Hi friends, this already a stale topic, but I post for those yet to understand, how to bet 16numbers of pies and stars numbers in matrix.
stringbeanpc said Nov 28, 2013...
Something I notice is when the spins are arranged in rows of 6 is that
the frequency of total pies/stars represented per row resembles a bell curve.
Example, find a group of about 600 spins and organize them into rows
of 6 spins each.
For each row of 6 spins
1) Most frequent are rows with a total of four separate pies
2) Next with approximate equal frequency is rows with either three or
five separate pies
3) Next, but with much lower frequency is a rows with either two or
six separate pies
4) Lastly, very rare is to have rows with only one pie represented
I notice approximately the same statistics with the stars.
xxxxxxxxxxxxxxxxxxxx
and Kimo AGREED with the finding of stringbeanpc,
see below.
xxxxxxxxxxxxxxxxx
KimoLiRoulette said Nov 28, 2013
Hi Stringbeanpc,
You are on the right path. Consistency is in line with the law of
averages and probability.
Kimo Li
xxxxxxxxxxxxxxxxxxxxx
stringbeanpc said Dec 02, 2013
Here is some of my analysis from a set of 1200 European spins (200
rows of 6) with the green zero removed
Rows with 4 pies = 105
Rows with 3 or 5 pies = 38 & 51 respectively
Rows with 2 or 6 pies = 4 & 2 respectively
Rows with only 1 pie = 0
thoug it just a small sample, the most important hypothesis, is...
Rows with 4 pies = 105
Rows 5 pies = 51 .
Rows with 6 pies = 2
...
thus
4pies/6spins=105vs51vs2
=105/158
approximately 2/3.
thus we track 4hit in 5spins, that has independent pies and stars, and bet 16numbers only, hoping that both pies and star will stay 4hit/6spins
we could also bet only 20numbers, if pies has 5hit/5spin and stars has 4/5spins, (or vice versa)...
stringbeanpc said Dec 02, 2013
Here is some of my analysis from a set of 1200 European spins (200
rows of 6) with the green zero removed
Rows with 4 pies = 105
Rows with 3 or 5 pies = 38 & 51 respectively
Rows with 2 or 6 pies = 4 & 2 respectively
Rows with only 1 pie = 0
xxxxxx
think hard!
Rows with 3 or 5 pies/stars = 38 & 51 respectively
Rows with 2 or 6 pies/stars = 4 & 2 respectively
Rows with only 1 pie/star = almost 0.
you think in term of pies/stars combination.
Lets say, after 5spins, theres only 1pie/star hit, then how you bet the risk that POSSIBLITY, 6/37, of losing?
How about only 2pies/stars hit after 5spins???
you bet only the numbers that overlap by the horizontal/ vertical pies/stars...
we bet the possibilities, of both pies/ stars high probabilities of hit.
eg, all six pies and all six stars hit in next 6spins, has low probabilities.
only 1pie and only 1star hit in next 6spin, has low probabilities.
only 2pies and 2stars to hit in next 6spins, has low probabilities, thus how to bet?
The RISK, still there.
Cover the zero, in long term, everything, will hit math expectation!
You cant find those Stringbeanpc and Kimo Li conversations, Kimo Li had erase them since Stringbeanpc had inadvertently disclose the core math probabilities...
I copied them 10years ago, and only now understand ...
you can bet pies and stars in form of PINWHEELS..
you need two pinwheels.
one for pies and one for stars.
then OVERLAP...
Now, you understand, how Kimo Li walk to a table, bet straight away, by just looking at past few previous spins, as he said, "THE EVENT IS THERE!
But people just not see them..."
how to bet the pinwheel?
Kimo Li wont tell..
His students wont tell...
People who understand wont tell...
I tell....
it has 1/27 odd of losing....
very very surprised thats theres no interest in this...
Can you help me understand where the probability of losing only 1/27 comes from? Thank you
Kimo Li's pies and stars matrix for the reference :thumbsup:
why bother with a picture just use the real thing by crystal
here you go
I bothered with the picture because I didn't know a tracker existed.
Thank you for sharing. Beautiful visual layouts.
your welcome
This tool is a good starting point. But hard to figure out how to use it. Every spin has information attached to it.
Process it like computer & you should win at less one bet some how.
Quote from: number25 on Sep 23, 10:29 PM 2023This tool is a good starting point. But hard to figure out how to use it. Every spin has information attached to it.
Process it like computer & you should win at less one bet some how.
1st re read what sweet has told you...apply that as a starting point,,,this tool lets you do that
@ 6th-sense
Need pen & paper in casino. The Matrix is a good tool.
Sry about the confusion
in next six spins, except covered green, one of these permutations will hit.
eg.
1star and 1pie
1s/2p
1/3
1/4
1/5
1/6...
2/1
2/2
2/3...so on
3/1
3/2...so on...
till
6/1
6stars/2pies..
so on.
so...
the least be
after 6spins..
only..
1star/1pie
after six spins...
max...
6stars and 6pies.
.after 6spins.
and you see, there multiple way strategy to bet
.
so after we track FIVE SPIN,
after five spin,
we see
ONLY
1STAR AND 1PIE.
OR
1PIE AND 1 STAR.
how to take the risk?
or
after 5spin,
we see
FIVE STARS AND FIVE PIES,...
HOW TO BET?
OR AFTER...any ideal permutation...
Then how you bet
the SIXTH SPIN!?
xxxx
Thanks to stringbeanpc.
quote:
in 200rows of six spins...
Rows with 4 pies = 105
Rows with 3 or 5 pies = 38 & 51 respectively
Rows with 2 or 6 pies = 4 & 2 respectively
Rows with only 1 pie = 0
unquote.
xxxxxx
think hard!
in stringbeanpc's
small sample of,1200spins,
hypothesised,
200rows of six spins...
pies,
or stars hit
Rows with 6=2
Rows with 5=51
Rows with 4= 105
Rows with 3 =38
Rows with 2=4
Rows with 1=0
total=200rows.
xxxxxx
tracking 5spins...
At the FIFTH SPIN,
only 1pie hit after 5spins,
at the SIXTH SPIN...
It may,
STAY at 1pie,1/6,
or hit another,
to BECOME TWO pies in six spins.
After 5spins,
if it had hit only 2pies,
it may STAY. ..
at 2pies/6spins,
or hit another to become
3pies/6spins
After 5spins,
3pies may stay 3/6, or become 4/6
After 5spins, 4pies may stay 4/6, or hit 5/6.
After 5spins, 5pies, may
5/6, or hit 6/6...
So you must understand,
that only four pies in six spins, though 105/200rows,
mean...
it was 3/5 before hit 4/6...
OR...
it was 4/5spins, to become 4/6spins...
thus bet when 3/5,
to become 4/6...
and,
bet 4/5, to become..
4/6...
Sweet you've just spelled it out in basic form from your 1st explanation which was itself was self explanatory..
They just had to re read like I said ..
:smile:
thanks 6th-sense for your advice.
xxx
this just for understanding only, for for betting.
Let say, you see number 32 hit, and hit again,
that star1 and pie4 hit twice,
and you know that it highly unlikely that star1 and pie4 will hit
6/6 in six spins,
now you cover zero, and bet the other stars and pies for the next four spins.
If pie4 and star1 keep hitting for next 4spins, then you lose your underwear!
Now I hope you understand how to bet...
betting with matrix, is betting TWO game in a single bet, pie is a game, star is a game, if both goes in their winning probabilities, then you win, but with "LESSER STAKING",
thus if you understand that,
only 4stars and only 4pies,in next 6spins...
are most likely to occur,
then if both win, then you win with less staking...
hope you understand...
is it really TWO game in one? Or any possibility, three in one??
There are numerous games that can be played just from those stats amongst the pies and stars
Quote from: SWEET on Sep 25, 03:29 AM 2023xxx
this just for understanding only, for for betting.
Let say, you see number 32 hit, and hit again,
that star1 and pie4 hit twice,
and you know that it highly unlikely that star1 and pie4 will hit
6/6 in six spins,
now you cover zero, and bet the other stars and pies for the next four spins.
If pie4 and star1 keep hitting for next 4spins, then you lose your underwear!
Now I hope you understand how to bet...
definatley lose your underwear and house and life with this example....
@ SWEET
You understand the madness.
Will you play roulette now?
Now you need to pick what side of the wheel is this going to happen. This way you can bet less numbers.
GOOD JOB!!!
betting
1star/1 pie
to hit 2...
bet 25number.
bet 2stars/2pies..
to hit 3....
bet 16numbers
bet 3stars/3pies...
to hit 4/6...
bet 9numbers.
bet 4stars/pies,
to remain at 4/6.
bet 16numbers...
and to bet lessor numbers, you may follow last, red only, hogh or low, or even pinwheel :xd:
Thanks 6th-sense and thanks number25,
The risk is always there, we need good mm, to recover losses too.
I think,
lets say, we bet 3stars/3pies to hit 4, need to bet last three spins, with marthy, labby, or flatbet
personally, I think,
bet,
after,
2STARS AND 2PIES,
would better.
You just bet
16numbers,
hoping that the
2stars/pies,
would not remain as two, before or at the sixth spin.
Both stars and pies need to hit 3,
thou, or you dead meat
maybe something a bit more logical would be better than hope and a super massive progression....?
brute force isn,t the answer...if your going to use brute force narrow the numbers down a bit to something more managable...
I have had some unbelievable days betting just 9 numbers. Playing low red, low black etc or low even, high odd numbers etc. I had gotten away away from this kind of betting. Maybe I should go back to concentrating on this.
Play 9 inside numbers is a good option, also 6 or 12. Flatbet
See 5 possible systems and the tracker for them. Always bet temporal hot groups, can jump from one system to other....5Switch 9 Tracker.xls
System 1
A 32,15,19,4,21,2,25,17,34
B 6,27,13,36,11,30,8,23,10
C 5,24,16,33,1,20,14,31,9
D 22,18,29,7,28,12,35,3,26
System 2
E 19,4,21,2,25,17,34,6,27
F 13,36,11,30,8,23,10,5,24
G 16,33,1,20,14,31,9,22,18
H 29,7,28,12,35,3,26,0,32
System 3
I 21,2,25,17,34,6,27,13,36
J 11,30,8,23,10,5,24,16,33
K 1,20,14,31,9,22,18,29,7
L 28,12,35,3,26,0,32,15,19
System 4
M 25,17,34,6,27,13,36,11,30
N 8,23,10,5,24,16,33,1,20
O 14,31,9,22,18,29,7,28,12
P 35,3,26,0,32,15,19,4,21
System 5
R 34,6,27,13,36,11,30,8,23
S 10,5,24,16,33,1,20,14,31
T 9,22,18,29,7,28,12,35,3
U 26,0,32,15,19,4,21,2,25
Thanks 6th-sense , Irish88.and Kattila...
When we heard,
"Bet 16numbers",
We thought,
"wow",
that MASSIVE STAKING!"...
Actually, betting 16 numbers, plus a covered zero, thats 17numbers total, is just as about SAME,
as betting EVEN CHANCE...
RED,ODD, HIGH, IS
betting 18numbers, without zero...
I mean massive staking doing a Marty on 16 numbers ..the example you gave still falls within loss expectation..
A better example is needed I think
Nice idea kattila
SWEET, thanks for sharing your idea.
One important thing, though: The statistics you provided (1200 spins) treat pies and stars as separate events. And both the pies and the stars reach these expectations (probabilities) somewhat separately, as it very often happens that (for example) in 4 spins we have 4xstars and 3xpies, or 4xstars and 2xpies, or even 4x stars and only one pie. The same thing the other way around. Playing them separately at this point we have, of course, 24 numbers bet + 0 = 25 numbers bet. In your approach, when we wait for 4 simultaneously on the pies and stars side and play overlapping numbers, these statistics may be different.
Of course, it can be programmed. 1200 spins is not a lot.
Just a little note from me. I'm thinking out loud while playing with the tracker.
Thanks 6th-sense.
Thanks duchobor for your thought.
Yes, you're right, the stars and pies have separate probabilities, and permutation and combination of stars and pies event.
Sweet, a little ....riddle,
Where are two might be the third, maybe the 3rd it s not in P :( but is in S :) , maybe the 3rd it s not in S :( but it s in P :) , maybe the third it is not in both >:( , maybe the third it is in both ;D
:( :) :( :) >:( >:( ;D ;D
समानांतर समानांतर
cheers
Maybe = house edge
Of course :) , but at least is flatbet, maybe no big lose , maybe no big win , maybe little win , just maybe ....
Thanks Kattila for your riddle.
let say 32hit, that's s1/p4...
then 17hit,
that's s2/pie6...
then if we bet 16 numbers,
(thats not s1,s2,p4,p6...)
then we only win when next number, not land on s1,s2,p4,p6..
stars will also,
move to third,
if the next number land on pie4/p6..
same for that 2pies, move to third if ball landing on s1,s2...
and we lose, even star and pies move from two to three.
According to
Stringbeanpc's small sample,
200rows of six spins...
pies,
or stars hit
Rows with 6=2
Rows with 5=51
Rows with 4= 105
Rows with 3 =38
Rows with 2=4
Rows with 1=0
total=200rows.
it seems that
1,2,3 hit
42/200rows,
and 4,5,6 hit
158/200rows...
then THRESHOLD OF...
3stars /pies
to move to 4,
seems interesting.
at the FIFTH SPINS,
THE
2 move to 3 AND/OR
3 STAYED at 3, at SIXTH SPIN,
both event COMBINED, are
38/200...
Thus at fifth spin, either,
2 move to 3, OR 3/5,
not moving,
stayed at 3/6
=38/200...
SO,
to bet,
we track the THREE, the soonest,
at third-spin,
then fourth,
or fifth spin,
(3 hit at third, fourth spin, or fifth spin), and at the sixth spin, or fifth spin, or 4th spin, the chance of 4stars hit=105/200,51/200,2/200.
That's
4,5,6../200
respectively
thus the chances, of 1,2,3 NOT moving, and stayed, after SIX SPIN,
=0+4+38..
/200...
4,5,6materialised, at or before SIXTH SPIN
=105+51+2..=158/200...
So how to bet?
So, in OVERSIMPLIFIED EXAMPLE.
If you sit at the table, then wait for THREE SPINS to hit, then bet straight away, that 4 will hit,
AT or before,
SIXTH spin,
Then the chances to hit 4,5,6 =158/200.
losing chances=42/200...
SO?
HOW TO BET?
To bet STARS ONLY, OR PIES ONLY, OR BET BOTH PIES/STARS?
thus,
we need to bet ...
marthy..
that's
1,2,4...
so if that small sample, is as good as trillions spins sample,
then,
bet marthy...
1,2,4=losses=6.
losing 6=?/200.
winning 1=?/200
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?
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
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!
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
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
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...
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.
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
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.
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.