PhP Pair-Completion

[TRD]

A request to coders ..


I would like to get the results of a simple test, regarding the % of the distribution inthe number of spins.

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PhP states that in  n  numbers (natural integers), there will be at the latest in  (n/2)+1 appearances at least two numbers that sum up to n+1.

Or practically speaking;
number 11. on this link
mindyourdecisions.com/blog/2008/11/25/16-fun-applications-of-the-pigeonhole-principle

11. If you pick five numbers from the integers 1 to 8, then two of them must add up to nine.

Every number can be paired with another to sum to nine. In all, there are four such pairs: the numbers 1 and 8, 2 and 7, 3 and 6, and lastly 4 and 5.

Each of the five numbers belongs to one of those four pairs. By the pigeonhole principle, two of the numbers must be from the same pairâ€"which by construction sums to 9.

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In E-roulette;
n=37


18 pairs + 1 standalone number (with its repeat making a pair) that sum up to 36
36+0
35+1
   ...
19+17
18+18

the pairs do not sum up to (n+1) but n, since Zero.



18 spins is the threshold limit by when at least two outcomes sum to 36; providing we bet SU numbers.



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Let's say that at each number appeared we place a bet on it + its counterpart (summing to 36); in case of 18 only that --so we are effectively betting

Obviously, in roulette, we have a possibility of a repeat. If we had only unique integers, we would need to only bet the pair counterpart, having a hit or =pair completion by ((n/2)+1) numbers out.

Thereof the overall mechanics & stats change.


What I would like the test result, 10.000.000 spins, to show by to show in spin-by-spin format
spin1
spin2
spin3
  ...
spin18
spin19


till either a) or b) happens, then restart the cycle using the last outcome of the previous cycle as the first of the new cycle with one pair positioned already;
   a) pair-completion realized (appearance of both counterparts)
   b) a repeat of the previous appearance

in percentage for each spin against the total cycles in 10.000.000 spins.


Also, further break each spin down by repeats & completions.

[algojack]

TRD,

11. If you pick five numbers from the integers 1 to 8, then two of them must add up to nine.

So in roulette terms, you want to know the least number of numbers which have to be spun such that any two of them add up to 36?

e.g.

1. 23
2.  9
3.  5
4. 26
5. 36
6. 11
7. 27

spin #2 and #7 add up to 36. End of cycle. 

[TRD]

TRD,

So in roulette terms, you want to know the least number of numbers which have to be spun such that any two of them add up to 36?

e.g.

1. 23
2.  9
3.  5
4. 26
5. 36
6. 11
7. 27

spin #2 and #7 add up to 36. End of cycle.

Correct, in essence.
The applications of this can be of two variants, & on many districts (=ds,st,sp,su),
although ds & st can appear all 6 or 12 respectively unique - meanwhile the su has 29/30 spin practical limit to the first repeat. Don't know about this on sp?




Read this thread, first five posts being the intro, the rest one the first possible variant - application on streets - betting the pair counterpart only & if the number repeats then the positions cover the whole pair.

Code Select Expand

roulettelife.com/index.php/topic,3302.msg66652.html#msg66652




This is the second possible variant, betting both pair counterparts as soon as one of them appears.

Code Select Expand

roulettelife.com/index.php/topic,3480.msg66611.html#msg66611

This is foremost being aimed at being an enclosed system (Dycksexlick type - meanwhile it incorporates the Vaddi pair principle as well), but for this to stand it has to be figured out how to close (=new high) with the repeat hit (=by spin 30 the latest), meanwhile the counterparts hit (by spin 19 the latest) would carry the game onwards .. without the progression exploding (having potentially more completion cycles till the first repeat hit).
So it would be the flat-bet pair counterparts games, overlayed with the 'repeat' game somehow (Dycksexlick implied betting various amounts per position .. further implying more hits fill the first 'I need one repeat').

The version in the thread in more Vaddi-style;
without keeping the rolling block, but resetting the positions at each hit in accordance to the core idea - so covering all the pairs on the marquee .. from the last spin number till its repeat  or complementing pair counterpart further back.

Look at the questions at the bottom of the posts directing the research into a potential comprehensive solution. There might be other enrichments - implementing a dynamic stream .. to further reduce the 19-spin threshold, ideas?




Btw, @algojack are you a coder?

[TRD]

The first three tests would be

•   threshold for a SP to repeat, with spin-by-spin % distribution (SU= 29or30 spins)
•   threshold for a PhP Pair-Completion, with spin-by-spin % distribution
          betting the SU 'x' pair-counterpart upon 'y' appearance (x+y=36; exception x=18 â†' bet 18 only)
              ( so, if 'y' shows â†' bet 'x' )
          if 'y' shows twice or repeats, the whole pair is covered;
              ( then onwards - first 'x' appearance  or  the next=3rd 'y' appearance = end of cycle)
•   spin-by-spin % distribution for PhP Pair-Completion
          betting both SU counterparts (x+y=36; x=18â†'18 only) as soon as one shows, till any of them hits
         

[stringbeanpc]

TRD, point number 11 from the document cht provided is quite interesting. Thanks for bringing it up

Just my thoughts, right or wrong

Interpret the numbers from 1 to 8 (and 9) as the root sum. Roulette has 4 groups of 9 numbers.
So each root sum has a subset of 4 numbers (example)

01 10 19 28
02 11 20 29
03 12 21 30
04 13 22 31
05 14 23 32
06 15 24 33
07 16 25 34
08 17 26 35
09 18 27 36

Furthermore,
each subset of 4 numbers has 2 Low and 2 High
each subset of 4 numbers has 2 Odd and 2 Even

Perhaps there is a way to use this to our advantage ?

[algojack]

Hi TRD, I didn't understand some of what you wrote in your previous longer post (I don't know Dycksexlick and Vaddis systems), but in answer to your question, yes I am a coder and will attempt to code your requests so far as I understand them over the next few days (probably not until the weekend though).

The first three tests would be

•   threshold for a SP to repeat, with spin-by-spin % distribution (SU= 29or30 spins)

What do you mean by "SP"? Does "SU" mean "Straight Up"?

[TRD]

EC  --  equal chance, 18-numbers
DZ  --  dozen, 12 numbers
CL  --  column, 12 numbers
QT  --  quarter, 9-numbers
DS  --  line = double street, 6-numbers
Q   --  quad, 4-numbers
ST  --  street, 3-numbers
SP  --  split, 2-numbers
SU  --  straight-up, single-number

[TRD]

Perhaps there is a way to use this to our advantage ?

Yeah, I had an intuition of running a two-layered PhP (as in pair completion),
on two districts (e.g. STs, SUs) concurrently, but haven't quite been able to crystalize in solid terms yet ..

Right now I am at the point of simply exploring seeing how various templates,
& mapping things out to solid facts like extremes in the performance etc.
to have a better grasp of, orientation & bearing in what I am dealing with;
in accordance with MoneyT101 4-steps

Code Select Expand

rouletteforum.cc/index.php?topic=26317.90
[Sep 27, 08:50 AM 2019]

(although cht recently conveyed to me alomg the lines that the second part of the linked post is equally, if not more, important; namely quantifying entropy)

Step 1 - Create your template
Step 2 - Limit your session
Step 3 - Breakdown your stats
Step 4 - Create your system

What I am saying is that, take the material in my posts related to PhP Pair-Completion as the findings of my initial exploration.


I began using two-counterparts pair completion, due to first testing the betting of one counterpart only -
the opposite of the appearance that lead to my first posted roulettelife link. Can be as well repeat=x & its counterpart=y  (x+y=36).
Although I later expanded the research on betting both counterparts, since its more in line with point 11. - with only the possibility of repeat being the discrepancy to the principle.

Now, about putting more than two numbers together, as in three & four in one layer only
well, its just a question of defining what the pigeonholes are, but to be in-line with the point 11. those have to sum to the same amount, & figure out the formulas for at least e.g. numbers to appear

but I am questioning the viability of this type .. the things, seems so, get more much more complicated

[number25]

01 10 19 28
02 11 20 29
03 12 21 30
04 13 22 31
05 14 23 32
06 15 24 33
07 16 25 34
08 17 26 35
09 18 27 36

Furthermore,
each subset of 4 numbers has 2 Low and 2 High
each subset of 4 numbers has 2 Odd and 2 Even

Perhaps there is a way to use this to our advantage ?
[/quote]

Yes you can beat the wheel with these root numbers.

You have so many options to play.  You can track multiple things. I played today using this exact numbers.

1. Track streets

2. Rows

3. Quads 1 to 9  10 to  18 19  to 27  28 to 36

4.  Root numbers 

I played two games today & made profit.

Number 25

[TRD]

OK ..

I see you made four groups (pigeonholes?) made of QT or quarters
[I reserve the terms Quads to 4-number positions]
You allegedly track 4x different parameters.

. but how exactly is this related to PhP, & furthermore PhP pair completion? Explain.


Also, my aim in roulette is to design an all-winning system with which every single individual game ends in nominal profit =new high, a chain of them a session.
Two games .. might be two of always won; but I have no interest in systems than win more than loose overall. Just to clarify.

[number25]

OK ..

I see you made four groups (pigeonholes?) made of QT or quarters
[I reserve the terms Quads to 4-number positions]
You allegedly track 4x different parameters.

. but how exactly is this related to PhP, & furthermore PhP pair completion? Explain.


Also, my aim in roulette is to design an all-winning system with which every single individual game ends in nominal profit =new high, a chain of them a session.
Two games .. might be two of always won; but I have no interest in systems than win more than loose overall. Just to clarify.

I was just sharing my experience with those root numbers.   

Stringbean said maybe we can use them to are advantage! 

Good luck with you quest.

Number25

[algojack]

I've made a start on the coding. The average number of spins before a repeat is 8.306, and the average number of spins before the sum of the last spin and a previous spin = 36 is 8.665. This is the result after 10 Million cycles.

So it seems that you're better off betting on a repeat rather than a sum to 36. If you think about it, this makes sense because for a repeat it doesn't matter what number comes up next - there's an equal chance for a repeat. But this isn't the case for a sum to 36 because the situation isn't perfectly symmetrical as it is for repeats.

[Ross]

Why do I detect an overpowering smell of voodoo here?

[TRD]

@algojackk check PM

OK, averages .. great. In addition, I would like to see the three specific tests requested above.

[algojack]

Why do I detect an overpowering smell of voodoo here?

You're probably right. But where's the harm?