How to calculate roulette advantage using ball jumps chart and prediciton data?

[anonymous3]

How to calculatte roulette standard deviation, the best answer

To answer this question I will first calculate SD for each number hit. [Standard%20Deviation] From that we get: Probability to win for a single number on roulette would be 1/37 Probability to lose for a single number on roulette would be 36/37 or (1 - Probability to win = 1 - 1/36 ) Trails is 120 spins. From that we get SD = SQRT (1/36 x 36/37 x 120) SD = SQRT (3.15) SD = 1.77 The Mean, (expected hits on each number for even distribution) for 120 spins on 37 numbers is 120/37 = 3.24 L…

Here we calculated possible advantage based on ball jumps scatter. That would be truth if we can spotlessly predict each time number on which the ball would drop. In other words it would be a maximal advantage we can get based on ball jumps distribution without altering the jumps based on any other factors.

https://rouletteplace.s3.dualstack.eu-central-1.amazonaws.com/uploadsrp/original/1X/1ba30e3626c319bcf6af8a4d82c8b7b794f3d52c.png

On the chart is ball jumps distribution for 120 spin.

And if for 100 spins we get ball drops in relationship to our prediction on this way;

Which advantage to expect when combining jumps chart with prediction and how to combine it?

[bebediktus]

@anonymous3

Which advantage to expect when combining jumps chart with prediction and how to combine it?

We can in some way combine these two charts, but if that will be what we can expect in reality i not know. Most chances - that not , because jumps from predicted diamond is not the same as jumps from not predicted diamond, jumps from DD after predicted is not the same as jumps from before predicted diamond.Here if simply we add 12 jump to every prediction, but as i said - not think that it will be near to reality

Best variant is when we predict actual number and mark how far real result is from prediction point

[anonymous3]

Then do we need jump charts for ball hits predicted D, chart ball hits earlier D and the chart when ball hits later D? If we predict ball drop using visual prediction we can’t have enough accurate information about ball speed.
If we predict using a roulette computer and if we have 4 vertical diamonds on the wheel we can have better information on which diamond the ball should drop and with which velocity.

@bebediktus

because jumps from predicted diamond is not the same as jumps from not predicted diamond, jumps from DD after predicted is not the same as jumps from before predicted diamond.

What if on some wheels they are the same?

[anonymous3]

@bebediktus

Better information, than what ?

It was compared to visual prediction, just a sentence earlier. VB has accuracy ~ 1 rotation, prediction often is based on overlap as 3/4 of rotation while RC in such case can definitely better decide. Point of comment is not what is better and for who, but is it worth using different scattergram for different ball speeds hitting the diamond.

[forester]

It may not be the truth but it may be truth in most of the cases that the ball from each drop would behave with similar jumps scatter.

For that reason each ball drop point needs to have applied jumps chart multiplied by amount the ball drops at that point. Than add all results together for each jump divide by amount of spins in jump chart and we get chart as on the picture.

Black chart indicates total advantage for the player based on jumps and quality of predictions defined on 2 white charts on the left. It shows about 20% advantage over the casino.

The chart for prediction was from -18 to +18 but the data in the chart in XL is used as from 0-36 therefore for the exact answer all data on the black chart needs to be shifted by -18 pockets.

Here I will attach XL file you can use to add your own data.

You can add data only to cells light blue and green color, other cells are locked for protection.

Prediction-Jumps-mul.xlsx (39.7 KB)

[bebediktus]

@forester

For that reason each ball drop point needs to have applied jumps chart multiplied by amount the ball drops at that point. Than add all results together for each jump divide by amount of spins in jump chart and we get chart as on the picture.

This part is not clear. Looks like you repeat the same jump chart from each ball hit place, so is like sum of 100 charts data. If so then not clear why

divide by amount of spins in jump chart
What gives this operattion ?

Ok i look to all that more simple - possible simply mark final result, from some point - from these do chart or do some math elements and know which place is best to bet.

Prediction must point to that “best place” . There are no other variant to win only, bet that place which will be best in future.
Enough for that to use something from past or not… who knows … but anyway aim must be - predict place which will be best in future.

[bebediktus]

@anonymous3

but is it worth using different scattergram for different ball speeds hitting the diamond.

Worth is all, nothing what you will use will make things worse, but i not imaginge how you can use

scattergram for different ball speeds hitting the diamond

I think that you too much want.
Firstly you must understand which edge can be in some ideal situation. Maybe that maximal edge is not enough for you

[forester]

@bebediktus

Looks like you repeat the same jump chart from each ball hit place,

Yes if the ball dropped say +4 pockets 5 times then you apply jump chart from that position 5 times.

@bebediktus

divide by amount of spins in jump chart

It gives you chart based on spins you predicted not collected amount of spins for ball jumps jumps.
That way you can also see the advantage.

[bebediktus]

@forester

It gives you chart based on spins you predicted not collected amount of spins for ball jumps jumps.

@forester

add all results together for each jump divide by amount of spins in jump chart

Still not clear what you add and what divide. For me looks more logical to add all variants of jumps for every predicttion…

Looks like you try shift all based to predicttion point, but something here is not good, because predicttion must be diamond and number under it say in hit moment.

So if ball hit correct so 0 and it was 5 times we must on that 0 add 5 times all chart meanings - so how long was jump, but you do something other…

[bebediktus]

@anonymous3

If we predict using a roulette computer and if we have 4 vertical diamonds on the wheel we can have better information on which diamond the ball should drop and with which velocity.

Better information, than what ?
I think are much cases, when without RC have better information. Say 7 vertical diamonds and 90% hits to two adjacent. In such situation can be that RC only will rise problems…

I often see that you about RC talk like about some professional player. RC is not professional player and it do only what YOU enter to it. Remember - YOU.

So in most cases RC play the same like play player - if player play good - RC helps much, but if not…then sorry, but it can do much damage…

[forester]

@bebediktus

So if ball hit correct so 0 and it was 5 times we must on that 0 add 5 times all chart meanings - so how long was jump, but you do something other…

No, I do as you described.

here i made only few predictions as 6,6,6,6,6,6,1 when jumps are applied you see orange line at 1 , it is an average of 37 spins used, that way you can see how much you are above expected.