Hi,

I have tried countless times to test a simple 6 step system but I keep doing something wrong.

If someone wouldn't mind helping me, it would be appreciated. I want to test a limited progression

based on the last color result.

The betting starts at $10 on the outcome of the last color. If win the bet stays the same.

If lose, the next bet is $16 on the outcome of the last color.

If that bet wins the next bet is $12, if lose the next bet is $22, all bets are made on the last color.

If lose the $22 bet is lost, increase to $28, if the $22 bet is won, decrease to $18.

If lose the $28 bet is lost, increase to $34, if the $28 bet is won, decrease to $24.

If lose the $34 bet is lost, increase to $40, If the $40 bet is won, decrease to $36.

Stop betting anytime the is a $150 total lost.

I'm sorry, the easiest explanation would be is to start betting $10, On every lost

increase $6 and on every win decrease by $4, Stop betting once you lose

a $40 bet! All bets are on the outcome of the previous color. If zero or double zero

appear it's treated as a lost and the next bet is the same color prior to the "O or OO."

I want the half the bet returned if any O or OO come out.

The bankroll is $150 for the 6 steps and I know the odds of losing 6 consecutive steps

in a row are 1 in 64. Just curious with this step ladder betting if it will produce a positive

result in the long run doing the testing over and over with a large number of trials.

Any help would be appreciated, I just can't seem to get the steps correct of the test

your strategy page. Thanks

I have tried countless times to test a simple 6 step system but I keep doing something wrong.

If someone wouldn't mind helping me, it would be appreciated. I want to test a limited progression

based on the last color result.

The betting starts at $10 on the outcome of the last color. If win the bet stays the same.

If lose, the next bet is $16 on the outcome of the last color.

If that bet wins the next bet is $12, if lose the next bet is $22, all bets are made on the last color.

If lose the $22 bet is lost, increase to $28, if the $22 bet is won, decrease to $18.

If lose the $28 bet is lost, increase to $34, if the $28 bet is won, decrease to $24.

If lose the $34 bet is lost, increase to $40, If the $40 bet is won, decrease to $36.

Stop betting anytime the is a $150 total lost.

I'm sorry, the easiest explanation would be is to start betting $10, On every lost

increase $6 and on every win decrease by $4, Stop betting once you lose

a $40 bet! All bets are on the outcome of the previous color. If zero or double zero

appear it's treated as a lost and the next bet is the same color prior to the "O or OO."

I want the half the bet returned if any O or OO come out.

The bankroll is $150 for the 6 steps and I know the odds of losing 6 consecutive steps

in a row are 1 in 64. Just curious with this step ladder betting if it will produce a positive

result in the long run doing the testing over and over with a large number of trials.

Any help would be appreciated, I just can't seem to get the steps correct of the test

your strategy page. Thanks

Ad

Hello Chingy711,

every amount is without any unit, meant as "pieces", you can imagine the worth as you like! $ € ¥ 元

If i understood your description right, this should be the steptree of your strategy:

The notation is: step Red/Black amount - when there is ony a number (the step) a old step is used again.

The bankroll would be set by the

Here is your strategy: last SC win-4 loss+6

without stops, for unlimited tests

As you see: its always a good idea to write down a strategy

Off course this can take a while, but its more comfortable than testing an idea only on paper.

Usually the result is more precise...

every amount is without any unit, meant as "pieces", you can imagine the worth as you like! $ € ¥ 元

If i understood your description right, this should be the steptree of your strategy:

The notation is: step Red/Black amount - when there is ony a number (the step) a old step is used again.

1 R0

+/ \-

2 R10 3 B10

+/ \- +/ \

2 4 B16 3 5 R16

+/ \- +/ \-

6 B12 7 R22 8 R12 9 B22

+/ \- +/ \- +/ \- +/ \-

2 10 R18 11 B28 3 12 B18 13 R28

+/ \- +/ \- +/ \- +/ \-

14 R14 15 B24 16 R34 17 B14 18 R24 19 B34

+/ \- +/ \- +/ \- +3 \- +/ \- +/ \-

2 20 B20 21 R30 22 B40 23 R20 24 B30 25 R40

+/ \- +/ \- +/ \- +/ \- +/ \- +/ \-

4 26 R26 27 B36 1 5 28 B26 29 R36 1

+/ \- +/ \- +/ \- +/ \-

7 30 B32 1 9 31 R32 1

+/ \- +/ \-

11 32 R38 13 33 B38

+/ \- +/ \-

16 1 19 1

The bankroll would be set by the

**loss limit**in the test configuration.Here is your strategy: last SC win-4 loss+6

without stops, for unlimited tests

As you see: its always a good idea to write down a strategy

**step by step**- in the end you know what to do :-)Off course this can take a while, but its more comfortable than testing an idea only on paper.

Usually the result is more precise...

I totally forgot to mention:

If you would only use one of the two colors as start condition,

it would save the half strategy .

But if you run it on a long test it would have the same results!

same results with the half work! ;-)

I just wrote it up completely because i was curious if there would be "jump overs" of the two trees.

(there were none...)

If you would only use one of the two colors as start condition,

it would save the half strategy .

But if you run it on a long test it would have the same results!

same results with the half work! ;-)

I just wrote it up completely because i was curious if there would be "jump overs" of the two trees.

(there were none...)

Hi Trizero,

I have been testing the strategy you wrote up for me and was

curious if it was all possible to test the number of consecutive color

streaks to the number of single even money events.

One of the arguments I have had with Houston on the forum is that

the Laws of distribution and appearance state that 3/4 of all trials are

series compared to single even money events. Can your tester show

the number of streaks (two or more consecutive colors RRR, BB, RRRRRR)

and the number of single events ( R, B, R, O, B, OO,) in a chart form.

I want to test how many series of event are streaks and how many are single

events based on color outcome. Louie/Chingy

I have been testing the strategy you wrote up for me and was

curious if it was all possible to test the number of consecutive color

streaks to the number of single even money events.

One of the arguments I have had with Houston on the forum is that

the Laws of distribution and appearance state that 3/4 of all trials are

series compared to single even money events. Can your tester show

the number of streaks (two or more consecutive colors RRR, BB, RRRRRR)

and the number of single events ( R, B, R, O, B, OO,) in a chart form.

I want to test how many series of event are streaks and how many are single

events based on color outcome. Louie/Chingy

Hi Chingy711

was it possible? what do you mean with "even money events"?

No, this tool is for testing and not for analysis.

But you could use the same permancence and run different tests over it,

whose strategy just "counts out" those figures.

It wouldnt be in a chart, but at least facts... ;-)

Hope i understood you so far!

Double Zero is not included here at all - because of the higher bank profit!

PS: forget Houston, this annoying guy is just not worth talking about..

Chingy711:

... if it was all possible to test the number of consecutive color

streaks to the number of single even money events.

was it possible? what do you mean with "even money events"?

Chingy711:

Can your tester show the number of streaks (two or more consecutive colors RRR, BB, RRRRRR)

and the number of single events ( R, B, R, O, B, OO,) in a chart form.

I want to test how many series of event are streaks and how many are single

events based on color outcome. Louie/Chingy

No, this tool is for testing and not for analysis.

But you could use the same permancence and run different tests over it,

whose strategy just "counts out" those figures.

It wouldnt be in a chart, but at least facts... ;-)

Hope i understood you so far!

Double Zero is not included here at all - because of the higher bank profit!

**Boycott 00!**:-)PS: forget Houston, this annoying guy is just not worth talking about..

trizero:

Hi Chingy711Chingy711:

... if it was all possible to test the number of consecutive color

streaks to the number of single even money events.

was it possible? what do you mean with "even money events"?Chingy711:

Can your tester show the number of streaks (two or more consecutive colors RRR, BB, RRRRRR)

and the number of single events ( R, B, R, O, B, OO,) in a chart form.

I want to test how many series of event are streaks and how many are single

events based on color outcome. Louie/Chingy

No, this tool is for testing and not for analysis.

But you could use the same permancence and run different tests over it,

whose strategy just "counts out" those figures.

It wouldnt be in a chart, but at least facts... ;-)

Hope i understood you so far!

Double Zero is not included here at all - because of the higher bank profit!Boycott 00!:-)

PS: forget Houston, this annoying guy is just not worth talking about..

Hi Trizero,

What I meant by even money chances was Red/Black, Odd/Even, High/Low.

Thanks again and yes the "00" should be boycotted but here in the United States

it's on most casino wheels unfortuntely. I'll do as you suggest, I'll just run the test and

count all the series of two or more reds and blacks and compare the totals to all the

single event spin results. Have a good day, Louie

ching I advise you to describe your strategy always in units not in dollars or pounds.The testprograms can only compute in units.

Dobbelsteen

Dobbelsteen

dobbelsteen:

ching I advise you to describe your strategy always in units not in dollars or pounds.The testprograms can only compute in units.

Dobbelsteen

Hello dobbelsteen.

I dont understand the meaning of your posting.

If you calculate with a real currency - for example €uro - how could you divide 1 ¢ent for a zero split?

The Casino would take this 1 ¢ent in that case.

Thats exactly the way, that this simulation works.

Help me understand your point please.

Have funny tests!

trizero