• Welcome to #1 Roulette Forum & Message Board | www.RouletteForum.cc.

News:

Odds and payouts are different things. If either the odds or payouts don't change, then the result is the same - eventual loss.

Main Menu
Popular pages:

Roulette System

The Roulette Systems That Really Work

Roulette Computers

Hidden Electronics That Predict Spins

Roulette Strategy

Why Roulette Betting Strategies Lose

Roulette System

The Honest Live Online Roulette Casinos

BET KIMO LI'S PIES&STAR MATRIX

Started by SWEET, Sep 17, 10:22 AM 2023

Previous topic - Next topic

0 Members and 3 Guests are viewing this topic.

SWEET

then, so...

we need to wait for the 3,
to hit first..

3/6,
or
2 move to 3, at sixth spin...

=38/200..

that 38, mean, mixed of 3 ,stayed at 3, and 2 move to 3, at sixth spin.

thus it should less than 38, because, 2 move to 3, is quite huge...

then...
4,5,6 hit AT or BEFORE sixth spin,
=158/200.

then what the RISK?

SWEET

so in OVERSIMPLIFIED EXAMPLE,
in that VERY SMALL SAMPLE,

MARTHY1,2,OR 4=may bet three, two or one spin..

may hit in 1st, 2nd or third spin,
 WIN 1=158/200...

LOSING CHANCE,
1,2,4=
less than 38/200


duchobor

Quote from: SWEET on Sep 29, 10:04 PM 2023so in OVERSIMPLIFIED EXAMPLE,
in that VERY SMALL SAMPLE,

MARTHY1,2,OR 4=may bet three, two or one spin..

may hit in 1st, 2nd or third spin,
 WIN 1=158/200...

LOSING CHANCE,
1,2,4=
less than 38/200

Good morning SWEET,
it's -202 units after this 200 rows (1200 spins).
If you wait to have 3 different pies or stars and then bet for it to move to 4, you bet 3 remaining pies or stars which is a 18 numbers bet + 0 = 19 numbers bet. With a payout of 36, you gain 17 units on every hit.
158 x 17 = 2686
Every lost attempt using Martingale costs you 4 x 19 = 76.
76 x 38 = 2888
2686 - 2888 = -202
Martingale is not a solution here, I'm afraid.
All the best!
 

6th-sense

Quote from: duchobor on Sep 30, 05:01 AM 2023Good morning SWEET,
it's -202 units after this 200 rows (1200 spins).
If you wait to have 3 different pies or stars and then bet for it to move to 4, you bet 3 remaining pies or stars which is a 18 numbers bet + 0 = 19 numbers bet. With a payout of 36, you gain 17 units on every hit.
158 x 17 = 2686
Every lost attempt using Martingale costs you 4 x 19 = 76.
76 x 38 = 2888
2686 - 2888 = -202
Martingale is not a solution here, I'm afraid.


All the best!


excactly..falls within expectation as i said

6th-sense

but at the end of the day its not as straightforward as sweets excited example...and kimo always said its about money management ie progression...the key is on what?...his lessons are mainly to help you memorise the different layouts...and not to use pen and paper...and to track hot zones/areas etc and use them to your advantage,,,,

after all hot is hot ...cold is cold

SWEET

Thanks 6th-sense and Duchubor,
for you thoughts.

We take Stringbeanpc's SMALL sample, and temporarily...REGARD them as statically truth.

Rows with 6=2
Rows with 5=51
Rows with 4= 105
-----------------versus
Rows with 3 =38
Rows with 2=4
Rows with 1=0
total=200rows.

NOW,
we basically, bet the
CROSSOVER...

of

 THREE stars
to FOUR stars,

before or at sixth spin.

FOUR stars,
 4/6,
 in 6spins,
may hit in...

FOURTH spin,
FIFTH spin,
SIXTH spin...

of course,
THREE must
hit first,
before FOUR...

THUS,
Three may hit soonest,

at THIRD spin,

or at FOURTH SPIN,
or,
at FIFTH spin,
or,
at SIXTH spin.

all that 3rd, 4th,5th and 6th, show,
in that small sample,

=38/200rows.

mean,
not all required 3-step marthy...to bet.

2STEP,
 at 5th bet,
and 6th bet,


1step, at sixth spin...



We cant bet, when 2stars hit at 5th spin, (and then, 3stars hit at 6th spin)...

thus, betting. THIRD TO CROSSOVER TO FOUR, losing chances less than 38/200rows...

BUT...
(at that small sample thou...)
FOUR STARS, AND MORE..( before 5 and 6stars, hit,... FOUR  must hit first)...
the chances =158/200.

thus, in that 1200spins 200rows.
SMALL SAMPLE.

If bet marthy,
we hit absolute 158win,
but less than 38losses...


Why?
because, 3stars in 6spins, do not mean we bet all that 38, and not all bet placed,
 lose THREE STEP MARTHY.


We do not bet when...

only 2stars hit in 5th spin...
because no 3star, then 3stars hit at 6th spin.

We may only bet at 6th spin, that 1step bet...
when 3stars hit at 5th spin

at 5th bet,
that's 2step...
when 3s hit at 4th spin

at fourth spin,
that's 3step..
when 3s hit at 3rd spin.

and
not bet at all, when only 2stars hit at 5th spin,

and 3stars hit in 6th spin.

Thus , in that SMALL SAMPLE 200ROWS,
losing bets,
are
less ,
than 38bet.
in 200rows

Maybe a kind soul here could do a few millions spin test to see that stringbeanpc's small sample hold true...

SWEET

My thought,

it just basically bet CROSSOVER

 from 3stars, or pies to 4stars,
AFTER 3stars had hit.

We may bet at
 4th spin, after 3s had hit in 3rd spin.
That marthy 1,2,4 bet

We may bet in 5th spin, AFTER, 3s hit in 4th spin...that marthy1,2..bet

We may bet in 6th spin, when 3s hit in 5th spin.
that flatbet 1only

We CANNOT bet, when 3s hit in the 6th spin...
But, 3s hit in 6th spin, ALSO part of that 38/200rows STATISTIC...

THUS, losses is less than 38/200, because we cant bet when 2s hit in 5th spin, and 3s hit in the 6th spin.

and not all losses is 1,2,4...because we may lose 3step, 2step, only 1step, and no bet.

SWEET

if we bet 3s WILL NOT crossover to 4s,in that small sample 1200spins, 200rows...

 that statiscally dangerous, because...

you lose absolutely...
158/200,

and win less than 38/200...

why lose 158/200, because, 4s,5s,6s hit after 3s had hit.

Why win less than 38/200,
because we cant bet at 2s at fifth spin

-