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Paradigm Shift ? series of bets VS a single outcome

Started by Arteinvivo, January 30, 2009, 09:50:11 AM

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Arteinvivo

Think of a Paradigm Shift as a change from one way of thinking to another.

A flat bettor will concentrate on one thing, ie, the next bet only.

A progressive bettor will concentrate on one thing, ie, a series of bets.

Which way of thinking is more solid and why ?

P.S. this post is open to all except Gizmotron/Spike. I don't want to see their comments here since they have nothing to share anyway as they keep repeating.

MATTJONO

hi arteinvivo

maybe combinding the both a system ste of numbers and a random choice of number at the same time. dont know whever or not it would be better or worst just a thought.


mattjono

Arteinvivo

Quote from: John on January 30, 2009, 12:29:20 PM
A series of bets is more solid, because any single bet will always be 50/50.  If you focus on a series of bets you are generally not that concerned with losing the first or sometimes beyond.  Also treating things in a series can yeild a greater than 50% expectation.

So what you say is consider many decisions as a single event. Can you give an example how a series of bets can be considered as a single event ?

Spike

Which way of thinking is more solid and why ?>>

Because a series of bets changes with every single bet, whats the difference? I don't have any idea what the fifth bet in a series will be, or the third or even the second. I can only go one bet at a time. So can you really seperate the two, single bet and series, in any practical way?

VLSroulette

Quote from: Arteinvivo on January 30, 2009, 08:16:00 PM
So what you say is consider many decisions as a single event. Can you give an example how a series of bets can be considered as a single event ?

Hello Arte,

For instance, you could be betting this event: "any number hitting 3 times within a 37-spin cycle".

You would then split your actuals in sets of 37 spins and assign a "w" (win for the event) if the cycle had a 3-timer at any point of it, if not, you assign "L" (lose for the event).

Series of 37 consecutive spins are transformed into a single binary events on which you can find groupings, etc.

Kind regards.
Victor

TwoCatSam

I bought a system from a fellow who tried in vain (me, too) to create events from EC coming three at a time.  Basically, here was his logic:


RRR
RRB
RBR
BBB
BBR
BRB

Considering colors only, this is all the ways a series of three can come.  There are six ways, and 4 vs all six would be the law of the thirds.  He said if four of these events had come, there was a much greater chance the other two would come than a repeat of the four that had come.  Get it?  He called them Triads.  When there were four different Triads, he would wait for the first to colors of the fifth Triad to come and if those and one other color would made a new Triad, he bet for it to happen

Example:

RRR
RRB
RBR
BBB
----------------------at this point we wait for either BB or BR and then bet for R in the first and B in the second.

BBR
BRB

He became disgusted with his study and abandoned it.  Sent back my 40 pounds and thanked me for my help.  I still have his system in my files, but I've told it all.  His name is Adrain and if you're out there, answer my e-mails!

He also used the Fibonacci in a very unusual way with this system.

Sam

The color tag no longer works.


Arteinvivo

this is all the ways a series of three can come

2^3 = 8 not 6 ???

BBB
BBR
BRB
BRR
RRR
RRB
RBR
RBB

VLSroulette

QuoteThe color tag no longer works.

It will work after upgrade.




...as for the combinations at Adrain's system as highlighted by Arte, yes they are 8. Law of the third theory is right for 6 elements as Sam exposed. As Law of the third runs throughout all the roulette game, in a cycle of 6 spins for lines 1/6,7/12,13/18,19/23,24/30,31/36 the natural observation is only 4 will appear. Of course, allowing dispersion at data (i.e. cases where all 6 will apear, and less than 4 too), but the "belly" of shows taking 6-spin slices worth of data is right at 4.

Arranging 6 elements in "thirds" = 2 , 2 , 2 , so two of these "thirds" are 2 + 2 = 4 elements.

When arranging 8 elements in "thirds" = 2.66,2.66,2.66... two thirds is 5,32... I'm a bit worried about using a decimal of a pattern, what is .32 out of RBR... "R"? Also unlike single "solid" elements (i.e. line 7/12) combinations of R/B patterns can overlap (i.e. RBRRBR contains [RBR]RBR, R[BRR]BR, RB[RRB]R, RBR[RBR]), then I think the paradigm needs to be rethought a bit. I'm sure there's a way to extrapolate law of the third to those, but needs some more thinking (encouraging anyone to take it from here).

Kind regards,
Victor

TwoCatSam

The more I looked at that series, the more I knew I was making a mistake and could not see it.

I need to quit posting for a while............

Sam

VLSroulette

QuoteI need to quit posting for a while............

Absolutely not Sam!

We are all here to learn from each other. I have been told many things I slipped to notice in the past and it is okay. Nobody is perfect, we are always improving.

Besides: No posting no learning! ;)

Best regards.
Victor

VLSroulette

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