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How to find a consistent bet the Enrique way

Started by enrique malou, December 03, 2008, 07:28:48 PM

0 Members and 1 Guest are viewing this topic.

enrique malou

Hello  :)


I am going to write something out and I think if some of you look at it and study it, then maybe you will be nicely rewarded for your hard work.  You can accuse me of playing silly games but there is no such thing as a free lunch and I hope anyone who plays and wins money once they have discovered this does donate some money to a worthwile charity.

17. 2B, (THIS IS DOZEN AND COLUMN) 1C,   3C.
--
28. 3A,   2C,   1B.
--
15. 2C,   1A,   2C.
--
24. 2C,   2C,   1C.
--
35. 3B,   2B,   1B.
--
16. 2A,   2C.   2C.
--
In the first section going down of dozen and column, the first dozen is missing.
In the second section going down of dozen and column, the 3rd dozen is missing.
In the third section going down of dozen and column, the first column is missing.

So from the number 17 you can discount all of the 1st dozen=1,2,3,4,5,6,7,8,9,10,11,12.
You can also discount from the second section all of the 3rd dozen=12,35,3,26,0,32,15,19,4,21,2,25.
you can also discount from the third section all of the A column=18,21,24,27,30,33,36,2,5,8,11,14.

Now that only leaves these numbers left to back=13,16,17,20,22,23,28,29,31,34. (10 numbers in all)

The winning number was 28. You have 36 chips.

Now from the number 28, once again you can discount all of the 1st dozen=1,2,3,4,5,6,7,8,9,10,11,12.
You can also discount from the second section all of the 3rd dozen=24,16,33,1,20,14,31,9,22,18,29,7.
you can also discount from the third section all of the A column=29,32,35,1,4,7,10,13,16,19,22,25.

Now that only leaves these numbers left to back=0,15,17,21,23,26,27,28,30,34,36. (11 numbers in all)
You won 36 chips from the last bet, so now you put three each on these 11 numbers and you have 3 chips left.

The winning number was 15. You have 111 chips.

Now from the number 15, once again you can discount all of the 1st dozen=1,2,3,4,5,6,7,8,9,10,11,12.
You can also discount from the second section all of the 3rd dozen=9,22,18,29,7,28,12,35,3,26,0,32.
You can also discount from the third section all of the A column=16,19,22,25,28,31,34,0,3,6,9,12.

Now that only leaves these numbers left to back=13,14,15,17,20,21,23,24,27,30,33,36. (12 numbers in all)
You have 111 chips. So now put 9 chips on these 12 numbers and you have 3 chips left.

The winning number was 24. You have 327 chips.

Now from the number 24, once again you can discount all of the 1st dozen=1,2,3,4,5,6,7,8,9,10,11,12.
You can also discount from the second section all of the 3rd dozen=17,34,6,27,13,36,11,30,8,23,10,5.
You can also discount from the third section all of the A column=25,28,31,34,0,3,6,9,12,15,18,21.

Now that only leaves these numbers left to back=14,16,19,20,22,24,26,29,32,33,35. (11 numbers in all)
You have 327 chips. So now put 29 chips on these 11 numbers and you have 8 chips left.

The winning number was 35. You have 1,052 chips.

Now from the number 35, once again you can discount all of the 1st dozen=1,2,3,4,5,6,7,8,9,10,11,12.
You can also discount from the second section all of the 3rd dozen=33,1,20,14,31,9,22,18,29,7,28,12.
You can also discount from the third section all of the A column=36,2,5,8,11,14,17,20,23,26,29,32.

Now that only leaves these numbers left to back=13,15,16,19,21,24,25,27,30,34,35. (11 numbers in all)
You have 1,052 chips. So now put 95 chips on these 11 numbers and you have 7 chips left.

The winning number was 16. You have 3,427 chips. (if they were £5 chips, that is a lot of money ;) )

That used to be as far as I could go using a positive progression because the table limit was £100 per number. So now if I wanted to continue I had to stick to a base unit of £100 per number. That only took me 5 steps to get there starting with £10.

**This has got me barred from one major chain of casinos in the U.K. do not take this lightly**

On this particular session I had another 2 winners and walked out of the casino that night with over £20,000.


All you need to do is work out the code for the second and third sections and if you study the appearance of the numbers it is not that difficult. To be honest I was never going to write any of this up but then again I don't owe the casinos any favours in fact I would rather see them all close down. They are greedy parasites. I have made my money from them. I hope this helps some of you.

good luck and best wishes

enrique.

p.s. If you think you have got it and want to make sure let me know at enriquemalou@ymail.com
If you don't hear back then you know you need to go back to the drawing board  ;D

enrique malou

I will try my best to explain. If you look at the two pictures on my website of the two wheels, you will see that they are marked.

The 0 is pointing at the 17 on wheel 1 because that was the starting number in the example above. The next number out is 28 so that is in the 3rd dozen and it is column A.
The next number out was 28. So looking around on wheel 1 at number 28 that is in the 2nd dozen and it is column C.
The 0 is pointing at the 17 on wheel 2 as well. So because the next number out is 28. looking around the wheel at number 28 that is in the 1st dozen and it is column B.

That gives me  28=3A,   2C,   1B.

On the two wheels the smaller disc in the middle revolves. So now because the new number is 28, you need to turn the smaller disc around both wheel 1 and wheel 2 so that the 0 on both wheels is now pointing at 28.
The next number is 15 so that is in the 2nd dozen and it is column C.
The next number out was 15. So looking around on wheel 1 at number 15 that is in the 1st dozen and it is column A.
The 0 is pointing at the 28 on wheel 2 as well. So because the next number out is 15. looking around the wheel at number 15 that is in the 2nd dozen and it is column C.

That gives me 15=2C,   1A,   2C.

So putting the 3 numbers together so far gives me
17=2B,   1C,   3C.

28=3A,   2C,   1B.

15=2C,   1A,   2C.

So what I am looking for is a run where a dozen or column is sleeping. Very often if you are looking at dozens or columns you will see the following.
1221122 (so the 3rd dozen is absent) and ACCCAACA (so the 2nd column is absent)
I look at the 3 groups above and wait for a dozen or column in each of them to be absent.
So looking at the first group going down, I can see that there is 2,3 and 2. So the 1st dozen is absent.
Also in the 1st group there is B,A, and C. but that is no good because there is one of each. So at least the 1st dozen is absent.  (So for betting purposes I will hope for that to continue and rule out 1,2,3,4,5,6,7,8,9,10,11 and 12 on the next spin.)
Now looking at the second group going down, I can see that there is 1,2 and 1. So the 3rd dozen is absent on wheel 1.
Also in the 2nd group there is C,C and A. So the B column is absent on wheel 1.
So now for betting purposes and hoping for that to continue, I can rule out the 3rd dozen on wheel 1 and also column B on wheel 1.
(so from the number 15 on wheel 1  *with the 0 at pointing at 15* I can leave out 9,22,18,29,7,28,12,35,3,26,0 and 32 and 4,25,6,36,8,5,33,14,22,7,35, and 0.)
Now looking at the third group going down, I can see that there is 3,1 and 2 but that is no good because there is one of each.
Also in the third group there is C,B and C. So the A column is absent on wheel 2.
So now for betting purposes and hoping for that to continue, I can rule out the A column on wheel 2.
(so from the number 15 on wheel 2  *with the 0 pointing at 15* I can leave out 16,19,22,25,28,31,34,0,3,6,9, and 12.)
So taking all these numbers out of the reckoning leaves me with just the following numbers 13,15,17,20,21,23,24,27, and 30. (that is only 9 numbers.
The next number out was

24=2C,   2C,   1C. That is a winner.

This is really complicated and will really take some getting into for anybody who wants to give it a try. First you will have to replicate the wheels, then you will have to write out the codes on them and understand them and give it all a good trial before you can take it to the casino and give it a try.
When you can play it, you will see the power of it. The absences come for long enough periods of time so you can catch a really good win. Like in my earlier example above of just 5 wins netting nearly 3,500 chips from a 10 chip start.

regards

enrique

xman1970

Many thx for explaining how Enrique  ;)


No doubt it will save my head from exploding..... :o


I will look into some more after some sleep..... 8)

MXkid77

Hi Scooby,

You want to be reading this thread, all is explained there as well as templates for the wheels.

nolinks://vlsroulette.com/bet-selection/how-to-find-a-consistent-bet-the-enrique-way/

pighead


Hi Enrique,

I do not understand why you adjust the wheel by pointing 15 to 0.. Can you plz give me some clues?

"(so from the number 15 on wheel 1  *with the 0 at pointing at 15* I can leave out 9,22,18,29,7,28,12,35,3,26,0 and 32 and 4,25,6,36,8,5,33,14,22,7,35, and 0.)"

thanks


enrique malou

Hello Pighead.

Once a number comes out you point the 0 at that number so that you can work out your bets for the next spin. Every spin you are re-setting the wheels to work out the next bets.

So just as an example pointing the 0 on the inner disc to number 15 on the outer disc on wheel 1. Say you wanted to leave out all the numbers on the B colum from the inner disc, that would be 4,25,6,36,8,5,33,14,22,7,35, and 0.
Whatever number comes out next, say number 17. Then you would point the 0 on the inner disc to the number 17 on the outer disc and then work out any new bets you had to place.

I hope that explains it for you.

regards

enrique

pighead


hi Enrique,


thank you for your prompt reply.

Just a matter of curious, how do you get the idea of using two wheels and also how successfully have you been using the system?

happy new year!

PH

enrique malou

The idea came because I was looking to find a way to bet less numbers than I was betting at that particular moment in time. Only one number can appear at any one time and the more you have out on the table, then the less you get back. I know it is all about odds and what not, but because of the nature of roulette and the deviations you get, I can get some good winning runs betting just a few numbers using the wheels.
To use them in a casino takes some getting used to and you can get some strange looks but they have helped me win many sessions in the past few years.

I will say however that my new method  "1320 combinations" looks really promising and will be a lot more practical for use in the casino with the aid of the charts where you can just glance for the info you need.

regards

enrique

pighead

Your method reminds me the ancient chinese bagua, another way to predict the future and to tell the past..

nolinks://en.wikipedia.org/wiki/Bagua_(concept)

I will study  your 1320 system for sure..

pighead

Enrique,

God father of roulette in SA, I tested the system:

3 sessions, 130 spins@dublin,turned $500 to $2000.. No wonder Casinos had to throw you out of their doors.. ;D ;D ;D ;D ;D

PH

enrique malou

They can not stop my playing, I have my ways  ;D

regards

enrique

JHM

Enrique,

Good to see you back. I'm looking forward to your new charts. I hope you'll finish theim (off course when you have the time).

bobbybobby

wow pighead, are you serious??




Quote from: pighead on January 06, 2009, 12:34:52 AM
Enrique,

God father of roulette in SA, I tested the system:

3 sessions, 130 spins@dublin,turned $500 to $2000.. No wonder Casinos had to throw you out of their doors.. ;D ;D ;D ;D ;D

PH

pighead


bobbybobby

hi to all reading this thread, does anyone really understand how to use this method already?

i am actually quite confused....





BobbyBobby

bobbybobby

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