The American wheel

Started by John1234, December 10, 2009, 07:57:52 PM

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John1234

There seems to be very little discussion on the possibility of a flaw in the American wheel. Over the last month I looked into a lot of possibilities.
Here is a quick system that I came up with

Use these numbers only. Look at the wheel that I posted and look at the board.

0,00, 7,8,9, 10, 11, 12 and 25,26,27,28,29,30 ON THE AMERICAN WHEEL. these numbers are not circled on the board. Now look at where they sit on the wheel.

I am not exactly sure about the most effective way to play these numbers. One Idea that I tried with a bit of success was this:

Wait for one of the numbers to hit. the number will be the trigger. Bet the splits. So you will cover all 12 numbers plus the split of 0 and 00. You will bet until you lose. When you lose you wait for a new trigger.

Progression is 1,1,2,3,4 then Idk what. I suck at designing progressions.

The letters on the wheel are as follows:
A= column 1
B = column 2
C= column 3

The two pictures are attachments.

elmo

give me 5-10 minutes john1234 and I will show you a good way to play.

elmo

o.k. here we go!

Wait for the 7-12 or 25-30 to come out 4 times at least.
What must NOT come out is the 1-6 or 31-36.
If the 13-18 or 19-24 come out, just ignore them, so as an example.

11
13
19
7
14
20
25
13
30

o.k. there you have had 9 spins and the 7-12 and 25-30 have come out the required 4 times. The 1-6 and 31-36 have went missing which is excactly what we were looking for.

Now we will use an 8 step progression but in a unique way.

put 1 unit on the 1-3 street.
put 1 unit on the 4-6 street.
put 1 unit on the 13-18 double street.
put 1 unit on the 19-24 double street.
put 1 unit on the 31-33 street.
put 1 unit on the 34-36 street.

Now to make a profit, we are hoping the 1-3, 4-6, 31-33, or 34-36 street appears but if the 13-18 double street or 19-24 double street appear, we get returned our chips. So in effect we are either going to win 6 or draw even.
The only thing that can beat us is if the 7-12 double street or 25-30 double street appear, in which case we go to step 2 in the progression of a total of 8 steps.
This is a good way of playing because even if we dont hit our 1-3, 4-6, 31-33 or 34-36 street for a profit, we will many times just get our stake back and not have to move up 1 in the progression.
What this all means is that for us to lose completely, we are going to have to witness a dozen (which our 4 potential winning streets are) not coming out for something like over 20+ spins. Remember just for the qualifying in this instance, it took 9 spins and we saw no 1-3, 4-6, 31-33 or 34-36. And now we can play an 8 step progression where even if the 13-18 or 19-24 show up a lot, we are always going to be returned our chips and we do not have to move up 1 in the progression. So it is possible you could go through 25-30 spins and still only be at level 3 or 4 in the progression when you hit your winner. It is fair to say that you will very seldom ever lose a game playing this way and you will build up some really nice profits.

John1234

Thanks for showing me a way to play using these numbers. I am not too familiar with designing systems for roulette so it is always nice to have someone like you help out. I am actually focusing on baccarat right now but I always like to have a roulette system in my back pocket for when I get board playing baccarat. I"ll have to look more into what you suggested, it looks very good.

Here's an idea for you John1234.  When a number hits on the second column (trigger), bet columns 1 and 3 with a 1, 3, 9 progression.  Wheel-order, columns 1 and 3 are connected on the wheel-order while column 2 is almost totally seperate.

kattila

Elmo, which would be  the next 7 steep progression? Thanks.

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