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A dozens system for the grinders.

Started by flukey luke, May 23, 11:19 AM 2012

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flukey luke

Hello.

I posted a system on another forum about a year ago which I will post here as well.

It is a bit of a grinder but the results from my testing showed a lot of promise. Maybe some of you guys can take a look and tweak it if you wish to come up with something stronger.


The best way to explain it will be to go through it step by step using an example. Hopefully that will give you an idea as to what is going on.

The actual bet itself is a dozens bet. You will be betting on two six lines.

The six lines comprise of the following.

1-6 = 1.
7-12 = 2.
13-18 = 3.
19-24 = 4.
25-30 = 5.
31-36 = 6.

The progression for the bet is as follows.


Step 1.   bet 2 (1 on each six line) win +4  /  lose -2.
Step 2.   bet 2 (1 on each six line) win +2  /  lose -4.
Step 3.   bet 4 (2 on each six line) win +4  /  lose -8.
Step 4.   bet 6 (3 on each six line) win +4  /  lose -14.
Step 5.   bet 8 (4 on each six line) win +2  /  lose -22.
Step 6.   bet 12 (6 on each six line) win +2  /  lose -34.
Step 7.   bet 18 (9 on each six line) win +2  /  lose -52.
Step 8.   bet 28 (14 on each six line) win +4  /  lose -80.
Step 9.   bet 42  (21 on each six line) win +4 /  lose -122.
Step 10. bet 62  (31 on each six line) win +2 /  lose -184.


So let's get started with the example.

36 is the first number out. What you need to do is keep a count of all the six lines and keep a rolling count of how many times they are missing.

So because 36 was the first number out, your count would look like this.

1 (1) 2 (1) 3 (1) 4 (1) 5 (1) 6 (0) The 6 sixline is the first one to hit and so all the others are missing after the first spin.

OK, let's move onto the next number.

35  You also need to check if the new six line that has just appeared is an alternate from the last one.
In this instance, it is not. So I like to mark it like this.

35 (N)  The N stands for (no)

How does the rolling count look now. 1(2) 2 (2) 3 (2) 4 (2) 5 (2) 6 (0) The 6 sixline repeated and therefore all the other sixlines have went missing for 2 spins.

next number.

22  Now this is an alternating six line. So I mark it like this.

22 (Y) The Y stands for (yes)  I also need to keep a running count of how many times the six line does alternate from the previous one. So you could mark it in the following fashion....

22 (Y) (1)
The rolling count now looks like this. 1 (3) 2 (3) 3 (3) 4 (0) 5 (3) 6 (1)

Next number.

16 (Y) (2)      ROLLING COUNT. 1 (4) 2 (4) 3 (0) 4 (1) 5 (4) 6 (2)

Next number.

27 (Y) (3)       ROLLING COUNT. 1 (5) 2 (5) 3 (1) 4 (2) 5 (0) 6 (3)

Next number.

28 (N) (0)  The 0 is here because the 25-30 sixline just repeated.
    ROLLING COUNT. 1 (6) 2 (6) 3 (2) 4 (3) 5 (0) 6 (4)

35 (Y) (1)      ROLLING COUNT. 1 (7) 2 (7) 3 (3) 4 (4) 5 (1) 6 (0)

3 (Y) (2)       ROLLING COUNT. 1 (0) 2 (8} 3 (4) 4 (5) 5 (2) 6 (1)

19 (Y) (3)      ROLLING COUNT. 1 (1) 2 (9) 3 (5) 4 (0) 5 (3) 6 (2)

35 (Y) (4)      ROLLING COUNT. 1 (2) 2 (10) 3 (6) 4 (1) 5 (4) 6 (0)

3 (Y) (5)        ROLLING COUNT. 1 (0) 2 (11) 3 (7) 4 (2) 5 (5) 6 (1)

12 (Y) (6)      ROLLING COUNT. 1 (1) 2 (0) 3 (8} 4 (3) 5 (6) 6 (2)

3 (Y) (7)         ROLLING COUNT. 1(0) 2 (1) 3 (9) 4 (4) 5 (7) 6 (3)

36 (Y) (8}       ROLLING COUNT. 1 (1) 2 (2) 3 (10) 4 (5) 5 (8} 6 (0)

Here is where you have the first bet. Once you see the current six line has alternated at least 8 times, that signals there could be a possible bet.
What you then need to do is go over to the rolling count and check for the furthest back six line.
Here it is the 3 which has missed 10 times. (The qualifying target for a possible bet  is 8+. If there are more than one with 8+, always play the one which has missed for the longest.)

So both qualifying targets have occurred.

How do you bet? I am going to bet for the current six line to repeat. (In this instance, it's the 6)
I am also going to bet for the longest missing six line in the rolling count to appear. (In this instance, it's the 3)

So betting the 6 six line and the 3 six line combines as a dozens bet. I am going to use the 10 step progression.

16 (Y) (9)         ROLLING COUNT. 1 (2) 2 (3) 3 (0) 4 (6) 5 (9) 6 (1)

That was a winning bet on the 1st step of the progression. The missing 3 from the rolling count came in.

You will notice that there is also another bet straight away.  Why?

Because the six lines are still alternating (currently standing at 9 which qualifies for a bet and the 5 six line in the rolling count is also at 9. This also qualifies.

So now I am betting for a repeat on the 3 six line and also betting the 5 six line.

28 (Y) (10)          ROLLING COUNT. 1 (3) 2 (4) 3 (1) 4 (7) 5 (0) 6 (2)

That was a winning bet on the 1st step of the progression. The missing 5 from the rolling count came in.

There is now no bet on the next spin. There are no qualifiers in the rolling count.

8 (Y) (11)           ROLLING COUNT. 1 (4) 2 (0) 3 (2) 4 (8} 5 (1) 6 (3)

There is a bet now because we have a qualifier in the rolling count.
I am looking for a repeat of the 2 six line and also playing the missing 4 six line.

25 (Y) (12)       ROLLING COUNT. 1 (5) 2 (1) 3 (3) 4 (9) 5 (0) 6 (4)

That was a losing bet. I will now move up to step 2 in the progression and I am betting for a repeat 5 six line and the missing 4 six line.

6 (Y) (13)        ROLLING COUNT. 1 (0) 2 (2) 3 (4) 4 (10) 5 (1) 6 (5)

That was another losing bet and I will now move up to step 3 in the progression. I am betting for a repeat 1 six line and the missing 4 six line.

7 (Y) (14)       ROLLING COUNT. 1 (1) 2 (0) 3 (5) 4 (11) 5 (2) 6 (6)

That was another losing bet and I will now move up to step 4 in the progression. I am betting for a repeat 2 six line and the missing 4 six line.

22 (Y) (15)       ROLLING COUNT. 1 (2) 2 (1) 3 (6) 4 (0) 5 (3) 6 (7)

A win there on the missing 4 six line.

There is no bet on the next spin.

27 (Y) (16)         ROLLING COUNT. 1 (3) 2 (2) 3 (7) 4 (1) 5 (0) 6 (8}

There is a bet on the next spin. I am betting for a repeat on the 5 six line and also looking for the missing 6 six line.

20 (Y) (17)         ROLLING COUNT. 1 (4) 2 (3) 3 (8} 4 (0) 5 (1) 6 (9)

That was a losing bet on step 1 of the progression.

I am betting again on the next spin and looking for a repeat of the 4 six line or the missing 6 six line.

31 (Y) (18)         ROLLING COUNT. 1 (5) 2 (4) 3 (9) 4 (1) 5 (2) 6 (0)

A win there on the 2nd step of the progression.

There is another bet on the next spin. I am looking for a repeat of the 6 six line or the missing 3 six line.

25 (Y) (19)         ROLLING COUNT. 1 (6) 2 (5) 3 (10) 4 (2) 5 (0) 6 (1)

That was a loss on the 1st step of the progression. I will be betting again looking for a repeat of the 5  six line or the missing 3 six line.

30 (N) (0)           ROLLING COUNT. 1 (7) 2 (6) 3 (11) 4 (3) 5 (0) 6 (2)

That was a win on the 2nd step of the progression. Finally the street repeated. There is no bet on the next spin and it will take a while now for one of the qualifying rules to work it's way back up to at least 8 again.

So there were 5 bets there in all never going past step 4 in the progression.

***This was the original example I wrote out last year***

What I later found out is that it works MUCH better if you have the time to wait for the alternating dozens to hit 12 before looking for bets. You can still leave the rolling count at 8+. I suppose it depends how much patience and time you have. You also could come up with a more gentle progression instead of the one I propose here.

I am just throwing it out there. Play around with it if you like and see what you can come up with.

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