Reverse-engineering the HG

[mickavelli]

I'm not a math head mate just trying to look at things logically.
Wondering if there might be a way to capitalize on that neat little trick

[falkor2k15]

I don't know mate that looks too much like a guessing game to me i believe the only randomly drawn order you would not want to encounter is a perfectly drawn order from start to finish or vica verca in relation to your beginning sequence.
For example 123456 or 654321 drawn ramdomly in those orders exactly. Highly unlikely - and so in all other instances pigeonhole kicks in and a position must repeat. So unless you are drawn a sequence in perfect order never can your positional sequence ever reach 6 uniques numbers.
Well that's the basic takeaway i get from outside the box thread. I know Priy said something along the lines of as you go up and up with streets, splits etc the chances of the repeat coming on the lower half gets higher and higher but I feel there is more to it than just playing for a repeat. There is a remainder. How to make use of this?

The positions stream is equally-likely and behaves exactly the same way as the standard stream. Standard cycles and position cycles of, say, lines or dozens both carry the same stats. So when we create custom streams we are simply creating additional random streams that have some kind of connection (based around pos/order 1 and the repeat) but no obvious exploit when used together. It's an interesting discovery, but it doesn't bring us closer to the HG.

[mickavelli]

Maybe entering the game with that stream at a specific betting point. Not sure mate

To me that positional stream could be seen as pigeonhole for repeats.

I had an idea a while ago of playing roulette like dice. Think of the game craps and rolling a 7.
Basically you can do it with any number.
For example 9

Partition 1,2,3,4....9
So.. (1,9) (2,8) (3,7) (4,6) (5)

Clearly the sum of all the PAIRS is 10...

So there are only 2 ways to have a length of 6 subset...
Either it contains a 5, or it doesn't..!!
If it does contain a 5...
There are only 5 more spaces in your subset to fill...

You can't possibly fill those 5 spaces without a pairing that sums to 10...

To me - that's pigeonhole for uniques!

[mickavelli]

The positions stream is equally-likely and behaves exactly the same way as the standard stream.

This is the thing. 
Standard stream has the possibilty for the seq to go through every option only to end in a deadlock leaving us with 1 option - to guess which number repeats

Positional stream will repeat before the options are exhausted avoiding the deadlock each and every time

Maybe not with dozens,  the likelihood of drawing 1,2,3 or 3,2,1 is there

But as you go up and up,  the likelihood of drawing for example 12 numbers from 1-12 or 12-1 in perfect order is gone

[falkor2k15]

Positions can still deadlock just as easy as the standard stream because they are equally-likely and have the same stats as the standard stream.

Anything is possible with lines or street positions - the same as normal lines or streets:

1234566 = cycle length 6 order 6.

Doesn't matter lines or positions of lines. They can both deadlock at exactly the same time too.

Perhaps you are getting mixed up with the position values on a repeat - or you are taking only the position value when there is a repeat? That would not be position cycles/repeats - but defining position cycles.

[mickavelli]

Perhaps you are getting mixed up with the position values on a repeat - or you are taking only the position value when there is a repeat? That would not be position cycles/repeats - but defining position cycles.

Yes, the values of the positions. Sorry mate i thought that part was pretty self explanitory using pigeonhole principle because how else are you going to use the positional stream - it guarantees a maximum length!
Meaning - It will be defined before a deadlock each and every time - Unless drawn in perfect order

[mickavelli]

I think if you were to play for a numberset of say 6 , the key might lie in constructing a game around a divisor of that number
Yes 6 uniques are possible in 1 stream, but impossible in another

Sorry I shouldn't say impossible - highly highly unlikely

[falkor2k15]

Not sure if we are any closer to resolving this...

If I am right: you take the standard stream and only take the position value on a repeat, ignoring the position values when a standard unique shows? You then take the repeat position only and make another cycle from it?

Sure - cycles on the repeat position will show less deadlocks. There are many examples using the same concept that I have posted such as Quadruplets and Six Dozen Options. With those kind of cycles it's possible to reduce deadlocks to about CL11 - but it still doesn't result in edge and can still break the bank at CL11/12 even.

If you use a divisor then you simply reduce those 12 outcomes down to just a couple - similar to "Same" or "Different" or "Odd" or "Even"; like what standard cycles does it just reduces the permutations down to even lesser combinations. However, it doesn't result in any exploit or edge.

[mickavelli]

All im saying is either a randomly drawn number repeats, or a position repeats!
Its either or - both can not go to deadlock simultaneously.. How do you figure that?

[mickavelli]

Take 6 uniques (6,1,2,3,4,5)

Starting with seq    123456

6 is in position 6 > 612345

1 is in position 2 > 162345

2 is in position 3 > 216345

3 is in position 4 > 321645

4 is in position 5 > 532164

If the random stream is to stay unique our positional stream will repeat

If the positional stream stays unique the random stream will repeat.

So how can both result in a deadlock?

The length of 1 of these streams is limited

[mickavelli]

 

[Herby]

(1,9) (2,8) (3,7) (4,6) (5)

Clearly the sum of all the PAIRS is 10...

So there are only 2 ways to have a length of 6 subset...

Hi mickavelli ,
thanks for your interesting posts. Most of them I can follow, but above:
"Clearly the sum of all the PAIRS is 10...  "     -  still clear, but here I miss something:

"So there are only 2 ways to have a length of 6 subset..."

maybe a little example helps. 

[mickavelli]

No worries herby,

Either it contains a "5", or it doesn't !

So if your playing a game of 9 options on the layout and you have 6 uniques,  either there will be a 5 and a pair that total 10, or no 5 and 2 pairs that total 10 - either way 1 side will have minimum 3 hits...
All i was meaning from it was,  "5" could be a constant bet - to help split things into 2 sides

[RouletteGhost]

I've finally glimpsed the HG and now know what form it takes!!! 

This topic is dedicated to RouletteGhost... 

You are so full of shit

I don’t even attempt to understand what the hell you are saying

I feel sorry for the people that follow you.

[falkor2k15]

Take 6 uniques (6,1,2,3,4,5)

Starting with seq    123456

6 is in position 6 > 612345

1 is in position 2 > 162345

2 is in position 3 > 216345

3 is in position 4 > 321645

4 is in position 5 > 532164

If the random stream is to stay unique our positional stream will repeat

If the positional stream stays unique the random stream will repeat.

So how can both result in a deadlock?

The length of 1 of these streams is limited

Well, dozens and quads certainly can deadlock at the same time.

As for lines - you are right - I can't seem to find an example of when both the standard stream + pos stream deadlock at the same time. However, I doubt this is due to any mechanism - but rather just a rare sequence - as there's no reason why it should not happen on lines and only quads, dozens and EC other than lower probability.

I found CL6 vs. CL5 quite easily: