BET KIMO LI'S PIES&STAR MATRIX

[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]

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

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]

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!

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

[SWEET]

I am dumb in math.
Nevertheless,
if,
we bet Stars and pies,
as two game together in pinwheels,
then math wise, the best and worst shall like ...

bet 16numbers FLAT, when two different stars, and pies, hit in last two spins,  so the numbers to bet, as lapped=16 albeit zero.

The math 66 vs 33% for DD.
66/33 for stars
66/33 for pies.
lapped 16numbers bet.

THUS,
OVERSIMPLIFIED!

OVERSIMPLIFIED, mind you..

the best PROBABILITY expectation...in next 100rows QUALIFIED BET...=
BEST AND WORST(IMPOSSIBLE!)
100HIT=100x (+19)=
100Lose=100x(-16)=

2nd best...
as
NORMAL math expectation...
66HIT=+19
33LOSE=-16.

since we play TWO GAMES IN ONE, play the lapped 16number albeit green, can anyone do the best and worst math expectation?
Thanks in advance.

[SWEET]

the bane is, both stars and pies, must hit at the same bet.
if one side hit, and other side lose, or both sides lose, then the losses=-16.

When both side hit,
=+19.
thus if both side indeed hit 66%,then, 66+66=132-100=32..
The
worst "only two side stars and pies win together"=32hit.