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System based on Law of the third-need comments

Started by Drazen, Apr 26, 12:39 PM 2011

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0 Members and 2 Guests are viewing this topic.

albertojonas

Made a test on this one, on fresh real spins...

start bank 500

::)

(added part2)

A few things I must consider:
Retrack 37 spins no matter what happens.
at the end if I get sleppers=
12-play to 6 hits (not 6 sleepers, since any of them can hit more than once)
13-play to 6 hits
14-play to 7 hits
15-play to 8 hits
16-play to 8 hits
17-play to 9 hits
18-play to 9 hits
...
when game is over I just retrack new 37 spins.

i am using a simple martingale progression. Any suggestions on a more conservative / efficient progression?

i know that by the law we would have 12 sleepers but what does really happens?
anyone knows how many sleepers we should have, average, after 37 spins?
can anyone post that info here?
thanks in advance

it is nice to be back



Drazen

Quote from: albertojonas on May 10, 10:18 PM 2011
Made a test on this one, on fresh real spins...

start bank 500

::)

(added part2)

A few things I must consider:
Retrack 37 spins no matter what happens.
at the end if I get sleppers=
12-play to 6 hits (not 6 sleepers, since any of them can hit more than once)
13-play to 6 hits
14-play to 7 hits
15-play to 8 hits
16-play to 8 hits
17-play to 9 hits
18-play to 9 hits
...
when game is over I just retrack new 37 spins.

i am using a simple martingale progression. Any suggestions on a more conservative / efficient progression?

i know that by the law we would have 12 sleepers but what does really happens?
anyone knows how many sleepers we should have, average, after 37 spins?
can anyone post that info here?
thanks in advance

it is nice to be back




I can't belive what I see!!! You take all sleepers together and than playing marti on all of them together? Jesus.... I don't have enough words to say how dangerous that is...  And bank only 500 units? I can't belive that realy succed to you.
I am not sure that you understand this system realy... Let me tell you. Sometimes in my tests my first sleeper for all of 3 dozens apeared in 9th spin. First sleeper at all! It is not unusal to have 6-7 (even more) spins between hit any sleeper. Your way is hilarious considering that.  I would like to see how would your marti handle that. Recommended bank is 2700 units. Sometimes we  are betting 3 diferent progressions at the same time.  Be aware of that. One wrong step can be deadly and with progression like that you don't want to make wrong steps, belive me on that. Very thankful for testing but you have to do it on the right way mate...

You have detailed progression explained. I don't understand. Why are you playing  like that?
I personaly don't like progression, but in the safest variant (1 hit in dozen with most sleepers) it doesn't go so high. (if you don't have enough high bankroll) Progressions for this system sound very scary, but in this case, hits with 1/2 sleepers are quite good controled with explained progression so it shouldnt be problems.
Regards
               Drazen

GLC

What if we say that according to the law of the third not only will the sleepers hit at a 2/3s rate, but also the numbers that did hit will hit again at a 2/3s rate.

Track for 37/38 spins. take all the numbers that hit and separate them according to 3 sectors on the wheel.  You can divide the wheel however you want, but the sectors need to be in consecutive order.  Pick the sector with the most numbers hit in it.  If you want, you can just pick any consecutive 12 numbers with the most hits in them and start betting on each of these numbers that have hit. 

Let's say we had 9 hits in our 12 numbers, we can expect to have 6 of those 9 numbers hit in the next 37/38 spins.  We bet these numbers for 1 unit each.  Play for 1 win.  If we have a new high bank, we continue to play for 1 unit each number.  Play for a 2nd win.  If we have once again reached a new high bank, we stay at 1 unit but we re-track for a new sector to bet on.

We can start from scratch after our 2nd win with a new 37/38 spins tracking, or we can backtrack however far we want and use those 37/38 numbers.  I prefer to backtrack no more than 24 numbers and continue tracking for the next 13/14 spins.  Then I determine the numbers I'm going to play.

Anytime we win and are at a new high bank balance, we stay at 1 unit per number.  If we win and we are in the hole, we add 1 unit to each number.  Every time we win and are minus, we continue to add 1 unit to each number until we finally hit and it brings us to a new high water mark.

This tweak is based on the idea of playing hot numbers, wheel based, instead of cold numbers, table based.  By picking the hottest section of the wheel, we are giving ourselves the chance to take advantage of any dealer footprint.

Drazen, if you don't want this idea in your topic, delete it and I will open a new topic with it. :thumbsup:

GLC
In my case it doesn't matter.  I'm both!

Drazen

Delete??? No mate. Very good idea. I was saying in my first post about that posibilites for hot numbers also. And i exposed my idea for sleepers on roulette board as you can see. But that should effect on the wheel maybe even better... I am doing some tests considering this applying to wheel. I am investigating many new ways of this system, but for me doing this all alone will take some time. Maybe you can run some tests considering that you just explained?

albertojonas

i showld plat 3 different progressions? one for each dozzen?

how i calculate the progression?

should i take a sleeper out once it hits and the recalculate progression?

should i aim for half of the sleepers hit in each dozen?


thx in advance for your attention. i am sorry to ask all this. :o

Drazen

Quote from: albertojonas on May 11, 12:30 PM 2011
I showld plat 3 different progressions? one for each dozzen?

how I calculate the progression?

should I take a sleeper out once it hits and the recalculate progression?

should I aim for half of the sleepers hit in each dozen?


thanks in advance for your attention. I am sorry to ask all this. :o

Observe and bet all sleepers from one dozen individualy. So that means that you will have 3 progressions at the same time. Whenever you have a hit cross of that number and recalculate progression with the rest of numbers in that dozen. So with every hit in dozen your progression for that dozen will have more steps because you have one number less betting on.  Aim 1/2 of sleepers for each dozen.  You have recalculated progression for some numbers earlier in the post, for the rest you can use calculator on loothog.com
I cant explain more and easyer to figure than i did. Read whole thread carefuly and you will figure out. And then please come back with some tests on real numbers only.
Regards
                Drazen

albertojonas

I got it know
re did the first test with real spins,
first game +169
2nd game +131

not bad at all

what bankroll do you suggest for attack each dozzen?
meaning, 3 separate banks..

best regards

albertojonas

shouldn't we aim for 1/3 of the sleepers in each dozzen?

or am i missreading the law?

testing with 500 bankroll for each dozzen.

has not sunk yet.
;-)

Drazen

Quote from: albertojonas on May 11, 01:37 PM 2011
Shouldn't we aim for 1/3 of the sleepers in each dozzen?

or am I missreading the law?

testing with 500 bankroll for each dozzen.

has not sunk yet.
;-)
You are missreading the law. It should be 2/3 but sometimes we have fluctuations  about that so 1/2 is the best. Just use 1/2.
800 units per dozen for each progression, about 2500 total. Just to be secure

Regards
            Drazen

albertojonas

Quote from: drazen_cro on May 11, 02:11 PM 2011
You are missreading the law. It should be 2/3 but sometimes we have fluctuations  about that so 1/2 is the best. Just use 1/2.
800 units per dozen for each progression, about 2500 total. Just to be secure

Regards
            Drazen
would not be safer?

albertojonas

Quote from: Fripper on Apr 28, 07:23 AM 2011
As I said in one of my posts, I don't chase 2/3 hits on the unhit numbers, I play around 1/2 hits on them.

Here's my chart:

Unhit numbers: 2, 3, 4, 5, 6, 7, 8, 9
I bet on:              1,  1, 2, 2, 3, 3, 4, 4      numbers

Example: If I have 5 unhit numbers, I look for 2 hits then take those numbers away.
If I have 6 unhit numbers I bet them until I have 3 hits then don't bet anymore.

I don't take away numbers after a hit.

Here is my first results:

* = highest progression

Session 1 : +202, 68 spins. (-164) *5
Session 2 : +145, 53 spins. (0) *1
Session 3 : +143, 64 spins (-55) *3
Session 4 : +138, 56 spins (-34) *4
Session 5 : +91, 60 spins (-53) *4
Session 6 : +109, 52 spins (-15) *2
Session 7 : +82, 53 spins (-68) *4
Session 8 : +86, 71 spins (-169) *6
Session 9 : +91, 71 spins (-229) *10
Session 10 : +145, 70 spins (-207) *10

:thumbsup:

this sessions are quite revealing...
how many sessions do we have to win to double our bankroll?
the progression is frightening
too much risk for my fainted heart

Drazen

Quote from: albertojonas on May 11, 06:26 PM 2011
would not be safer?

1/2 didtn fail by now. You can play 1/3, 1/5 if you want....

Drazen

Quote from: albertojonas on May 11, 11:54 PM 2011
this sessions are quite revealing...
how many sessions do we have to win to double our bankroll?
the progression is frightening
too much risk for my fainted heart

You dont have to play all 3 dozens at the same time. Take one for example with most sleepers and take 1/3 or 1/5

catalyst

Hi forum members
i downloaded the following system couple of months before. it seems to me drazen system has been reinvented with the progression. the following system was played in flatbetting. great roulette mind people such as GLC, F_LAT_INO, Albertojonas, twisteruk etc could find it useful and could guide us for longterm win. HERE IT GOES: :smile:

"Hey everyone, ::)


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






 

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? :twisted:


THANKS
CATALYST


Drazen

Thank you Catalyst for this. Personaly I have seen this system already and have been testing in the past for a while...I consider it good also because every time you will be in some plus at least, but not always in the end. So it is important to predict when to stop. That is my opinion.
But I am sure that for someone can be very interersting.

Regards
             Drazen

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