TurboGenius Advantage of Repeaters explained - part 3 - 2020

That looks familiar

 

Ty Denzie

 

[QUOTE = "Denzie, publicación: 96561, miembro: 6493"] Cada número tiene 1 unidad.
Cada vez que aparece un número con 1 unidad, lo hacemos 2u y eliminamos cualquiera de las apuestas 1u.
Si una apuesta 2u gana (repetidor), no hacemos nada.
Si aparece un número sin apostar (eliminamos previamente un número que ahora apareció), coloque 2u en él y elimine cualquiera de las apuestas restantes de 1u. (estamos corrigiendo nuestro error de eliminarlo)

Entonces, después de 51 giros, no me quedan números con 1u y estoy jugando 23 números
que TODOS han aparecido. NO estoy jugando un solo número que no ha aparecido.



Ahora podemos repetir el proceso con nuestro bonito conjunto de 23 números.

Si aparece uno, va a 3u y se elimina cualquier número de 2u.
Si aparece uno en el que no se apuesta, obtiene 3u y se elimina cualquier número de 2u.
Si aparece un número 3u, no haga nada.

Entonces, después de 73 giros, no me quedan números con 2u y estoy jugando 16 números
que TODOS han aparecido. Todavía no estoy jugando un solo número que no ha aparecido.




Aún no hemos apostado en ninguno de los números del tiempo
El viajero dijo que no apostara, a pesar de que no sabíamos cuáles eran esos números.


De todos modos, antes de arrastrar esto para siempre, nunca necesitamos ese viajero del tiempo,
los giros nos dicen qué números jugar, también nos dice qué números NO
para jugar, [/ CITA]
With effort and work, I have also come to that conclusion and others that are also very good, but you are not playing the winning horses, this is something else, just like others that also work, Mr. Denzie.
My tenure talks about whether you can win with the winning horses!
I know there are other ways to overcome any roulette, which you have mentioned is very good, but I am waiting to see the session that I publish with the winning horses win, not with a different bet.
surely this publication people will be grateful for its publication, but he did not tell me anything new.
That session does not win with the winning horses, I only got a few hits and no positive progression, nor game regarding the winning horses that I know, exceeds those 104 balls.
Another way, of course I do, I know!

 

Hi Denzie!! Could you be more specific or give some more clues?

Thanks in advance!
Regards!

 

Hi CDN, I don't understad this.
Could you explain any more please?
For example, for colum A when we have A1,B1 and C1, only we can bet for A1.
When we have A2,B2,C2. If A2 has more hits than A1 we only bet A2 for column A.
When we have A3,B3,C3, only play one number between A1 and A3. This number would be which with more hits.
This is ok?
And what progression is fine for this?

Thanks in advance!
Regards

 

This reminds me of a post by TG in which he said that he chooses the left seat to be closer to the table which in turn could mean that he defined the first dozen as his framework of repeaters.

Are you suggesting aggressive progression even though we discard numbers and start anew as soon as the next stage kicks in thus we should up the progression even though we didnt win (having too many we dicarded some) or raising only on a win?

Limiting the framework as you already said will require double play time (theoretically) especially if the the defined set is less active than the rest of the sum of numbers.

 

TY TurboGenius

 

I wouldnt play this on a single dozen though. Knowing they can sleep for 40(ish) spins . I prefer a hit every...1-20 spins(ish)

 

I think he means...
Play every 1x with 1u
On a hit play every 1x with 4u
On a hit play every 1x with 16u
We do this till all our 1x are hit OR a 5th 1x comes . We then stop our progression and reset to 1u for our 2x (there might even already be a 3x but thats ok . We work line by line)

Play all 2x with 1u
On a hit play every 2x with 4u
On a hit play every 2x with 16u
On a hit play every 2x with 64u
Again...we stop betting if we got all our2x OR when a 5th 2x comes

Etc etc etc
(Personally i would start at 2x or even 3x )

Correct me if im wrong please @CDN

 

Yes but you wont have as many betting opportunities as if you had the whole table thus less numbers, equals less drawdown.
Thats why I said double the time especially if you cover 1-18.
A dozen can sleep indeed but Im talking about already hit numbers inside the dozen (frame) not the sleeping ones(unhit).

 

To Quos

In my example, instead of one race we have 3 races, each for 1 column
A1,A,2...An = race1
B1,B2,...Bn = race 2
C1,C2,...Cn = race 3

Let's take the A1.. An example
We count all numbers on that column until one is+1 ahead of the herd.
Let's assume the A1 An column ( first to 1x,2x, nx)looks like this:

10
1
3
10 bet 10 2x compared to the rest of the herd
1 stop betting 10 as 1 and 10 are both at 2x
1 bet 1 as 1 has 3x +1 more than the others.
5 continue betting 1
4 continue betting 1
1 win

This happens all the time in roulette, 1 will eventually be at least 2 hits higher than the rest.

What we overlook are the rest of the columns, in one of which a number will always break through.

While the A1. .An column can and will lose, what happens in the rest if columns B1.. Bn column /C1 Cn, D1..Dn column will have a number that will be +1 hits more than the local herd. That HAS to happen.
If it can be +1 higher it will certainly be +2 higher than the herd. If you think about it is exactly the classic example of a 37 spin cycle: you will ALWAYS have a 2x (+2 more than the sleepers).
We are not betting on the hottest of the hottest, but the warm becoming hot.

The A1..An is just one race. That by itself will not win this game.
Instead, think about all columns that have from 2x upwards hits, and bet only on those numbers that are +1 ahead of the individual (columns).
Any 1x has the potential to become a 2x.
Not a sleeper. Common sense.

Progression wise i would increase the progression depending on the number of hits above the +1 (a number that has +1 shows more than the heard step 1, 2 shows more step 2, 3 shows more step 3, and so on.
The progression will depend on the distance from the followers. If they catch the leading horse, progression goes to 0 no bets, and so on, another horse will be at plus +1 , step 1 progression, plus 2 step 2 progression, plus 1 step 1 progression.

Think of the big picture.

 

Unless im not getting something right you are essentially aiming for the +2 condition to happen to win.
If you start betting at +1 and other numbers catch up and you have equals, you stop betting - so you are betting that a +2 will happen before other numbers catch up and "force" you to pause.

 

in the example above put in columns, we would have the following:

***A B C
1x 10 1 3
2x 10 1
3x 1
4x 1

When you stop betting number 10 in column A only we have number 10. Just at that moment the number "1" is the 2x of column B not of A.
I thought that we only had to take into account the numbers that appear in column A. And the same independently for column B and C.

Thanks again!!

 

Hi Smitridel
In that example, anything +2 and above. The maximum numbers played would be the number of columns used. That is highly improbable to happen in a real situation as you can not be +1 on all of the columns. Some leaders are caught up, others rise. Its a moving scenario with different numbers in play.
TG matched in the initial thread the columns into first numbers to go to 1x, first to 2x and so on(column A) second to go to 1x 2x and do on (column B) third numbers to go to 1x,2x and so on (column B). But that does not necessarily mean to stop at three columns. Add any number of columns you want- column D fourth to get to nx, Column E fifth to go to nx, and so on. The more columns you have, thet more betting opportunities you will get.
BECAUSE of random, there will be cases where the fifth shows will have a champion 3 steps ahead of the pack. It does not need to be the most hit number overall, just the best in its column. Each column will have its own progression adjusted on the distance to the followers.
If one column doesn't hit, another one will.
And that is the column race.
If you like a 1 number bet only, use all the matrix and start at A1 and finish at Zn. Count all of the numbers and bet the one that has +1 dhows more than all.
A1 B1... Z1
A2 B2 ....Z2
................
An Bn.....Zn



21 14 7 27 19 36 8 7 -bet 7
5 8 stop betting
7 bet 7
5 5 stop betting
7 bet 7
13 5 stop betting
7 bet 7
13
7 win
5
19
7 win
5
19
7 win
5
18
7 win
5
13
7 win
5
13
7 win
5
13
7 win
5
7 win
7 win
7 win

OR

Bet the last number that repeated in each column
21 14 7 27 19 36 8 7 -bet 7
7-5-8-7 win continue with 7
7-5-5 bet 5 lose
5-7-13-5 bet 5 win
5-7-13-5 bet 5 lose
5-7-13-7 bet 7 win
7-5-19-7 bet 7 win
7-5-19-7 bet 7 win
7-5-18-7 bet 7 win
7-5-13-7 bet 7 win
7-5-13-7 bet 7 win
7-5-13-7 bet 7 win
7-5-7 bet 7 win
7-7 bet 7 win
7-7 bet 7 win

 

Basicly the're few ways to play this columns...

Horizontal ? Going for 3 matches ? Or 2 ? Using a 1/5 or 1/5/25 progression?

Vertical? Using a 1/5/25 progression?

Eugene his way ?

CDN his way ?

Your own way?

Or....

The complete method TG shared that i copied here above ? ( of course tweak it so its actually playable)

Using law of the third? Each cycle keep going with the hits of previous cycle(s) with a progression? ( 24ish, 12ish, ...)


I suggest you start testing . 100 sessions each. This should already tell you something right? And while you test you might see where to improve it....

Lots of solid info here...time to do some work

 

I hate to say it but I got more confused about your way even with your examples. Especially using TGs matrix.
Can you provide a small session to clarify when do you decide to bet and when to stop or is it too much to ask?

 

Who; of these 37 do we bet?

41 spins nothing; but we are shown who we can bet. Obviously not 11-12-24-32.
32 of the starting 37 hit by spin 60. just 3 repeats in 1st 20 spins, not my #'s

 

Nice

 

Excellent post !

 

Predictable.