as the so called even chances are not really even I wonder if the betting results would differ if we use the two "artificial" even chances (see attachments). by placing the chips like this you would get 9 red, 9 black, 9 high, 9 low, 9 odd and 9 even numbers. so it's totally balanced.
would the results be any different from just betting on the even as offered by the layout? ???
There is a small error in the first jpg
You must bet 31,32 33 instead of 34 35 36
:)
ludo8400
Quote from: HansHuckebein on March 17, 2010, 02:11:07 PM
would the results be any different from just betting on the even as offered by the layout? ???
Your artificial EC bet covers 18 numbers.
Therefore it has the same 0.473 hit probability.
Therefore the results will behave similarly to outside EC's.
thanks ludo - correction done. :)
noble savage,
if you just look at even chances as "isolated" ones you're absolutely right.
but if all even chances show similar results, would then the attempts to find an advantage within the r/o and b/e imbalance be useless?
cheers
hans
I'm afraid so.
There might be some "imbalances" in terms of how these bets are spread on the layout (and wheel) and how they overlap with each other, but there are no imbalances in terms of payout. It's always based on how much you're covering.
First you take 300 trails then put them into groups of 8 and count how many changes every group of 8 has.
Probability dictates that at least one chance in 8 trails.
But we all know nothing is due to happen as getting 8 reds or 37 blacks.
Now the only thing that is constant regarding even money bets is "present change"
So it does not matter if there is red or black, forget about colours or odd/even and so on.
All you have to know is that change is constant present and that you only play when it is present.
Other wise you dont play and just observe.
There is many ways to explore change and here is one way.
First you divide the game into 8 patterns of three.
Then you pick all the patterns that begin with the same colour, like red, exampel.
Now when you get at least three patterns of three that begins with the same colour in an row - you will have 9 trails.
When you find does continue to track untill it gets broken with one of the oppisite patterns of three that begins with an black out come.
When you have this situation put them into an horizontal line.
RBBRRBRBR
BRBRRRRBB
If you look at the formation.
00x xx0 xx0
5 changes in 9 trails.
lw lw lw
That would produce non change would be XXX XXX XXX or 000 000 000
There is other ways then i explain above this is just me wanting you to look at the issue with new mind and eye.
Take care
@cookies
thanks for your advice. I still try to work it out. :)
cheers
hans