37 back to basics

[6th-sense]

You have to bet 2. Because that is now in position 3.

it doesn,t mean you have to bet it,,,its just for reference...

[Herbyx]

"an upsequence" provided for any a.i and a.j in the subsequence we have

i < j ⇒ a.i ≤ a.j ;
in the case that we have
i < j ⇒ a.i ≥ a.j ;

it is called "a downsequence".

should look like this:

upsequence:
i < j ⇒ a_i ≤ a_j ;

downsequence:
i < j ⇒ a_i ≥ a_j ;

where i and j are indices
anyone who is my opinion ?
 

[6th-sense]

CarpeDiem refers to it as balance

just going to quote this herby from mels posts.....what does this actually mean

Dyksexlic called it twins he also referred to it as sunset/sunrise


and these
Pri mentioned something about parallel universes

[Herbyx]

Thanks 6th, I have to sleep it over.

[ᶦ ᵃᵐ|Ä-łëx]

youtube.com/watch?v=Ova4BHJbtcw

[TRD]

Funny, this ↑, in principle, works as VWD theorem ..
at least length of x=4 increasing & y=6 decreasing in z=16 numbers.

I wonder if number z can be modified, to the payouts & its cycle lengths, observed simultaneously & as your terminology would go stitched, & what the x,y are then?

[TRD]

Also, there must be a version for repeats .. latest n+1
& another for uniqyes .. at least n

in a certain number of spins.

used, both combined.

[TRD]

Ideas, practical?

[Herbyx]

Thanks alexlaf,

Theorem from the film:
Any sequence of m*n+1 distinct (real) numbers either contains an increasing subsequence of length m+1 or a decreasing subsequence of length n+1.

Remark: the numbers of the subsequences don't have to be neighbours (but must be seen from left to right)

distinct = uniques as TRD remarked

[6th-sense]

if i could give a thumbs up for the video and comments i would do... wait i can

[TRD]

Great to know we are thinking in the correct direction ..

[TRD]

Hmmm, "either" .. so we are back at the hedge between both.

& although the initial number gives some type of constant, as a starting point for higher or lower, thereof subsequence (up/down-sequence), effectively eliminating a portion of the board as bs ..


.. what real value is in this? From the main stream & su, none.

Perhaps combined with the derived stream, might complete the picture into something sensefully, advantageous?

[TRD]

Another detail;

m*n+1


Where the guy in video is using an even number (16), as a starting point (z=m*n).

Given that 16 can be made of either
1x16
3x5
4x4

Question; within the theorem, must be m>n -- or does one subsequence have to higher than the other?

Otherwise, we would have multiple completing options.
n(dec)= 5+1, 4=1
m(inc)= 4+1, 5+1



Also, in roulette given 37 numbers, the probability of an upsequence being longer than downsequence is higher .. when the z≤18, linearly more so the more z gets towards & closer to Zero.

However we may have a long string as a streak highs -- when based firstly on the fact that there is a hedge primarily between betting on up/down-sequence, & despite the higher probability -- none it, the formulation, matters.

Say I have 18 as the first string number, the kind of constant, then 15 high repeats .. the theorem does not hold. There is not 1 (n+1), actually none additional numbers in the string, bein only n itself.

Even we took 16x1 option, we do not arrive at 2 (=n+1), the theorem does not complete.

!?!



True, the probability of 15xEC streak is low, very low,so we might dismiss that.

But we have also the possibility of having only 18, 17 & 14x High repeats -- or 6, 3 & 14x of numbers higher than 6 .. much higher probability, more likely.

So where are we at!?!



Given that the theorem does not hold (how not!?, am I mistaken on something), & there is no constant to override theprobability --

the only one more possibility I see is a repertoire of multiple streams on multiple payouts, where z might be somewhere in the middle, taken as a starting point (of the up/down-sequence).

Duh!?

[TRD]

What am I missing?

Anyone, 6th?

[MoneyT101]

What am I missing?

How do you know something is missing?

Only way to know that it's missing is for you know that something was already there.  That's a stranger question that only real friends would know the answer to...

Good luck