Random Thoughts

[RouletteGhost]

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.

this is EXACTLY what I have been trying to exploit 100%

[RouletteGhost]

Priyanka

can you confirm to me if this is something you are doing?

when the last 3 results were all different, example: 132 bet for the sequence to repeat

also when last 3 are the same 222, bet for that to repeat

please confirm

edit: also, if the last 2 results were different, such as dozen 1 and dozen 3, bet those 2 dozens for 3 spins as they are destined to repeat? trying hard to grasp

thanks

[3Nine]

3nine - It is clear to me that Priyanka is a man, is a senior, and resides in the UK. All this can be gleaned quite clearly (to me at least) from this thread alone. I mean no disrespect to you my friend - but it really is irrelevant. Just make your choice and write accordingly. Do not allow trivialities to distract you from what is really important.

Scarface - while I applaud your efforts I do not think you are on the right track with your research into the stats. While these are useful to know, and perhaps important later on, I 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.

all the best
Rog

Ok, then integrity becomes an issue, doesn't it? 

I know what's important, thanks, and it has nothing to do with this thread. 

Good luck chasing 2 rabbits.

[RMore]

Fine. I wasn't trying to be insulting - shame you have to interpret my reply that way. I'm not understanding why you are posting here at all if this is unimportant to you. Look - people post with anonymous handles all the time - says nothing about integrity at all - just a desire for privacy. I can't fault that, and in fact I applaud it. We should defend our privacy rights as a part of the rights of all free people. THAT'S what is important. Anyway, troll away my friend - I will say no more to you until and unless you suffer an attitude change.

[RMore]

Folks - need some help here. I'm not understanding what Pri said here in post #7.

The above example of 9 spin finite cycle non-random behaviour can be done for any set of numbers. For example for a dozen, this limit is 27. We cannot not have an A.P of length 3 of same dozens in a set of 27 spins. It is debatable whether to call it non-random or limits of random. You can device a way to play even chances or dozens using these. The lower 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.

Look 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?

thanks
Rog

[3Nine]

Fine. I wasn't trying to be insulting - shame you have to interpret my reply that way. I'm not understanding why you are posting here at all if this is unimportant to you. Look - people post with anonymous handles all the time - says nothing about integrity at all - just a desire for privacy. I can't fault that, and in fact I applaud it. We should defend our privacy rights as a part of the rights of all free people. THAT'S what is important. Anyway, troll away my friend - I will say no more to you until and unless you suffer an attitude change.

Ok, cool - you're right.  Love you buddy - let's get back to the program.

[3Nine]

Folks - need some help here. I'm not understanding what Pri said here in post #7.
Look 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?

thanks
Rog

EC is the HIGHER of the 2.  18 #s vs 12.  Is that helpful?

[RMore]

Sweet - 3Nine and I are friends again! Long may it continue - I wish to be friends with everyone.

OK - next question for all you dudes and dudesses. As I said yesterday I am now reviewing the first 6 pages in detail. See above question - hope somebody can help out with that. But next up - what about this quote.

W â€" 256 times
L â€" 48 times
LW â€" 104 times
LL â€" 32 times
LLW â€" 36 times
LLL â€" 16 times
LLLW â€" 10 times
LLLL â€" 10 times

Look at it from an "each play" point of view.
On the first opportunity we have 256 chances of a W and that means 256 chances of a loss. Play the predicted or the opposite - no difference. If we don't play and it would have won then nothing ventured, nothing gained, just start again.
If it does lose then what's next? If I am reading this right then we have a W chance 104 times leaving 152 potential losses. But 48 times there is no other opportunity. So again it is 50/50 - 104 potential wins versus 104 potential losses.
If we lose then what? Next up - W 36 times, lose 36 times, no opportunity 32.
Finally we are left with 10 wins, or 10 losses, 16 no opportunity's.
Net 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?

[RMore]

Oh - so you think that the number being referred to is the number of numbers (so to speak) involved in the bet? Mm - hadn't thought of it that way - obviously. You may be right. Will try and expand my brain to encompass that concept. Thanks.

[3Nine]

A sample 12 spins. 133 323 123 323
133 â€" 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 â€" 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.

If you're looking for 1 dozen to repeat here, why isn't 3 a play after the first spin?  Thanks.

[Nickmsi]

Hi 3Nine . . .

133 is called a Set of 3 that has a dozen that repeats, ie the #3 Dozen is repeated.  So this set is called a Repeating Dozen.

There are 3 Types of Sets:

1.  Repeating Dozens (133, 221, 112, etc)  one of the dozens in the set repeats, there are 18 of this type

2. Same Dozens (111,222,333)  all dozens in the set are the same.  there are 3 of this type

3.  Different Dozens (123, 321, 213 etc) all dozens in set are different. There are 6 of this type.

Hope this helps.
Nick

[3Nine]

Hi 3Nine . . .

133 is called a Set of 3 that has a dozen that repeats, ie the #3 Dozen is repeated.  So this set is called a Repeating Dozen.

There are 3 Types of Sets:

1.  Repeating Dozens (133, 221, 112, etc)  one of the dozens in the set repeats, there are 18 of this type

2. Same Dozens (111,222,333)  all dozens in the set are the same.  there are 3 of this type

3.  Different Dozens (123, 321, 213 etc) all dozens in set are different. There are 6 of this type.

Hope this helps.
Nick

Thanks, but in the example the actual dozen that repeated is irrelevant.  So, the first 3 of the 2nd set could have repeated there and would have fit with the 'repeating dozen' of the last set, correct?

[Nickmsi]

133  323

The first set of 133 is a "Repeating Dozen" set.

We are playing that the second set is also a "Repeating Dozen"

The first number of the second set is "3".

We do not know if the first Dozen of the 2nd set (3) will be the repeating dozen or the second Dozen of the 2nd set (2) will be the repeating dozen so we wait until both numbers are spun.

Now we bet that either the 3 or 2 will form a "Repeating Dozen"

Yes you could bet for the 3 to repeat after the first spin and if wrong you would have to make a 2nd bet.

More efficient to make only 1 bet.

Hope this helps.
Nick

[3Nine]

133  323

The first set of 133 is a "Repeating Dozen" set.

We are playing that the second set is also a "Repeating Dozen"

The first number of the second set is "3".

We do not know if the first Dozen of the 2nd set (3) will be the repeating dozen or the second Dozen of the 2nd set (2) will be the repeating dozen so we wait until both numbers are spun.

Now we bet that either the 3 or 2 will form a "Repeating Dozen"

Yes you could bet for the 3 to repeat after the first spin and if wrong you would have to make a 2nd bet.

More efficient to make only 1 bet.

Hope this helps.
Nick

It does, thanks - so, then the question becomes what if it did repeat on spin 2?   33X.  Now you need to wait and see without a bet, correct?

[Nickmsi]

First spin a 3, then

If 2nd spin is 3 then you would play 1 and 2 to complete a Repeating Dozen Set.(332,331)

If 2nd spin is a 2 then you would play 2 and 3 to complete a Repeating Dozen set.(323, 322)

If 2nd spin is a 1 then you would play 1 and 3 to complete a Repeating Dozen set.(313, 311)

Nick