Best of Six System

[buffalowizard]

Through some searching, I came to the conclusion that most of the time, all three dozens will appear in each 6 spin frame.
So to exploit this simply wait for 2 dozens to appear within six and then go for the 3rd dozen that hasn't hit.
These are all done within 6 spin frames.

Example:

3
1 <both dozens have shown so bet for dozen 2
3
3
2 <win on third

From Wiesbaden today:

2
2
1 trigger
3 win
3
2 trigger
3
2
1 win
2
2
1 trigger
2
1
1 loss
3
2 trigger
2
3
1 win
2
3 trigger
1 win
0
1
2 trigger
1
2
3 win
1
1
3 trigger
2 win
2
2
2
3 trigger
1 win
1
3 trigger
3
2 win
2
2
1 trigger
3 win

If for example you have

2
2
2
2
1

Then still keep to the method and bet dozen 3 once

So again, bet for all three dozens to appear within a six spin frame. If lost then move to next and retrack

I thought mild 11,22,33,44 progression

BW

[Colbster]

The odds of not getting the other dozen in 6 spins is (2/3)^6, 64/729, or less than 1/11.  With the 2-1 payout, there is a decent chance here to work with a very slow progression for dozens to give you lots of chances to win.  The question becomes how do we optimize your method?  After 1-2, do we chase 3 for the next 4 spins?  Last 3 of 6, 2 of 6, 1 of 6?  Each has to be weighed with the likelihood of getting a hit vs. the potential cost of betting 4 spins instead of the last 1 or 2.  Also, waiting for 4 or 5 spins to get the better risk-to-reward ratio comes at the cost of fewer betting opportunities.

I like the idea but am curious how you would specifically apply it.

[buffalowizard]

Colbster, great reply, thankyou

The way I have looked at this is to stick rigidly to the six spin structure regardless how many bets are left. If we have 2 bets to go, then bet twice and then move on.

I'm sure this could be tweaked better and hopefully improved. The correct MM would also do with some altering

[warrior]

Colbster, great reply, thankyou

The way I have looked at this is to stick rigidly to the six spin structure regardless how many bets are left. If we have 2 bets to go, then bet twice and then move on.

I'm sure this could be tweaked better and hopefully improved. The correct MM would also do with some altering

cubano had a system like this.

[buffalowizard]

Do you have a link to the Cubano system warrior? thanks

[warrior]

Do you have a link to the Cubano system warrior? thanks

Search cubanopro stanard deviation.

[ego]

 
I write this so i can find this topic later as i am at work now.
I did post the same idea at notepad.
Will get back and add my 2 cent to this topic.

Cheers

[buffalowizard]

cheers warrior/ego


Just goes to show - not many new ideas out there - but maybe doesn't mean that some shouldn't get given a second chance.

[warrior]

cheers warrior/ego


Just goes to show - not many new ideas out there - but maybe doesn't mean that some shouldn't get given a second chance.

All you can do is try.

[chiricahua]

I think this is the cubano system link:

link:://vlsroulette.com/index.php?topic=16973.msg120222#msg120222

[chiricahua]

Thanks for share Buffalo

[Colbster]

The problem with this sort of selection is that 2 of your 6 are spent determining where we are going.  Those are already in past, and we need to remember that past spins don't have a bearing on future spins.  If 1 and 3 hit, they are past.  They have no bearing on us expecting 2 to hit in the next 4 spins.  The odds of 2 hitting in the next 4 is (1-(2/3)^4), 65/81, or about 4 in 5.  Even with a safe progression like 1,1,1,2, we can only expect to break even on a no-zero table, meaning we fall prey to the house edge.  The only way it gets good is by waiting for the right circumstances to appear, specifically meaning that 1 and 3 hit in some order during the first 2 spins of any given 6 spins with a rigid 6-spin cycle.  This essentially becomes a 0,0,1,1,1,2 progression.  It happens quite rarely that the spins fall exactly this way, meaning there would be a lot of tracking and not much betting.

[amk]

I don't know Colbster,

You have strong stats....

"The odds of not getting the other dozen in 6 spins is (2/3)^6, 64/729, or less than 1/11."


I only disagree with the "or less than 1/11" stat. Should be a little more than 1/11  (1/11.39)

Could it go 1000 spins/15 sessions of 11 games, each game consisting of 6 spins without seeing an overall average of 1/11??

Statistically after max X amount of spins we should always see a 1/11 average appear.  I don't think this can be more than 2000/3000 spins max? My calculations for the average units won is advantages but there is always a possibility they are wrong.



Sorry Buffalowizard, : )

Thank you for the thread! Really creative!

[buffalowizard]

The problem with this sort of selection is that 2 of your 6 are spent determining where we are going.  Those are already in past, and we need to remember that past spins don't have a bearing on future spins.  If 1 and 3 hit, they are past.  They have no bearing on us expecting 2 to hit in the next 4 spins.  The odds of 2 hitting in the next 4 is (1-(2/3)^4), 65/81, or about 4 in 5.  Even with a safe progression like 1,1,1,2, we can only expect to break even on a no-zero table, meaning we fall prey to the house edge.  The only way it gets good is by waiting for the right circumstances to appear, specifically meaning that 1 and 3 hit in some order during the first 2 spins of any given 6 spins with a rigid 6-spin cycle.  This essentially becomes a 0,0,1,1,1,2 progression.  It happens quite rarely that the spins fall exactly this way, meaning there would be a lot of tracking and not much betting.

Colbster,
It wouldn't take too long to wait for two different dozens to appear one after the other, and then bet 4 times, and then retrack for another two different consecutive.
It will take longer, but would definitely be a stronger version. As an example:
1
3 <now bet 4 times for 2
1
2 <win. Retrack
2
2
3
1 <bet 4 times for 2


Worth looking into

[buffalowizard]

amk,


No worries! My heads a little lost in the calculations but glad you're interested mate