Artur (translated):Hi, id like to test the Jagger Formula, but I don't know if this can be done.

You can find the rules on page 67 of the manual:

www.gamblersbookcase.com/JaggerFormulaCan someone help?

I estimate, you don't know how to build this roulette strategy?

The answer is simple:

With the **first bet**, lets take the colors as opposite EC. *One shall play a color as long as it wins.*

Let's build a switch to detect the color first.

1. Step: Red, Amount 0: on win: step 2, on loss: step 1

Comes Red, then the strategy jumps to the 2nd step,

with Black (and zero) it remains in the 1st step.

For the moment I ignore the game on black, so let's go on the "red string".

Its still that thing: *One shall play a color as long as it wins.*

But now its clear, we play the red string.

1. Step: Red, amount 0: on win: step 2, on loss step 12. Step: Red, amount X: on win: this, on loss: step 3

So we bet on Red (repeat step 2), until we lose, then we go to step 3.

(the amount of the bet is not important here)

After a loss, *one shall play the opposite chance, as long as its lost again*:

1. Step: Red, amount 0: on win: step 2, on loss step 1

2. Step: Red, amount X: on win: this, on loss: step 33. Step: Black, amount X: on win: this, on loss: step 2

As far as I understand this rule, its valid for both "strings",

so we change to Red on loss.

Remember the switch in the first step?

Now the Black string is also ready to be played.

Change the switches (1st step) "on loss" to step 3 (Black).

1. Step: Red, amount 0: on win: step 2,on loss: step 3

2. Step: Red, amount X: on win: this, on loss: step 3

3. Step: Black, amount X: on win: this, on loss: step 2

Now the roulette strategy plays a color until its lost,

then it changes to the other color.

And so on.

Now everything changes:

When it has lost two times in a row, the last color shall be played on, instead of changing again.

This means: two losses in a row doesn't change the color

For this I add two steps, this makes the 3rd step to the 4th.

1. Step: Red, amount 0: on win: step 2, on loss: step 3

2. Step: Red, amount X: on win: this, on loss: step 3

3. -->new4.step: Black, amount X: on win: this, on loss: step 2

5. -->new

The new strategy:

1. Step: Red, amount 0: on win: step 2, on loss:step 3

2. Step: Red, amount X: on win: this, on loss: step 3

3. Step: Black, amount X: on win:step 4, on loss:step 4

4. Step: Black, amount X: on win: this, on loss:step 5

5. Step: Red, amount X: on win:step 2, on loss:step 2

The 1st steps "on loss" is changed from 3 to 4, as the step was moved down.

On loss in step 4 is changed to step 5.

The two new steps bet on the opposite color as the steps above them do.

On loss and on win point to the same steps.

This keeps the rule to keep the color after two losses in a row.

Example:

(step 1 R, R comes: on win --> 2) Switch

step 2 R, B comes: on loss --> 3

step 3 B, R comes: on loss --> 4

step 4 B

*The Zero shall be ignored and the Color shall be kept.*

For this we add a new Chance to every step: Single, Zero, Amount 0

1. Step: Zero, amount 0; Red, amount 0: on win: step 2, on loss: step 3

2. Step: Zero, amount 0; Red, amount X: on win: this, on loss: step 3

3. Step: Zero, amount 0; Black, amount X: on win: step 4, on loss: step 4

4. Step: Zero, amount 0; Black, amount X: on win: this, on loss: step 5

5. Step: Zero, amount 0; Red, amount X: on win: step 2, on loss: step 2

This makes the march complete.

For the honor of the thread starter, I call this strategy: ** Artur Jagger**.

Here are the flowcharts, that represent the course nicely.

Have Fun with Strategy,

trizero