Gentlemen,
Albalaha's posting ,
have give me an idea.
[maybe you already knew it]

He says, after millions spins investigation,
the virtual limit for,

dozen=250hits/1000spins.
EC=405hits/1000spins.

Thus now we move away,

from,  trying to bet in tens of spins,
or a 100spins, which, after 300years of casinos open shop,
that none of any human being can figured out how to beat the wheel!

We try to  in 1000spins!
===================
dozen=250hits/1000spins.
EC=405hits/1000spins.

the trillion dollars question, is ,
how to have a progression, that can win 1u, with only,


405hits/1000spins.
for EC.???
[There may never less than 405hit, in 1000spins].

250hits/1000spins,
for dozen bet???
[There may never less than 250hit, in 1000spins].

column/dozen: 250 hits in 1000 spins

then, if you can,

250x2=+500u
losses=1000-250=-750.
losses=750-500.=-250

if you can figure out, how to recoup,
-250losses, by 250hits,


or win, with only 250hits/1000spins...
then you got a way to win...constantly.

Greenguy,
3700, is  assumed,
that all bet may around BREAKEVEN.

If you bet flat, u may win, or lose,
accordingly to your ladyluck.

Say, if we see, that a single number sleeps for 500spins, then you may bet that single, continuously, to 3700th spin.
and hopefully, as, or if it breakeven earlier, then you win, even before the 3700th spin.

Hi Albalaha,
you say,
------------------------------------
------------------------------------------
Virtual limit of persistent bad doing bets, in my observations(in european roulette):

straight up:only 4 hits in 1000 spins
split: only 15 hits in 1000 spins
street: only 40 hits in 1000 spins
corner: 60 hits in 1000 spins
line bet: 110 hits in 1000 spins
column/dozen: 250 hits in 1000 spins
EC bet: 405 hits in 1000 spins
double dozens/columns: 575 hits in 1000 spins

unquote.
----------------------

Thus, if we could have a progression, that beat, say,
EC bet: 405 hits in 1000 spins,

win=+405u,
lose=595u,
losses=-190...

then we need to have a progression,
that recoup -190losses,
by 405hits.
=====================================
or win, at least 1u, with only 405hits/1000spins!!!

that's a grail!!!

Hi Albalaha,
in your blog, you say...
===========================
Every bet selection reaches at its break even point minus house edge in long run


I do have my own set of trackers and bots. I have extensively tested all bet selections over millions of spins. In long run, say in 3700 spins (100 cycles of all numbers) or more, every bet selection gets to its break even point minus house edge. This applies very accurately to outside bets.
               This fact ensures profit for casinos unless a player get lucky streaks and he could capitalize that too.

RG,'
and gentlemen,
I tell you an stale open secret.
If you cut the wheel into three almost equal portion,
whether you bet one portion, or double,
after long spins, you will notice, in long spins, they'll always almost equal,

BUT, they tend to CLUSTERED,
AND NOW YOU KNOW WHAT TO DO!

Say,
An OVERSIMPLIFIED EXAMPLE.


You bet marthy for 10losses,
[helooo..this just an OVERSIMPLIFIED EXAMPLE, OK?]

You wait, and wait, and wait,
for 10streak of losses to happen.

Then VIOLA!! It happen!!!

10 streaks of red,
then wait for a black to hit,

and the you bet that TEN RED won't HIT AGAIN.


Hope you understand.

The roulette is casino's HG,
the only way to win,
is to avoid LOSSES,

or losses to happened,
that the losses avoided!

POSSIBLE losses that depleted your profit.

What you gentlemen think???

Bet from different angle.

Ok,
You have a wonderful system,

and you see that ,

say in every 10season,
YOU ARE very CONFIDENT,

your system will lose , say 3times,


Darn!, that THREE session,
will cause,
the other 7season winning,
DEPLETED to NEGATIVE!

Thus how you ensure profit long term???

EASY,

don't play with real money,
play your seasons, on paper,

then only play real when a losing,
or two losing season losses, on paper, happened!

say, 2winning, and 2losing seasons.

AND YOU BET THE REST OF THE SEASON WITH CONFIDENT.

of course, YOU MUST, very CONFIDENT,
THAT YOUR SYSTEM, ALWAYS 'averagely',
LOSE 3, OUT OF 10SEASONS.



What you , gentlemen, THINK???

wait for the losses to happen, then bet the rest, will save chips.

the 12 street hit in next 12spins,
will highly impossible,

and if 8hit/12, is the accepetable averages max,

then wait for 8/12,
and bet it will not 8/12 again.

may better risk/reweard.
What you think.
====================
Wade,
6/6 of DoubleStreet, will hit,
then wait 6/6, then bet,
for it won't happen, will better.

Instead of waiting for 18nos, why not wait for 24,
and bet the 24nos???
Since the law of third, says, always 24???
isn't this better?
Anyone???

They 'say', single will always, hit 100times, in 100cycles, that 3700spins.
Thus if, we bet the last hit single, after x cycles, then we need to scrutinize, the rest of 100cycles-minus-x-cycles, so to speak...and expect around 100hits, to profit...
Anyone???
I think Denzie, has something to say here.
Denzie???
Where art thou???

That 13, is your expectation, of unhit 13cycles.
24, is your expectation, of hit 24cycles.

------------------
say if we wait for last two unhit numbers,
which may always happen around 200spins ,
[I think.]...

Then we bet that two for eternal, or hit.
for profit.
I don't think, two numbers will sleep together for 1000spins?

Anyone here knows, how many spins, the virtual limit for two-numbers to sleeps??

Maybe Mr J could enlighten us, as he swears that 4-number, and 2-numbers bet, as his near grail?
Mr J???

To calculate the total sum of lost bets is easy:
13 times multiplied by 37 (1 x 37) equals minus 481 units.
-------------------

If we see, say, number x, had sleep, for 37spins
then it has 12more cycles to lose...
If we see, say, number x, had sleep, for 74spins
then it has 11more cycles to lose...
If we see, say, number x, had sleep, for 3cycles
then it has 10more cycles to lose...
--------------------------------------------
Then number X, if have sleep for 3cycles,
has 10more cycles possiblity to lose,
thus 10x37=-370u.

And we need to ensure, the theory that it will hit,
in 24 cycles,out of next 34cycle,
[13+24, already slept for 3cycles]

Thus we need to ensure, that the 24 cycles hit must win more than 370u.
[370/24=+15u]

To calculate the total sum of lost bets is easy:
13 times multiplied by 37 (1 x 37) equals minus 481 units.
-------------------

If we see, say, number x, had sleep, for 37spins
then it has 12more cycles to lose...
If we see, say, number x, had sleep, for 74spins
then it has 11more cycles to lose...
If we see, say, number x, had sleep, for 3cycles
then it has 10more cycles to lose...
--------------------------------------------
13cycles multiplied by 37 =481spins


and we may see, say, number v,y,z,
that sleep for the entire spins of say 100spins.

and a few number will stubborn, and thus, we aim for, say LAST FOUR numbers that has sleep for the whole spins,not yet hit,

then the simplest is to wait for the LAST FOUR unhit numbers, for they have sleep, for 1,2,3 or more cycles, thus the averages 13, now reduced to 12,11,10,9..so on,

and then bet them till profit, or cut loss...
AGREED?
I don't think FOUR numbers, together, will sleep for 1000spins?