ANOTHER REALLY GOOD SYSTEM!!!!!!!!!!! The Perfect Pattern...

[Newlight]

14 17 20 23  Was another Pattern... then I split 27-30 33-36  25-28 31-34  also split 3-6 9-12 1-4 7-10

Another Pattern I got was 13 14 16 18 12.... That Makes a X....

In Regards To the Above  when I got my win, I would Start a NEW SHEET....

YOU CAN Start a NEW SHEET ANYTIME IF YOU WANT TO... 

I RECOMMEND starting a new SHEET so you don't get confused.

This is ONLY my word but the last 24 hrs I made 290 on a reputable online casino. I won't Mention the name for people will accuse me of advertising for some one else.

Last Night FOR REAL 80% of the Time I would Win in the 1st Spin after the Pattern.

I was betting like so Low Like as cheap as I could.  I logged on and off cashed out 4 times...

What I'm saying I was prepared to progress Bet but Never got the opportunity.

Again you can Try this on FREE Roulette Maddness on Facebook it's a app I use to Practice on and it's FREE... So again not Promoting nothing. 

[chrisbis]

This is ONLY my word but the last 24 hrs I made 290 on a reputable online casino. I won't Mention the name for people will accuse me of advertising for some one else.

Last Night FOR REAL 80% of the Time I would Win in the 1st Spin after the Pattern.

Its OK friend, U can mention where U played Ur game.


We know the difference between Advertising, and genuine "Information".


Cheers.

[Smee]

This is really interesting Newlight....but im afraid i really dont get it. Can you explain it please spin by spin?

Cheers!

[Newlight]

Also just want to add... When I was doin Online I was betting $1 on RED AND BLACK to get the spins to get the pattern...

Got 0 twice and $2... It's up to the Moderators if they take the Post but I was Using Party Casino...

I got on to them playing Texas Holdem and I've always been able to cash out with no worries...

Party Poker.... Again if you USA you won't be able to get on and this system is based on European roulette only...

But On Party Poker if you PUT EVEN MONEY BETS and ZERO comes out YOU ONLY LOSE HALF your BET... Not all....

[Smee]

So....we start crossing off numbers as the ball lands on them. For example we have 6,5,4,2 and 8 crossed off - forming a + in the first dozen.

We then bet that we dont get a + in the 2nd or 3rd dozen.....but how do we actually do that?

By betting on every number except 18,17,16,14 and 20 (if we were going for the  + not to happen in the 2nd dozen) ?

So we put 1 unit on every number except those 5....and one of those 5 comes up so we lose that bet. Do we then put 36 units on every number except those 5 to get our bet plus our money back as in a martingale? At the live wheels I play on they wont let any bet higher than 35 units on the single numbers so I dont know if that will work - it wont let you go to a second progression due to table limits.

I really like your idea but Im sure im missing something here. Thanks for the explanations tho!

[catalyst]

hi newlight
its a interesting strategy from the raindrop analogy. but you have messed up with your poor explanation. please be advised to be organized in explanation. for the forum members i lay down the raindrop analogy:

i explored the following system couple of months before. it seems to me drazen system (see the system by Drazen-co) has been reinvented with the progression. the following system was played in flatbetting.



====================================

Clercx posted an excellent analogy concerning raindrops and probability.
Mostly concerning sections of the wheel that have failed to show.
If you compare this to rain, you can assume that both are random - but the largest dry area
will probably be receiving raindrops in the very near future since we know the eventual outcome
is complete coverage of the area (wheel).

I'm looking at it another way, although simillar.

Let's look at the 3 dozen sections in front of us at the table as 3 squares side by side (a sidewalk - for this analogy)
All three will receive raindrops randomly, but we can be almost 100% assured that all 3 will never have the same identical pattern between them.

This applies to the dozen sections when looking at the table layout in graphical layout.

Let's assume that after 37 spins we have sleepers in dozen 1 of #3, #8, #9 and #12
What are the odds that dozen 2 and dozen 3 both have those same identical sleepers in their layout ?
for dozen 2 it would be #s 15,20,21,24 and for dozen 3 #s 27,32,33,36
The odds of all three sections having the same pattern is nill in my opinion.
(I'm sure a possible "rare" event though of course)

Now let's assume we have sleepers in the dozen 1 section again of 3,8,9 and 12.
From the above we can "assume" that 15,20,21,24,27,32,33 and 36 "have" hits on them,
probably many but no doubt about average.   Will these locations show a higher hit rate than
expected since their dozen 1 neighbor has these same locations as sleepers ?

Let's run off a quick test from actuals and see where the merit in this is (if any)

I chose a random s/h day, in this case Jan 1 2005

We can record the first 37 spins, and have 14 sleepers (as normal).
The sleepers on the image below are left on the layout, the numbers that hit
have been removed.




What we see is that (as expected) the pattern between all 3 dozens is different.

If we look at the corresponding numbers to the sleepers, we can get the following info :

dozen 1 sleepers :
1,2,3,6,8,11,12
the same locations in dozens 2 and 3 :
13,14,15,18,20,23,24,25,26,27,30,32,35,36
(there are 14 numbers here, during the last 37 spins they had 18 hits on them )

dozen 2 sleepers :
13,19,22,24
the same locations in dozens 1 and 3 :
1,7,10,12,25,31,34,36
(there are 8 numbers here, during the last 37 spins they had 8 hits on them )

dozen 3 sleepers :
28,34,35
the same locations in dozens 1 and 2 :
4,10,11,16,22,23
(there are 6 numbers here, during the last 37 spins they had 8 hits on them )

in summary to this point -
we have 14 sleepers total, 28 numbers that are corresponding to those sleepers,
and a total of 34 hits on them.

===

The advantage here ?
Some people will say "so what ? of course the sleepers had 0 shows and the others had
more than average, that's common sense"
also, "what good is this info ?  It's from the past and doesn't effect the future spins where
we are going to be playing"
This isn't the case in my opinion.

We can assume a lot with this game,
above we assumed that all 3 dozens wouldn't have the same pattern of sleepers. .  and they
don't (and probably almost never will)
Now, we can assume the the next 37 spins will not be identical to the 37 that just happened. . .
(leaving the exact same sleepers in the exact same locations)

What does this tell us if anything ?

Let's look at the dozen with the largest number of sleepers.
This will be our base dozen.
I'll assume that during the next 37 spins there will be hits on these sleepers (somewhere)
and I don't need to know which ones or when actually.
Now, since all 3 dozens won't follow the same pattern, I can assume the the "opposite"
if these sleepers in the other 2 dozens will also show during the next 37.

This leaves us with a bet selection that is very simple.

We are going to bet on the sleepers in the dozen that had the most of them,
in this case Dozen 1
We are also going to bet in the opposite of these numbers in the other two dozens.

From this day - Jan 1 2005 it means the following :

Bets :
Dozen 1 : 1,2,3,6,8,11,12
Dozen 2 : 16,17,19,21,22
Dozen 3 : 28,29,31,33,34

17 number total flat betting $5. 00 each
On any win, we simply add to the winner $5. 00 (there's no sense in removing the bet after it wins if we believe that it will show again)

let's look at the next 37 spins in order :

19 win #1
24
3 win #1
26
15
21 win #1
31 win #1
2 win #1
24
32
25
30
35
12 wn #1
20
9
2 win #2
35
19 win #2
21 win #2
16 win #1
22 win #1
17 win #1
17 win #2
3 win #2
31 win #2
32
18
23
12 win #2
5
10
13
22 win #2
2 win #3
12 win #3
22 win #3

= end of second 37 spin cycle =

Type your title here.



+ $1,315. 00 for only 37 spins.

So far live I've had one +5K day and one +4K day of play using this approach.

Would I call it the holy grail ?
Of course not.

But a better form of bet selection that most systems use, yes.

The probability guys should be able to back up this method also,
although in the end it may fair no better than any other method of play.



==================================

So, what does everyone think about this? 


THANKS
CATALYST

[Bayes]

Very nice. 

As you say, the odds are practically nil that the exact same sleepers will occur in the corresponding dozens over the next 37 spins. This is a fresh approach, good job!

How much testing have you done?

[ZeroBlue]

This is cool!


will test asap.


thx

[akuuka]

Hi Catalyst ...


Very interesting System and clear explanation. 17 Number for 37 Spin. IMHO, there is bankroll that we should to consider to play this system.


I've been testing using live result from casinosbo.com.

I play just only on Dozen:
Bets :
Dozen 1 : 1,2,3,6,8,11,12
Dozen 2 : 16,17,19,21,22
Dozen 3 : 28,29,31,33,34
Here the result :


[reveal]


1   30   l   -17
2   31   w   36
3   14   l   -17
4   29   w   36
5   7   l   -17
6   34   w   36
7   14   l   -17
8   10   l   -17
9   24   l   -17
10   11   w   36
11   14   l   -17
12   29   w   36
13   32   l   -17
14   11   w   36
15   3   w   36
16   14   l   -17
17   28   w   36
18   19   w   36
19   33   w   36
20   0   l   -17
21   18   l   -17
22   27   l   -17
23   17   w   36
24   11   w   36
25   12   w   36
26   4   l   -17
27   36   l   -17
28   1   w   36
29   8   w   36
30   14   l   -17
31   24   l   -17
32   25   l   -17
33   11   w   36
34   15   l   -17
35   30   l   -17
36   6   w   36
37   33   w   36
38   0      325 <---------- total win after 2nd round of 37 spins
39   28   w   36
40   30   l   -17
41   35   l   -17
42   6   w   36
43   23   l   -17
44   7   l   -17
45   36   l   -17
46   27   l   -17
47   27   l   -17
48   9   l   -17
49   10   l   -17
50   19   w   36
51   34   w   36
52   16   w   36
53   13   l   -17
54   23   l   -17
55   11   w   36
56   11   w   36
57   26   l   -17
58   17   w   36
59   22   w   36
60   27   l   -17
61   15   l   -17
62   28   w   36
63   18   l   -17
64   35   l   -17
65   22   w   36
66   36   l   -17
67   25   l   -17
68   36   l   -17
69   5   l   -17
70   15   l   -17
71   19   w   36
72   19   w   36
73   10   l   -17
74   5   l   -17
75   8   w   36
76   14      113 <---------- total win after 2nd round of 37 spins[/reveal]


This is great but with considering the bankroll 

[Fripper]

Hi catalyst and thanks for sharing, it's a nice idea and I like it already.


Testing is needed as with everything but shouldn't the method be posted in it's own thread?


Cheers

[MrJ]

As I have always said.....this is EVENT betting and I love it. Event betting has little to do with the actual numbers, per say.

Ken

[Hermes]

OK, it is playable this way but much finer approach is with the original raindrop strategy mapping on wheel.  Why mess with this mimic incomplete strategy when there is one already better suited?
Finally somebody explained it properly.
Thanks Hermes

[Smee]

ok...now its making sense!

So we have 37 spins, find the dozen with the most sleepers and bet them, and also bet the opposite of those sleepers on the other 2 dozens. We flat bet those same numbers for 37 spins - is this correct?

Wouldnt this be difficult to play anything but RNG? You would have approximately 25 seconds after the 37th spin to work out the bets and place them on the board....

So I tried this on RNG and after the 37 spins I only had 2 sleepers in doz 1, 3 in doz 2 and 3 in doz 3.

That means I bet on 3 numbers in doz 2 or 3, i picked 2 - the sleepers - and 9 numbers in doz 1 and 3. I guess its correct as per the system but the bet selection isnt quite what was in the example.

[Smee]

I just played 5 games RNG real money.

As long as the rules I posted above are the correct system -

Game 1 + 50 units
Game 2 + 159 units
Game 3 - 54 units
Game 4 + 160 units
Game 5 - 124 units

Overall tho up 191 units in 185 spins. Not too bad i guess - but it still kinda seems as tho im placing the chips on random numbers....Ill give it another 5 games tomorrow and post results.

[monaco]

i paper-tested this on 4 different sessions (3 smartlivecasino live dealer, 1 random.org) - i tried to do it exactly as per the original instructions by clercx, using 5u as base bet, adding 5u to a hit number.

i had 1 winning session (+400u) & 3 losing ones (-520u, -1245u, -420u)

i also tried just flat-betting, 3 sessions ended in the minus, 1 positive.


I would say though that all sessions fluctuated about 3-5 steps either way - all 4 were in the positive at some point - so it might be worthwhile applying some kind of parameters along the lines of, if 3 or 4 steps up, quit, if 2 or 3 steps down, aim to break even or 1 step up.. doing it this way, 1u flat betting only, could’ve come out with 142u profit in 34 spins total across all 4 sessions.