A request to coders ..
I would like to get the results of a simple test, regarding the % of the distribution inthe number of spins.
-------------------------------------------------------------------------
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.
-------------------------------------------------------------------------
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.
-------------------------------------------------------------------------
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.
TRD,
Quote11. 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.
Quote from: algojack on Oct 13, 04:08 AM 2021
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.
roulettelife.com/index.php/topic,3302.msg66652.html#msg66652This is the second possible variant, betting both pair counterparts as soon as one of them appears.
roulettelife.com/index.php/topic,3480.msg66611.html#msg66611This 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?
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
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 ?
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).
QuoteThe 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"?
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
Quote from: stringbeanpc on Oct 14, 06:06 AM 2021Perhaps 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
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)
QuoteStep 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
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
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.
Quote from: TRD on Oct 15, 07:52 AM 2021
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
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.
Why do I detect an overpowering smell of voodoo here?
@algojackk check PM
OK, averages .. great. In addition, I would like to see the three specific tests requested above.
Quote from: Ross on Oct 17, 04:55 PM 2021
Why do I detect an overpowering smell of voodoo here?
You're probably right. But where's the harm?
Quote from: TRD on Oct 13, 08:01 PM 2021
The first three tests would be
• threshold for a SP to repeat, with spin-by-spin % distribution (SU= 29or30 spins)
I'll take them one at a time. What do you mean by a spin-by-spin-distribution? A file with 10M cycles showing every spin would be far too big. What do you mean by "Threshold"? Is it the maximum number of spins before a split repeats?
Quote from: algojack on Oct 18, 03:16 AM 2021You're probably right. But where's the harm?
No harm at all!
Quote from: algojack on Oct 18, 03:20 AM 2021
I'll take them one at a time. What do you mean by a spin-by-spin-distribution? A file with 10M cycles showing every spin would be far too big. What do you mean by "Threshold"? Is it the maximum number of spins before a split repeats?
What I mean is like redhot's post, just spins -- in essence this output
format.
rouletteforum.cc/index.php?topic=26049.msg228070#msg228070Total cycles
spin1 ...... Cycles ..... % relevant to Total Cycles
spin2 ...... Cycles ..... % relevant to Total Cycles
spin3 ...... Cycles ..... % relevant to Total Cycles
...
spin16 ...... Cycles ..... % relevant to Total Cycles
spin17 ...... Cycles ..... % relevant to Total Cycles
spin18 ...... Cycles ..... % relevant to Total Cycles
Threshold.
We know that practically DS & ST can hit all unique/different in a row.
For SU, Straight-Up, we know r]that practically the first repeat treshold is 29th or 30th, test done over 10.000.000 spins -- so there's a window of 7-8 spins till 37 total numbers.
TEST1: What's the window for SP, splits two-number position, -- if any? Bet any appeared SP, add another on each spin,
at repeat restart betting the last outcome as the first bet.
Populus: 10.000.000 spins, even better cycles if not too much of a ask
This is the table for Splits.
NO OF CYCLES AS %
------------------------------------------------------------
First Repeat Occurred @ Spin no. 2 555753 5.558%
First Repeat Occurred @ Spin no. 3 1049581 10.496%
First Repeat Occurred @ Spin no. 4 1399462 13.995%
First Repeat Occurred @ Spin no. 5 1554231 15.542%
First Repeat Occurred @ Spin no. 6 1512373 15.124%
First Repeat Occurred @ Spin no. 7 1310145 13.101%
First Repeat Occurred @ Spin no. 8 1016764 10.168%
First Repeat Occurred @ Spin no. 9 712423 7.124%
First Repeat Occurred @ Spin no. 10 444302 4.443%
First Repeat Occurred @ Spin no. 11 247508 2.475%
First Repeat Occurred @ Spin no. 12 120768 1.208%
First Repeat Occurred @ Spin no. 13 50952 0.510%
First Repeat Occurred @ Spin no. 14 18600 0.186%
First Repeat Occurred @ Spin no. 15 5566 0.056%
First Repeat Occurred @ Spin no. 16 1310 0.013%
First Repeat Occurred @ Spin no. 17 239 0.002%
First Repeat Occurred @ Spin no. 18 21 0.000%
First Repeat Occurred @ Spin no. 19 2 0.000%
These figures are a little too high because for simplicity I didn't include the zero. So the first row % should be closer to 2/37=5.41%, whereas they are actually closer to 1/18=5.56%. So as you can see, all splits arrived in the first 18 spins in only 2 of 10M cycles.
I'll code the system next (bet first split and then add each one as they appear).
@alogojack; great, thanks - that ticks one unknown off the list.
= there no open window on SPs to the first repeat related to the pool of all splits (18/18),
similar to DS (6/6) & (ST 12/12) all can show in a row; & unlike the SU district where the
practical window is (29-30/36; 10mil spin test).
Still, the test shows that expanding the bet by an additional SP position on each & every spin, it only makes sense to bet till including perhaps spin 12, but later no more.
Why - to keep the max cycle drawdown low, & rather recover the outstanding debt with consequent cycles till in nominal profit â†' restart next individual game in the session.
-1-2-3-4-5-6-7-8-9-10-11-12-13-14-15-16-17-18= 9*19= -171
-1-2-3-4-5-6-7-8-9-10-11-(12)= 6*13= -66 (-78)
Big difference, ≈3x between (171) & (-66) = keeping drawdowns recoverable.
With first 5 spins being positive.
â†'
This relative to the Php Pair completion (version1), betting the pair-repeat or pair-counterpart only on each outcome appearance.
Of course the above ain't totally precise, just the first of the parameters to regard, still narrowing the extremes we play within a bit further.
Why - for when both pairs numbers are out .. the whole pair is bet, so on that particular spin the same number of SP positions as a bet are repeated.
although,
1) since all my system's betting is continuous - it would have been great if you have used the previous' cycle repeat =end of cycle SP as the first position used â†' thus the 'spin1' parameter would also show as relevant
2) include 0, for this is integral part of the game & approximations derived from neglect do not work in our favor, especially when designing a system
3) when opting for actual split positions I would determine that upon the numbers already out in recent spins, current or previous cycle - top & bottom columns outcomes have (except numbers 36 & 34) three possible positioning options & middle column each 4 options (except number 35); effectively if number 9 is the previous cycle & in current cycle 8 shows SP8-9 would be placed if placing the repeat counterpart, or it pair-counterpart first
4) the SP repeat bell curve shows peaks at spins4,5 (5,6 in the chart), so it may make sense to bet for the repeat-counterpart first & then complete it with
eg. Number 8 shows =splitNâ¿5 = split8-10
(x+y)=19 â†' splitN°5 + splitN°14 as pair conterparts
Can you first re-run the test two times, each enriched by the next of 1) & 2) points; disregard 3) & 4) for now. No betting amounts, still only distribution percentages to get really precise results.
Quote from: TRD on Oct 20, 11:40 AM 20212) include 0, for this is integral part of the game & approximations derived from neglect do not work in our favor, especially when designing a system
Ok. Are you also betting on the zero when it appears, or only the splits?
Quote from: TRD on Oct 20, 11:40 AM 2021eg. Number 8 shows =splitNâ¿5 = split8-10
(x+y)=19 â†' splitN°5 + splitN°14 as pair conterparts
Hi TRD!!
what is split nº5 and split nº 14?
Thanks
Although you, playing SPs only, can get a bit more creative â†' the splits being arbitrarily labeled differently in each of the cycles ie. as the board fills up, when opting for a specific SP placement option (vertical & horizontal) & the formation derived from that as the board fills up, based on previous cycle outcomes coupling them with the already shown (to repeat, as a refinement that may exclude unnecessarily betting the cold numbers, pushing them towards the cycles' end =full board )
the most simple categorization per the screenshot, construct pair's complementing SP counterparts according to (x+y)=19, zero excluded
Otherwise, if you are essentially playing the game on SU Pairs, betting both counterparts simultaneously; but rather than playing SUs from the beginning opting for a parachute-style attack that closes many games much quicker in nominal profit & moves from 2x DS (spin1, 1st pair), then 4Q (spin2, 2 pairs), 6ST (spin3, 3 pairs), 8SP (spin4, 4 pairs), 10SU (spin 5, 5 pairs) -- I suggest that you, irregardless of the position type, couple the pairs according to (x+y)=36 for the coupling to be consistent all the way through. After all the goal is to have 2 numbers shown within a threshold of numbers shown producing a complementing pair, thus a hit. Also, the attack varries ad-hoc cycle-by-cycle, since with fewer positions all the pairs shown can be already covered
eg.
spin1 #16
16+20=36
no need for 2DS, one DS covers the pair (-1)
spin 2 #16,19
16+20=36, 19+17=36
no need for 4Q; with 2DS both pairs are already covered, resulting in an exposition
that allows for another DS-based spin pushing the need for Qs a spin onwards ..
as per the nominal profit requirement (-2, total-3)
spin3 #16,19,25
.. 22+14=36
no need for 6ST; 2Q does the job (-2, total -5)
spin4 #16,19,22,25
25+11=2=36
no need for 8SP, nor 6ST;
you wanna break even & continue with 2DS theonwards anew ? --
bet 3Q (+6) = Q11-14, Q16-20, Q22-25; or Q7-11 instead if 7,8,10 is a recent show ..
(-3, total -8)
spin5 #16,19,22,25,13
5ST does the job covering all 5 pairs;
ST10-12, ST13-17, ST16-18, ST19-21, ST22-24, ST25-27 (-6, total -14)
if spin5 hit (+6, total -2)
no need to continue with SU,SP,ST, or Q -- 2DS does the job (+3, total +1)
take the last two outcomes & start the pair-completion anew
spin6 #1
1+35=36
DS1-6, DS31-36 (-2, -4)
spin7 #1,13
1+35=36, 13+23=36
4Q = Q1-4, Q11-14 (since of 11&13 show), Q22-25, Q32-36
spin8 #33 Q32-36 hit (+5,total +1); game done â†' restart