Gizmo

You can't really do much more than what you have already done.

I for one appreciate the software and explanations you have gave.

No player can go from step 1 to step 10 in my opinion without going through the intermediate stages.

All the best gamblers I have ever met were not young! 

One of the reasons I particularly like this idea is because I find it easy to follow for the most part.

Here are 2 play sessions.

The S represents the furthest back staying missing on both the table layout and the wheel layout.

The D represents one of the furthest back appearing from either the table layout or the wheel layout.

Session 1.

D
S
S
S
S
D
S
S
D
S
S
S
S
S
S
S
S
S
D
D
S
D
S
S
S
S
S
D
S
S
S
S

Session 2.

S
D
D
D
D
D
S
D
D
S
S
D
D
D
D
S
S
D
S
D
D
D
D
D
D
D
S
D
D
D
S
D
S
D
D
D
S

As you can see, it can be very streaky.

Some type of consistency has to be the key if you don't have a physical edge.

You can then work your way through a session confident of coming out ahead.

cheers 

LoL ralph!

Neigh onions on my burger thanks!

As a further example carrying on with the numbers from above....
　
19 2A
12 1C 3C
19 2A 2A
32 3B (1C) 3B (2)
18 2C (1) 3A (2)
31 3A (1) 3A (2)
02 1B (2C) 2B (C)

You can see that the 2C is furthest back for the table layout.

C is the furthest back for the wheel layout.

This is a situation where the bet is not costing much.

Leave out all of the 2C from the table layout.

13,14,15,16,17,18,19,20,21,22,23,24,3,6,9,12,27,30,33,36.

Leave out all of the C from the wheel layout.

34,10,29,26.

This only leaves 13 numbers left to bet on.

1,2,4,5,7,8,11,25,28,31,32,35,0.

The next number out was 32.

19 2A
12 1C 3C
19 2A 2A
32 3B (1C) 3B (2)
18 2C (1) 3A (2)
31 3A (1) 3A (2)
02 1B (2C) 2B (C)
32 3B (2C) 3B (C)

So you can see that the 2C from the table layout and the C from the wheel layout are still missing.

That was more or less just a dozen bet betting only 13 numbers.

You can get times where the furthest back tie (ie. 2C in this example) can also go on a long run of not appearing.

I was monitoring this at a live casino the other day and saw the furthest back missing from both the table layout and wheel layout for an astonishing 37 out of 45 times!

That's incredible if you ask me.

It would be like watching RED come out 37 out of 45 times.

I am sure it happens and I may just have witnessed a freak occurence. However it's well known for dozens and columns to sleep for paticularly long periods of time.

This method which combines both the table layout and wheel layout seems to throw up some nice runs.

I am going to try out a few of Stetson Baily's progression ideas with this one using a bit of (hopefully) good timing.

Here is something which you may find interesting.
　
It is based on finding the furthest back Dozen or Column using both the table layout and the wheel layout.

The best idea will be to show you an example of what I mean.

The first number out is 19.

This is represented by 2A. (2nd dozen and column A)

So far so good!

The next number out is 12.

This is represented by 1C.

So I have...

19 2A
12 1C

Now I am using the table layout to arrive at these.

However I also want to use the wheel layout as well.

So the first number out was 19 and it then went to 12.

I need to assign some dozen and column markings from the number 19.

It would look like this going around the wheel.

19 - 0
04 - 1A
21 - 1B
02 - 1C
25 - 1A
17 - 1B
34 - 1C
06 - 1A
27 - 1B
13 - 1C
36 - 1A
11 - 1B
30 - 1C
08 - 2A
23 - 2B
10 - 2C
05 - 2A
24 - 2B
16 - 2C
33 - 2A
01 - 2B
20 - 2C
14 - 2A
31 - 2B
09 - 2C
22 - 3A
18 - 3B
29 - 3C
07 - 3A
28 - 3B
12 - 3C
35 - 3A
03 - 3B
26 - 3C
00 - 3A
32 - 3B
15 - 3C

I hope you get the idea.

So the number 19 went to 12.

Looking at the 12 on that chart and you can see it's 3C.

So my marking card would look like this....

19 2A
12 1C 3C

I put the appropriate dozen and column from the table layout first.

Then I put the dozen and column from the wheel layout second.

The complicated part is that the wheel layout is going to change from spin to spin.

The number out after 19 was 12.

So the dozens and columns from number 12 are as follows....

12 - 0
35 - 1A
03 - 1B
26 - 1C
00 - 1A
32 - 1B
15 - 1C
19 - 1A
04 - 1B
21 - 1C
02 - 1A
25 - 1B
17 - 1C
34 - 2A
06 - 2B
27 - 2C
13 - 2A
36 - 2B
11 - 2C
30 - 2A
08 - 2B
23 - 2C
10 - 2A
05 - 2B
24 - 2C
16 - 3A
33 - 3B
01 - 3C
20 - 3A
14 - 3B
31 - 3C
09 - 3A
22 - 3B
18 - 3C
29 - 3A
07 - 3B
28 - 3C

So you can see the dozens and columns are always starting from the new number that appeared.

The 12 went to number 19.

19 2A
12 1C 3C
19 2A 1A

The 19 went to number 32.

19 2A
12 1C 3C
19 2A 1A
32 3B 3B

Now here is where I can find out what the furthest back dozen and column is for both the table layout and the wheel layout.

Looking at the first set from the table layout...

2A
1C
2A
3B

The furthest back is a tie between dozen 1 and column C.

Looking at the second set from the wheel layout...

3C
1A
3B

The furthest back is dozen 2. In fact it has not even appeared yet.

So the marking card now looks like this...

19 2A
12 1C 3C
19 2A 1A
32 3B (1C) 3B (2)

The furthest back is located in the brackets.

The 32 went to number 18.

19 2A
12 1C 3C
19 2A 1A
32 3B (1C) 3B (2)
18 2C (1) 3A (2)

1C was the tied furthest back from the table layout.

The C appeared and so now the 1 is the furthest back on its own.

The 2 is still the furthest back from the wheel layout.

Now here is the interesting part if you have got this far!

The furthest back dozen or column in both sets can go missing for long periods of time.

So how do you take advantage of this fact?

I will show you from the last number which was 18.

First of all, leave out the entire dozen 1.

That is the furthest back using the table layout.

Now it's a case of leaving out the furthest back 2 from the wheel layout.

Let's take a look from the last number out which was 18.

18 - 0
29 - 1A
07 - 1B...Left out.
28 - 1C
12 - 1A...Left out.
35 - 1B
03 - 1C...Left out.
26 - 1A
00 - 1B
32 - 1C
15 - 1A
19 - 1B
04 - 1C...Left out.
21 - 2A ...Leave out the 21.
02 - 2B...Already left out. (Belongs to dozen 1)
25 - 2C...Leave out the 25.
17 - 2A...Leave out the 17.
34 - 2B...Leave out the 34.
06 - 2C...Already left out.
27 - 2A...Leave out the 27.
13 - 2B...Leave out the 13.
36 - 2C...Leave out the 36.
11 - 2A...Already left out.
30 - 2B...Leave out the 30.
08 - 2C...Already left out.
23 - 3A
10 - 3B...Left out.
05 - 3C...Left out.
24 - 3A
16 - 3B
33 - 3C
01 - 3A...Left out.
20 - 3B
14 - 3C
31 - 3A
09 - 3B...Left out.
22 - 3C
So that is 20 numbers that are getting left out.

That only leaves 17 left to bet.

They are 18,29,28,35,26,0,32,15,19,23,24,16,33,20,14,31,22.

The 18 went to number 31.

19 2A
12 1C 3C
19 2A 2A
32 3B (1C) 3B (2)
18 2C (1) 3A (2)
31 3A (1) 3A (2)

The 1 is still missing from the table layout and the 2 is still missing from the wheel layout.

A session from Smartlive Casino last night produced a session where the 3 was the furthest back.

It was furthest back for both the table layout and the wheel layout and went on a 15 spin winning run.

I like the idea because the table layout and wheel layout are independent from each other.

You can get long runs of the furthest back stay missing for both.

Playing it this way equates to just slightly less than an even money chance.

There are a few things to take note about.

You could have 1C (a tie) missing in the table layout and 3 missing on the wheel layout.

You would need to leave out the entire 1C first which is 20 numbers.

Then you would leave out the 3 from the wheel layout.

This just means you end up betting less numbers.

It is possible you could have 2 ties for both the table layout and wheel layout.

This would result in very few numbers actually bet.

You also have the option of playing for the alternative to this bet.

That would mean looking for either the furthest back to APPEAR.

Say the 1 was furthest back for the table layout and the C was furthest back for wheel layout.

You would then bet all the first dozen and the C numbers from the wheel layout.

This works out to roughly about a 20 chip bet.

I have noticed after doing some research that it is not that difficult to follow the patterns with this.

I much prefer it as an alternative to an even money bet like RED or BLACK.

Superman

I am just gathering some data at the moment and then I am going to have a look at it. There are no real concrete ideas just yet.

cheers

One thing this shows me is that playing lots of numbers is a disaster waiting to happen. You are going to get hit by some bad variance when your group of selected numbers matches the majority of these underperforming numbers and there are a lot of them as you can see.

However there does always seem to be a small group of numbers that are performing at a rate of 50% better than average. The problem as you have likely figured is these numbers can change quickly. A hot number can go to sleep and a cold number can wake up.

So maybe one way of catching that consistency is to look for numbers that have appeared regularly in the previous cycles (see the red dots in the pics) You could also keep an eye on this for waking up numbers.

[attachimg=1]

[attachimg=1]

One thing I noticed looking at the pic is how six of the numbers have appeared in every cycle. I put a red dot next to those. Could be something or nothing.

I am breaking it down cycle by cycle because I want to take a good look at things. The danger as always with roulette is in making quick assumptions. I need to try and find something which happens consistently over all these cycles or maybe it's a case of trying to find an extreme situation which only happens at certain points.

[attachimg=1]

So just over half of all the numbers have not appeared on average 4 times over the 4 cycles.

[attachimg=1]

After the second cycle and there are 17 numbers that have not shown up the average of 2 hits over the 2 cycles.


[attachimg=1]

Here is the first cycle from a set of 370 spins I have.

There are 15 numbers that have not appeared.

I am going to go through 10 cycles just to see what happens!


[attachimg=1]

The best way to go forward with this in my opinion is to go through some cycles and see how many numbers can keep close to the average at each stage. I want to find out if there is a point where a higher proportion of numbers start lagging behind. That might then give me something to work with.