzz!
It turns out 2 types:
1-starts and ends before repeating (length 2-3)
2 starts after repetition and ends with repetition (length 1)
I understand this ttak ...

[ati]

It's so hard to find the proper application of all the statistics. But I'm sure once you find it, it becomes obvious.

Sequences like the below are the worst. If I play every spin in both streams, I can easily lose 15+ units before the two cycles end.
There also seem to be a contradiction on the 5th and 6th spins, because one of the statistics says that there is over 90% chance that one of the last 3 lines will repeat. The other stat says there is over 60% chance the repeat will come from the low position.
Once you said you can't win playing for just repeats or playing for just uniques, I'm not sure if you still think that's true. If it is, then the question is, when to switch bets?

[ati]

.

[Person S]

Thank you boyzzz!
It turns out 2 types:
1-starts and ends before repeating (length 2-3)
2 starts after repetition and ends with repetition (length 1)
I understand this ttak ...

Or so?
33 length 1
33 - length1 (starts after repetition and ends at repetition)
322 - length 2 (started after repetition and ended before repetition)
For repetition, I take the defining dozen.

[MoneyT101]

It's so hard to find the proper application of all the statistics. But I'm sure once you find it, it becomes obvious.  YES

Sequences like the below are the worst. If I play every spin in both streams, I can easily lose 15+ units before the two cycles end.
There also seem to be a contradiction on the 5th and 6th spins, because one of the statistics says that there is over 90% chance that one of the last 3 lines will repeat. The other stat says there is over 60% chance the repeat will come from the low position.
Once you said you can't win playing for just repeats or playing for just uniques, I'm not sure if you still think that's true. If it is, then the question is, when to switch bets?

First understand the statistic and then you will find you’re answer.

In the chart you posted that’s only 7 spins.   The 93% chance didn’t have enough spins to kick in.  Right there is your 7% chance.

Do atleast 50 spins so you can see the percentages better.  100 spins is good and 200 spins clears it up.

1000 spins or more is just so you can see it holds no matter what

10000 is to much and unnecessary because you will see the same percentages just with more spins

You can do 100000 spins, but it’s no point percentages will hold!

[redhot]

First understand the statistic and then you will find you’re answer.

In the chart you posted that’s only 7 spins.   The 93% chance didn’t have enough spins to kick in.  Right there is your 7% chance.

Do atleast 50 spins so you can see the percentages better.  100 spins is good and 200 spins clears it up.

1000 spins or more is just so you can see it holds no matter what

10000 is to much and unnecessary because you will see the same percentages just with more spins

You can do 100000 spins, but it’s no point percentages will hold!

Easy to test and confirm this using the cycle length stats for double streets:

Cycle Length 1: 166288 --- 16.6288%
Cycle Length 2: 277432 --- 27.7432%
Cycle Length 3: 278163 --- 27.8163%
Cycle Length 4: 185184 --- 18.5184%
Cycle Length 5: 77330 --- 7.733%
Cycle Length 6: 15603 --- 1.5603%

These results are from 1 million cycles.

"93% of the time the repeat will be from the last 3 double streets" - So immediately we can include all CL1, CL2 and CL3 as they have no option other than to repeat from the last 3 so that gives us

166,288 + 277,432 + 278,163 = 721,883 cycles

For CL4 we have 1,2,3,4,(R). The repeat (R) could be from the last 3 results (2,3 or 4) OR it could be from the first result (1). The odds of it being from the last 3 are 3/4, the odds of it being from the first result are 1/4.

There were 185184 cycles that ended CL4, 3/4 of these would have the repeating number coming from the last 3 results which is:

185184 * 0.75 =  138,888

Applying the same logic to CL5 and CL6 we have:

CL5 = 77330 * 0.6 = 46,398
CL6 = 15603 * 0.5 = 7,802

In total:

721,883+138,888+46,398+7,802 = 914,971 cycles where the repeat was from the last 3 results.

914,971 / 1,000,000 = ~ 92%

The percentage holds over 1 million cycles

[MoneyT101]

Easy to test and confirm this using the cycle length stats for double streets:

Cycle Length 1: 166288 --- 16.6288%
Cycle Length 2: 277432 --- 27.7432%
Cycle Length 3: 278163 --- 27.8163%
Cycle Length 4: 185184 --- 18.5184%
Cycle Length 5: 77330 --- 7.733%
Cycle Length 6: 15603 --- 1.5603%

These results are from 1 million cycles.

"93% of the time the repeat will be from the last 3 double streets" - So immediately we can include all CL1, CL2 and CL3 as they have no option other than to repeat from the last 3 so that gives us

166,288 + 277,432 + 278,163 = 721,883 cycles

For CL4 we have 1,2,3,4,(R). The repeat (R) could be from the last 3 results (2,3 or 4) OR it could be from the first result (1). The odds of it being from the last 3 are 3/4, the odds of it being from the first result are 1/4.

There were 185184 cycles that ended CL4, 3/4 of these would have the repeating number coming from the last 3 results which is:

185184 * 0.75 =  138,888

Applying the same logic to CL5 and CL6 we have:

CL5 = 77330 * 0.6 = 46,398
CL6 = 15603 * 0.5 = 7,802

In total:

721,883+138,888+46,398+7,802 = 914,971 cycles where the repeat was from the last 3 results.

914,971 / 1,000,000 = ~ 92%

The percentage holds over 1 million cycles

I said I’m only speaking facts and thank you for showing numbers to back it up!

I think that’s enough info in the forum!  With that everyone should be a able to come up with some ideas on how to improve their games or create a winning method.

Please don’t post anymore Facts!  Let everyone do their own work. 

Enough info has been posted!  Anyone wanting to learn and win can go with facts and find other facts!

Facts will help you win this game!  You just learned something new that you didn’t know before and was hidden within the stats.

If you look at things differently you might find gems!

With this gem alone you have an edge! Work in private and good luck

[MoneyT101]

Dyksexlic attempted to share similar facts but his riddles were good. I’m still having trouble figuring some out.  But you should understand his word roulette a bit better try going back to that post