Collecting roulette ball jumps information

Roulette ball jumps scatter
39

STD is: SQRT(NPQ) where N=103, P is probability of winning and Q is probability of loosing.

I not know now exact name of this value in english must look , but the same calculate many players . STD is counted in sigmas and show probability that result is not random…

3STD ( sigmas ) in front is 99.9% that result is not random , that are some reason for ball stoping or falling here , 1 STD ( sigma ) chance is about 66% that result is not random…

You can count for himself what you want from chart to back side …To front we can only estimate, so calculations which are in chart are acordingly mine estimation…

So what will be usually , if i will have such results to back . Because results to back itself say nothing usually.
Must be - prediction / result this way we can calculate STD for prediction point.

40

Thanks for the explanation but you always explain something else and not what you quote.

Perhaps you could say yes or no.

But that is not SD for 18 pockets.

You write explanation what the SD is.

I am not an expert in data and probability but it is logical to me that if you look SD across 18 pocket SD value on same amount of spins would increase, it may look as you have less SD but in reality it may not be the truth, because and the mean increases more same as the amount of hits from the mean.

41

Yes i that understand - what for me is very clear - often i think that is clear also for all and some such moments in talk i simply skip…
Because of that i simply writed formula which i use for calculations and with which is possible calculate STD in any sittuation. Only marked that must remember that STD is calculated in front - so firstly prediction, then data where ball landed. Not first - collect data then look for best point and for it calculate STD . This way we get i not know what , but not what we want…

42

Thanks @forester.

This edge is if I play for 105 spins. but in reality I may play 15spins, would I get the same edge betting 15 offset?

Another thing is conditions change on the wheel, with dealer change, rotor speed… The 105 spins were taken from different dealers with varying rotor speeds. But the peak is up tbere, so long term is consistent but what about short term.

43

Finally find one article i will copy past it from other forum
Here is about finding bias but the same all is when to apply in detecting best distance.
Someone ask about detecting two most biased numbers and gave 6*37 =222 spins data . Where are in these two numbers 22 hits.

Chi square is an essential part in all statistical analysis, including tracking results in VB. Having said that, the correct way to bias spot, is to find the bias first, visually, and then from this assumption measure the Std/Chi Square. If the the Std is peak picked from random numbers its not quite valid. Laurance has recently done a very good explanation on advantageplayer.com heres a copy and paste, you will then as a bonus get the formula for
** calculating the standard deviation.**

Laurance writes:

The formula for STDEV is: SQRT(NPQ)

In this case, N = 6 * 37 = 222

I assume you are playing a single zero wheel, so the probability of winning § = 2/37, and the probability of losing (Q) is 35/37. P = .054, Q = .946

SQRT(222.054.946) = SQRT(11.34) = 3.37**

So, 3.37 hits represents one Standard Deviation (STDEV) from expectation given 222 trials and two numbers bet.

So, we now need to know what expectation is. Expectation = N * P (Number of spins * Probability of winning).

E = 222 * .054 = 11.988. Let’s call it 12. Your expectation is 12 wins in 222 trials.

You have 21 hits, so you are 21 - 12, or 9 hits above expectation.

9 / 3.37 = 2.67 STDEV from expectation.

But wait … there’s more.

The 2.67 STDEV only has meaning IF you predicted the two numbers in advance of the 222 trials.

If you just took 222 trials, and then “Peak Picked” the top two numbers, the traditional STDEV does not apply. It only applies in the case where the outcome is predicted IN ADVANCE of the sample taken.

So, how rare is it to find two numbers which, combined, have a 2.67 STDEV over 222 trials? To answer this question, we need to run a simulation. I have a simulation program, and these are the results:

I ran 10,000 games of 222 trails each on a single zero wheel. I counted all of the games where there were at least two numbers, which combined, had a STDEV >= 2.67 (e.g. there were 21 hits or more).

8,025 of those games passed. In other words, if you don’t predict the numbers IN ADVANCE, there is an 80% chance you will encounter two random numbers that, combined, will have 21 hits over 222 spins.

But, what if the two numbers are contiguous? The 80% figure applies if you pick ANY two numbers on the wheel. If you are looking at a contiguous 2 number section, this becomes a rarer event, but not that rare. The chances of a 2 number contiguous section rising to the 2.67 STDEV level is 28% (again, determined via simulation).

So, if you predicted the number IN ADVANCE of the 222 spins, the chance of those two numbers having 21 hits is about 1 in 50 (98% = 2.67 STDEV). If you peak pick the top two numbers after the fact, the chances of randomly winding up with two numbers hitting 21 times is 8 in 10. The chances of randomly finding two adjacent number >= 2.67 STDEV is about 1 in 4.

Hope this answers your question.

44

That looks correct, but the question is did you understood my point.
You calculate individual numbers SD but you want to play 18 numbers.

Calculate for 18 or 9 then you will understand why the chart is not a coincidence.

45

For 18 i calculate groups of 18 , for 9 i calculate groups of 9, so all is good . Main is that I recalculate all what will be if i will predict that group in advance

The 2.67 STDEV only has meaning IF you predicted the two numbers in advance of the 222 trials.

If you just took 222 trials, and then “Peak Picked” the top two numbers, the traditional STDEV does not apply. It only applies in the case where the outcome is predicted IN ADVANCE of the sample taken.

46

Yes , you can have 4 x same number by luck but if you look 18 and have 3SD there is so small chance for coincidence. Do your chart with 18 and you will see.

47

Otter i want to share one theoretical solution that might improve you way looking at scatter patterns.
But i can not do it in the open section because there is a RAT among us and i don’t want to teach or give him any hints about visual ballistics solutions, because later on he will claim he is the inventor of such solution.

I will add link to private section

48

Can you help me calculate scatter please?

49

Just took an average of 3 positions,
There is not much there, 11-19

50

Here’s my opinion on scatter. Collect three plots over three different time periods (different days and different dealers) if you’re looking at a weak wheel.

300 spins…minimum. If you’re not sure by then, find a better wheel.

1 Like
51

Main problem is time, when we aim 5-6 digits number as win, we can spent as much time as we need. When we aim 4 digits number also much is possible , but when our aim is 3 digits number… ok understand not aim , but possibilities :slight_smile: - our possibilities are much smaller .

If player know that his “army” can win 500 he can spent for that day , but sure not three or more.
Many of us here , not have possibilities play where we want, we must play where we can… and that of course will be weak wheel and we must look to all from this point - how to take best and much as possible , from “nothing”.

I put for himself aim , to understand, how with minimal data get near to good decission… usually that is enough to be in safe possition…

52

Main problem is time, when we aim 5-6 digits number as win, we can spent as much time as we need. When we aim 4 digits number also much is possible , but when our aim is 3 digits number

Bebediktus,

You and I tend to aim a bit higher than 4 digits though. :wink: Especially since we both travel. Most of the time three or four digits wouldn’t even cover our travel expense. That’s probably why we spend more time on the details.

53

So I posted about main problem. When aim is one, but abilities can be other. When we start 2-5$ we simply cant reach sum about what you talk.

To start from 10 times bigger sum, at least for me, looks very often - too risky. When start low and increase - have problems with dealers , if game is positive to me…

54

I think we’re kind of arguing the same but also different points.

We both travel to play, so when we play we want the highest edge possible in order to be efficient and win the most money. But when playing for small amounts of money we tend to be less thorough ( less detailed…collecting less data.)

55

I like that thought.

56

Maybe not tend, now about me - very often I know that I simply not have enough time I would like that will be some mine partner which will do initially this job. Nobody wants that do…
Really what I met - much more like to talk, than use all in practice.
So when I travel I have say payments for a day - travel itself + hotel+ food + something else. Now say I observe some wheels on the first day and detected that best wheel can give me 6% ADVANTAGE. If I spend observing it may be one more day say that advantage can become 8% if one more day maybe 9% but say I totally have 4 days. So easy calculation. Say I play the first day 10000$ bets with +600 if all is good maybe next day I can play 20 000$ bets with 1 200 and if all is good, say the third day I can play 30 000 bets with 1 800 in total 3 600$.
If I spent one more day this way I have +800$ and +1 600$ in total 2 400$ what is less than 3 600$.

And must remember that here I talk if all is good, but sometimes not all is good and I can spend the second day and on the third find that something changed !!!

Because of that I if only see advantage around 20-30% from 20-40 spins I immediately start to play, nobody knows what will be after even few hours…

20-30% initial edge usually equal 6% edge on the end…

57

So when I travel I have say payments for a day - travel itself + hotel+ food + something else. Now say I observe some wheels on the first day and detected that best wheel can give me 6% ADVANTAGE. If I spend observing it may be one more day say that advantage can become 8% if one more day maybe 9% but say I totally have 4 days. So easy calculation. Say I play the first day 10000$ bets with +600 if all is good maybe next day I can play 20 000$ bets with 1 200 and if all is good, say the third day I can play 30 000 bets with 1 800 in total 3 600$.
If I spent one more day this way I have +800$ and +1 600$ in total 2 400$ what is less than 3 600$.

For me it’s opportunity first. I travel to the ends of the earth always looking for the absolute highest edge I can find rather than attempting to grind out a win. Unfortunately I sometimes find myself doing the later when the weather grounds flights or traffic. The longer we spend on tough wheels, the more we fine tune our craft. Consequently it’s of no surprise that Bebebiktus is one of the best technical players out there. As for myself I just can’t stand playing slow games and tough wheels. Slow games for me is purgatory. If it’s a very fast paced game, then I can tolerate a slightly lower edge wheel.

My suggestion to the player earlier in the thread stands, find a better wheel if he’s doubting the scatter on his wheel because there are tons of wheels out there. Network, line up minions in other cities to help keep tabs on wheels.

58

The main problem is that I not have so big opportunities like you. I have a mind, that in Europe wheels are more new and other problem, that we here have not enough players, say wheels works not all day and I can collect data only if somebody play. That is a big problem. And maybe this is the biggest reason why I look for fast starting ways.