Another idea for using groups.

[flukey luke]

Hello guys,

Here is another idea I had for the groups that I use.

Just to recap about the groups using RED/BLACK as an example.

Group 1 = RED/RED.

Group 2 = RED/BLACK.

Group 3 = BLACK/BLACK.

Group 4 = BLACK/RED.

So suppose you see the following three groups appear.

1
2
3

This leaves Group 4 as the furthest back group. If Group 4 continues to be absent, you will
see one of either Group 1, Group 2 or Group 3 appear.  So by backing RED on the first spin and BLACK on the second spin, you are guaranteed a winner should the Group 4 continue to be absent.

So am now going to present my new idea.

This time I will be working in groups of three.

Here is the first group.

1)RED/RED = 1. (this means the first two spins to come out were red/red)
2)RED/BLACK = 2. (the next two spins to come out were red/black)
3)RED/RED = 1. (the next two spins to come out were red/red)

I will also keep a track of the groups as they appear.

1 = 1.
2 = 2.
3 = 1.

Next group.

1)RED/BLACK = 2.
2)BLACK/BLACK = 3.
3)RED/RED = 1.

1 = 1. 2.
2 = 2, 3.
3 = 1, 1.

Next group.

1)RED/RED = 1.
2)BLACK/BLACK = 3.
3)RED/RED = 1.

1 = 1, 2, 1.
2 = 2, 3, 3.
3 = 1, 1, 1.

Next group.

1)BLACK/RED = 4.
2)BLACK/BLACK = 3.
3)RED/BLACK = 2.

1 = 1, 2, 1, 4 (3)  so the furthest back group is the 3 which I have put in brackets.
2 = 2, 3, 3, 3.
3 = 1, 1, 1, 2.

Next group.

1)RED/BLACK = 2.
2)RED/BLACK = 2.
3)BLACK/RED = 4.

1 = 1, 2, 1, 4 (3) 2 (3) the 3 is still the furthest back group.
2 = 2, 3, 3, 3, 2.
3 = 1, 1, 1, 2, 4 (3) so the furthest back group here is the 3. I am waiting for there to be a furthest back group in all three rows before I can place a bet.

Next group.

1)RED/RED = 1.
2)RED/RED = 1.
3)BLACK/BLACK = 3.

1 = 1, 2, 1, 4 (3) 2 (3) 1 (3)
2 = 2, 3, 3, 3, 2, 1 (4) so the furthest back group here is the 4.
3 = 1, 1, 1, 2, 4 (3) 3 (1) the furthest back group has now changed to the 1.

You can now see that there is a furthest back group in all three rows.

The furthest back groups are the 3, 4, 1.

So now on to how to work out the bet.

Here is a simple way to work it out.

Remember the groups.

1)RED/RED = GROUP 1.
2)RED BLACK = GROUP 2.
3)BLACK/BLACK = GROUP 3.
4)BLACK/RED = GROUP 4.

So I will list those groups again and also put in their mirror opposite.

1)RED/RED  =  BLACK/BLACK.
2)RED/BLACK  =  BLACK/RED.
3)BLACK/BLACK  =  RED/RED.
4)BLACK/RED  =  RED/BLACK.

If Group 1 is the furthest back, you just bet it's mirror opposite for the next two spins.. You will win as long as the Group 1 remains absent. You do this for all the groups.

So the Group 3 is the furthest back in the first row. This means I will be betting the mirror opposite of Group 3 for the next two spins. So I will bet RED/RED.

Obviously I will be stopping on any win and restarting again. I am using the following betting progression over 6 decisions which will hopefully produce a win somewhere along the line.

1 2 4 8 16 32. (total risk = 63 units.)

So what happens if I lose the first two bets?

The furthest back group in the second row is Group 4. This means I will be betting the mirror opposite of group 4 for the next two spins. So I will bet RED/BLACK.

What happens if I lose the third and fourth bet?

The furthest back group in the third row is Group 1. This means I will be betting the mirror opposite of group 1 for the next two spins. So I will bet BLACK/BLACK.

What happens if I lose the fifth and sixth bet?

I am down 63 units and curse my rotten luck.

The furthest back group would need to appear in all three rows to lose the 6 bets.

So you should get some nice winning runs out of this bet.

[flukey luke]

The only slight variation for this would be to track all three e/c at the same time.

You would play whichever e/c was displaying the furthest back group.

For example: I am looking for my bet for spin 1 and 2.

The furthest back for RED/BLACK is group 2 and has not appeared in 10 groups.

The furthest back for ODD/BLACK is group 4 and has not appeared in 5 groups.

The furthest back for LOW/HIGH is group 3 and has not appeared in 2 groups.

Therefore I would play the RED/BLACK option because that group 2 is pretty cold.

This way could be played for all the possible 6 bets. I am not sure if it would randomize things a bit better.

My personal view is you need some kind of  perceived edge to win in the long run. At least typing this up saved me from going to the gym.