Author:Mark Anthony Howe  1994

Many people have attempted to beat the game of roulette Physics integrated into computers. The task is a little more insurmountable than one may first assume, in that not only do you have to take human errors of timings taken into consideration, but also the mathematics involved for varying degrees of tilt of the wheel. Early attempts tried the use of primitive Polynomials, that although gave reasonable accuracy using exact data as the seed for the calculations, taking timings of the ball and rotor and using human intervention would introduce errors that needed to be overcome. Air pressure and temperature was another factor.

In 1983 I designed a program based on Roulette Physics that ran on a PSION Series 1 Pocket Computer with less than 32 K of memory I was able accurately predict with an increased edge of over 38% consistently. To this day I still have an habitual tendency to dabble with alternative methods of roulette prediction and have several different methods of successfully gaining an edge in the game of roulette. I also have an abundance of technical gadgetry that I have designed myself in order to pull off the perfect heist time and time again covertly and under the ever scrutinising eyes of the pit bosses. From completely in the ear canal hearing aids to transmitter pens, pocket vibrators, we have a full arsenal of weapons in the war against the casinos. I am currently still involved with helping people beat the game of roulette and now reside in the UK.


What was needed was a method where the equations could be set up as you play, therefore being flexible and coherent to the actual game in play that could be constantly changing every 20 spins or so.


Firstly a study was made by myself and Formulas invented and tried to replicate the exact data that was observed empirically for level and tilted wheels.


Here is some early experimental roulette physics, some of the formulas are invented, but lend themselves to the problem at hand.


Some equations describing the ball behaviour on a tilted wheel have been invented.


I am very content that the virtual experimental results resemble what we can see in the real world!


This research has been carried out with a purpose to understand what causes the tilt-like behaviour and what we can expect from this in Roulette Physics.

Enjoy it!

Roulette Physics

Roulette Physics and Mathematics for beating Roulette

The following mathematics deals with the X,Y and Z axis within the confines of a Roulette Wheel environment.


Y axis¦N1¦*COS(a)-(mg)*COS(a)=0


X axis¦N2¦+¦N1¦*SIN(a)+¦mg¦*sin(@)*COS(Y)=m*¦@ centre)=m*V^2/R=m*[Y’)^2]*R


V=Linear Velocity

R=Ball Track Radius

@=Centripedal acceleration


Z axis¦Ffr¦+¦Air Drag¦=m*¦@tan¦=m*Y”*R


Friction Force a This is negative as it is opposing the Z axis


Air Drag is the force that is equal too:


¦Air Drag¦= – 0.5*CD*P*TT*r^2*V^2 (TT is pie) this is also a minus value!


CD is Drag Coefficiaent

P is AIr density

r is the balls Radius


Z axis is always tangentially directed.




After some very simple Algebraic Transformation and incorporating the above formulas we get the next differential equation:

More Roulette Physics


Y”=(a+air*R)*(Y’)^2=b*SIN(Y)+c*COS(Y)+d (*)




a is the determining friction factor(Ie 0.004)


Air =-[0.5*CD*P*TT*r^2*V^2]/m








The ball movement sters to this equation only till the moment when it loses the contact with the vertical side of the ball track or:




So the Drop off condition is:


[(Y’)^2]*R+g*COS(@)*tg(a)-g*SIN(@)*COS(Y)=0 (**)


Now lets introduce some real values into the equations and see the predicted results:





a=16.7.TT/180 inner slope of stator


r=^-3 Radius of ball


m=9.10^-3 Mass of Ball

(a)= 0.004 Friction factor for rolling between the ball and the track

@=0.8 grad Tilt Angle


t0= 0 sec

t1=30 seconds


These values determin the time interval of 30 sec since the start of spinning!


I also calcualted the time the ball loses contact with the vertical wall of the ball track, this is when(**) becomes true!


Time till drop off is 17.04 seconds


By this time the ball passes 4935 Grad or 13.7 revolutions from start point!


At this moment the ball has a velocity of 2.7 Rads/Sec or 0.43 Revs per/Sec!


Now lets build a graph Y=Y(YO).


If we change the point from where we spin the ball, in ideal level wheel conditions (When the tilt is absent) the ball travels the same velocity, no matter where the ball is spun from along the circumfrence of the wheel, if it is spun with the same initial velocity, hence:


Y=YO+some constant value


Thats if the tilt angle in degrees is the same and we change the YO from Ograd to 360grad at a given initial velocity 2*pie/0.7, then we get the next picture of Y expressed in Gradients.



Y=Y+4870.53 Grad


Now lets tilt the wheel with a severe tilt angle of 0.8 grad, then we would have the graph shown below. It is obviously non-linear and also has a big leap up between points B and C.

In general they are all showing just a transformation of any point from which we spin the ball into a point where it drops off. We can see that the majority of spins will result in falling into a narrow sector of the wheel. This is clearly illustrated in Figure#4.

Look at Figures #1- #4


The probability of the ball dropping off in the drop off sector is:


P=100%*Grey Area/(White Area+grey Area)


in this case;



The steep gradient of the graph(on the interval of the initial angles lying between B and C, where there is a sharpe prolonging of the distance which the ball travels) is due to some interesting behavior of the ball around the point where it is ready to drop off.


Lets spin the ball from the point with an angle coordinate equal to;


YO=312grad(Around the middle between B and C)


Then we have the picture of the angle velocity shown in the graph. When the ball is nearly losing its velocity up to the drop off velocity and its contact with the track is almost lost, it manages to pass the “Peak of the hill” and begins to accelerate due to traveling downhill from the Apex.


Its velocity increases which in turn contributes to increase centrifugal forces against the ball track side until it approaches the Peak of the hill yet again, where it loses its orital velocity finally and proceeds to head towards the rotor after leaving the tracks banking.


This can produce increases of distances of ball travel in the region of 0.7 revolutions and greater.





The modeling has shown that the greater the tilt, the narrower the resulting sector.


The narrowing drop off sector results in hitting some particular diamonds which are located AFTER the sector (as the ball still has to pass some distance from the ball track to the diamond), spiralling inwards towards the rotor.


The modeling also allows investigating the influence of different parameters on various tilts of wheel. One important note to make is that we can reach the same degree of tilt behavior bu either increasing the tilt angle, or by decreasing the friction arounf the ball track. This also means that roulette wheels with less frictional surfaces need less tilt in order to produce very good narrow sectors which can be used in the prediction of the outcome.


I have also found that A ‘Critical Tilt Angle’ exists on most wheels, which greatly increases the narrowing of the final predicted sector, that is the intervals between B and C sharply decreases. I found in my research that the the ‘Critical tilt angle’ was around 0.35 grad.






We are only going to deal with physical size,but other parameters are the elasticity of the composite, frictional surfaces.

“Some secondary thoughts about the influence of the ball size on the real distance offset on the rotor.”

Let’s simplify the movement of the ball along the rotor pocketed surface to a simple moving of a body on which some braking or friction force acts. We can assume that the main part of this force is due to the Nx (the horizontal part of the vector {N}). {N}-reaction force from the pocket divider.

So, the braking force: Fbrk=k*Nx, k-is just secondary coefficient showing the proportional dependence between Nx and Fbrk.

Using general formulas for deceleration movement we get:

dS=-d(V^2)/(2* k*Nx/m)

after taking integrals:

S=V0^2/(2* k*Nx/m),

V0-initial relative velocity between the ball and the rotor when the ball enters the rotor,

S-distance which the ball manages to travel till it finds rest. (as we have simplified the situation so in real it would be the most frequent on average distance which determines the peak on the real distance offset histogram)

Nx=|{N}|*sin(Y), the more Y the more hitting energy redistribute in favor of horizontal braking of the ball.

R*cos(Y)=R-H gives cos(Y)= (R-H)/R and sin(Y)= (2*R*H-H^2)/R^2



If we increase the ball radius R at given H(pocket divider height) then we see that the distance S will increase.

Say, we have two balls: 21mm and 18mm diameter. R=10,5mm and R=9mm

The height of the pocket divider is 5mm(as my wheel has). H=5mm

Then finding the relative change of the averaged travelled distance with changing the balls (leaving V0 and H unchanged) we obtain that the bigger ball travels about 9% farther than the smaller one.

I think it’s worth taking into account when we define the offsets.

Contact : Mark Anthony Howe

Author :  Mark Anthony Howe Copyright 2013 ©

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