If we take scatter when ball hits rotor until ball stops.

Then if we take data, prediction vs ball hits rotor.
And each amount for any position hit on rotor we multiply with all possible scatter from that position we should end up with data of advantage.

Example,

If prediction €“ rotor hit looks as follows. Each of this x has chance to end up on any of pockets, but chance to end up on some distances is higher then average. So if for position -18 we apply 4 x all scatter result then for position -17 we apply 8x but we shift it by 1 and last cell we transfer to start€¦etc , it would be interesting what we would get. Any volunteers good in XL?

If we take scatter when ball hits rotor until ball stops.

Then if we take data, prediction vs ball hits rotor.
And each amount for any position hit on rotor we multiply with all possible scatter from that position we should end up with data of advantage.

Example,

If prediction €“ rotor hit looks as follows. Each of this x has chance to end up on any of pockets, but chance to end up on some distances is higher then average. So if for position -18 we apply 4 x all scatter result then for position -17 we apply 8x but we shift it by 1 and last cell we transfer to start€¦etc , it would be interesting what we would get. Any volunteers good in XL? :-*

-18 xxxx
-17 xxxxxx
-16 xxxxxx
-15 xxxxxxx
-14 xxxxxxxxxxxxx
-13 xxxx
-12 xxxxxx
-11 xxxxxxxxxxxx
-10 xxxxxxxxxxxxxxxxx
-9 xxxxxxxxx
-8 xxxxxxxxxxx
-7 xxxxxxx
-6 xxxxxxxxxxx
-5 xxxxxxxxxxxxxx
-4 xx
-3 xxxxxxxxxxxxx
-2 xxxxxxxxxxxx
-1 xxxxxxx
0 xx
1 xxxxx
2 xxxxx
3 xxx
4 xxx
5 xxxxx
6 xxxxxxxx
7 x
8 xxx
9 xxxx
10 xxx
11 xxx
12 xxx
13 xx
14 xx
15 xxx
16 xxxxxx
17 x
18 xxxx

Email

[color=blue]I’m an quantative analysis professional and I command better analysis tools than Excel, for really serious statistical analysis. So I could probably help you if you need some calculations on your data.

However, I don’t quite understand what you need… Anyway, this is as I understand it, confirm or clarify if you want:

1. Your x-graph is a frequency diagram based on data recorded from experiments.

2. The column marked -18 to 18 signifies the distance, in number of pockets, between the winning number and the number closest to the ball-to-wheel impact point.

3. The number of x:es shows how many times the number that many pockets away won. (I.e. the ball hit the wheel at pocket marked “0”).

[This spread caused by the jumping of the ball is called “the scatter” and is genuinely random and unpredictable due to chaotic behavior]

1. You then say something about “shift by 1 and last cell we transfer to start” and “Multiply by all possible scatter” which I don’t really understand.

Do you want to smooth out the observed scatter to a long-term statistically stable distribution? That I could do. That’s a common kind of task (and a whole science too!)

It is surpricsng that the “local differences” in frequency are so big:
-14 has thirteen hits.
-13 only has four hits.

That must be coincidental (and even that seems unlikely, are these data really true observations?). No pocket can anywhere close to three times as probable as its neighbour. In the long run they might differ one or two “x:es”, no more. If you that is what you want, I could calculate the long run probability distribution based on any set of observed data.

Hello,
You should post this on forum, I want some action going.

Yes I did not put it well all together.

My objective is to multiply (prediction €“ ball hits rotor) with scatter (rotor hit €“ ball stops)
That graph is prediction-ball hits rotor.

To any of those hits we can give chance of scatter and we will get final result by adding to every rotor hit same possibility of scatter. I have somewhere in boys club data of scatter (ball hits rotor- ball stops). I can probably dig it out if you willing to do it. So if scatter is done on 50 spins on position -18 we would add it 4 times. For position €“ 17 we would also add it by amounts of hits but shifted by 1 position€¦etc. On the end final result would be a graph with about 2500 chances. I think that would give us better picture then looking only 50 spins prediction-ball stops.

1. Your x-graph is a frequency diagram based on data recorded from experiments.
Yes it will be, this one is only sample. Real one looks even better.

2. The column marked -18 to 18 signifies the distance, in number of pockets, between the winning number and the number closest to the ball-to-wheel impact point.

No it is distance from prediction to position where the bal hits rotor.

1. The number of x’s shows how many times the number that many pockets away won. (I.e. the ball hit the wheel at pocket marked “0”).

It shows how many times the ball hits rotor at position distanced by so many pockets from predicted number.

1. You then say something about “shift by 1 and last cell we transfer to start” and “Multiply by all possible scatter” which I don’t really understand.

Yes, if we have chart in XL we need to apply scatter from each of positions on graph that I gave, by amount how many x’s is on that position.
So if we start with -18 we apply it 4 times. With -17 we do it 6 times (amount of x’s) where -18 of scatter will be at -17 of my graph and position of +18 of scatter will be multiplied by -18 of my graph. Further more on position -16 we have also 6 hits so we would add 6 times scatter from that position.

1x
2xxx
3xxx prediction rotor hit
4xx
5xx

1x
2xx
3xxxx scatter from rotor hit to final stop
4xxxx
5xx

1x€¦…oooooo€¦zzzzzzzzzzzz
2xx€¦.ooo€¦zzzzzz
3xxxx€¦.oooooo€¦zzz
4xxxx€¦.oooooooooooo€¦zzzzzz
5xx€¦.oooooooooooo€¦zzzzzzzzzzzz

Simple example, “o” is where on first graph we add second graph but from position 2. “Z” is from position 3.

If we do 50 or 100 spins prediction final result we do not get much.
For each rotor hit number we get 2-3 possibilities of scatter. But from same amount of spins we can reasonably well format scatter graph and if we multiply it with rotor hits we can have better picture.

[b]Ok, this is result for 148 spins, if I did not miss any.
Used DVD recorded spins, live clocking should be better
System set for accuracy 1
Rotor clocking = full rotation
System missed to predict `10%
Spinners included, but few spins where ball spins without hitting diamond are excluded

Result= distance in between prediction and position where the ball hits rotoron leveled wheel

[/b]

-18 xxx €¦€¦€¦. 3
-17 €¦€¦€¦. 0
-16 €¦€¦€¦. 0
-15 xx €¦€¦€¦. 2
-14 x €¦€¦€¦. 1
-13 x €¦€¦€¦. 1
-12 x €¦€¦€¦. 1
-11 €¦€¦€¦. 0
-10 x €¦€¦€¦. 1
-9 €¦€¦€¦. 0
-8 x €¦€¦€¦. 1
-7 x €¦€¦€¦. 1
-6 x €¦€¦€¦. 1
-5 €¦€¦€¦. 0
-4 xxxx €¦€¦€¦. 4
-3 xx €¦€¦€¦. 2
-2 xxxxxx €¦€¦€¦. 6
-1 xxx €¦€¦€¦. 3
0 xxxxxxx €¦€¦€¦. 7
1 xxxxxx €¦€¦€¦. 6
2 xxxxxx €¦€¦€¦. 6
3 xxxxxxx €¦€¦€¦. 7
4 xxxxxxxxxxx €¦€¦€¦. 11
5 xxxx €¦€¦€¦. 4
6 xxxxxxxxxx €¦€¦€¦. 10
7 xxxxxxxxxxxxx €¦€¦€¦. 13
8 xxxxxxxxxxxxx €¦€¦€¦. 13
9 xxxxxx €¦€¦€¦. 6
10 xxxxxxxx €¦€¦€¦. 8
11 xxxxxxxx €¦€¦€¦. 8
12 xxxxxxxx €¦€¦€¦. 8
13 xxxx €¦€¦€¦. 4
14 xx €¦€¦€¦. 2
15 xxxx €¦€¦€¦. 4
16 xxxxx €¦€¦€¦. 5
17 €¦€¦€¦. 0
18 xx €¦€¦€¦. 2
Total €¦€¦€¦. 148 151 Positions, 6,7 and 8 i counted wrongly by 1 x when i made total 148

This is probably the best as it can be, when I find time I will do same but in accuracy 3 mode. But first I want to apply to any position hit, chances of scatter across the wheel.

Wow Forester

I think that calculates out to a 1:19.4 spin hit rate and a 59% profit on Turnover.

Mike.

Yes, something as that, but we still do not have jumps across rotor.
I am confidant that hits before pick point mostly will bounce more then hits after.
But if we apply average to all we still can have idea.

This is good hit rate and when scatter is applied it will probably drop down or maybe will keep narrower pick point.
I am trying to demonstrate how important is to have common drop point on rotor.
Without that if any advantage shows it is probably only coincidence.

Please click on image to enlarge it.

On left side is same graph from before but I moved data in center.

On the right are same spins but different kind of prediction.
First one was where prediction is projected after the diamond.
Second one where it is up to the diamond.
What is better it is hard to tell.
Chart one is more stable while chart 2 can be very unbalanced.
And that is what was happening when I was clocking spins.

First 100 was really narrow pick point with huge advantage.
But next 50 spins half were hitting same spot but another half was missing and going around position +9 on graph. Before those 50 spins there was almost nothing around position 9 on graph 2. That kind of prediction would require fixed clocking point.
It is easy for me at home to see what is going on but in casino it would be hard to follow. Spins can divide in between 2 points 9-12 pockets distanced, but if scatter also has 2 picks this may produce huge advantage.

Previous 2 graphs are showing how well prediction is in relationship to where the ball will hit rotor.

Next graphs are showing prediction to final result where the ball stops.
How much of advantage we will have after accurate prediction is dependant on how wide is scatter across rotor. Data on the right side is something new for me and I will definitely look more in to it.

-4 xxxx €¦€¦. 4 -3 xx €¦€¦. 2 -2 xxxxxx €¦€¦. 6 -1 xxx €¦€¦. 3 0 xxxxxxx €¦€¦. 7 1 xxxxxx €¦€¦. 6 2 xxxxxx €¦€¦. 6 3 xxxxxxx €¦€¦. 7 4 xxxxxxxxxxx €¦€¦. 11

Forester,

I am still confused by your wording.
In the above excerpt from your ‘graph’ are you saying that you predicted accurately ‘7 times’ the precise point where the ball will first meet the rotor before it starts its scattering?

It is correct, but it is not the best way to look that graph.
Position zero is only where prediction is but the ball can hit anywhere from that position.
What counts is where the ball will most likely hit from prediction on particular wheel where system I applied.
On the graph positions 7 and 8 are most important.
It means that the ball most likely will hit rotor 7-8 pockets from number that system gives as prediction.

-8 x €¦€¦. 1
-7 x €¦€¦. 1
-6 x €¦€¦. 1
-5 €¦€¦. 0
-4 xxxx €¦€¦. 4
-3 xx €¦€¦. 2
-2 xxxxxx €¦€¦. 6
-1 xxx €¦€¦. 3
0 xxxxxxx €¦€¦. 7
1 xxxxxx €¦€¦. 6
2 xxxxxx €¦€¦. 6
3 xxxxxxx €¦€¦. 7
4 xxxxxxxxxxx €¦€¦. 11
5 xxxx €¦€¦. 4
6 xxxxxxxxxx €¦€¦. 10
7 xxxxxxxxxxxxx €¦€¦. 13
8 xxxxxxxxxxxxx €¦€¦. 13

Example of perfect prediction

-2
-1
0
1
2
3
4
5xxxxxxxxxxxxxxxxx
6
7
8
9

Here the ball will hit every single spin 5 pockets from prediction. Notice there is zero hits where prediction is but system still produces huge advantage. All that is needed is to add 5 pockets from whatever system predicts, if there is ball bouncing then add average of it on top.