Hello to everyone! This is my first post on this forum and that is why is so long. Sory about that. First of all i have to apologize for my not so good english, but i hope you will understand me. Please correct me on anything i do or say wrong.
I will later in the post explain my system, of course that is why i choosed this subject. But please, allow me to tell you in short my story about roullete. It is imoprtant.
I am 23 years old student of economics faculty in Croatia. I am relative newbie in this thing, beacuse i started to study roulette a few months ago. But in that time i was spending up to 7-8 hours a day studying all mathematical and statistical facts about roulette. . (And of course they were all saying that roulette is in long term unbeatable). That game was magical for me and i was infatuated by this game. ( So please dont underestimate me about knowing all that facts about roulette). But deep down i felt the desire to beat him. I have red almost anything i could find about roulete and i have tried a hundredths of systems. Most of them i tested,were based on some kind of progression. . Of course all of them failed at some point, because of the table limits. Some of them were more successful than the others, but that wasnt enough for me. I wanted more, i wanted to beat him every time. Although I never knew how I felt that it is possible. . . I was looking for something I could use against him. . and now i think i found that. . . It is based on the Law of the third. Although of course I knew about it, and some systems are based on that, I have never seen anyone is using all advantages of this, earlier. Sorry I have not read plenty posts on this forum, but if there is this system like this please tell me. Any suggestion or correction is welcome. And now my idea.
As you know The law of third affects on every random occurrence in nature, and of course on randomly throwed roulette ball. As it effects on the wheel of the roulette it should also affect on the board too, precisely on the positions of the numbers. It is all conected thanks to that law. . . . Let me try to explain you how. . . Huh this is the difficult part but i hope you will understand me. I ll do my best. (because of my average english).
Many variations can be done with this, but i will explain to you how did i figured this out on original example. Lets take an example on first 37 roulette spins. Folowing the law of the third you all know how many approximatly numbers will show up, and how many will be sleepers. Aprox. 1/3 of course will be sleepers. And if we look at the table through the thirds ( i mean dozens), we will see that numbers in the dozens are also hit by that principle. You all know that in 37 spins is imposible that in any of dozens wont be sleepers. . . it cant be that in 37 spins all numbers in at least one dozen are hit. (I will explain all of this on real numbers later. ) Bearing in mind that the law of the third act constantly, I wondered if that would not be expected to act in the next 37 spins. For now I am focused on sleepers only, but i think that applies to repeaters also. . . Just focused on sleepers for now.
Now i will give an example with real numbers taken from “ random. org „
Ok the first 37 spins are:
14 29 31 15 19 32 13 34 8 7
10 9 13 32 16 9 31 1 26 21
18 12 8 31 19 0 21 36 1 8
36 6 29 10 4 15 34
We record the sleepers:
1. dozen sleepers: 2 3 5 11
2. dozen sleepers: 17 20 22 23 24
3. dozen sleepers: 25 27 28 30 33 35
We have 15 numbers that didnt show up. . . And if the law of the third acts all the time, these numbers should behave similary in next 37 spins. Wright? So now I know that 2/3 of these sleepers will show up in next 37 spins. When we look in dozens, that means that we must have 2/3 hitten sleepers in every dozen also. Wright? Example: if we have 4 sleepers in the first dozen we can expect that that number of sleepers will occur in ratio 2/3. So 4*2/3= 2. 4. we expect to have at least 2 hits in next 37 spins for sleeper numbers in first dozen. We do the same thing for every dozen. 2nd dozen: 5 sleepers *2/3= 3 so here we expect 3 hits minimum. 3. Dozen: 6*2/3=3. 6 and that is 3. Of course in some dozens will be much more hits. I ll explain that also, but later. We dont know which dozen is that going to be, and in that we are not interested, but the thing that protect us from unnecessary risking is that we know that 2/3 of the sleepers will have a hit in every dozen. Maybe now you will say what is new in that, we can bet on sleepers, eventualy they will be hit. . . But, since we expect a certain number of hits in every dozen we can take the numbers in each dozen individually and bet on them like they are only numbers we are betting on in 37 spins. Example of 1. Dozen: we have 4 unhit numbers, and we expect at least 2 hits in 37 spins. So that means that we can use much better (safer, longer) progression, than we bet on all sleepers together. If I use profit calculator, with minimun profit (this was tested on unibet, min. Bet 1 max 25 per number) i can get 32 steped progression, and i know that first sleeping number in my first dozen must show up in 2/3 of spins. So 25. Spin is the latest spin in which should first number appear. And i have 32 steps for that. Much more, you see. ïÂÅ When i hit the first of expected numbers, i reset progression or recalculate it. And when i hit last expected in dozen, I stop. No matter on which spin i am. And continue playing like that with the remaining numbers in dozens which are not hitted as many times I expected. I did my best to explain this, but still i owe you real example.
After we have tracked first 37 spins, wrote sleepers, and calculated their progressions. (i use tools on loothog. com for that) we can run next 37 spins and start betting.
Now i will write next 37 spins and explain progression and try to explain how this works in practice.
You have to follow all sleeping numbers individualy, but within dozens. (i hope you understand)
28 33 4 12 0 9 22 36 2 17
28 1 24 33 3 20 0 3 0 35
33 9 16 30 21 14 4 34 18 25
7 1 34 30 13 16 8
In the first dozen i have 4 sleepers and I expect 2 hits.
I already said that i can make 32 step progression whith 4 numbers
In the 2nd dozen, there are 5 sleepers, I can use 25 stepped progression, but appearance of the first number has to be for 1/3 earlier than if there are 4 sleepers.
For the third, we have 6 sleepers, we can use 21 stepped progression max, but the time of the first number appearance is 1/3 earlier than with 5 sleepers, and that means that first number must appear latest by 12th spin and of course diference between 2 hitted number cant be more than 12 spins. . Now lets see. Down i will write, sleepers and next to them spin when some of them from that dozen was hitted. As the next 37 spins were going, i was just waiting and carefuly watched the progression. . .
I ll explain each dozen and number individualy so you can better get it. , ok DOZEN 1:
We know which are our sleepers and we wait them to be hitted. Follow the 2nd spin cycle from the above
2 --number that has first hit from this dozen and that is at 9. Spin (+32 in my progression) than i reset progression and wait for the second hit. . . . that happened at 15. Spin. . . i hit number 3 and beacuse i reset the progresion that was 6. step of progression (15-9=6) you get it? And that is +6 units more at that step of progression . All together 32+6 =38 ,and here is the point where i stop for this dozen
3
5
11
As you can see from the numbers later, number 3 was hitted again in 18. Spin but at that time i have already stopped, because i got 2 expected hits, and I didnt want to unnecessarily risk. Numbers 3, 5, 11 werent hit at this cycle, but law of the third is respected. I hope you see it.
Now lets see 2nd dozen. Sleepers were 17 20 22 23 24 – we expect 3 hits here nad we have 25 step max progression.
Ok, first hit from this dozen happened on 7. Spin, number 22 ( 7th step in progression, +1 unit, we reset progression,start from begining or we recalculate it)
2nd hit was number 17 on 10. Spin and by the progression for this dozen that means step 3 (diference between 7th and 10th hitted spin is 3) (profit +21 unit)
For the 3rd hit we didnt had to wait for too long. As you can see it happend on spin 13. and with this hit we have got our 3rd expected hit, and we are removing all our bets from this dozen. . . In progression this is also step 3 (2nd hitted spin 10. and 3rd on 13. spin so diference is again 3, that means 3 step i progression and again profit +21) Total profit for this dozen is (1+21+21= 43 units)
As you can see and here we have one hit more than we have expected,(16. spin number 20) like in 1st dozen, but this is not our concern, beacuse, by the law of the third we couldnt guarante more than 3 hits.
Ok, last dozen. 6 sleepers, expect to hit 3, and i have said that first number has to show up in max 12 spin , and the diference between 2 most farthest hitted number is of course also 12, logical. I hope you understand all of this by now. . . just pure following of The law of the third as i explained before. . .
And let see the numbers and hits. . . I ll try to explain one more time, what i think is the most important when you follow this law. . . This is the dozen in wich we have the most sleepers. . 6 of them (25,27,28,30,33,35) When we put 6 numbers in our progression calculator, it gaves us max of 21 step progression
Now lets see. . . .
1st hit for the 3rd dozen came up on 1. Spin ïÂÅ (number 28, 1st step of progression +30 units. )
2nd hit we have on 2nd spin. Number 33, (after restart on the first win, this is also 1. Step so +30 units again)
3rd hit happened on 11th spin (number 28 again- that is 9th step in progression and that is +24 units) 3 hits accomplished. Stop.
Total of 30+30+24=85 units for this dozen
Total profit from all 3 dozens: 38 units(1st dozen) + 43 units(2nd dozen) + 85 units (3rd dozen) =166
Now something interesting, if you look at the rest of the numbers for this dozen. . . . You will see that we had 9 hits in this (3rd) dozen. And I stopped at the third hit. . . Why? Because distrubution like this could happen in any of dozen. . . I cant tell in which dozen is going to be the most of the hits, and that is not important. But after all of this i can tell you one more thing. We completed our game in just 15 spins. You get it how? We got all of our needed hits by that spin. . . And sugar for the end of explanation of this system. At the start we had 15 numbers (sleepers). Now look at these numbers after 74th spin. . . .
1st dozen numbers 5 11 stays unhit
2nd dozen numbers 23 27 stays unhit
3rd dozen number 27 stays
4 out of 15. Hm. . isnt that close to 1/3? Yap! How are they distributed? I hope you see this, this law is magical. . . And apply this in the next 37 spins, and you will hit again 2/3 of these sleepers that have left. . . That means at least 3 numbers out of these 5. Apply appropriate progression and bum. Do i need to say that i continued as i just suggested, and The Law of the Third just did the rest? :)
Similar variant of this system is that you can do this with dozen, double dozens. . . But about that later. . . There are many ways of betting and many ways of using this method.
Wow, finaly the end. . . Once again i apologize for my english, i ll try to make it better, i promise.
One little confession at the end. . . I already said that i am a newbie in this, at least in terms of playing for real money. . . I have never played this system for real money, and i tested it most on RNG-s, but i made a few tests with numbers from live casino, and all passed smoothly too.
And dont laugh at me, but i have some basic questions about some thing, that i dont understand.
Why are there some systems that are not for RNG? Does that means that RNG-s are cheating at some point?
Any kind of suggestion, question, lesson is welcomed. Test this, tell me if i am wrong at some point maybe?Be free to PM me at any time. I am new at this, but l want to learn, to improve my game. I want to help and to learn at this forum as much as i can. Is it okay to notice that members like Johnlegend, ScoobyDoo,Hermes for example know about roulette very very much? :) Regards
Drazen
