Evolution of roulettes advantage play


plz Example 5 and Example 2 are not same what is 1;24 advantage plz give me more explain how to play this i like to play

I prepared here five examples about different possibilities at roulette table. With each example we play with an advantage.

Example 1

We play 66 spins, number 66 doesn’t have any particular meaning it could be any number, because I started explanation with advantage of 1 in 33 hit rate, so I used 66 spins so an average hit on a single number would be 66/33=2.

Also around that figure it may be amount of spins that a player may play in casino within 2-3 hours’ time frame.
In all examples we will play $9 per spin.

Example 1 and 4, we play $9 on a single number while in other examples we distribute $9 across 9 numbers sector, so we place $1 on each number.
If we play 66 spins total amount placed on table would be 66 x $9 =$594
with an average hit rate 1:33 we would expect to win from table 2 times

2 x 36 x $9 value = $648
Place $594 get back $648, profit $648-594=$54
As you can see it is close to 10% profit which is logical since we have an advantage of about 10%.
However, because of short sample and unequal distribution of results we can expect something else.

If we look possibilities what will happen by playing single number we would 95% of time end up in between minus 0.79 and 4.79 wins. Of course it is a mathematical limit, we can’t win 4.79 times, it would be 4 or 5 times and we can’t have limit lower than zero therefore minus 0.79 is zero wins.
After calculation using possibility limits 95% of times we will end up somewhere in between winning 0 to $1550 from the table we play.

It means we will end up losing every single spin or we may profit up to $1550-$594=$956.
Surely such high deviation is not very good for an advantage roulette player.
As you can see with 10% advantage we can play 66 spins without single win. The ball may stop 3 times next to number we played but it doesn’t make anything good to us.

Example 2

Here we have same situation but this time we distribute same amount of money $9 across 9 numbers and place $1 on each.
We expect to win 2 times per number played which makes an average of 18 wins.

And again, due to uneven distribution 95% of time we will end up from 10.76 to 25.24 wins in our 66 spins played.
With 10.76 hits we will still lose since we would get from table $387, and we placed $594.
As you can see this makes distribution more even, after 66 spins 10% advantage 95% of time we will end up in our hands with $387 to $909.

Example 3

To be 95% sure that we will profit when playing with 10% advantage we would need to play same as in example 2 but about 1500 spins in total.
To have constant advantage of 10% for 1500 spins using ballistic is extremely hard.

Some may say well dealers signature on long run may show us advantage of 5-10%.
The dealer may spin 50 spins 15,16 or 17 rotations most of the time.
Next 50 it may be 16,17,18. Next 5 mostly 17, next 50 mostly 15…etc
Rotor may deviate in speed, 1-2 pockets per sec, which on spin 15 sec long makes 15 -30 pockets differences. Ball may behave differently, and it does. 50 spins ball may be jumping differently.

The player without constant observation of ball drop point has no control and no idea if he is still playing with an advantage. Results combined with randomness of ball jumps are more pattern chasing, which we never know when will start or end, or how long it will last.
With small advantage for an AP everything is harder.

This graph is form page where I was explaining Genuine Winner system.

[i]Anything above green line is advantage. If we read values we can see that pick point is only 2.85 and green line is at 2.78 so the difference is 0.07. It means that for every 100 spins played we would win 35x0.07=2.45 units. Or if we round it up, we can say every 200 spins we should profit 3 units. Theoretically it will work; you can play for 10 hours and win few bucks.

The problem is that hit rate is only 0.07 above the profitable limit. Therefore we wouldn’t be able to detect it. It will take hours to define it but until then definitely the dealer will change. He may start spinning more often slightly different amount of rotations, the ball will be replaced or to be more dirty, polish on wheel may erode, temperature or air pressure will change, the wheel will slightly move from vibrations on the table, the tilt will change etc. It means soon we may play with negative advantage.[/i]

Example 4

It is same as example 1 but this time we have about 30% advantage (hit rate 1:24).
If we play only a single number we still can end up without single win.
We will place on table $594 and 95% of times we will end up with getting from table 0 to $1943.

Example 5

It is same as example 4 but this time we distribute $9 across 9 pockets ($1 on each)
Since advantage is significant ~30% and units are distributed across 9 pockets 95% of time we will win. We would place on table $594 and we would be getting back something in between $608 and $1174.

It is important here to understand that just playing more numbers doesn’t create same effect all the time.
Someone may come in conclusion that playing 18 numbers will give us better stability and certain wins. It is not the truth since created advantage is not across 37 numbers. With each spin one area of wheel will have less chance for ball to stop there, while the other area will have more. It depends how wide area with advantage is. In usual I play 5-9 pockets. If playing wider sector some numbers we play may have significantly reduced advantage or even negative.

This may be typical created advantage.

Everything marked with green colour is where the advantage is. After analysing data if someone was playing according to this prediction and covering 11 pockets from -5 to +5 from prediction, he would play with advantage of 53%. Since this graph are results from bench test, in casino we often can’t analyse results with such details. Therefore playing 5-9 pockets will give us some space for errors if we are not perfectly positioned in area where the best advantage is.
It is important to understand how advantage for ballistic AP player is created.

To be able to have previous graph the player will need accuracy of prediction to rotor point as it is displayed on this graph. We can see that most of spins the ball drops within 10 pockets of players expectation. After ball hits the rotor it jumps based on ball scatter law, for that particular wheel and ball. When a player has something so constant to observe, he has better control over the game.

From the graph we can see that the ball most of the time hits 10 pockets in front of predicted number.
If for example for next 5-6spins the ball hits 7-8 pockets after the predicted number the player instantly knows something is happening on the wheel.
If the player observes only results where the ball stops he would need at least 100 spins to be able to react and to make necessary adjustments so he can continue playing with an advantage. Playing 100 spins and losing is not fun.

That is why I was shocked when come across roulette computers that build data graphs on that way, or when watching some salesman making videos where computer can’t constantly predict where on rotor the ball will drop.

Anyway if you read this post with understanding you would also understand that making video of 50 to 100 spins in convincing someone to buy roulette computer or VB system is pointless. :smiley:

95% of times this is our chance to end up with any profit with each of previous examples.

Example1 … 53%
Example2 … 60%
Example3 … 99%
Example4 … 64%
Example5 … 100%


Example 2 and 5 are same playing 9 pockets sector but different winning hit rate 1:33 vs. 1:24
With advantage 1:24 after 66 spins played we are 95% sure to win 608-594=14 to 1174-594=580 units.


how to set 1;24 win hit ?






Congratulation Forester, this is a very good work… very instructive.

I tried Ds for a long time and i 've seen incredible series of same distance, the hardest part is to surmount the fluctuation. I think there is no “good or magical” selection to play Ds, except that we know we have to play less chips we can. I think that the real part of randomness stay in the final jump of the ball ( and chaotic spins).

The most important is to bet numbers in the area, i use to bet split or street pré selectionned for every area, it helps for psychological aspect. For example 16-13/5-8 are not stuck on the wheel but it allows the ball to jump a bit more or less.

I had recorded spins online that looks very similar with 5 min interval; same diamond, same deviation, same pocket distance from last winning number to the new winning number. So Ds is obviously not a caretaker 's gossip, but there is many concentration that are due to randomness too.


This is completly amazing.Big thumbs up for this.I have to admit that many people sell something similiar to this,and this is even better explained.This was written in 2010 and I know the guy who is crazy popular about this kind of play (won’t name him but you can go on youtube and you will get it) and he is selling it for a quite a bit of money and it is even explained worse,and it is 2014 now.This forum is the best thing that I came across in my long journey of roulette.It is for free and even better explained and has better support.BIG THUMBS UP!Congratz!


Every roulette visual prediction except VB2 is more than 25 years old.

On new roulette wheels visual prediction holds very slim chances and there is far less opportunities where it would have some chance.


Hello Forester,
I appreciate your hard work very much and Im as a
beginner want to thank you so much that you share
your knowledge in here.I want to ask few questions
about VB…
As you say;
“Since rotor moves 10 pockets per sec, if it is the most,
reasonably steady speed we can
divide 10p/s by 7.3 pockets and we get 1.36s .
Without any problem we can adjust the timer and use
Now if we apply that time at “any” moment during the
spin on left side of knee point we will always read same
number as we are predicting in a particular ball rotation
every time.”

Suppose that ı dont have vb2 but ı have vibrating metronome.when spin started ı first clock wheel with metronome.lets say 12 p/s.second ı calculate the reference time:%15 of 12 p/s =1.8 +5.8=7.6 12/7.6=1.57 lets say 1.6second.then ı clock metronome to 1.6 second.ıf ı have tıme still :))) ı wait till tenth maybe nineth rotation before the end of spin.then when ball is passing DD ı start to count 1.6 second.then ı read the number.wait.again when ball is passing DD ı count 1.6 sn again.then ı read number.ıf this two numbers are same thats the prediction.am ı wrong?

thank you again for your kindness…


That is not how you get the reference time!
There is an equation how to calculate it but your need to have ball deceleration time difference in between rotations.
The easier way is try as in your example 1.6 you try once you read number , you try right after that and you read numbers, since VB2 claims prediction can be in “any ball rotation” then if your reference time is right you should be reading same number.
By the way, I do not teach VB2 or any VB outside of VB2 forums section.


thanks for the reply forester…

ıf we observe one particular rotation which starts at DD and make one full revolution at exactly 1200 ms and instantly we start our timer how much remaining time will be (you can say at least average time) till drop?if its very changable because of ball deceleration times can significantly change between rotations and spins especially losing linearity later in the spin how can we predict fall of point with measuring two three or five rotation speeds early in the spin with such device?


by the way ım sorry for asking about vb2 but lets say as if ı asked about traiditional vb method :))


VB doesn’t use any device,

but if you timing 1200 ms rotation and when the ball takes ~1200ms for rotation you know that after that it will travel fixed amount of rotations. Fixed amount of rotations has it’s time to be completed, for example 10 sec.
Then 10s/rotor time will give you how much rotor will move in remaining time.
VB assumes the ball will drop at particular diamond so the drop point is constant.

Understanding roulette wheel and exploiting possibilities

[quote=“forester, post:53, topic:538”]VB doesn’t use any device,

but if you timing 1200 ms rotation and when the ball takes ~1200ms for rotation you know that after that it will travel fixed amount of rotations. Fixed amount of rotations has it’s time to be completed, for example 10 sec.
Then 10s/rotor time will give you how much rotor will move in remaining time.
VB assumes the ball will drop at particular diamond so the drop point is constant.[/quote]

Please more about that.
Well, how do you recognize the collision deflector? Say you have 4 vertical deflectors and 3 of them get hit.


Let’s say we take times of last full ball rotation and compare them with ball drop diamond in other words we clock last full ball rotation and predict drop point.
We may end up with following chart;

For example we measure ball time 2200ms and the ball hits diamond at dd 12 so it doesn’t travel any longer and we can say it hits diamond (DD) 0 . We can see same with shorter ball times until the ball is 2120ms when instead of diamond 0 it hit DD3. We can see it as a transition phase where similar ball speeds hit DD0 or DD3.

2070ms ball hits DD6 marked as red, however you can see only 2 hits surrounded by dd3 hits so it is very unstable hits to DD6. This is reality even if we measure the time in very last ball rotation. Such instability for us means that we simply cannot predict such isolated hits, and we might be better of predicting all spins with such ball speed range, as a diamond 3.
We do not predict in last ball rotation, we predict earlier, and the earlier we predict the ball is exposed longer to various factors which can affect the ball traveling distance.

Here you can see wider mistakes, for example green lines representing diamond 3, indicate that the diamond 3 gets hit more often even when the ball times match diamond 9 hits or diamond 0.
1135 to 1095 ball times resulting in almost equal amounts of hits in between diamonds 0 and 3. It means if we clock ball time 1110 it has same chance to hit DD0 or to travel ¼ rotation longer and hit DD3.
The Acrobat 4 program corrections define chart as this. With blue, green and purple triangles I marked the areas and approximate accuracy of predictions if such ball speed range is clocked.

Red collared DD6 speed range is not marked because most likely it would be predicted as DD3 or DD9 and that is the reality because most of the time such ball speeds did end up at DD3,9. Even only 2 spins from 40 hit DD6 in some occasions the system still may predict it as DD6, because it analyzed data taking many factors in consideration.
Explanation basically means even if we precisely measure time of last ball rotation as 2100 sometimes such ball may hit different diamonds. It happens more at sped ranges where there is a border in between diamonds. When we measure times earlier such mistakes increase, we do not measure time so precisely, and the ball simply doesn’t travel each spin on exactly same way.

Level wheel theoretically has equal distribution on all diamonds. If we look a wheel with 4 vertical diamonds we can say that each one gets 25% of hits. From perspective of borders in between diamonds each diamond has 2 borders. For example diamond at 6 o’clock has border with diamond 3 and diamond 9. If only 5% on each side is predicts wrong diamond (one earlier or later) it leaves us with only 25%-5-5=15% good predictions per diamond or 60%. (it is only an estimate)

On the picture above we have example data from a strongly tilted wheel. From 1050-1200 we have 100% accuracy. It is already 75% accurate. From 1050 to 1010 which is narrow ball speed range but we see instability and hits to various diamonds.

Now I will refer to Laurance Scott and his explanation in the book ” Professional Roulette PredictionVolume 2 Advanced Methods published in 2005.

Laurance calls it Compensating Rotor Speeds.

“This occurs when a wheel has multiple dominant fall-off points, which is fairly typical (usually two but sometimes three, fall-off points.)”

…….etc. Then continues…

“Assuming the secondary point is one quote wheel beyond the primary point and it takes the ball 1.5 seconds to travel to the distance of this one quote wheel. If the rotor speed is such that it will compensate by traveling three-quotes of the wheel in the same 1.5 seconds, the relative position of the wheel will be the same for the second fall-off point. In other words it doesn’t matter if the ball falls from the first or second fall-off point. The ball will still strike the same section of the wheel if the wheel is traveling at the optimum speed that fall-off pattern. “

So how that relates to our chart?
Say we clocked ball 1030ms in that unstable area and it is predicted to DD0, but the ball travels all the way to DD9 (3/4 rotation longer). Until the ball travels such distance it takes about 1.5s, for that time the rotor moves and if the rotor is slower, the rotor position will be correct even the ball hits DD9. From this we can conclude that actually there is no border in between DD0 and DD9 on one side, because regardless which diamond the ball hits the ball will drop on the predicted number.

There is only a border in between dd9 and DD12 which is same as dd0 but one rotation later. Based on above chart the player accuracy would increase when predicting the ball drop point about 75% plus ½ or 25% = 87.5% .

We can also look at it that the ball hits only 2 diamonds as n the following picture.

The border in between DDO and DD9 at 1100ms is not important if the rotor is slow.
To show the important border which creates mistakes I extended chart beyond one full ball rotations. You can see marked border in between DD9 and DD12 which is actually DD0 but one rotation later. That is the only place where the player would have mistaken in predictions.

If we apply same 5% rule we used when explaining level wheel mistakes since there is only one border it affects 5% dd9 and DD12 making predictions 90% accurate. That is the reason why playing tilted wheel with 1 or 2 dominant diamonds is much easier and results are better.

Just a small note to explain level wheel mistake a bit better. Level wheel is predicted on a different way. Tilted wheel is predicted more as a digital, one diamond or the other, but the level wheel is predicted more as an analogue with results in between. It helps to reduce significant mistakes.


Very interesting thanks for sharing ! I wish i could think like this :slight_smile:
I am a begin :slight_smile: ner casino player


This forum is impressive.


Manged to fix lost images in this post

However for this post

images were at imageshark and they removed them.
Only for the first post i recreated image.