## Bet Selection

Started by enrique malou, November 21, 2008, 10:50:53 PM

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#### enrique malou

In my opinion bet selection is the most critical part of enjoying a profitable time at the tables. It will be argued that each spin of the wheel is an independant trial completely unconnected from anything previous and therefore no bet selection can gain any kind of advantage.  Well would you be surprised if I told you I disagree . I thought not. The bet selection I am going to share here will allow you to bet on either 18 or 24 numbers. This can be done by placing 3 chips on three double streets or 4 chips on four double streets.
What I like about it is the number of long winning streaks it gives. Also you do not need to use any kind of progression. You can flat bet it all the way or once you build up some profits then you might want to let a few ride and accumulate at a quicker rate.

18=3
11=2
31=6
07=2
25=5
03=1
19=4
36=6
15=3

That much is easy to understand in regard to which number fits into which double street.
Now the above has to get translated and becomes the following.

18=3
-----
11=2   4   5
31=6   5   4
07=2   1   2
25=5   1   3
03=1   3   2
19=4   5   3
36=6   2   2
15=3   3   3

I have on a little piece of cardboard sitting next to me two lines of numbers. Going down the first line is
1.1
2.6
3.2
4.5
5.3
6.4
Going down the second line is.
1.1
2.2
3.3
4.4
5.5
6.6
The first number was 18=3.
Then 11=2   4   5
Where does the 4 and the 5 come from.
In the first line go to the 3 because that represents the 18. The next number after 18=3 was 11=2. So from the 3 you have to get to the 2. So counting from the 3 to the 2 you should count 4.
In the second line go to the 3 because that represents the 18.  The next number after 18=3 was 11=2. so from the 3 you have to get to the 2. So counting from the 3 to the 2 you should count 5.
Therefore you have 11=2   4   5.
With time you can do this very quickly and really does not require much effort.
Now that you know how to do the translation there is an extra step involved.

18=3
-----
11=2   4   5   (1,3,6)
31=6   5   4   (1,2,3)
07=2   1   2   (3,4,5,6)
25=5   1   3   (2,4,6)
03=1   3   2   (4,5,6)
19=4   5   3   (1,2,6)
36=6   2   2   (1,3,4,5)
15=3   3   3   (1,2,4,5,6)

In the brackets are the double streets that are missing.
In the 11=2   4   5  you can see the 2,4,5 double streets are there but the 1,3,6 are missing.
Some of you may be asking why have I come up with other double streets that do not exist.
I have created them myself You will understand why in my next post when I print out the chart which shows you what to bet and when.

regards

Enrique

#### enrique malou

#1
Hello.

Looking at this again

18=3
-----
11=2   4   5   (1,3,6)
31=6   5   4   (1,2,3)
07=2   1   2   (3,4,5,6)
25=5   1   3   (2,4,6)
03=1   3   2   (4,5,6)
19=4   5   3   (1,2,6)
36=6   2   2   (1,3,4,5)
15=3   3   3   (1,2,4,5,6)

If you do some testing with this one then you should start to notice something happening frequently in the missing double streets. That is the 6 will go missing for long periods of time.
To take advantage of this if you want to play an even money chance. In other words a single chip on each of three double streets, then here is the following betting chart.

No 6 on a 1. bet 3,4,5.
no 6 on a 2. bet 3,4,5.
no 6 on a 3. bet 1,2,5.
no 6 on a 4. bet 1,2,5.

Looking at the 11=2   4   5   (1,3,6) The 6 is absent. The double street that appeared on the wheel was 2.
So looking at the chart you would play the 3,4 and 5 double streets.
Next up was 31=6   5   4   (1,2,3) That is a loss. There is now no bet on the next spin.
Next up was 07=2   1   2   (3,4,5,6) The 6 is absent again. The double street that appeared on the wheel was 2. So looking at the chart you would play the 3,4 and 5 double streets.
Next up was 25=5   1   3   (2,4,6) That is a winner. The 6 is absent again. The double street that appeared on the wheel was 5. So there is no even money bet for this.
Here at your own discretion is where you can add an extra bet to the chart and it is as follows.
no 6 on a 5. bet 1,2,3,4. This would mean that you are betting a 1/2 chance.
If you applied this after the last number which was 25=5   1   3   (2,4,6) Then you would bet the 1,2,3 and 4 double streets.
Next up was 03=1   3   2   (4,5,6) That is a winner. The 6 is absent again. The double street that appeared on the wheel was 1. So looking at the chart you would play the 3,4 and 5 double streets.
Next up was 19=4   5   3   (1,2,6) That is a winner. The 6 is absent again. The double street that appeared on the wheel was 4. So looking at the chart you would play the 1,2 and 5 double streets.
Next up was 36=6   2   2   (1,3,4,5) That is a loss. There is now no bet on the next spin.
In the 5 bets you placed in total, you are up 1/2 a unit.

If you would prefer to play betting exclusively 4 double streets at a time, here is the betting chart.

no 6 on a 1. bet 2,3,4,5.
no 6 on a 2. bet 1,3,4,5.
no 6 on a 3. bet 1,2,4,5.
no 6 on a 4. bet 1,2,3,5.
no 6 on a 5. bet 1,2,3,4.

As I said earlier the 6 will go missing for the longest period of time, but I have a chart for all 6 numbers that can go missing. I will print this out tomorrow.

regards

Enrique.

#### enrique malou

#2
Hello. I will now print out the chart for all 6 double street numbers that can go missing.

no 1 on a 2. bet 2,4,6.
no 1 on a 3. bet 2,3,5,6.
no 1 on a 4. bet 2,3,4,6.
no 1 on a 5. bet 2,4,5.
no 1 on a 6. bet 3,4,5,6.
------------------------

no 2 on a 1. bet 1,4,5,6.
no 2 on a 3. bet 3,4,6.
no 2 on a 4. bet 1,3,4,5.
no 2 on a 5. bet 3,5,6.
no 2 on a 6. bet 1,3,4,6.
------------------------

no 3 on a 1. bet 1,2,6.
no 3 on a 2. bet 1,2,6.
no 3 on a 4. bet 4,5,6.
no 3 on a 5. bet 4,5,6.
no 3 on a 6. bet 1,2,4,5,6.
-------------------------

no 4 on a 1. bet 1,2,6.
no 4 on a 2. bet 2,3,5.
no 4 on a 3. bet 3,5,6.
no 4 on a 5. bet 1,2,5.
no 4 on a 6. bet 1,2,3,5,6.
-------------------------

no 5 on a 1. bet 1,2,3.
no 5 on a 2. bet 2,3,4.
no 5 on a 3. bet 1,3,4,6.
no 5 on a 4. bet 1,2,4,6.
no 5 on a 6. bet 2,3,4,6.
------------------------

no 6 on a 1. bet 2,3,4,5.
no 6 on a 2. bet 1,3,4,5.
no 6 on a 3. bet 1,2,4,5.
no 6 on a 4. bet 1,2,3,5.
no 6 on a 5. bet 1,2,3,4.
------------------------

#### enrique malou

#3
Here is an example from numbers from the casino this afternoon.

29=5
-----

2=1   3   2   (4,5,6) The 6 had missed for a few before as well so I decided to jump on board and start betting.
Here is the full betting chart for when the 6 double street goes missing.

no 6 on a 1. bet 2,3,4,5.
no 6 on a 2. bet 1,3,4,5.
no 6 on a 3. bet 1,2,4,5.
no 6 on a 4. bet 1,2,3,5.
no 6 on a 5. bet 1,2,3,4.
-----------------------------

So my first bet is 2,3,4 and 5.

28=5   3   4   (1,2,6) I win that one. my next bet is 1,2,3 and 4.

01=1   3   2   (4,5,6) I win that one. my next bet is 2,3,4 and 5.

15=3   4   2   (1,5,6) I win again. my next bet is 1,2,4 and 5. every time I win a bet I am making a profit of 2 chips. I am outlaying 4 to get back 6. So far I have won 6 chips.

01=1   2   4   (3,5,6) I win again. my next bet is 2,3,4 and 5.

17=3   4   2   (1,5,6) I win again. my next bet is 1,2,4 and 5.

04=1   2   4   (3,5,6) I win again. my next bet is 2,3,4 and 5.

18=3   4   2   (1,5,6) I win again and now my total profit is up 14 chips. my next bet is 1,2,4 and 5.

04=1   2   4   (3,5,6) I win again. my next bet is 2,3,4 and 5.

20=4   5   3   (1,2,6) I win again. my next bet is 1,2,3, and 5.

32=6   2   2   (1,3,4,5) This is a loss and now I stop. the good news is that I made 16 chips profit very quickly.

This explains this component of the winning edge system.

regards

Enrique

#### enrique malou

#4
Hello, Here is another example from a live online casino this evening.

03=1
-----

06=1   6   6   (2,3,4,5) The 2 had missed for a few spins so I thought I would have a go.
Here is the full betting chart for when the 2 double street goes missing.

no 2 on a 1. bet 1,4,5,6.
no 2 on a 3. bet 3,4,6.
no 2 on a 4. bet 1,3,4,5.
no 2 on a 5. bet 3,5,6.
no 2 on a 6. bet 1,3,4,6.
------------------------

So my first bet is 1,4,5 and 6.

01=1   6   6   (2,3,4,5) I win that one. My next bet is 1,4,5 and 6.

04=1   6   6   (2,3,4,5) I win that one. My next bet is 1,4,5 and 6.

23=4   5   3   (1,2,6) I win that one. My next bet is 1,3,4 and 5.

30=5   4   1   (2,3,6) I win that one. My next bet is 3,5 and 6.

35=6   4   1   (2,3,5) I win that one. My next bet is 1,3,4 and 6.

16=3   3   3   (1,2,4,5,6) I win that one. My next bet is 3,4 and 6.

09=2   4   5   (1,3,6) I lose that one and stop betting. In total I won 10 chips.

regards

Enrique

#### wiggy

#5
Quote from: enrique malou on November 21, 2008, 10:50:53 PM
The first number was 18=3.
Then 11=2   4   5
Where does the 4 and the 5 come from.
In the first line go to the 3 because that represents the 18. The next number after 18=3 was 11=2. So from the 3 you have to get to the 2. So counting from the 3 to the 2 you should count 4.
In the second line go to the 3 because that represents the 18.  The next number after 18=3 was 11=2. so from the 3 you have to get to the 2. So counting from the 3 to the 2 you should count 5.
Therefore you have 11=2   4   5.
With time you can do this very quickly and really does not require much effort.
Now that you know how to do the translation there is an extra step involved.

18=3
-----
11=2   4   5   (1,3,6)
31=6   5   4   (1,2,3)
07=2   1   2   (3,4,5,6)
25=5   1   3   (2,4,6)
03=1   3   2   (4,5,6)
19=4   5   3   (1,2,6)
36=6   2   2   (1,3,4,5)
15=3   3   3   (1,2,4,5,6)

Enrique - can you please redefine your explanation here... It really does not make sense to me...

This is how I understand what you have written...
In the first line go to the 3 because that represents the 18. The next number after 18=3 was 11=2. So from the 3 you have to get to the 2. So counting from the 3 to the 2 you should count 4.
I understand that double street Number 3 represents #18... and counting forward to the second double street we pass 4 double streets (4,5,6,1) before we reach Line 2... this is how you get 4?
Then straight after that you say:
In the second line go to the 3 because that represents the 18.  The next number after 18=3 was 11=2. so from the 3 you have to get to the 2. So counting from the 3 to the 2 you should count 5.
You're employing the same "count" yet arriving at a different number...