Recently someone asked me.
How can I win with advantage play?
Let's say I predict VB.
I play 5 numbers sector.
Let's say in average in 7 spins I can get it only once perfectly.
For example 4 times I predict spot on (the ball hits dominant diamond and drops on rotor where expected), and 3 times wrongly.
From that 4 well predicted spins, because of ball jumps distribution only once in 4 spins I win.
So to win I would need perfect hit on rotor and perfect jump.
For example the ball hits 10 pockets in front of number I predicted and jumps 10 pockets. But it happens only every 4 spin that the ball perfectly jumps.
If all together happens once in 7 spins I am still not winning.
Playing 7 spins by 5 units is 35 units, we can say close to zero wins.
What we need to understand in this example is that when player wins he did not win with 5,10,or 50% advantage but he placed 5 units and won 35-5=30 which is 600% of what he played.
In 4 spins that he predicted correctly, he wins only one and loses 3 times. He can't win on 3 spins because the ball jumped wrongly so it did not end up inside of his 5 pockets sector. If we distribute winnings for that 4 spins he still plays with 75% advantage.
Important point here is that remaining 3 spins badly predicted are not 100% loses but it is a play with negative advantage.
Instead of -2.7% house edge the player may be getting on them -10% or -30%. It is negative bu it is still something.
For example when with perfect hit he gets ball drops at rotor 5 pockets in front of predicted number and chance 1 in 20 that the ball will jump 5 pockets from there.
If the spin was wrongly predicted and the ball dropped at rotor 15 pockets in front of prediction he may have chance of 1 : 50 that the ball will jump 15 pockets.
Surely, 1:50 may be any value, it could be anything from 1:20.. 1:30..1:40..1:100 etc
It is defined by scatter law.
An average distribution would give us 1:37 therefore 1:50 is a significant negative advantage.
Since the player plays 5 pockets sector, 1:50 hits will mean a hit in every 10 spins.
Distributed across 3 spins it means an extra 0.3 hits in 7 spins.
Now players average win every 7 spins is 1+0.3 times which makes him to profit
10.5 units, it also mean he is playing with 30% advantage.
[info border=green float=right width=502 height=352]
[center][img width=500 height=350]http://img35.imageshack.us/img35/2653/rouletteballscatterleve.jpg[/img]
The advantage is in light blue areas
In usual there is a reason why player gets spin wrongly predicted.
For example when playing traditional VB prediction may be in wrong rotation.
It could mean 10-15 pockets difference. (it depends on rotor speed)
With specific conditions skilled roulette player may optimise his play and take advantage of 2 pick point. In such case even his bad predicted spins would produce positive advantage. If we look the graph we can see that all the way from about position -8 up to +15 the player doesn't have negative advantage. (only small area around zero).
(It means if the ball drops on area across he still has chance of at least 1:37 to win)
If his VB prediction is floating in between 2 ball rotations distanced ~12 pockets it really doesn't matter to him if he predicted in one or the other ball rotation.
If the player predicts in 6th rotation to the end.
His prediction is zero the ball hits number 19 (EU wheel, ball direction CW)
The ball most of the time jumps -3 or +9 pockets (graph).
If the ball jumps -3 (1:20 chance based on data from the graph) he wins
If he plays 5 numbers sector it will happen every 4 spins.
If the player misses to predict in 6th ball rotation but instead gets prediction in 5th.
Instead of predicted number zero he will get predicted number 13, based on graphs scatter law he has a chance od 1:25 that the ball will jump 9 pockets. ( from number 19 to 13)
Based on this the player will win 2 times every 9 spins.
He will place 9x5 units=45 and get back from table 2 x 36 =74
It is 64% advantage. (perfect prediction would produce max 75% advantage)
Easier said than done.
In this example 64% advantage would be top limit. (4 spins predicted in 6th and 5 in 5th ball rotation)
Reality is that pick points with advantage are relatively narrow and prediction will never be so accurate therefore the player will not always take the best from pick point.
The point of this explanation is to show how predicting where the difference in between 2 ball rotation is equal to distance in between two pick points on scattergram can actually give benefits to a skilled player.