RANDOM THOUGHTS BY PRIYANKA – A CONCISE REFERENCE (VERSION 2) ![]()
Priyanka’s current position on the multiplayer roulette leaders board![]()
Priyanka’s past position on the multiplayer roulette leaders boardAfter reading some of that thread I decided to follow Priyanka in the Roulette game. I've followed her five times and the results is very similiar as this. It's starts off as a slow game with small wins and small losses and then boom, out of nowhere, she makes 1-5 high wins before she quits for the day.
Honestly, I don't know what to make out of it. The way she's playing in this doesen't resemble the way she's playing in the video, W/L wise.
I can't see how she bets, only knowing what numbers she gets profit from. I'm pretty sure at spin 117 and forth she's playing with dozens with 50$ bets but other than that, I have no clue. Maybe others might see and finds it interesting. If not, than alright.
Spin # BR Profit Info
1. 4
2. 19
3. 34 33609 -3
4. 28 33606 -3
5. 13 33603 -3
6. 17 33636 +33
7. 15
8. 18
9. 1
10. 29
11. 11
12. 19 33634 -2
13. 29
14. 35
15. 26
16. 0 33631 -3
17. 33 33630 -1
18. 14 33621 -9
19. 11 33618 -3
20. 3 33616 -2
21. 30
22. 25
23. 26
24. 14
25. 10
26. 34 33615 -1
27. 16
28. 16 33613 -2
29. 18
30. 20
31. 15
32. 1
33. 19 33610 -3
34. 5
35. 25
36. 26
37. 25
38. 23
39. 3
40. 4
41. 32
42. 17
43. 13
44. 1
45. 5
46. 23
47. 32
48. 26
49. 36 33606 -4
50. 30 33603 -3
51. 15 33599 -4
52. 22 33593 -6
53. 15
54. 23
55. 8
56. 26
57. 24
58. 9
59. 31 33591 -2
60. 32
61. 11
62. 18
63. 20
64. 0
65. 34
66. 2 33588 -3
67. 14 33585 -3
68. 19
69. 3
70. 14 33618 +33
71. 26 33609 -9
72. 19 33600 -9
73. 35 33591 -9
74. 36 33582 -9
75. 20 33573 -9
76. 31 33564 -9
77. 3
78. 7 33558 -6
79. 20 33588 +30
80. 0 33583 -5
81. 30 33578 -5
82. 21 33573 -5
83. 21 33569 -4
84. 24 33566 -3
85. 14 34166 +600
86. 28
87. 20 35666 +1500
88. 21 38954 +3288
89. 34
90. 13 38966 +12
91. 27 38942 -24
92. 5 38918 -24
93. 3 38894 -24
94. 5 38870 -24
95. 15 38846 -24
96. 19 38894 +48
97. 23 38942 +48
98. 27
99. 2 38930 -12
100. 5 38918 -12
101. 3
102. 32 39506 +588
103. 18 41006 +1500
104. 3 40706 -300
105. 14 41306 +600
106. 6 Comment: She left
107. 33
108. 36 Comment: She came back
109. 31 41303 -3
110. 3 41302 -1
111. 36 41102 -200
112. 8
113. 11 42502 +1400
114. 13
115. 36 45898 +3396
116. 33 45873 -25
117. 31 45973 +100
118. 11 46073 +100
119. 24 46023 -50
120. 30 45973 -50
121. 9 46073 +100
122. 27 46023 -50
123. 11 46123 +100
124. 29 46073 -50
125. 35 46173 +100
126. 0 46123 -50
127. 16 46073 -50
128. 35 46173 +100
129. 6 46123 -50
130. 23 46073 -50
131. 12 46173 +100
132. 21 46123 -50
133. 2 46073 -50
134. 3 46023 -50
135. 32 46123 +100
Quit
Start BR: 33612
End BR: 46123
I started my research by reading everything Priyanka has written since joining. I've observed an apparent learning curve at the beginning, someone learning unusually fast. What i'd like to know is if Priyanka is also "Grandpa" (not sure i spelled that right). Anyways, the Priyanka that went to New Zealand to bungie jump seems a different Priyanka that came back this fall, two years later.
http://www.rouletteforum.cc/index.php?topic=15938.165 (page 12)
Priyanka, your writing style is usually very coherent, but this is rather cryptic to my limited comprehension skills... I wish you could be more explicit. Are you under oath or something?
I wish. But yeah am under oath, hence I can't.
http://www.rouletteforum.cc/index.php?topic=15938.690 (page 47)
1.1 EDGE / GENERAL INTRODUCTION OF PRINCIPLESEDGE – Four letters that every gambler looks for. Whether it is roulette or baccarat or blackjack. Whether it is real life or casino life. Whether it is advantage play or system play. Every gambler looks for an edge. Unfortunately it is Casino who has kept it locked in a safe and every gambler plots to break that safe. There are some who has the access to the key to this safe, but prefers to keep it a secret so that you don’t kill the golden goose. There are some who doesn’t even know how a key looks like but quite good in theoretical plots around designing it. There are some who takes pleasure in weaving stories just like the captain flight in Planes, where in actually doesn’t have a clue. In between all these are a confused set of individuals who has the majority to win an election if there was a vote. What a strange world.
So where is the key. How do we locate this thing called edge. Let us restrict our discussion only to roulette, apt to the title of this forum.
To start with let us assume that there is no edge. Is that a right assumption? Can we prove this assumption with an evidence.
If someone assumes that I am going to explain that I have an edge that everyone can play, read back what is stated in bold.
http://www.rouletteforum.cc/index.php?topic=15938.0 (page 1)
The edge that I called out is the edge for the player.
http://www.rouletteforum.cc/index.php?topic=15938.15 (page 2)
How can we prove either way that edge exists for the player or there is no edge (only for the house)? Tackling house edge becomes better once we have found an edge that will overcome the expectation. If am talking about edge in even chances am talking about getting more than 50% wins always(YES, always 100% guaranteed, but am not promising I have the solution) and every time over a finite number of spins. Once we establish this edge it is easier to attack the house edge equation.
http://www.rouletteforum.cc/index.php?topic=15938.15 (page 2)
Roulette is a game that can give you endless possibilities to play. And there lies the beauty of the game and the beast. It is easy to get oneself lost into the complexities of the game. But if you are able to breakdown those complexities into simple principles, then you will be able to effectively play it with a better understanding of what to expect at the end of every session. Sure one or two odd session may turn out to be exceptional, but you will figure out that a 98-99% of the games will fall within your expectation (win or loss!).
There will come a day when you will be able to see past whats happening on the surface and free from the wheel, felt and the statistics that the pit bosses want you to keep your attention to.
It seems to me that way you analyze or dissect this game is kind general, and can be applied to any part or odds in the game? One just needs to understand it right.
You are so right there. Unless you dissect the game into simple parts irrespective of bet placement and odd, you are not going to understand game.
http://www.rouletteforum.cc/index.php?topic=15938.60 (page 5)
At times things that defy logic at first seems so. It just waits for a proof before it becomes logic. It's like you can pause time just by observing uranium. It never decays when observed. Something like the question between virtual play and personal permanence. It's like Suns atmosphere is hotter than the sun even though sun generates the heat. It just waits for a proof to become logic. Keep an open mind general.
http://www.rouletteforum.cc/index.php?topic=15938.495 (page 34)
Yes Drazen there are 27 combinations possible
(Dozens in 3 spins) and you cannot use 18 and leave the other 9 around. Simply because that is the reality. You cannot play a waiting game waiting for your favourable event to occur.
http://www.rouletteforum.cc/index.php?topic=15938.75 (page 6)
My typical betting method is ECs with usage of straights to complement them. I see fun in using quads and lines.
http://www.rouletteforum.cc/index.php?topic=15938.300 (page 21)
RMoreAs for the lines, I think Pri plays both lines (6 numbers) and quads.
it’s possible that her quads system also observes the linesMy favourites are double street/lines/6 numbers. But I agree with you, that the more you push your boundaries and come out of your comfort zone, you will be able to take the learnings back to your favourite playing position and play a completely different game.
see section on “creating your own playing sections”http://www.rouletteforum.cc/index.php?topic=15938.60 (page 5)
Let me be very clear, no hints, no puzzles, and above all no claims on HG - just an attempt to see whether there is a way to rewrite the facts and beat the house edge.http://www.rouletteforum.cc/index.php?topic=17014.0 0 and 0/00 are the biggest problems. Now lets assume that they are removed. Can someone give me a winning method? No. It might win, it might lose, if we play a random game. So even without 0, we are having a problem at hand.
http://www.rouletteforum.cc/index.php?topic=17014.15 2.1 OVERVIEW (RANDOM)There are two main ways to think about roulette and its outcomes.
1. Conditional probability, Odds and random – This is the common if not traditional way to approach the game. There is nothing wrong with any of these methods unless you are fighting to beat the random. Repeaters, variance, 3SD, playing the last, playing the opposite, playing for streaks and chops - whichever method you use to deliver your bet selection, what you will finally select is a random selection. You are just trying to see whether you can align the random to the laws of probability and you will not get a 100% correct selection. Knowingly or unknowingly, you are trying to fit things within a distribution pattern.
2. Not everything is random – This is the most uncommon way of looking at roulette outcomes. Again there are two interesting sections here.
2A. Physics – This is a way to approach the game where the physics of roulette play a major role. The speed of the rotor, the position in which the ball is released, the speed in which the ball is released, the abnormalities with the ball and the wheel a lot of physical aspects of the game come into play here. This is not random. The accuracy of prediction is greatly improved with the random variables coming into play being very limited like the air pressure in the room, sweat from the dealer hands impacting the speed in which the ball rolls, dirt falling in the roulette table impacting the wheel friction – it goes on and on. But in summary, this is another way to play the game.
2B. Maths – It is a little bit more complex to explain (especially as it is not the common way to play). Of three spins that yielded red or black numbers, there will be at least two red or two black. Hmm! This is not random right. This is an absolute result. The difficulty in this is the practical applicability. And hence very uncommon way to play.
It is very very important to know when your selection is random and when it is not.
http://www.rouletteforum.cc/index.php?topic=15938.0 (page 1)
Random: Now you are left to the mercy of deviations, variations and statistic reality to either fail or win. This is the reason I was pointing back to find out finite, non-random methods within the bet selection process.
http://www.rouletteforum.cc/index.php?topic=15938.60 (page 5)
Now you can think about
(after formulating a Non-Random game) statistics and progression in that sequence. Not before and not in a different sequence of progression and then statistics. Typically we tend to focus on these two subjects first, leaving ourselves buried deep into the big hole.
http://www.rouletteforum.cc/index.php?topic=15938.75 (page 6)
Statistical relationship will come in handy first followed by progressions. I am not trying to steer towards all 27 sets will look the same or on average we will have similar sets. All I was pointing to was there are imbalances here which could be utilized for your selections and progressions.
Seems to be hinting that we should opt for one bet selection over another and also vary our progressions based on the scenario (see section below on combining Non-Random…)http://www.rouletteforum.cc/index.php?topic=15938.75 (page 6)
Regarding your question around other principles, yes I do use others. But using only the concepts so far I have mentioned you can play with an edge.
http://www.rouletteforum.cc/index.php?topic=15938.75 (page 6)
3.1 NON-RANDOM / LIMITS OF RANDOM (THE BASICS)People always say Roulette is a random game. But they do forget that it has its limits. They do forget that non-randomness is part and parcel of this game and embedded in it. There are numerous situations which are really finite in roulette.
http://www.rouletteforum.cc/index.php?topic=15938.60 (page 5)
The first and foremost thought process should be how can I make it finite rather than making it a game of chance. In other words, how can i reduce the non-predictability aspect of the game and move closer to predictability.
(appears to be a reference to Non-Random) http://www.rouletteforum.cc/index.php?topic=15938.75 (page 6)
The direction I am trying to steer you is towards thinking away from statistics based selection as the primary selection. Thinking towards selection that focus on events that definitely happen. There is no variance involved in here. In this case such an event happen every 12 spins.
http://www.rouletteforum.cc/index.php?topic=15938.75 (page 6)
The point am trying to prove is
unless you remove the randomness from the game there is no way to beat the monster. This might not be the only thing that we need to do to overcome,
but this is the basic. http://www.rouletteforum.cc/index.php?topic=15938.90 (page 7)
Mathematically, there is only one way to beat the roulette and that is through seeing the game with a non-random lens.
http://www.rouletteforum.cc/index.php?topic=15938.255 (page 18 )
A non-random lens or a VdW lens?
http://www.rouletteforum.cc/index.php?topic=15938.720 (page 49)
All am saying is there are more non-random ways rather than just exploit the wheel. The basic assumption people have taken is everything is random in the game of roulette. I am just saying that, that assumption doesn't hold good in certain aspects of roulette outcomes. When that assumption is shaken, all the proof we had so far doesn't hold good. Law of large numbers gets shaken when that assumption is shaken. Proof based on randomness and convergence gets shaken when you shake that assumption.
It is always possible to obtain certain non-random events withing any random stream. http://www.rouletteforum.cc/index.php?topic=15938.525 (page 36)
Regarding removing randomness, all I am trying to advocate is try to play some of the steps in your sessions or some of your moves which are not random.
http://www.rouletteforum.cc/index.php?topic=15938.120 (page 9)
A Random System is what we all have played, such as, FTL (Follow The Last), Bet Black after 4 consecutive Reds, bet that a Pattern will form, etc.
A Non Random System is one based on Math or Statistics, ie, the Van de Waerden theorem, the Pigeon Hole Principle which the 12 spin Dozen cycle is based on.
I (Nick) have tested a Random System (FTL) verses a Non Random System (12 spin Dozens cycle).
Both were tested with identical 3,170 spins from BVNZ table.
Both were tested Flat Betting of 1 unit each bet.
Both bet every spin until a Profit Target of 1.
See the results below.
![]()
Random System
![]()
Non-Random System
http://www.rouletteforum.cc/index.php?topic=15938.405 (page 28 )
3.2 DOZEN TRIPLETS OVER 12 SPINS – ONE TYPE HAS TO REPEAT111
112
113
121
122
123
131
132
133
211
212
213
221
222
223
231
232
233
311
312
313
321
322
323
331
332
333
Three possible outcomes. Three dozens in three spins, two dozens in 3 spins and 1 dozen in 3 spins. So If you take a set of 12 spins, you will have one of these combinations to definitely repeat. Limited. This has to happen. It is not random. It will happen always. That is the key. Identifying events that will always happen.
A sample 12 spins. 133 323 123 323
133 – There is one dozen that is repeating here. Our basic premise is in 4 sets of 3 numbers one combination has to repeat. So we will play for the second set to have 1 repeat.
323 – You start playing after 32 has spun. For one repeat to happen you have to have either 2 or 3. So you play the double dozen (2,3) and you win.
Second sample 111 131 111 122
111 – All dozens are same. Again based on our basic premise. We will play for this to repeat.
131 – You start playing after the first spin here. You will be playing for all dozens to be the same. Second spin is 3. Loss. Now you have two outcomes. Three dozens in a row or one dozen to repeat.
111 – You start playing after the first spin. You will be playing for either three dozen in a row or one repeat to happen. So you play for dozen 1. Win.
Third sample 321 311 223 312
321 – All dozens different. We will play for this to repeat.
311 – Start playing after the first spin. For a repeat of first combination to happen, the second spin can be either 2 or 1. So we play double dozen. Win. Now here I pause. One can play every session until a win happens or until the combinations repeat. For those who want a win to happen can stop playing here this set and start fresh with a new set. For those who will want a combination to repeat will go for the next spin. For the combination to repeat the next dozen has to be 2. Play 2 and lose. Two combinations are available for us to replicate. All dozens to be different and only one dozen to repeat.
223 – We cannot play after the first spin here. We will not be able to make a decision after the first spin as for one combination to repeat the second spin can be any of 1,2 or 3. So we play only on the third spin. As we have seen 2 and 2, we know that this is not all dozens different. So we play for two dozens in three spins. So our choice for next spin is 1 and 3 and we win.
Fourth sample 132 112 123 111
132 – All dozens different
112 – Start playing after the first spin. We play double dozen 2 and 3. Loss.
123 – We cannot play after the first spin. We cannot play after the 2nd spin. This is a deadlock and we exit out of this sequence and look for the next 12.
So what did we do. We did not leave our destiny to the hands of chance. We are playing for something that we know will definitely happen. You are building a game based on limits to the randomness of roulette or the non-random aspect of it.
http://www.rouletteforum.cc/index.php?topic=15938.75 (page 6)
But if you read through i have played the complete non-random sequence in this sample. The second bet on this sample was a loss and the sequence repeat as we expected did not happen. So we went ahead and played the third set of spins as well. There we had a repeat in the form of 311 and 223 and that completes our sequence.
http://www.rouletteforum.cc/index.php?topic=15938.90 (page 7)
ScarfaceNot sure what advantage it is to bet for the previous cycle to repeat. I understand at least 1 in 4 cycles has to repeat, but if it gives no advantage, then why bet it?
Sample cycles:
132 222 322 121
132 - 3 different dozens
222- 1 dozen
322 - 2 dozen
121 - in this cycle, something has to repeat. Bet double dozen 12 after the 2 here. A cycle containing 2 dozens will hit 18 out of 27 times. So this is a winning bet
Basically, if you see the last 2 cycles are different, and one of them contain 2 dozens, then bet for the 2 dozen to repeat after the second spin of the 3rd cycle
We know that there are 27 different combinations for dozen cycles. 3 will be 1 dozen. 6 will be 3 dozen. And 18 will contain 2 dozen.
I'm wondering what would be the statistics on winning cycles. Obviously, 2 dozen cycles will win most of the time. But how can we use this to gain an advantage? Maybe wait till the first 2 dozen cycle appears, then bet on it repeating?
http://www.rouletteforum.cc/index.php?topic=15938.390 (page 27)
ScarfaceWould the law of a third qualify as a non random system?
113
333
231
322
212
122
331
213
323
133
131
221
132
312
122 Repeat
211
322 Repeat
332
322 Repeat
331 Repeat
121
312 Repeat
333 Repeat
122 Repeat
233
221 Repeat
Based on these 27 sets, there were 8 sets that repeated. A third of the cycles should repeat out of 27, amiright?
Take a look at the first repeat 122. We're betting only the last spin of each cycle. Looking at the prior spins, there is only 1 set that begins with 12, so bet the same on the last number which is a winner.
There has to be x amount of repeat cycles (8 or 9?). There has to be a way to take advantage of this!
( I guess the Law of the Third can be represented as a finite cycle - providing we see each cycle through till there are 12 repeats?)http://www.rouletteforum.cc/index.php?topic=15938.420 (page 29)
3.3 GUT (Great Universal Theory) Regarding the question of whether “Crossing” is a non-random event or not, it is difficult for me to answer as am not able to make up my mind on either side. For an event to be non-random there has to be a limit that need to be defined and the event has to happen within that limit. In a single zero table, if you say “there will be at least one crossing between 0-1 in 37 spins”, this is definitely a non-random event. But the way Professor explains crossings and plays, am not 100% sure.
However it is good that you brought GUT for the discussion. The most important learning that I have learnt from Winkel is an adoption of Parrondo’s paradox. In GUT, if you keep betting on the same crossing you will ultimately lead to a -2.7% expectation. However switching between crossings, and betting different crossings is a different beast altogether. The answer to your question around dozens and ECs lie there
(optimum combined play – see section on “Combining Non-Random”) .
http://www.rouletteforum.cc/index.php?topic=15938.15 (page 2)
3.4 CYCLES / WHEN A SECTION HAS TO REPEAT / PIGEON HOLE PRINCIPLEHow many of us have wondered why a few systems always work well at the start and then the graphs grow towards the south? If you are not one of those who has experienced this, then you have not played enough roulette. The law of large numbers always catches up. This is why when some one tests thousands of spins, you always get a southward graph. So what is the issue?
Your playing sessions are not short enough to stay ahead of the curve for forseeable future. Unfortunately, playing the game as is will always lead to the session being long enough to catch up on the game edge. For some it could happen in a minutes. For some it could happen after building a solid bankroll over a year or two. However, if you see roulette as a
game made up of a number of finite non-random events, it can help you constructing your sessions short. Short not in its literal sense of minutes or seconds or few spins, but
short enough to avoid the game edge catching you forever.
http://www.rouletteforum.cc/index.php?topic=15938.60 (page 5)
Also how can you make your sessions short enough (not in number of spins, but in terms of elements of play) so that house edge doesn’t catch you and you are able to ride on those imbalances or variances.
This appears to be a reference to cycles as well as possibly combining Non-Random with Random (see next section) Of three spins that yielded red or black numbers, there will be at least two red or two black.
http://www.rouletteforum.cc/index.php?topic=15938.0 (page 1)
As usual, we will ignore the zeroes throughout until we get to a place where we have managed to explore an edge. Let us consider that we are playing dozens. Can you predict the next dozen? If I bet on the negative, the odds will be better than what i will get from playing roulette. However, what we can say for sure is there will be at least 1 repeat of a dozen in 4 spins. Hmm! Is that random? Or is it a finite characteristic and hence non-random?
See the following spins. Construct them into sets of 4.
21 - Dozen 2
17 - Dozen 2. At least one repeat of a dozen
24
12
36 - Dozen 3
18 - Dozen 2
29 - Dozen 3. At least one repeat of a dozen
2
17 - Dozen 2.
17 - Dozen 2. At least one repeat of a dozen
19
10
16 - Dozen 2
7 - Dozen 1
11 - Dozen 1. At least one repeat of a dozen
20
How can we take advantage of this non-randomness. Now here is where Probabilistic and non-probabilistic approach has to go hand in hand
(see next section on “Combining” Non-Random). http://www.rouletteforum.cc/index.php?topic=15938.60 (page 5)
Everythign that happens in roulette happens in a cycle. A cycle starts and ends when a number repeats.
For the dozens, lets see that it will be like this.
19
25
18 – This is a dozen cycle of length 2
19
20 – This is a dozen cycle of length 1
18
31
1
30 – This si a dozen cycle of length 3.
http://www.rouletteforum.cc/index.php?topic=15938.210 (page 15)
You have said you used cycles. Isnt that patterns?
Other way to look at it is they are repeaters.
http://www.rouletteforum.cc/index.php?topic=15938.540 (page 37)
Can someone please confirm that the spin to end/define a cycle is included as the first spin of the next cycle ( A ) or is a fresh new/next spin the first for the next cycle ( B ) as the stats for both are totally different.
(A) is the right approach
http://www.rouletteforum.cc/index.php?topic=15938.285 (page 20)
![]()
rrbbThis can best be visualized by a "tree" Dozen Cycles
You start at the top with the first spin.
You have two possibilities for the next spin:
1. A repeat. Out of 3 possible dozens you can only pick one (the first spin)-> 1/3 this ends our spincycle of length one!
2. No repeat. Out of 3 possibilities you can choose from 2 (no repeat of the first spin)-> 2/3.
For the next spin we again have 2 possibilities
A. A repeat. Now we have 2 possibilities to choose from (2/3). But remember: to reach this point, we first had to choose the second spin to be no repeat!
The total probability of a repeat on the second spin equals 2/3 x 2/3, which equals 4/9.
![]()
http://www.rouletteforum.cc/index.php?topic=15938.450 (page 31)
Cycles of length 1 - Very straight forward. The dozen that started the cycle has to finish the cycle.
Odds of this happening = 1/3Cycles of lenght 2 - A little bit more complex. For this to happen the second dozen that appears in the cycle has to be different than the first dozen (2/3 odds) and the third dozen has to be one of the first two (again 2/3 chance).
Odds of this happening = 2/3*2/3 = 4/9Cycles of length 3 - Again, following what we did in cycles of length 2, the second dozen has to be different than the first dozen (2/3). But the third dozen has to be different from first and second. Hence only one dozen is a possibility or 1/3 chance.
Odds of this happening = 1/3*2/3 = 2/9http://www.rouletteforum.cc/index.php?topic=17014.0 Likewise, this can also be converted into a finite cycle as well:
111 131 111 - Cycle 1
111 122 .... - Cycle 2
(Dozen triplets – see section above)
http://www.rouletteforum.cc/index.php?topic=15938.675 (page 46)
I (Nick) have coded a system that uses a 4 spin cycle. Excel Tracker attached.
1st Spin you bet the last Dozen(FTL). If it wins, then No Bet the next 3 spins.
2nd Spin you bet the last 2 Dozens, win or lose No Bet the next 2 spins.
![]()
http://www.rouletteforum.cc/index.php?topic=15938.60 (page 5)
making the sessions short enough to capture the variations. How on the earth do we do that?
As usual let us take a simple example. Going back to the dozens.
![]()
If you see the attached picture, let say you are tracking for 1 repeat of a dozen to happen. Quite often you will find that you will have to track all 3 dozens before a repeat can happen.
Now look at the same thing for 2 repeats of dozen to happen. You will find that you are starting to track lesser number of unique dozen for the second repeat can happen. The bigger the number of repeats you are tracking you will find that the number of unique dozens that you will track on an average will reduce. Translate this to a betting position that offers more options like double street, street or numbers. What do you see? Does this ring any bells?
It reminds me personally that 1 number should repeat before spin 25, so perhaps 2 numbers should repeat way before spin 50. Also see how stitched bets can be “translated” to different betting positions in the “stitching bets” section below. How about 4th repeat? Is there an optimal number that you can think about, which can help you elongate the session? Is this somehow related to the number of positions (3 in case of dozens, 6 in lines) that you are tracking? What is the relation? Can the relation be utilized to your advantage? After all the ultimate aim is to make the playing session short. How are you able to achieve it?
This is just one way of making your sessions short enough to capture variations. However, I hope this gives a fantastic view of how you can make your playing sessions shorter and take advantage of variance.
http://www.rouletteforum.cc/index.php?topic=15938.105 (page 8 )
3.5 VdW (Van der Waerden) / AP (Arithmetic Progression) If the payout is less than odds, how do AP players or ones who is using computers win. By increasing the accuracy of predictions. So technically by increasing the accuracy of predictions you can win the game (beat the game = beat the house edge). One way we know is AP. One other way we know is computers. One way you seem to know is the math beat math game you claim. One way I seem to know is something based on certain events which breaks the basic philosophy of equally likely outcomes as spins are equally likely but certain events are not. This can be proven that all events are not equally likely. . So in summary what you have written is not what I was looking for as it just proves that odds are greater than the payouts. It doesn't prove that roulette cannot be beaten as one can increase the accuracy of prediction by any means that works and still come ahead of your house edge equation.
http://www.rouletteforum.cc/index.php?topic=16972.0While using AP, am assuming the house payout will always be short of odds, regardless of how you treat or observe random data. Is that assumption correct? If that assumption is wrong, are you assuming that you are increasing odds through AP? If that is true then increasing odds = increasing the accuracy of prediction?
http://www.rouletteforum.cc/index.php?topic=17014.30BayesYou can't get better odds in roulette like you can by shopping around the bookies in sports betting. Roulette odds (the payoffs) are fixed, so the only way to win is to get a better probability of winning (increase the accuracy of predictions).
http://www.rouletteforum.cc/index.php?topic=17014.45 “in 9 spins of roulette yielding black and red, there will be one arithmetic progression of 3 integers holding the same colour”
“in 27 spins of roulette yielding 3 dozens or columns, there will be one arithmetic progression of 3 integers holding the same dozen/column”Lets play the game using a template and let the casino catch us rather than we going after predicting the casino. It is paradoxical to note that even though we are playing based on what we see as previous spins, we are not making any guesses here, but playing to a fixed template. The casino is trying to predict and win over us rather than we predicting what the next spin is. We are just playing to prove the (VdW/AP) theorem right
The term is “arithmetic progression”. It is nothing but a sequence of numbers where the difference between consecutive terms is constant.
1,2,3,… is an arithmetic progression
1,3,5,7… is an arithmetic progression
1,4,7,10.. is an arithmetic progression
2,6,10, 14… is an arithmetic progression
Now coming back, lets take a set of 9 spins.
RBR BRR RBB
123 456 789
See the arithmetic sequence of spins 1,3 and 5 – I get RRR. There are 512 combinations that can happen, but none of them will fail to have an arithmetic progression of 3 integers holding same colour.
Now look at the following 8 spins.
BRRBBRRB
There are two choices:
1. B - because you have 1,5,9 as a possibility.
2. R - because you have 3,6,9 as a possibility.
We cannot play the ninth spin. Of all the 512 combinations, this is an example of a combination where you will be left with a loss if you choose to play based on this theorem.
http://www.rouletteforum.cc/index.php?topic=15938.0 (page 1)
Yes, always no bet.
(multiple sequences/clashes) There is no magic to speak against the mathematical truth.
http://www.rouletteforum.cc/index.php?topic=15938.60 (page 5)
no instincts that will come into play here. Instincts will appear in random selection. Here it is a strict mechanical rule. When in dilemma we don't bet
http://www.rouletteforum.cc/index.php?topic=15938.15 (page 2)
No, it won’t give you any advantage on its own. It will give you a loss, if you play it ditto as I have explained in a step by step manner. The examples that I depict are for explaining the principles for better understanding. This is only a part of the puzzle. Let me explain why.
Without 0(yes even if there is no house edge), just consider only R and B as an example. There are 512 combinations of 9 spin sets possible. Out of this 512 combinations, 406 combinations will give you a win and the rest will give you a loss. Sure a high win ratio inching towards 80%. But, the risk of losses will outweigh the impact of wins. See the following possibilities out of 512 combinations.
W – 256 times
L – 48 times
LW – 104 times
LL – 32 times
LLW – 36 times
LLL – 16 times
LLLW – 10 times
LLLL – 10 times
http://www.rouletteforum.cc/index.php?topic=15938.15 (page 2)
This equates to:
50.00%
9.38%
20.31%
6.25%
7.03%
3.13%
1.95%
1.95%
http://www.rouletteforum.cc/index.php?topic=15938.30 (page 3)
If we play all 512 combinations the way the example suggests, in terms of individual outcomes, we will get 406 wins and 406 losses. 50-50, nothing more nothing less. Not any different from the 50-50 chance of next spin being red or black. Unless you can find a way to make this 50-50 tilt towards one side for a set of all the possible combinations, this doesn’t have an edge on its own and it’s a failure. Some runs will give you profit, some will give you loss and if you add Zero to the mix you will get -2.7% equating to a single zero house edge.
http://www.rouletteforum.cc/index.php?topic=15938.15 (page 2)
it's 50/50 like betting R/B normal, but it has a certain stability about it, so I don't think the 50/50 fluctuates in the same way; however, the most important thing is that the 50/50 comprises several different outcomes instead of just 2.
http://www.rouletteforum.cc/index.php?topic=17014.0 I don’t think the example
(VdW/AP) need a simulation
(on it’s own it’s 50/50) , unless you are planning to study the simulation and observe the principles and cycles. If you are using it for latter, I will be very happy to answer any questions as always.
http://www.rouletteforum.cc/index.php?topic=15938.15 (page 2)
rrbbAt the beginning you had a great discussion with Priyanka about VdW remember? She gave a list of occurences of L, W, LW etc.
Her question was: how can you tip the balance? Think about that question. if i would tell you, you would go like "duhh". Just write down the list she gave, but without the numbers or percentages. The answer is there.
The answer will put many of the things Priyanka said in a new light!
http://www.rouletteforum.cc/index.php?topic=15938.720 (page 49)
there will not be any predictive advantage. The advantage there is knowing that in random everything is defined not in "equal to" but "at least". Vdw says there will be at least one AP, which just means there will definitely be one and there could be more. There is no direct applicability that am aware of.
http://www.rouletteforum.cc/index.php?topic=17013.60 Three small modifications I would suggest.
1. Play only thrice in a set. Don't place the fourth bet ever.
2. Play always opposite. Don't play for AP to complete.
Play a progression 1-2-4. But only within a set. Once the set is finished irrespective of where u r start from 1.
You might find a slight edge.
Personal communication, September 20, 2015 could it be to do with converting the outcomes into their opposite components and then playing just for 3 wins max?
W
L
LW
LL
LLW
LLL
LLLW
LLLL
L
W
WL
WW
WWL
WWW
WWWL
WWWW
http://www.rouletteforum.cc/index.php?topic=15938.720 (page 49)
So might we gain some advantage by keeping a tally of each outcome based on the set (1 cycle?) - and change our bet selection based on whatever outcome(s) are trailing behind the above percentages?
I like the way you are thinking.
I would encourage you to think a bit harder. You are again getting into distribution and probability area where things are left to chance. Unless you are able to increase the Win%
(in VdW/AP on Red/Black) which is currently standing at 50% in the above set (406 wins and 406 losses), whatever variance based methodology or progression that you use will drive you down.
OK, I reckon we might gain EDGE if we bet on the opposite colour for first bet (if the Ws are greater than 50%)?
Why only first bet? You are getting there mate.
This might be worth exploring: playing different variations based on whether we are falling below or above expected ratio as they seem to complement each other from the below graph. (see also start of section on “Combining Non-Random”). ![]()
http://www.rouletteforum.cc/index.php?topic=15938.30 (page 3)
maestroBBRRBBRR...9
RRBBRRBB..9
BBRBBRRB..9
RRBRRBBR..9
BRBBRBRR..9
RBRRBRBB..9
BRBRRBRB..9
RBRBBRBR..9
BRRBBRRB..9
RBBRRBBR..9
SO NONE OF THE ABOVE spins have arithmetic progression and also spin 9 where we got 2 choises so we can get it as loss...so say if you look for sequence with say 4 outcomes and dont have AP we can bet against that pattern to form without arithmetic progression...just a thought
sorry maybe i was not clear enough...say we get spins like....RBBR now i will bet 4 times FOR arithmetic progression to form so i will bet BRRB AND IF NOT WIN LAST SPIN IS 9 SO I WILL TAKE THE LOSS..hope it is clear
http://www.rouletteforum.cc/index.php?topic=17013.0MoneyT101Trust me write it down... You will see when you have to bet against!
http://www.rouletteforum.cc/index.php?topic=17013.0 RMoreNet result - no benefit in switching, i.e. play as predicted or switch to the opposite - no benefit accrues, so I am struggling to understand why, when Falkor suggesting switching on the first bet, that Pri said "why only the first bet?". I see no benefit in switching at all, for any bet - according to the stats that is, and we know that Pri uses the stats and therefore believes in their truth - so why would he bend an elbow towards switching?
http://www.rouletteforum.cc/index.php?topic=15938.375 (page 26)
RMore I don't think switching the bet is the answer. Look at my post #382 in this thread where I break down each possible way through the 512 combinations that can occur. It ends up 50/50. In this situation switching makes no difference to the final result. It has to be something a little more subtle I think.
http://www.rouletteforum.cc/index.php?topic=15938.720 (page 49)
rrbb Do not think to complex like: "maybe when i do a VdW on the results of an VdW and then multiply it by..." Do not forget what you try to achieve!
http://www.rouletteforum.cc/index.php?topic=15938.750 (page 51)
If you lose then switch to playing opposite? If you lose again then switch back to normal? Something like that maybe? ![:question:]( "Question")
rrbbWay too complex falkor. You are thinking in solutions. Do you have a nephew or niece of say preschool age? Replace the L's by apples, and W's by pears. Then think about the right question to ask, and they will give you an answer right away.
The answer is not a solution nor a betting scheme. It is a guide that, combined with all info Priyanka gave, can give some nice results.
http://www.rouletteforum.cc/index.php?topic=15938.720 (page 49)
RMore Nice response - rrbb. "Think of the right question". That is so often the best approach. It sounds so simple but can be so difficult to twist the brain around to do that. But that is where I am going to try to go. Apples and pears. Don't want an apple - want a pear. I can't rearrange them – but I could ignore the apples until I reach a pear and then take the next one. I dunno - just rambling.
So what is the question? In my mind's eye I see that table with apples instead of L's and pears instead of W's (no bananas - sorry Turner). And then I try to become a pre-schooler. Shouldn't be too hard - they say that as you get older you revert more and more to a child-like state. How can I go straight to the pears? or perhaps how can I make it more pears than apples? What are we trying to achieve here? Mix up the pears and apples somehow? Jeez - I dunno.
http://www.rouletteforum.cc/index.php?topic=15938.735 (page 50)
rrbb "But uncle, why do you eat so much pears when you do not like them?"
http://www.rouletteforum.cc/index.php?topic=15938.735 (page 50)
Now, lets see the VdW. As I said already the Vdw doesnt give us any advantage as it doesnt help in predicting what is going to happen next spin. What it did give 512 possibilities with 406 wins and 406 losses. Now is there a way that we can tilt this in favour of Ws? Like apples and pears. You have 5 pears in one hand and 5 apples in other hands. How do we make apples more than pears when we dont have the possibility of having more apples? By losing pears.
What does this translate to here. If we can potentially find a way of not playing even 1 spin that leads to a loss, the advantage can trip to Wins. But how the heck do we do that? I dont know yet. So unless we find a way to do this, this information is useless.
http://www.rouletteforum.cc/index.php?topic=17014.15 MoneyT101, rrbbdo the results carry over after the 9 spins for vdw? If the reply gives away to much info you can pm me...
ex
R
R
B L
B
B W
R L
B W
B
R L ..........does this L carry over to the next 9 spins to read LLLL? or is it considered just L?
R
B
B
R L
R
B
R
B L
R
B
B L
B W
R
R L
B L
R L
R W
R W
Whatvis it that you try to achieve? Think about that i would suggest
http://www.rouletteforum.cc/index.php?topic=15938.750 (page 51)
MoneyT101Go through a couple of spins and after any W reset and start a fresh set of VDW. Manually is easier to see it as you go along.
VDW is to give you an idea and information...once you have the info you can do with it what you feel is best.
http://www.rouletteforum.cc/index.php?topic=17013.0rrbb re: VdW’s application in Priyanka’s videosNo, I'm not using a tracker. If you know what you are looking for it is easy to see! What was the subject that Priyanka started these posts with??
The tracker question was for Priyanka. Thanks though.
Same here. Initially I needed pen and paper but not anymore. It is quite easy once you have grasped it.
http://www.rouletteforum.cc/index.php?topic=15938.435 (page 30)
When you look to complete the AP there can be a gap of 0, 1, 2 or 3 between the integers used to form the AP, so I just bet when there was a gap of 2:
1,4,7 and 2,5,8 and 3,6,9
Here it made +18, but it was just a fluke...
![]()
http://www.rouletteforum.cc/index.php?topic=15938.705 (page 48)
VdW/AP: Just by looking R and B as R and B, will not help the cause. You could play, four or five games here. If you refer to my earlier posts, i was pointing to play multiple games before a session is complete. As an example, you could play every alternating spin to be part of the 9 spins and hence 18 spin as one game instead of just 9 spins. You could play single and series formations to complete an AP. You could play three sets of alternating spins with one set for completing AP and two sets for not forming an AP. The possibilities are endless, but the key is finding that set of games where 1+1 <> 2. Let me take another example of a game, to illustrate a different game you can play. You can play the fastest colour to reach 3 to complete an AP. Take a set of spins that we saw earlier.
Spin R/B Fastest to 3
18 R
19 R
19 R Red wins. Outcome 1
9 R
31 B
21 R
17 B
25 R Red wins. outcome 1. Now play for AP to complete on red to become fastest to achieve 3.
26 B
27 R
36 R Bet red.
31 B Black appears. Loss. Bet red.
17 B Loss. Outcome 2. Our outcomes read 112
34 R
13 B
0 0
12 R
26 B
12 R Red is fastest. Our outcomes read 1121
12 R
10 B
36 R
12 R Red fastest. Outcomes read 11211. Outcome 1 on next set will complete the AP
18 R
23 R
0 0 Loss.
1 R Win. End of set. It read LLLW for this set.
10 B
6 B
30 R
Am not suggesting you play the ways i have played here as is. All am trying to get at is look at non-random possibilities that has a limit, as opposed to random variables (AP is just one example), create multiple games that can be played at a frequency can lead to a potential edge you could look at.
(see also the section below on Parrondo’s Paradox)http://www.rouletteforum.cc/index.php?topic=15938.60 (page 5)
The video is to explain different ways of play and playing multiple games within a single game. The game I depict in the video is not necessarily the game am playing.
http://www.rouletteforum.cc/index.php?topic=15938.30 (page 3)
How about this? I play 9 number sets to catch an AP on R and B. Within the same 9 numbers, I know that either red or black has to appear at least 5 times (no zero), so I'd bet on a color that has already appeared 4 times. Would this be a better way of playing multiple games?
Unfortunately this is not viable. Try playing a couple of games and you will realise. Eventhough I have explained AP in a simple form, if you have observed carefully, you will notice that it is nothing but playing dominant in case of ECs.. If the selection for AP is not dominant, then you would ideally get a dead-lock.
http://www.rouletteforum.cc/index.php?topic=15938.105 (page
![8)]( "Cool")
Couple of things, if you go back to my post. Playing just R and B will always result in that 50% scenario. If we need to get ahead of that scenario, we need to figure of ways of staying ahead and play multiple games within a game. An example of playing multiple games here could be
1. Instead of 9 spins, take 27 spins. You could play 3 alternating games. Play 1st, 4th.....25th spins as one game, 2nd, 5th...26th spin as one game and so on.
2. One game is a straight RB game. Other could be a series and singles game on RB (single is one outcome and series is another outcome, these two we can look for AP). Play these games alternatively.
Personal communication, September 24, 2015Now look at the following 8 spins.
BRRBBRRB
If we play based on the theorem, what will we play for the 9th spin? Black or Red? Leaving you with these thoughts.
Does the clash here appears because we have a possibility of betting both black and red. What if we tie our hands that we cannot bet black and we can bet only red. Does this clash happen. Does this handicap situation of betting only one colour makes this theorem more workable from a VdW perspective. Does this handicap really a handicap or is it a boon in some form making us lose less?
http://www.rouletteforum.cc/index.php?topic=15938.255 (page 18 )
rrbb"what other bet selections would encompass a certain win within 9 spins while just playing either red or black for on average ±2.3 times per cycle"
Priyanka seems to hint that this could be connected with my suggestion of the possible conversion of VdW into cycles based on “approach (A)” (see previous section on Cycles) as opposed to 9 number cycles – needs to be tested.RBRBR - cycle 1
RRR - cycle 2
RBB... - cycle 3
http://www.rouletteforum.cc/index.php?topic=15938.660 (page 45),
http://www.rouletteforum.cc/index.php?topic=15938.675 (page 46)
If it is possible to get a pattern of 8 spins that did not form an AP and that would only allow one color (R or B) to finally form an AP in the 9th spin that would be a sure bet. Even if a long wait was needed.
http://www.rouletteforum.cc/index.php?topic=17013.30 Scarface I don't think there is ever a situation where just one color has to come up to form an AP. In my opinion, AP is useless when it comes to gambling since there always will be 2 choices
Could Scarface be the devil’s advocate?http://www.rouletteforum.cc/index.php?topic=17013.45 ScarfaceMaybe Pri and rrbb may be dropping these clues to lead Falkner on to mess with him? And others perhaps? There is no basis for an edge in anything in this post. I'm no expert at VDM Theory, but it's just a non random fact no different than saying a dozen will repeat at least once in 4 spins....doesn't help us choose which one to pick.
http://www.rouletteforum.cc/index.php?topic=15938.750 (page 51)
MoneyT101There is an edge if you wait for it.. *The longer the AP takes to form the higher your chances of hitting
Also you can play two games of 9(18 total spins, 9 each) and one game of pairs(18 total spins but 9 decisions) for a total of 3 games in 18 spins
needs to be testedhttp://www.rouletteforum.cc/index.php?topic=15938.690 (page 47)
rrbbI think that you are both right and wrong (hmm reminds me of Schroedingers cat ![:)]( "Smiley")
). What I mean with this is that if for example VdW is used on a two-coloring of numbers (like high low, odd even) you are right.
However, in case VdW is for example used for wins and losses (W, L ==> also a two coloring!) then not...
Please note: I do not claim to know exactly what Priyanka is doing, but the principles she is talking about are extremely versatile.
http://www.rouletteforum.cc/index.php?topic=15938.450 (page 31)
rrbbI did not suggest to use VdW on wins and losses! I just mentioned one could do that. VdW is extremely versatile: one could use it on number partitions (like high/lows), length of cycles, wins/losses, you name it.
http://www.rouletteforum.cc/index.php?topic=15938.705 (page 48)
rrbb You make a crucial thinking mistake: why dou you assume Priyanka uses VdW in numbers or wins? What other "numbers" are there? What did i propose to RayManZ???
http://www.rouletteforum.cc/index.php?topic=15938.705 (page 48)
![]()
Quad cyclesrrbb re: RayManZ’s analysis of the Quads video – with emphasis that this needs to be “seen” in the video – see videos sectionYou might want to add the number of unique numbers in the cycles. The betting scheme in the last section of your reversed engineered work then also becomes clear: "suddenly" a new bet is added... and a part of the "earlier" bet (consisting of two parts) also changed "suddenly"
http://www.rouletteforum.cc/index.php?topic=15938.420 (page 29)
rrbb re: RayManZ’s analysis of the Quads videoYou made a very good remark. Something like "notice she does not bet on a repeat on the first spin". This is true, but just write down the number of unique numbers in a cycle, and you will be able to observe 2 other things.
http://www.rouletteforum.cc/index.php?topic=15938.435 (page 30)
We have seen that VdW is not of any use of we use it in spins. I am making my write up on my attempt at using VdW in events and not spins.
http://www.rouletteforum.cc/index.php?topic=17014.60read carefully the replies from Rayman from rrbb. Perhaps the part where he talked about how many uniques. That holds the key between differentiating from spins to events and the dependency it creates. There is nothing new or not shared with you that I want to share.
http://www.rouletteforum.cc/index.php?topic=15938.690 (page 47)
rrbbThe essence of these kind of cycles is that they only contain unique numbers!. In your case it starts with a repeat, so the last number can be left out!
http://www.rouletteforum.cc/index.php?topic=15938.450 (page 31)
rrbbAnd concerning rayman: i think i said more: rayman noticed priyanka did not bet on cycles of length one. I mentioned something along the line that betting on 4 unique quads doesn't make sense. How many "lengths" are than left to bet on?
http://www.rouletteforum.cc/index.php?topic=15938.705 (page 48)
rrbbIndeed! It is not only quads! The betselection is also based on the first principle priyanka introduced. To answer 3nine's earlier question: in the nice worked out scheme of Ray you can quickly see the number of unique quads in a cycle! As i use cycles a long time, i do not need trackers...
because it makes no sense to bet on 4 unique quads, and because there is no bet on cycles of length 1 (as Ray observed) ...(you fill in the dots).
Is this a winning method? No, i do not think so. But it is an extremely clever showcase of how VdW (there you have it, i spilled the beans) can be applied. And as Ray indicated: it might just open a whole new way of selecting bets etc.
http://www.rouletteforum.cc/index.php?topic=15938.435 (page 30)
Can someone please explain how the VdW theorem can be used on double dozens?
I understand it on EC's and single dozens, but can it be applied to double dozens, when a double dozen bet can be 12, 23, or 13?
E.g. if the last number was 14, you cannot say it was the dz 12, it also could be dz 23.
Ati - there are many ways. I will explain one possible way.
12 - outcome A
23 - outcome B
31 - outcome c
For ease of explanation (only for ease of explanation!!!) wait for two independent dozen to form. Let's say they are dozen 1 and 2.
Consider the following spins.
3,15, 23, 2, 31, 21, 16, 34, 32, 23, 1, 15, 19.
3- dozen one
15 - dozen 2
Th double dozen is 12. So anything different from this we will mark it as different. Anything same as this we will mark as same.
23 - Same
2 - same. Possibility of ap in next spin. Play dd 12
31- different
21 - different. Possibility of ap in next spin.
16 - same
34 - same. Ap in next
32 - same.
Remember this is just one way.
Now the comment that ati you highlighted is different from playing double dozens. It is for an AP for 2 dozens in 3 spins. This will typically have one single dozen bet and one double dozen bet.
http://www.rouletteforum.cc/index.php?topic=15938.270 (page 19)
IMO as the outcomes are always 50/50, that 1 spin you skip that leads to a loss might as well lead to a win. You don´t know if you´re losing pears or apples until they have just hit.
what is the significance of this statement?You have made a beautiful point there psimoes. You will never know if you are losing pears or apples until they have just hit. If and if only there is a way. But here i would like to remind that Vdw is a versatile theory. It can be used in a number of ways. The simplified statement is if you are having two colours, then there is no way of colouring from 1 to 9 without creating an arithmetic progression of the same colour. As many have pointed out, it doesn’t increase the probability of the next spin to be a certain colour. So there is no usability there.
However, can we use it beyond colours? Yes. Let us explore some possibilities to understand how versatile this is without considering the usability of this theorem.
Example 1Consider the spins 15, 21, 23, 26, 15, 25, 33, 16, 28, 23, 14. Translating this to colours it will read B, R, R, B, B, R, B, R, B, R, R. Now let’s read the outcome as whether the colour was same as previous colour (S) or different from previous colour (D). The above sequence will read D, S, D, S, D, D, D, D, D, S. We know that within 9 of these events there will be at least one arithmetic formation with D or with S.
Example 2Same set of spins. Consider the outcomes as whether current dozen is different(D) or equal (S) to the previous dozen. The sequence will read S, S, D, D, D, S, D, D, D, S. We know that within 9 of these events there will be at least one arithmetic formation with D or with S.
Example 3Same set of spins. Consider the outcomes as where the dozens could be expressed in a clock with a clockwise movement taking us from dozen 1 -> dozen2 -> dozen 3-> dozen 1. The relation between two dozens could be expressed as either Clockwise(CW) or Counter clock wise(CCW), denoting the shortest distance to reach the next dozen. If both dozens are same then it is considered CW. The sequence for the same set of spins will now read – CW, CW, CW, CCW, CW, CW, CCW, CW, CCW, CW. We know that within 9 of these events there will be at least one arithmetic formation with CW or with CCW.
I know there will be lots of questions around so what? What is the applicability in roulette.
Sorry, I don’t have an answer. It is yet to be seen, but I have an inkling that this versatility could be put to use somehow when we are having two variables that do not essentially have a 50-50 probability appearing, but could or might give an advantage when lining up in a VdW sequence.
That´s creative thinking. So betting for Different than previous dozen to form an Arythmetic Progression will apparently have the edge over betting Same as previous dozen, as two dozens (D) have more chances of hitting than a single dozen (S). But like with every two dozen methods or similar, there will be losses and at a higher cost...
Exactly.
The applicability though is a big question mark. But what it does definitely teaches us is there is much more than what we already know and a creative application of most of what we already know is quite possible. It just needs an unconstrained mindset which doesn't lull over the same things again and again. As I said at this point in time this is one for the notebook which we will keep coming back to and see any possibilities of practical application.
http://www.rouletteforum.cc/index.php?topic=17014.75Psimoes BTW playing the vdw I go for the latest possible ap to form. I bet for 7-8-9 even if a furthest back 1-5-9 is in conflict.
http://www.rouletteforum.cc/index.php?topic=17014.msg156725#msg156725 pralineIf we know that probability for dozen that defined previous cycle to define the next cycle (S) is approximately 60%
And different dozen (D) approximately 40%
We can use vdw with S and D
Personally I bet only for AP for same dozen (S) with a great results.
SORRY FOR MY ENGLISH
AND DON'T COMPLICATE THINGS
S S D S S . . . .
80% WINNING BET
Your english is good Praline.
![:thumbsup:]( "Thumbs-up")
. And that definitely is a creative usage of VdW. :applauds:
However, I have to disagree with both of you on the approach here to use Vdw, as I am not able to see a clear mathematical advantage. If anyone can help with that it will be really great. HAving said that, I am eagerly looking forward to your simulations to see whether there is any empirical evidence of this advantage. Logically what you are saying makes absolute sense. More occurances of "S" defining dozen cycles with a dozen with a pay out of 2 to 1, seems like something that gives an advantage. But as I said, I cant figure out a mathematical working behind this, as it is always possible to get Ds in the AP before S does as VdW suggests only at least. I think we need more help here.
Psimoes - Should I read that as - You believe not betting "D" will make you lose some losses and give more wins than losses?
Not betting D to avoid the losses, yeah.
betting Same as previous dozen to form an AP
Psimoes - Here it is obvious because 2/3 of the times you will get a dozen that is different from previous. Hence the odds of Same to form an AP drastically reduces as the composition of D is 2 times more than the composition of S.
But what praline says is something different. He is suggesting the cycles. We saw that when in cycles, the same dozen to that defined previous cycle to define the current cycle is more than 60%
(see section on Cycles). So that means more S's than D's and the opporutnity to get an AP with S is significantly higher than the opportunity to get D. Betting on S's gives you 2 to 1 payout. So logically there seems to be an advantage. My point is I cant or to be exact not able see a mathematical advantage. I did a random test going back to my favourite wiesbaden spins from yesterday. 15Ss compared to 10Ds - closer to 60% of Ss. But we dont come ahead. They show why it doesnt hold an advantage. The problem is we dont get 2 to 1 payout in reality, it could be either 2 to 1 or 1 to 1 or just your money returned depending on the length of the same cycle.
12
13
36
26
18
36 - S
13
6
21 - D
8
15 - S
29
7
24 - S. AP formation in next possible.
26
36 - -2. D
8
5 - D. Conflicting AP formattion next. But we can still go for S.
27
19
24 - D. -5. AP formed. start tracking again.
28
30 - D
34 - S
23
36 - S. AP to be formed.
33 - -3. Start again
35 - S
9
9 - D
9 - S
19
2 - S. AP to form
19
14 - -5
30
5
32 - D. Again same situation as last.
26 - -3. Start tracking again.
33 - S
26 - S. AP to form.
4
11 - D. -5
27
26 - D
26 - S
29 - S
0
1
36 - -5.
praline My way of playing those numbers
lose and won are referenced to previous BET
1. 12
2. 13
3. 36
4. 26 d
5. 18
6. 36 s
7. 13
8. 6
9. 21 d
10. 8
11. 15 s next D possible NO BET
12. 29
13. 7
14. 24 s next S possible BET
15. 26
16. 36 d lose
17. 8
18. 5 d next S and D possible NO BET
19. 27
20. 19
21. 24 d AP formed (RETRACK FROM LAST SPINS WITHOUT AP, from spin #18). next D possible NO BET
22. 28
23. 30 d AP formed (retrack from spin #21)
next D possible NO BET
24. 34 s
25. 23
26. 36 s next S possible BET
27. 33 s won
AP formed (retrack from spin #24)
next S possible BET
28. 35 s won
AP formed (retrack from spin #27)
next S possible BET
29. 9
30. 9 d lose
31. 9 s
32. 19
33. 2 s next S possible BET
34. 19
35. 14 d next S possible BET lose
36. 30
37. 5
38. 32 d next S and D possible NO BET lose
39.26 s AP formed (retrack from spin #30)
next D possible NO BET
40. 33 s next S possible BET
41. 26 s AP formed (retrack from spin #40) won
next S possible BET
42. 4
43. 11 d lose
44. 27
45. 26 d next D possible NO BET
46. 26 s
47. 29 s next S possible BET
48. 0 - ignore
49. 1
50. 36 s AP formed (retrack from spin #47) won
next S possible BET
but then i took the wrong one and cant find the way out
in terms of Bets this session would be like this
LLWWLLLLLLLWLLLLW
13 L
4 W
For a flat bet result - 5
NOT GOOD AT ALL
PsimoesI see now. Thanks (I bet someone observing the abundance of Ds in the test will create a two-dozen method with 1-3-9 progression).
So the dozen that defined the previous cycle (and will start the next) has 60% chances of defining the next cycle. Shouldn´t it be more like 66% seeing that 100/3=33.3 BTW? What if you don´t start the next cycle with the defining dozn and wait a spin instead? Not that it will change the odds in any way, just curious...
http://www.rouletteforum.cc/index.php?action=profile;area=showposts;u=6798 4.1 COMBINING NON-RANDOM (WITH RANDOM/NON-RANDOM)may be absorbed into other sections for version 3While non-random is good, we often get into a dead-run. An example of a dead-run is below where you are trying to play for a dozen to repeat in 4 spins, you get sequences like 1231, 2311, 3121 etc. As Drazen and Turner rightly pointed out, there is still an opportunity to get these sequences over and over and over again that you can get into a deep hole. The key is how can overcome these dead-runs with a parallel bet or a parallel selection, which is the alternate game played on its own will give you a negative result, but played together will make this dead-heats into winning combination.
Possibly a reference also to Parrondo’s Paradox (see later section; also about when to enter and exit Non-Random games) http://www.rouletteforum.cc/index.php?topic=15938.90 (page 7)
i normally play a number of parallel games in a session.
presumably the same as the earlier references to “multiple” (VdW) games? http://www.rouletteforum.cc/index.php?topic=15938.180 (page 13)
rrbbWhen I played with it I just wondered: "what other bet selections would encompass a certain win within 9 spins while just playing either red or black?"
http://www.rouletteforum.cc/index.php?topic=15938.660 (page 45)
Before getting further into the world of random and non-random and how we can combine these two worlds, another question. As I touched upon dozens, “A dozen on the carpet, a dozen on the wheel, a selection of 12 numbers that changes constantly. Are they different? Do these bet selections result in changes to your predictions or the distribution?”
You can device a way to play even chances or dozens using these. The lower the number
(sic: the higher the number?) , the higher the complexity and difficulty to track and play. Try playing this for sets of 27 spins with both dozens and ECs and you will figure out a whole new way to play roulette. It doesn't seem humanly possible to keep track of the dozens without a tracker since there are many more arithmetic sequences and it seems that Priyanka actually adopts a Same/Different 2 integer AP approach with all.
http://www.rouletteforum.cc/index.php?topic=15938.0 (page 1)
RMoreLook at the bolded statement. "The LOWER the number .." What number? And any number I choose I find it LESS complex if the number is lower. For example, EC's with 2 chances versus dozens with 3. The EC's is a lower number but the complexity is LESS. What other number could he be referring to? The length of the series? But again, larger number (27) MORE complexity (than 9 for example). So help me understand - what number gives higher complexity when it is lower?
12 numbers (dozen) is lower than 18 numbers (EC); so does that mean there’s a 2 dozen VdW/AP method or we should change our position based on a trigger or something when playing VdW on ECs?http://www.rouletteforum.cc/index.php?topic=15938.375 (page 26)
Without getting into the complexities of money management lets adapt a simple 1-1-2 approach for EC which will suit our finite 9 spin cycles and a finite up 1 for 2 losses for dozens which will suit out 27 spin cycle.
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http://www.rouletteforum.cc/index.php?topic=15938.0 (page 1)
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http://www.rouletteforum.cc/index.php?topic=15938.15 (page 2)
The more the number of outcomes you are trying to fit the arithmetic progression, the higher this phenomenon of confusion. I have taken the approach of absorbing the loss if there is more than one possibility. Another way to play this is absorb the loss only if all the possible outcomes are possible. In case of ECs, it will be both the outcomes becoming possible. In case of dozens, it will be all three outcomes becoming possible. If one takes this approach then the game swings between playing single dozen and double dozens. Remember, there is no right or wrong way of doing things here. It is just to understand the concept that roulette can be played without getting lost into random.
http://www.rouletteforum.cc/index.php?topic=15938.0 (page 1)
One way of using this statistic is to bias towards one set when a conflict occurs for your bet selection. Other way of using this is application of VW theory as I explained earlier for the AP to form on 2 dozens in 3 spins. It is left to your imagination, your mood of the day or a mechanical way that you prefer.
Could this be referring to the following scenario: ![]()
For the next spin, considering we are presented with multiple sequences for both R/B and Dozens, should we bet only on Dozens since a single win game (W) has higher statistics compared to a LW game? http://www.rouletteforum.cc/index.php?topic=15938.75 (page 6)
What if instead of colours and dozens, you have lows and highs and dozens. Are we able to derive any relation?
http://www.rouletteforum.cc/index.php?topic=15938.105 (page 8 )
![]()
It’s possible to beat roulette using a mod of VdW/AP on Red/Black/ECs alone according to Priyanka (using a positive progression)! And the mod might possibly involve Parrondo’s Paradox (covered in a later section). http://www.rouletteforum.cc/index.php?topic=15938.15 (page 2),
http://www.rouletteforum.cc/index.php?topic=15938.30 (page 3),
http://www.rouletteforum.cc/index.php?topic=15938.45 (page 4),
http://www.rouletteforum.cc/index.php?topic=15938.90 (page 7)
rrbbI said spend some time on it, not trying every system in the world with VdW and posting why it will not work. You had this great conversation with Priyanka at the start of this thread. She did explain why it will not work as it is. Something else is needed. Otherwise, forget it. And no, you will never find it by chance!
Check out the video that RayManZ analyzed. Well actually you do not need to as he wrote down all the spins etc. Work with this for a couple of week, Priyanka even gave a nice spreadsheet, why not add the numbers of RayManZ in it? Print it out, and spend a weekend in the sun glazing over it: use RayManZ analysis and try to understand what happens, use a remark I made and see what comes of it.
i will guarantee you that you will find a possible way of using VdW. In doing so, you might also see what she is actually trying to convey. It is NOT a system, it is a mindset.
http://www.rouletteforum.cc/index.php?topic=15938.705 (page 48)
RMorePri said early on in the other thread that Dozens and EC's (and possibly 6-lines or was it quads) COULD be played with an edge using VdW and other non-random techniques such as repeating dozens in a cycle and so on.
http://www.rouletteforum.cc/index.php?topic=17014.90 RMore: I think - he has collected a couple of favourite non-random events not sequences (including of spins)! (as corrected by Priyanka) and then combined these with useful stats and appropriate betting plan not necessarily with MM! (as corrected by Priyanka) to create a synthesised approach that leaves room for subjective play that does not disturb the basic facts.
So the stats play no part in the basic strategy - only to assist decision-making when multiple options turn up. But I don't think that is all. I think there is another component that has to be added. Possibly another non-random sequence? One with a lower strike rate because of a smaller coverage perhaps? Consider what he hinted at with High/Low combined with dozens or perhaps six-lines. Or perhaps the quads (9 numbers as I understand it - personally I've always thought quads were 4 numbers, corners basically, but whatever) are the basic and the High/Low is used when a bet is ambiguous - the dead run situation. I see no reason why, when a dead run situation presents itself, that you can't, at that time, look back and create a sequence starting point that gives you a bet on this spin. After all, the VdW theorem says ANY 9 number sequence so why not create them on the fly as and when needed as the "other" bet?
RMoreSo -where does that leave us? Well, I can only assume that the references to PP are only done as similes - not for actual use. That you use 2 non-random events and that you alternate them in some fashion that is for us to discover. Actually, we also have to discover the 2nd non-random event don't we? I know it is probably on the 6-lines but it can't be VdW because this would be far too complex - even the dozens have a sequence length of 27 and nobody is able to accurately find all the AP's in that on the fly. Does anybody have any idea what the 2nd non-random event might be?
there is a line cycle ratio described further down but it doesn’t appear to be “non-random”. See also the section on Parrado’s Paradoxhttp://www.rouletteforum.cc/index.php?topic=15938.345 (page 24)
RMoreI really feel that the important things we need to focus on right now are, firstly, the second non-random event (assuming the VdW on the EC's is the first) and the way in which these 2 events are put together in the betting plan. Note the word EVENT. I believe that we need to think of the VdW theorem that Pri gave us as an example of an event to be also a good example of the meaning of the word EVENT in this context. We need to find another such event that fits with either the groupings of 9 numbers (which Pri calls a quad) or perhaps 6-lines, maybe even dozens although I personally don't think it is this.
http://www.rouletteforum.cc/index.php?topic=15938.360 (page 25)
5.1 OVERVIEW (ADVANCED PRINCIPLES)There are a few fundamental things I am trying to communicate… These are all things you can do and this is all things you need to know. There is nothing else.
1. One is exactly what Bliss is describing that you will be able to find out non-random events that has to happen in any random stream of objects. In roulette it is the random stream of numbers from 0 or 00 to 36.
Covered in previous sectionshttp://www.rouletteforum.cc/index.php?topic=15938.300 (page 21),
http://www.rouletteforum.cc/index.php?topic=15938.315 (page 22)
5.2 CONSTANTS / RATIO OF CYCLESThe fact is things do clutter. When they do clutter, repeaters do happen. When repeaters do happen the statistical relation between these finite cycles tend to lean towards and form a magical relation between two finite cycles.
(see section on playing cycles for increased odds)http://www.rouletteforum.cc/index.php?topic=15938.225 (page 16)
2. Second is the
constant explained by Drazen and the ratios of lengths. If you have 1000 spins, are you able to say with certainity that Red will be more or Black will be more? Are you able to say that number 36 will be more than any other number? No. But can you say that the number of repeating cycles of dozens will be more than number of different cycles of dozens. Yes, you can with absolute certainity. Leave aside winning every session for a moment. But lets say you keep a count of red and black. When red goes to 10, can you keep on betting black to balance that count, no. Keep a count of repeating cycles and different cycles. When there are 10 different cycles, can you use this count to get back the same cycles up? May be!
I mentioned you can bring in 2 or 3 constants together. What those constants that has to be brought together is your work. May be these two will work, but i dont know. You dont need to bring in more constants to gain edge. Even one constant is sufficient. To get a playeable method in a casino environment you might need to look at more opportunities.
http://www.rouletteforum.cc/index.php?topic=15938.315 (page 22)
While we talked about non-randomness, it is key that you dont forget statistics and what is a fact. We talked about cycles. Lets take the following dozen cycle as an example. Following is the statistics across various number of cycles for a set of few thousands of spins. The fact is the percentages defined there say something about the edge and they remain the constant irrespective of the set you will use.
500 cycles Dozen that defined the previous cycle same as the dozen defined the next cycle - 306 ~ 61%
Dozen that defined the previous cycle different from the dozen defined the next cycle - 194 ~ 39%
1000 cycles Dozen that defined the previous cycle same as the dozen defined the next cycle - 618 ~ 62%
Dozen that defined the previous cycle different from the dozen defined the next cycle - 382 ~ 38%
2000 cycles Dozen that defined the previous cycle same as the dozen defined the next cycle - 1241 ~ 62%
Dozen that defined the previous cycle different from the dozen defined the next cycle - 759 ~ 38%
No law of large numbers to curve the percentages? "Defined" - The dozen which caused the cycle or the dozen that repeated.
20
31
20 - the cycle was defined by the dozen 2
1
31
22 - The cycle was defined by dozen 2 again.
1
8 - cycle was defined by dozen 1
22
18 - cycle was defined by a different dozen - dozen 2.
Spreadsheet giveaway for showing Dozen, Line and Quad (9 Number) cycles based on above percentages when comparing one cycle to the next:
http://www.rouletteforum.cc/index.php?action=dlattach;topic=15938.0;attach=23017![]()
RMoreNatural? I guess it could be viewed that way. It simply depends on how you look at it. It's only "natural" when you choose to define a cycle by termination due to a repeating dozen. If I choose to close out a cycle by some other definition then the "naturalness" changes into something else. It all depends on your definition of the term - that was my main point really.
http://www.rouletteforum.cc/index.php?topic=17014.45 Scarface I noticed in all my testing cycles on Pri's thread, that it is rare to have several unique 3 dozen cycles in a row.
I was tinkering around with the idea of betting for a repeat after seeing 3 unique fall.
As far as your idea here, how would we bet a sequence with repeats in the middle like this:
1233312222133321123
Just wondering how common it would be to have a string of unique 3 dozens with several repeats in the middle.
http://www.rouletteforum.cc/index.php?topic=16937.30my favourite betting position the double streets, in three spins what is the probability of getting 3 unique double streets or the double streets not being the same? It is a over 55%. Surprising, but that is the truth. So chances of getting 134, 156, etc where all double streets are different are better than chances of getting 121, 555, 556, 322 etc. Can that be used to our advantage during the play where some steps are random and some steps are non-random. Yes definitely.
It is definitely possible to take advantage of birthday paradox
(refer instead to the Double Streets example above). Using it in conjunction with Pigeonhole principle and stopping when you are winning in an attacking session while you are progressing towards a non-random set will definitely give you the edge.
http://www.rouletteforum.cc/index.php?topic=15938.120 (page 9)
Whichever part of the table I analyzed, I am seeing certain strong ratios between cycles. And they are pretty much constant.
Apart from the ratio that I have highlighted, are there any other ratios that you are able to see. Do you think we will be able to use VdW theorem with which I started the thread in some form or other to bring a statistical concept and a non-random concept together.
Refers also to combining Non-Random with Random (see relevant section) http://www.rouletteforum.cc/index.php?topic=15938.255 (page 18 )
5.3 CREATING A BIASED GAME3. Can you bring 2 or 3 such constants together to
create a biased game, just like biased wheel readers who is constantly keeping on the look out for bias and look for the entry point. May be!
RMoreBut in this case Pri has said that the answer was A - remembering that A is the option where the last dozen in the previous group/cycle becomes the first in the next group. This will have a distorting affect on the results because you will be starting the next dozen with a built-in bias - there is a 1/3 chance that the last dozen WAS the defining dozen which will cause the next group to have an increased chance of a match. But that is the wrong way around isn't it? You would expect maybe 69% instead of 62% right? BUT - what if the previous dozen grouping was completed after only 2 dozens showed? This has a 2/3 chance. So the result Pri is showing is going to be distorted in all sorts of ways because of the way in which he set up the analysis.
http://www.rouletteforum.cc/index.php?topic=15938.330 (page 23)
ScarfaceThis may help. Based on the 97 cycles I tested:
1. 38 ended on the first spin
2. 43 ended in 2 spins
3. 16 ended in 3 spins
4. Based on this, 39% hit on the first spin. If playing last 2 dozens on the second spin, I would've got a 73% rate.
Stats show better than expected returns. Not sure if it's due to varience or edge until further testing
I forgot to mention that 15 out of 16 cycles that ended with 3 spins were would've been won betting the last 2 dozen hit within that cycle.
http://www.rouletteforum.cc/index.php?topic=15938.345 (page 24)
rrbbyou can calculate the probability that a repeat on dozens occurs on the previous "defining" dozen: 17/27 (which is of course Priyanka's 63%)
![]()
http://www.rouletteforum.cc/index.php?topic=15938.450 (page 31)
rrbb I think that what Priyanka shared is extremely important:
The probability on any dozen is of course 1/3. But Under the condition of a repeat, the probability is "suddenly" 17/27
Btw: there are many other imbalances. Priyanka showed you some...
http://www.rouletteforum.cc/index.php?topic=15938.450 (page 31)
Nothing strange or new. Going to bayes site it is
http://www.roulettician.com/probability1.html#mozTocId598749winkel: http://www.vlsroulette.com/index.php?topic=13004.0http://www.rouletteforum.cc/index.php?topic=15938.795 (page 54)
[color=red[Scarface[/color]Betting on the last dominant dozen is basically betting on the last dozen that repeated. And, testing is showing no statistical edge in doing this. Counting the same dozen twice on paper may change the stats, but its artificial so if it shows any edge, it will be artificial too. Hopefully Pri can help us out on this ![:)]( "Smiley")
I wonder if we take it a step further, could we find some sort of edge. Maybe, always bet the last 2 dominant, or repeating dozens. But only play the most recent hit 3 streets from each one (total of 6 streets). Seems like a good way to catch hot sections being hit. May test this later with the same data
http://www.rouletteforum.cc/index.php?topic=15938.330 (page 23)
ScarfaceWhat if we bet the last 3 hit lines in the dozen, instead of the whole dozen. If a cycle ends with dozen 2 as dominant, bet the last 3 hit lines in dozen 2 on the first bet. If dozen 1 hits next, bet the last 3 hit lines in dozen 1 and 2.
Seems like there is always 1 line in a dozen that stays cold. Thought this might increase the odds of a hit better.
http://www.rouletteforum.cc/index.php?topic=15938.495 (page 34)
ScarfaceDon't waste time testing this one. I thought by skipping the first spin on a new quad cycle would give an advantage, but it doesnt. Wins and losses still come up as they should statistically
http://www.rouletteforum.cc/index.php?topic=15938.780 (page 53)
RMoreBut the fact that there is no edge being demonstrated here is no surprise. That is not what Pri was trying to say - I believe anyway. He has never said that the stats would provide any sort of advantage, in fact quite the opposite - only that in certain circumstances they could be used as part of an overall strategy.
http://www.rouletteforum.cc/index.php?topic=15938.345 (page 24)
There is an important thing here around statistical advantage of same element defining the next spin. What if we remove cycles of length 1, do we see any difference in ratios. Can cycles of length 1 be exploited? Can cycles greater than length 1 be exploited?
In her video it has been shown that Priyanka avoided cycles of length 1. http://www.rouletteforum.cc/index.php?topic=15938.255 (page 18 )
RMore:You refer to combining the stats. And I recall that it is the - what did you call it? - the "dead runs" that stops the simple non-random component giving an edge by itself straight out of the box if you know what I mean, and so these have to be dealt with in some way. A sort of PP has been discussed but as I see it what you are suggesting is just a similarity - it is not a true PP per se but rather just a similar thing where a complimentary strategy is combined to mitigate the situation where the dead run turns up. But could this also be handled, where appropriate, by the use of suitable stats? For example, in the early dozen example where Turner rightly pointed out that the win wasn't really a win - just the probability asserting itself - to which you wholeheartedly agreed - then as the next dozens appeared we could change our attack from length 1 to length 2 when the opportunity presented itself because, and here is the stat, length 2 is statistically more prevalent that length 1 and so is the better choice when a dead run possibility appears, or even when you have both on review waiting for an opportunity. Right? There are only 3 length ones, 3 length threes, and 12 length twos. So it is better to swap your game to the 2's if that opportunity appears rather than hang out for the completion of a 1.
see previous discussions on dead-runs, dead-heats, multiple/parallel games and combining non-random as well as, in the next section, Parrondo’s Paradox).http://www.rouletteforum.cc/index.php?topic=15938.315 (page 22)
You see those cycles of dozens. Imagine each of those set of unique numbers within a dozen has a statistic quality associated with it. What if those statistic qualities give us an advantage something along the lines of below.
Dozen 1 is no longer 12 numbers but it is 14 numbers. Dozen 2 is no longer 12 numbers but 16 numbers. Dozen 3 is no longer 12 numbers but 6 numbers. But the payouts don't change. All the dozens still give you 2 to 1.
That's the target you need to work on. Sorry can't get more explicit than this.
One more thing. In the tree that you are publishing from the excel that I have shared, is there any elements that can be swapped around so that it gives you the view above? In the hurry before sending the excel I forgot to modify the title of one column. That is the reason there are two columns with cycle lengths. Does that give away anything?
http://www.rouletteforum.cc/index.php?topic=15938.690 (page 47)
5.4 INCREASING THE SPAN OF THE BIASED GAME4. Can you
increase the span of that biased game, by making the limit of that cycle larger that you will always find a bias and the law of large numbers will never come into picture. May be!
Why do house edge catch up with you. Because of the law of large numbers. Simply put, lets say you constantly bet on red. If it is 10 spins, you might win, you might lose. If it 10,000 spins, then most of the times you will be losing. 100,000 spins, you will definitely be in negative as the variance decreases with a larger sample size. This is because the cycle limits of even chances is only 3 spins excluding zero. However imagine you have defined a cycle with a very large limit. Then you can play such that the law of large numbers will take longer to catch you, and hence you will always have variance to take advantage on.
It’s not clear what a “biased game” is but it sounds like playing a game per cycle that increases a bias towards the defining dozen – the longer the cycle length the better to increase the span before being caught by the law of the large numbers – all based on the “constant”(?) ratios. 5.5 CREATING YOUR OWN PLAYING POSITIONS AND STITCHING BETS TOGETHER5. Can you increase that edge further by not using a hook to catch fish but using a net as Turner would put it by
stringing together your bets. May be!
Lets say you are tracking a biased wheel which is biased towards the 0 pocket. Odds of the game do not change. But the number of times you hit a winner will increase if you are not just targeting zero but pockets around 0 as well. Thats increasing the accuracy. If you follow a betting plan such that this increased hit rate is giving you a higher edge, why not.
Sounds like we should change the focus/scope of our bet slightly if we have more information/support re: accuracy of predictionhttp://www.rouletteforum.cc/index.php?topic=15938.300But stringing together ECs we can create an odd placement that we like like quads, dozens, so on and so forth. We don’t even have to look at the numbers or wheels. How is this possible. See this example below on Red and Black.
Instead of playing one position of just R and B, what if we play RR, RB, BR and BB. Instead of giving odds of 1/1 we have converted ECs to give odds of 3/1. An example play is below. For simplicity, what we will be looking to play is for getting the outcome RB.
25 - 1 unit on red. Win.
27 – Place both units on blck. Loss.
7 – 1 unit on red. Win.
29 – 2 units on black. Win. We got the win at odds of 3/1
4 – 1 unit on red. Loss
18
27 – 1 unit on red. win
10 – 2units on blck. Win. We got 3/1 odds
14
28 – Won this sequence
34
27 – lost this one
6
16 - lost
12
20 - won
This is not a progression. This is not letting it ride. This is an example of stitching together simple EC components to create an odd that is better than even return. Now the possibilities are endless and everyone can create opportunities based on their comfort and style of play. You can create dozens, quads, splits, all possible odds through stitching together these components.
Now when it comes to the topic of stitching together bets, it is also important to understand which combinations are profitable and which ones are not. The combinations which might seemingly give better odds at first sight may not be the ones that will be profitable and vice versa. Taking a simple example.
Red and Odd. If we need to stitch together these two, will you place one bet on red and one on odd or one bet on red and 8 bets on the black odd numbers? Any creative ideas and view points? http://www.rouletteforum.cc/index.php?topic=15938.60 (page 5)
I think 1 bet on red and 8 bets on black odd numbers is better.
Ati - You are right and perfect. Now the follow up question that one should ask is we can clearly see there are imperfections here. Is there a potential for us to modify the bet sizes across these positions instead of 1 unit bet uniform to create an edge? I will let you ponder on that.
Could be a reference to progressions otherwise sounds like a new concept of converting/”translating” a stitched bet into different sections/”positions” of the board (EC > Numbers) and then placing different unit sizes across those numbers. http://www.rouletteforum.cc/index.php?topic=15938.105 (page 8 )
It always worries me that a set of numbers could be given for your whole video that meant you didnt win once...or didnt win enough and that set of numbers have the same odds of showing as the ones that happened.
Turner - You are right. There is no need to worry. There will always be a set of numbers that could be given for the whole video which will mean that we dont win or dont win enough. However the other part is not right, as this is not really based on numbers, it is based on combinations and number cycles.
It is like this - the odds of 2 blacks out of 3 spins is different from the odds of black happening only in the first and third spin. It is based on combinations and as I have been saying from the start of the thread, roulette is not necessarily random. It does have a limit. Some of the plays in this game are based on non-random occurances and we are covering every combination possible. When you are covering every combination possible and you are getting a positive result, irrespective of whether you lose 1 or 2 or 3 or for that matter 100 sessions, eventually the edge will prevail. Just like the casino prevail on the house edge.
http://www.rouletteforum.cc/index.php?topic=15938.195 (page 14)
So if we betting two different dozens in a row not to become 3, we are winning on 12 patterns and losing on 6.
6 combinations against 12 combinations. Making it simple once against twice. On the outset it looks as if 12 combinations are better than 6 combinations hence this looks like a very good selection. But what is the return on those 12 combinations. You will have to play 2 dozens and hence the return is 1/2 (Apologies for people who don’t understand English odds, it is 1$ returned on a win for every 2$ staked). If you see it that way, you are essentially going back to 6 against 6 (1/2 multiplied by 12). Now you are left to the mercy of deviations, variations and statistic reality to either fail or win.
This is the reason I was pointing back to find out finite, non-random methods within the bet selection process. You have taken the step in the right direction, but I would really encourage you to look deeper.
What is that you are finding common in those 12 combinations that you have selected?
Why have you not taken into consideration that other combinations where the dozens have repeated in the first two positions?
Is there a way you can stitch these dozens together?
Are you able to find out a common theme between the first and second spins or first and third spins?
Is three spins sufficient for you to derive that commonality?http://www.rouletteforum.cc/index.php?topic=15938.60 (page 5)
Thinking about statistics now. Out of the 27 combinations that is possible
(3 dozens in 3 spins) , 3 will be one dozen in 3 spins, 6 will be 3 dozens in 3 spins and 18 will be 2 dozens in 3 spins. It is like drawing a ball from a bag of 3 red balls, 6 green balls and 18 blue balls, then putting it back in and repeating this whole process. Your chances of drawing a blue ball is higher. There is an irregularity and the statistically speaking the 12 spins (4 sets of 3 spins), there is a higher probability of 2 dozens in 3 spins to come through.
http://www.rouletteforum.cc/index.php?topic=15938.80 (page 6)
RMoreAnd besides, simply stitching bets together randomly is not going to produce the solution.
I suspect that a large part of the answer is going to be nailing the principles first and then working on the bets and the bet structures afterwards - for a specific purpose. For example, let's say we are following a dozen cycle. There are different bets at different points in the cycle, right? After the first dozen, let's say it is a 1, then we should consider a bet on 1 again because at this point there is only one choice to close out the cycle. But should we bet at this point? We should look for some support for this bet on the first dozen from elsewhere - some other non-random measure. For example maybe the VdW - although not sure how to do that. But the point is, can we find support for this bet? If not - no bet. So let's say that it is a 1 that comes out. Oh well, end of cycle, but that 1 is the start of the next so let's check again.
No Bet. Perhaps this time a 2 comes out. OK - then what next? Should we bet the 1 and the 2? Perhaps we need some support from what we are seeing from the quads - or the six-lines. Maybe there are some stats that are coming into play. But which? We can't measure everything that is going on on the wheel - way too complex.
The point? I suspect that we play a non-random cycle as a starter - a fundamental if you like. Principle 1. By itself - no advantage. But if we can solve the riddle of adding some weight towards one of the 2 possibilities in a dead run situation, or perhaps when we see more likelihood from the stats of one option versus the other, then and only then will we have a betting opportunity. Or perhaps if we can find a way to avoid a fairly certain loss, then we can improve our situation as well.
There are just too many possibilities. Cold numbers? Hell no! Stay away from those. Hot numbers (or sections)? Don't have a lot of faith in those either. Repeats? Maybe - but again, by themselves - no advantage. So we have to base our bets on a combination of factors as it is clear that any one of them, by themselves, is without advantage. This is key. So perhaps we should start bringing our discussion into more specifics. In order to do that let me propose that we use the dozen cycle as our base bet because we know that this is a non-random event. The objective is to close out a cycle with a repeating dozen. What can we say about the very first betting opportunity? It is, quite obviously, a simple repeat. What possible support can we get for this? Either positive or negative because if positive then we play the bet but if negative then we can consider playting the OTHER 2 dozens - but ONLY if there is strong suppport for either. Where can that support come from?
psimoesHere´s one way not to bet: if the defining dozen is dz1 or dz 3, bet for it to be next defining dozen until a win or until a different dozen defines the cycle. Bet as well the opposite EC (H or L) to reduce variance. It´s a bet on 30 numbers, I know! If Dozen 2 becomes the defining dozen, don´t bet.
I chose 2u on Dz and 3u on EC for a profit of 1u everytime it wins. Could have gone for 2u on EC and 1u on the Dz for 1u profit or break even. Less losses, but also less profit. It´s all linear.
Well here goes nothing. Sorry lost the file. But it had 81 wins and 22 losses. Nothing special really.
http://www.rouletteforum.cc/index.php?topic=17014.1206.1 PARRONDO’S PARADOX“There exist pairs of games, each with a higher probability of losing than winning, for which it is possible to construct a winning strategy by playing the games alternately.”
https://en.wikipedia.org/wiki/Parrondo%27s_paradoxLeaving that aside, lets take the two views here. First one states that PP cant work because the outcomes are independent. Second one states that no casino games change rules based on players bank roll (I wish they did, then things would have been easier for us to win
![:)]( "Smiley")
).
PP can't work because the outcomes are independentThe proof against this one is a little difficult to grasp.
First of all lets clearly understand the definition of independence. Two events are independent, statistically independent, or stochastically independent if the occurrence of one does not affect the probability of the other. Keeping this definition in mind, lets take the event of getting the spins.
First event - Spin 1 gets me 20.
Second event - Spin 2 gets me 24.
Both the above events are independent. Very much independent. Getting 24 in spin 2 is totally independent of getting 20 in spin 1. (Remove all physical factors that might cause dependence).
Now see the following two events.
First event - spin 1 gets me 20.
Second event - Sum of spin 1 and spin 2 gets me 44.
Are these two events independent? No. A big
NO. Actually, a better explanation of why PP can't work with casino games is because outcomes are independent, but PP requires some interaction between the current game and the previous one. In the above example, have we not created an interaction and made dependent events in roulette outcomes? As we have managed to create dependent events then the argument of why PP cannot work in roulette doesn't hold good.
Carefully creating those events to make them dependent is in our hands. We cannot achieve that just with spin outcomes, you have to find a way of stitching them together.
(see section on Stiching bets to increase odds) VdW and other non-random examples are ways and means to induce those dependencies and create and locate events that are dependent.
Casino doesnt change rules based on players bank rollThere is no flaw or nothing to prove here.
PP never says that you play based on your bank roll. That is just one example to explain it in a simple manner. WoV is true that constructing a PP based on your bankroll will not work. But what is PP? Is PP based on your bank roll. No. PP is exactly what you copied and pasted from wikipedia. It is creating a dependence between two of your playing streams so that you are more likely to enter one of the playing streams at the point where it will yield positive expectation. The dependency or the deciding factor of games doesn't have to be based on bank roll.
Let me explain one crude example which you might be able to relate to. One stream of play (Game A) is observing spins. Second stream of play is starting to bet(Game B). You are deciding to alternate between these two streams of play or games with a simple rule. Start playing Game A. Enter Game B if there are ten of an even chance. Exit Game B and start playing Game A on a win in Game B or after 3 spins on Game B. Repeat the process.
(This concept may be observable in one of Priyanka’s videos based on a follow up parlay bet or waiting for a virtual win or loss) What are we trying to do here. We are trying to enter Game B at a point where we believe it will most likely give a positive expectation. There is no dependency of bank roll. So as I said, nothing to prove against what has been said in WoV. It is the just that the basic premise of PP games has to be chosen based on bank roll is wrong. It can be created without bank roll coming into question. You will have to find out that tipping point that is most likely to give positive expectation.
If you remember the example of dozens we discussed the point where statistics comes in/progression comes in. There was an imbalance. One outcome was more likely than other. How we can enter the dozen game when that imbalance is in our favour and most likely to result in a positive expectation is the riddle that you need to crack.
(Priyanka always waits for virtual spins and has never shown a system where bets are placed every spin) http://www.rouletteforum.cc/index.php?topic=15938.90 (page 7)
I dont switch bets and there is no need to. The key is taking advantage of certain things which are non-random. However, yes, as Drazen rightly said, there has to be a when/where/what that can be defined for every entry point and exit point and that will be based on these non-random concepts.
http://www.rouletteforum.cc/index.php?topic=15938.300 (page 21)
However it is good that you brought GUT for the discussion. The most important learning that I have learnt from Winkel is an adoption of Parrondo’s paradox. In GUT, if you keep betting on the same crossing you will ultimately lead to a -2.7% expectation. However switching between crossings, and betting different crossings is a different beast altogether.
http://www.rouletteforum.cc/index.php?topic=15938.15 (page 2)
praline1. Cycles
2. VdW
3. PP related to triggers on CL
CL = Cycle Lengthhttp://www.rouletteforum.cc/index.php?topic=17014.135 7.1 PROGRESSIONS / MONEY MANAGEMENTThinking about progression now. Depending on how you chose to play, you can see the irregularities here and you can focus on tuning your progression to maximize your wins. Key is low drawdowns and achieving those low drawdowns using elements that are fixed and finite.
http://www.rouletteforum.cc/index.php?topic=15938.75 (page 6)
How about we change to the following units after each kind of set has finished?
This again goes into the equation of waiting for LLLL to increase your units so on and so forth. I havnt tried it, but my expectation is you will get into LLLL 3 times or 4 times in a row to wipe your bankroll or gains.
http://www.rouletteforum.cc/index.php?topic=15938.30 (page 3)
8.1 VIDEOSVideo 1: http://youtu.be/AIvAeaHzKVY - related to VdW/AP? What concepts are identifiable in this? http://www.rouletteforum.cc/index.php?topic=15938.0 (page 1)
Bet Amount Result Red 0.05 2 Black lose (107.95)
Red 0.05 4 Black lose
Red 0.05 17 Black lose
Red 0.05 7 Red win
Red 0.05 12 Red win
Red 0.05 14 Red win
Red 0.05 15 Black lose
Red 0.05 31 Black lose
High 9 29 Black win
1,19,20,21,22,23,24,25,26,27,28,30,32,33,34,35,36 0.5 each (8.5 in total) 21 Red win
Red 0.05 17 Black lose
Red 0.05 17 Black lose
Red 0.05 18 Red win
Red 0.05 11 Black lose
Red 0.05 35 Black lose
1,3,5,19,20,22,23,24,25,26,27,28,30,32,33,34,36 0.5 each (8.5 in total) 19 Red win
1,3,5,6,20,22,23,24,25,26,27,28,30,32,33,34,36 0.5 each (8.5 in total) 27 Red win
Red 0.05 35 Black lose
Red 0.05 16 Red win
Red 0.05 5 Red win
Red 0.05 19 Red win
Red 0.05 13 Black lose
Red 0.05 33 Black lose
20,22,23,24,25,26,28,30,32,34,36 0.5 each (5.5 in total) 36 Black win
Red 0.05 4 Black lose
Red 0.05 22 Black lose
Red 0.05 19 Red win
Red 0.05 30 Red win
20,23,24,25,26,28,32,34 0.5 each (4 in total) 35 Black lose
Same as above but 1.0 units on 20 20 = 1; rest are 0.5 (4.5 in total) 32 Red win
Red 0.05 34 Red win
Red 0.05 24 Black lose
Red 0.05 5 Red win
Red 0.05 28 Black lose
Red 0.05 22 Black lose
Red 0.05 23 Red win
Red 0.05 19 Red win
Red 0.05 17 Black lose
Red 0.05 36 Red win
Red 0.05 33 Black lose
Red 0.05 9 Red win
1,2,3,6,7,8,10,12,14,15,18,20,21,24,25,26,29,31 1,2,3,6,7,8,10,20,25,26 = 1; rest are 0.5 (14 in total) 9 Red lose
1,2,3,6,7,8,10,12,14,15,18,20,21,24,25,26,29,31 1 = 1.5; rest = same as before (14.5 in total) 18 Red win
? (cut) 6.5 in total 12 Red lose
? (cut) 7 in total 5 Red lose
? (cut) 7.5 in total 3 Red win
1, 20, 24, 25, 26 0.5 each (2.5 in total) 2 Black lose
1, 20, 24, 25, 26 1 = 1; rest = 0.5 (3 in total) 27 Red lose
1, 20, 24, 25, 26 1 = 1, 20 = 1; rest = 0.5 (3.5 in total) 26 Black win (180.80)
Can you figure out what Priyanka is doing in the above games? Which principles discussed herein are being applied?
Some analysis by Still, including for one of Priyanka’s earlier videos just prior to the Random Thoughts topic:http://www.rouletteforum.cc/index.php?topic=15938.165 (page 12)
Video 2: https://www.youtube.com/watch?v=g1RWS1Ar_YM (Parrondo’s Paradox) Bet Amount Result virtual 11 Black
virtual 7 Red
Low 5 24 Black lose
Dozen 1-12 5 16 Red lose
Double Street 7-12 5 9 Red win
Red 0.05 29 Black lose
Red 0.05 26 Black lose
High 5 19 Red win
High 5 11 Black lose
Dozen 25-36 5 14 Red lose
Double Street 25-30 5 5 Red lose
Double Street 25-30 5 28 Black win
Red 0.05 19 Red win
Red 0.05 35 Black lose
Red 0.05 19 Red win
High 5 21 Red win
Dozen 13-24 0.05 35 Black lose
Dozen 13-24 0.05 22 Black lose
Dozen 13-24 0.05 26 Black lose
Dozen 13-24 0.05 35 Black lose
High 5 18 Red lose
Dozen 25-36 5 12 Red lose
Double Street 31-36 5 13 Black lose
Double Street 31-36 5 19 Red lose
Double Street 31-36, Dozen 13-24, High 5 each (15 in total) 15 Black win
Low 5 9 Red win
Low 10 13 Black win
Low 5 7 Red win
Low 5 20 Black lose
Dozen 13-24 5 18 Red win
Red 0.05 8 Black lose
Red 0.05 6 Black lose
Red 0.05 19 Red win
Red 0.05 14 Red win
Red 0.05 33 Black lose
Red 0.05 26 Black lose
Red 0.05 17 Black lose
Low 5 17 Black lose
Red 0.05 3 Red win
Red 0.05 4 Black lose
Low 5 6 Black win
Red 0.05 33 Black lose
Red 0.05 26 Black lose
Red 0.05 7 Red win
Red 0.05 18 Red win
Red 0.05 27 Red win
High 5 23 Red win
High 5 17 Black lose
Dozen 25-36, Low 5 each (10 in total 15 Black broke even
Low 5 3 Red win
Low 10 13 Black win
Red 0.05 20 Black lose
Red 0.05 19 Red win
Red (accidentally missed a spin!? - but no spin was virtual) 0.05 17 Black lose
Dozen 13-24 5 36 Red lose
Double Street 19-24 5 30 Red lose
Double Street 19-24 5 24 Black win
Red 0.05 24 Black lose
Red 0.05 18 Red win
Red 0.05 14 Red win
Low 5 5 Red win
Low? (cut) 5 27 Red lose
Dozen 13-24 5 5 Low lose
Double Street 13-18, Low 5 each (10 in total) 16 Red win
![]()
Video 3: https://www.youtube.com/watch?v=dy_hSK4z-yIVideo 4: https://www.youtube.com/watch?v=3J4Lf7zxk4IVideo 5: https://www.youtube.com/watch?v=4dVbiXMIipINumber Quad Cycle quad W/L Bet Why?29 4
3 1
9 1 1 Bet 2 - 3 - 4 End of cycle: Bet all the other quads
26 3 W Bet 1 - 3 We won our first bet. Now we bet the last two quads
27 3 3 W Bet 1 - 2 - 4 End of cycle: Bet all the other quads
4 1 W Bet 1 - 3 We won our first bet. Now we bet the last two quads
27 3 3 W Bet 1 - 2 - 4 End of cycle: Bet all the other quads
32 4 W Bet 3 - 4 We won our first bet. Now we bet the last two quads
18 2 L No bet We lost. Wait for a virtual win.
1 1 No bet
7 1 1 Bet 2 - 3 - 4 End of cycle: Bet all the other quads
28 4 W No bet No bet. We wait for the virual win.
27 3 VL No bet Virtual loss.
24 3 3 Bet 1 - 2 - 4 End of cycle: Bet all the other quads
5 1 W No bet No bet. We wait for the virual win.
7 1 1 VW Bet 2 - 3 - 4 Virtual Win. End of cycle: Bet all the other quads
28 4 W Bet 1 - 4 We had our virtual win. Now we bet again the last two quads.
2 1 1 W Bet 2 - 3 - 4 End of cycle: Bet all the other quads
15 2 W Bet 1 - 2 We won our first bet. Now we bet the last two quads
31 4 L No bet We lost. Wait for a virtual win.
30 4 4 No bet
![???]( "Huh?")
No ideal why we dont make a bet here…
14 2 VW No bet Virtual win.
29 4 4 VW Bet 1 - 2 - 3 End of cycle: Bet all the other quads
31 4 4 L No bet Here we lost our bet. Now we wait for a virtual win.
36 4 4 No bet
35 4 4 No bet
5 1 No bet
11 2 No bet
20 3 No bet
23 3 3 No bet
23 3 3 No bet
1 1 No bet
9 1 1 No bet No bet. We wait for the virual win.
27 3 Bet 1 - 3 Virtual win. Bet all the other quads. This bet is still active.
19 3 3 W Bet 1 - 2 - 4 End of cycle: Bet all the other quads
7 1 W Bet 1 - 3 We won our first bet. Now we bet the last two quads
15 2 L No bet Lost
10 2 2 Bet 1 - 3 - 4 End of cycle: Bet all the other quads
16 2 2 L No bet Lost
12 2 2 No bet
10 2 2 No bet
4 1 No bet
26 3 No bet
16 2 2 No bet
15 2 2 No bet
22 3 No bet
31 4 No bet
25 3 3 No bet
9 1 Bet 2 - 4 Virtual win. Bet all the other quads. This bet is still active.
11 2 W Bet 1 - 2 - 3 Here we see a new trend. Our previous cycle was lenght of 3. Now we bet it will also be 3. Bet the 3 previous quads.
23 3 3 W Bet 1 - 2 - 4 End of cycle: Bet all the other quads
25 3 3 L No bet
14 2 Bet 1 - 4 Here we switch bet. We now bet the two missing quads because we bet for a cycle of 3.
2 1 W Bet 1 - 2 - 3 Our previous cycle was lenght of 3. Now we bet it will also be 3. Bet the 3 previous quads.
5 1 1 W Bet 2 - 3 - 4
![???]( "Huh?")
Why bet? We did not have a virtual win here.
29 4 W Bet 2 - 3 We now bet the two missing quads because we bet for a cycle of 3.
20 3 W Bet 1 - 3 - 4 Our previous cycle was lenght of 3. Now we bet it will also be 3. Bet the 3 previous quads.
2 1 1 W Bet 2 - 3 - 4 End of cycle: Bet all the other quads
24 3 W Bet 2 - 3 We now bet the two missing quads because we bet for a cycle of 3.
16 2 W Bet 1 - 2 - 3 Our previous cycle was lenght of 3. Now we bet it will also be 3. Bet the 3 previous quads.
12 2 2 W END
Video 6: https://www.youtube.com/watch?v=LKjvj4FQVuUIt is based on Iron steel and Turner's quads, but taking non-random into consideration.
To make things clear, firstly as there was lot of discussion about 0.05 bet, the 0.05 bet is used to place a bet to complete the non-random sequence when we cannot play. Secondly, as i mentioned already, the only position used are quads(group of 9 numbers). You can reverse engineer to figure out the method as it puts together all the concepts i have explained. However, this one game in itself gives an edge over the game slightly higher than 9% which should defeat the house edge of american roulette.
http://www.rouletteforum.cc/index.php?topic=15938.180 (page 13)
Video 7: https://www.youtube.com/watch?v=5VUUfwkFilIVideo 8: https://www.youtube.com/watch?v=T4KgiscwgRU Todo with cycles – designed to lose.Video 9: https://www.youtube.com/watch?v=C5rHlShQm1kVideo 10: https://www.youtube.com/watch?v=TwOiBydJJF09.1 PARALLEL UNIVERSESBefore I move my thought process into an interesting concept of Parallel universes, I would like to explore another aspect of non-randomness.
Was this covered in any more depth or was it a reference to Cycles or stitching bets?http://www.rouletteforum.cc/index.php?topic=15938.60 (page 5)
Now coming back to parallel universes, as drazen has asked about it, the whole thing of birthday paradox(problem) works because of these parallel universes. A person on its own will have a lesser probability of finding a birthday match as opposed to a group finding its match as there are more number of pairs involved.
could this be a reference to increasing the span of a biased game by catching with a net? (see prior section above)http://www.rouletteforum.cc/index.php?topic=15938.120 (page 9)
Linkback: http://www.rouletteforum.cc/index.php?topic=17047.0
As I looked at cycles , and dozens the one thing I noticed is that after a dozen has hit 3 x in row or more 24, 18, 13 (doz 2) then its broken by say (doz 3 ,,, 32) the next number is or will be a doz 2 close to 40% of the time in the 2000 spins I tested---hmmmm
No ideal why we dont make a bet here… 
