Is on the forum any willing mather (or having a great grasp of math on the roulette side of things)
prepared to take the matter at hand further, allowing me:
• add smth, you deeming me thereafter having an even better comprehension
of things specified & more, facilitating an even smoother application of it
• demonstrate the way of calculating the frequency of an event,
keeping in mind that its not constant, thereof precisely calculate
the 'worst' or most intense frequency of such an event being =1/q
.... so we are looking at getting some type of 'interval of frequency'
(very much the confidence interval results' format alike)
To begin with, before even applying MoneyT101's YN groups bs concept
(which by itself will definitely improve the avg & overall performance parameters anyway);
I've included here the detailed extract of stats of an all-winning method,
which has some very rare games going beyond the desirely acceptable
performance parameters .. those being:
(eg. -100u max drawdown + ≈100 spins max individual game to nominal profit length,
or -150u max drawdown + ≈100-150 spins max individual game to nominal profit length;
with bankroll vs session +100u profitable session ratio, nearly maximizing simultaneously
both, the base unit increase & thus compounding effect)
.. so for the purposes of generating an insight & assessment, I am playing/toying
with the idea of the stop-loss introduction = rather than driving those rare individual
games over +200 spins into the nominal profit, terminate the session based on the
fundamental premise of .. in short:
my method is so strong that I can afford to use Marti on it
or in a bit longer version.. my 'base sequence' (eg. up to 150u bankroll + ≈150 spins till termination = event in itself) is so great, with the extended bottlenecks games being so rare & far apart, that those can (potentially) be terminated as 'non-completes' -- & thereof the base unit Marti takes care of it ..
(link:s://i.ibb.co/PCHWHrF/Screenshot-20220517-162229-WPS-Office.jpg)
As you can see in this limited stats base, I've played/included:
3533 spins
555 games
11 sessions
.. where a session is a 'gamechain' .. of individual games till inevitable profit;
• it takes on average 50 games to form a session, ranging somewhere from 70 till 40,
even below that to 30 & less on faster games
• it takes on average about ≈300 spins to form a session, here specifically averaged to 321
made +1037u,
or 0.2935u/spin, 1.8685u/game
The last session, due to the extreme game & the sheer amount of spins has been terminated early ..(before the standard session goal of +100u, although if continued for another ≈150-200 spins this such is driven to +100u too) .. so basically just above session break-even .. with nominal session profit;
[which, btw, is some members' on this site a definition of a hg .. win or break-even --not mine though, I am a perfectionist]
Moreover, I've input all games played into the sheet, having two attributes --
individual game's .. max exposition amount, length (till new high, or profit);
then I've sorted all the games into two arrays categorized by either exposition or length,
broken down into multiple groups criteria for each --
ie.
max exposition, units | -1 till 9, -10 till -19, -20 till -29, .. etc. .., -100 till -150, -200 till -300, -300+
length, spins | 1 till 10, 11 till 20, 20 till 30, .. etc. .., 101 tiil -150, 151 till 200, 200 till 300, 300+
.. thereof got the number of games per each criterion, & the ratio of the criteria versus total games played .. or simply percentage
Hereof, let's say we trim the bottlenecks off ..
hypothetically applying only 150u bankroll.
The general stats change to:
(columns BC)
+923u
3343 spins
555 games .. 1x game terminated
11 sessions .. 1x session terminated
The terminated session already pocketed 43u (which are never re-staked, playing only with the allocated/designated bankroll .. in the big picture, contributing to the base unit increase & compounding effect) --
thus (+43 - 150=) resulting in -107u -- to be recovered.
Bringing us to the above-mentioned recently delving into the math side if things:
p = 1 - q = 0.9982
q = 1 - p = 0.0018
----------------------------
inv.p = 1/p = totality/p = 1/0.9982 = 10000/9982 = 1.00180224584
inv.q = 1/q = totality/q = 1/0.0018 = 10000/18 = 555.5~
[1x non-complete every 555 games]
------------------------------------------------------------------
var = p x q = 0.9982*0.0018= 0.00179676
n = number of games, at avg ≈50 games/session = 10000
[at avg ≈50 games/session ≈ 20 session]
---------------------------------------------------------------------
mean success = n x p = 150*0.9982= 998.2
[expected .. succes 998 in 1000 games played, or
9 in 10 sessions, non-complete 1 in 10 sessions]
variance for number of trials = n x var = 150*0.00179667= 1.7976
sigma, Σ = √(n x var) = √(150*0.00179676)= 1.340432766
std.error, se = 2 x sigma = 2 x 0.5191473= 2.680865532
-------------------------------------------------------------------------------------------
conf.int = p ± magnitude x Σ = p ± number of nines x se =
2 nines = 99.45 = se 2.68086553
3 nines = 99.75 4.02129823
4 nines = 99.9936 99.82 ±5.36173106
6 nines = 99.99994 6.70216383
9 nines = 99.99999998 8.04259660
p probability of an event happening successfully, reliability (=complete)
q probability of an event not happening (=non-complete)
inv.p inversed success probability, expected number of occurrences in which
success is to be seen
inv.q inversed non-success probability, expected number of event instances
non-success is to be seen
var variance, a measure of dispersion, or far a set of numbers (data) is spread
out from their average value (means the expectation is only ideal)
n instances of an event (can be a spin, game, session; here game)
mean success average rate of success
variance for number of trials a measure of dispersion within the specified number
of events
sigma, std.dev is a measure of the amount of variation or dispersion of
a set of values
se standard error, an estimate of how far the sample mean is likely to be from
the population mean
conf.int confidence interval, tells the possible range around the estimate & also
about how stable the estimate is .. meaning one that would be close to
the same value play was repeated.
REQUEST
1. Add more to the definitions --
if you deem smth will allow me to have a better comprehension of the dynamic between
the bold figures; secondly, at eg. variance within the number of trial decimal value .. what
does that actually tell me in the context, practically speaking, or its simply just used as a
factor in further calculation holding no other significance.
2. Moreover, & most importantly, given that the confidence interval is a vector + the
amplitude is the angle of the vector .. & we are looking for the frequency of the 1/q events ..
(in other words, practically -- the application of Marti, over/on the 'base sequence' =
game ↠bottleneck terminated as a non-complete of an event)
.. & since in roulette neither the amplitude nor frequency is constant .. specifically,
we are looking for the highest intensity of such frequency telling us essentially
either
• how many 1/q events in a row we can expect max .. in the desired interval of games
played, thereof the most likely frequency of these 2x, 3x, (4x?) in the row event appearing
• or, taking another angle, given the resulting frequency of 2x in row event is low enough,
what would the required 'play bankroll' amount be, ascertained from the fundamentally
required P (see image, top left array) .. which satisfies the premise of the one-step Marti
takes care of the most intense frequency manifested (potentially extended max to two-step)
what would the formula to get to this value be ?
Il faut avoir un prix nobel de maths appliqués pour comprendre ce fil
You have a drawdown. You may want to reconsider everything you're doing. Cheers.
@Blueprint;
debt creation, or drawdown, is inherent to roulette play -- from the point of view of an individual game to profit .. the first no-hit bet already creates an outstanding debt to be resolved (=essentially, 1x no-hit = recovery) .. so basically its unavoidableâ†inherent;
what's important are essentially two things -- interacting with variance ..
• how much drawdown the method self-creates (exposition increase rate, volatility, protrusion rate = is drawdown minimized & at any point recoverable, requiring only the shortest turn of variance in favor)
• how, or in what manner (which tools), is outstanding debt deal with -- does it get ALWAYS recovered (the answer is YES)
+
Regarding that I mostly play 2Q as a wide bet (1st hit) & a single ST as focused (combo hit)
in those ≈5% of the upper games with higher exposition & longer length (top left array) ..
the drawdown actually ain't that big at all.
Especially when only 2% go over (-49), 0.36% over (-99), & 0.18% over (-150).
So, the dickhead says .. NO.
======================
@toimeme .. & I need a translator to decode what you said; English, thanks.
Nonetheless, it really ain't that hard to comprehend = the frequency fluctuation of 1/q events â†'
telling how often 1/q happens in a row .. when the frequency is at its most intense
Quote
@toimeme .. & j'ai besoin d'un traducteur pour décoder ce que vous avez dit; Anglais, merci.
Néanmoins, ce n'est vraiment pas si difficile à comprendre = la fluctuation de fréquence des événements 1/q â†' dire combien de fois 1/q se produit d'affilée .. quand la fréquence est la plus intense
We obviously play very different games. Good luck in your quest!
You sound like casino dealer.
For those who find it useful, the above screenshot sheet;
manual input -- row 31â†", columns BCDE & Hâ†'
& on the second tab .. cells D5 (map to first tab column E, in the row of the bankroll amount)
D35 (number of event = games)
& I do not rely on luck.
Quote from: TRD on May 20, 03:49 PM 2022
You sound like casino dealer.
You sound like Falkor.
Quote from: TRD on May 20, 04:53 PM 2022
& I do not rely on luck.
Yup... you're in control, I see.
Quote from: Blueprint on May 21, 06:46 AM 2022You sound like Falkor.
Forum idiot
Turbo quote; take 5'000 to win 5'000
Sure, I am.
Erase that, you sound like a child.
Let's move on back onto the course.
Here I've included 13 contiguous sessions -- 620 games, till the first hypothetical non-complete;
game (to nominal profit) = event = 'base sequence' to potentially be Martied.
(link:s://i.ibb.co/9Vswdfc/Screenshot-20220521-171007-Browser.jpg)
(link:s://i.ibb.co/kDp6tJZ/Screenshot-20220521-170844-Browser.jpg)
(link:s://i.ibb.co/QX6fHbn/Screenshot-20220521-171317-Browser.jpg)
disregard the unit placement mistake on ≈spin210, 3ST 34u/position instead of 4u
(link:://%5Burl=link:s://i.ibb.co/hchpnmH/Screenshot-20220521-171836-Browser.jpg%5Dlink:s://i.ibb.co/hchpnmH/Screenshot-20220521-171836-Browser.jpg%5B/url%5D)
(link:s://i.ibb.co/CQPr6Xp/Screenshot-20220521-165223-Browser.jpg)
(link:s://i.ibb.co/t2BVQg0/Screenshot-20220521-165546-Browser.jpg)
(link:s://i.ibb.co/HdkyHTy/Screenshot-20220521-165143-Browser.jpg)
(link:s://i.ibb.co/YjJFDXq/Screenshot-20220521-170244-Browser.jpg)
(link:s://i.ibb.co/PFRp2bV/Screenshot-20220521-170417-Browser.jpg)
(link:s://i.ibb.co/r7BPJmW/Screenshot-20220521-173054-Browser.jpg)
(link:s://i.ibb.co/8Kq48Qn/Screenshot-20220521-182554-Browser.jpg)
(link:s://i.ibb.co/NTQzFpM/Screenshot-20220521-182721-Browser.jpg)
(link:s://i.ibb.co/5LrkqDf/Screenshot-20220521-182828-Browser.jpg)
(link:s://i.ibb.co/5F83kgM/Screenshot-20220521-183356-Mi-Xplorer.jpg)
& with this one makes, that was not added in the beginning
All these games are played with mechanical bs, last outcome;
so no advantage constituted from it (eg. trend, repeaters).
All done simply by mechanism constituted of the only two interactive things in control,
how many units bet & how those are spread .. or in short money management.
As you can see, 2x 'extreme games' went beyond 110 spins (0.3226%) over 200 spins;
one still resolved withing 150u bankroll, (-133) to be precise, & the other got to (-240),
so with 250-300u bankroll, ALL is resolved in positive.
The first-mentioned game, although it required
Although the variance in both games is mostly out of favor, in the non-complete game extraordinarily so, with an interval of over 100 spins without the focused single ST hit -- the exposition is kept in control, slowly descending meanwhile keeping the ratio between the combo hit (wide coverage bet first, with its gains reinvested sponsoring the focused bet = bot hitting close-by is a combo hit) & exposition proportional .. (not expanding the vertically unnecessarily);
so yeah, I am in control -- in control of my emotions, mind, focus, deliberation, resolve, exposition (keeping the drawdowns at all times recoverable) & thus the games in its entirety .. so that/& even when the variance is extremely out of my favor, simply patiently awaiting its (game's) inevitable resolve.
(link:s://i.ibb.co/XJDBtxZ/Screenshot-20220521-191346-WPS-Office.jpg)
(link:s://i.ibb.co/dDYmqhG/Screenshot-20220521-191622-WPS-Office.jpg)
Correction
QuoteAs you can see, 2x 'extreme games' went beyond 110 spins (0.3226%) over 200 spins;
As you can see,
4x/620 games went beyond 100 spins (0.3226%),
-240(266), -133(225), -92(110), -88(118);
of which 2x called 'extreme games went over 200 spins,
3x of those were still kept within a hypothetical 150u bankroll,
& 2x within hypothetical 100.
So, toying with hypothetical event termination,
taking the
150u hypothetical bankroll;
with probability p = 0.998387, or 99.8387% reliability of an event completing in positive,
makes 1 - p = q .. 1 - 0.998387 = 0.001613, or 0.1613% chance of event non-complete ↠terminated ..
the question is what's the avg & the most intense frequency of (1 -q)(=how many of those non-completes can happen in a row max).
in a bit longer version
Quotegiven that the confidence interval is a vector + the amplitude is the angle of the vector ..
& we are looking for the frequency of the 1/q events ..
(in other words, practically -- the application of Marti, over/on the 'base sequence' = event
= bottleneck game terminated as a non-complete of an event)
.. & since in roulette neither the amplitude nor frequency is constant .. specifically,
we are looking for the highest intensity of such frequency telling us essentially
either
• how many 1/q events in a row we can expect max .. in the desired interval of games
played, thereof the most likely frequency of these 2x, 3x, (4x?) in the row event appearing
• or, taking another angle, given the resulting frequency of 2x in row event is low enough,
what would the required 'play bankroll' amount be, ascertained from the fundamentally
required P (see image, top left array) .. which satisfies the premise of the one-step Marti
takes care of the most intense frequency manifested (potentially extended max to two-step)
Nevermind, that the above method has been already further improved by itself, further tweaked & optimized, moreover added the benefit of playing the repeats which further adds to its overall avg performance parameters;
& now evaluating the MoneyT101's high-performance bs concept in detail, further enriching something that already works consistently .. null-hypothesis: absolve the extreme games entirely;
making it a true masterpiece.
No matter what...
1/q will be the amount of numbers played divided by the amount of numbers available. At every single spin. No matter what you try, no matter what the averages are.
and odds are against the player. One cannot beat random. One cannot argue with maths.
You've probably had a lucky strike
Quote from: Bigbroben on May 22, 08:49 AM 2022
No matter what...
1/q will be the amount of numbers played divided by the amount of numbers available. At every single spin. No matter what you try, no matter what the averages are.
and odds are against the player. One cannot beat random. One cannot argue with maths.
Why do we go to work? We never get 100% return on our time spent at work, the governments take away 30% of our earnings.
The odds are against the worker to ever be able to get 100% of his earnings to take home.
Does that mean we stay at home and not work?
SUMMARY
My question to you is regarding the 'base unit increase' + Marti (reverse, halving
the base unit .. or neg, doubling the base unit) on the hypothetically terminated
game is --
are you able to calculate the fluctuation & thus the highest intensity of frequency
of such terminated events;
confidence interval is a vector (of standard deviation), amplitude is the change of vector (of its direction) .. amplitude & frequency are correlated (but not constant as in eg. cos & sin graph) -- thereof it should be in theory possible to get the result above out --
the result being the most intense frequency of the (1-q) event = in a row, or practically how many Marti-steps are required to surpass the worst variance;
where an event is one game, so how many session possibly terminated sessions in a row at worst, given that a session is about ≈50 on average (range ± 30-60)
REPHRASED QUESTION
bringing us back at the most intense point in the fluctuation of frequency, or every how many games the non-complete event happens at the highest congestion, or how close-by those non-complete (1-q) events in terms of games are at the worst variance
â†' =Marti-steps require -- & if this is still too high, how much does the 'p' or reliability of the 'event = base betting sequence' need to be (where q=1-p) for those (1-q) events to be .. to not exceed 2x in about 150-180 events or games
Lotsa varaiance there