Hello People.

I would like to share with you a lotto 6/49 prediction method, that can raise a bit the probability to guess the winning numbers on the next draw.

It is based on the intervals of the numbers, eg the number of draws between two appearances of the same number.

Suppose the number 1 appears after 7 draws, we write 7 as first number of the sequence, then same number 1 comes after 8 draws, we write 8 etc.

By this way we can build the sequence of intervals for the number 1, it looks something like: 7, 8, 30, 3, 10, 7, 5, 2 ...

The goal is to obtain a mathematical equation, so we can build the intervals curve, using a sequence of numbers, that we already know.

For example, using the sequence 1, 2, 3, 4, 5 ....

I spent a lot of time, analysing the databases of the most 6/49 lotteries, searching for suitable equation, to reproduce all intervals curves for the 49 numbers.

Below is the equation:

Y = a + a3*sin(a4 + c1*cos(b1*X+e1) + d1*sin(b2*X+e2) + c2*cos(b3*X+e3) + d2*sin(b4*X+e4)+c3*cos(b5*X+e5) + d3*sin(b6*X+e6)+c4*cos(b7*X+e7) + d4*sin(b8*X+e8)+c5*cos(b9*X+e9) + d5*sin(b10*X+e10)+c6*cos(b11*X)+e11 + d6*sin(b12*X+e12)+c7*cos(b13*X+e13) + d7*sin(b14*X+e14))+a5*cos(a6 + c9*cos(b17*X+e17) + d9*sin(b18*X+e18) + c10*cos(b19*X+e19) + d10*sin(b20*X+e20)+c11*cos(b21*X+e21) + d11*sin(b22*X+e22)+c12*cos(b23*X+e23) + d12*sin(b24*X+e24)+c13*cos(b25*X+e25) + d13*sin(b26*X+e26)+c14*cos(b27*X+e27) + d14*sin(b28*X+e28))

The parametters values are the following:

a 7.29968401551873

a3 -16.685835427847

a4 4.03362006820856

a5 11.7878901996141

a6 -0.929875140722455

b1 -2.18308812702256

b10 2.19257627739827

b11 0.646184039028009

b12 3.12875081362303

b13 -2.63990819911078

b14 -1.23445265954403

b17 1.68432237929677

b18 1.80681539787069

b19 -1.00807239478445E-02

b2 5.6223457630153

b20 1.8198683870071

b21 3.42192985805353

b22 1.86269712706211

b23 0.540543148349822

b24 1.86248490223944

b25 1.22919827028682

b26 1.88811410276383

b27 0.542228454843728

b28 1.85312655171971

b3 -2.93793519786925

b4 3.05516005231002

b5 4.15565748199625

b6 1.99914999103218

b7 1.42403484496882

b8 1.12315432067913

b9 0.58752842233569

c1 9.58219972192211

c10 444.826536028089

c11 -256.60103094578

c12 -590.091497107117

c13 173.815562399882

c14 -605.344333544192

c2 160.540667471316

c3 235.597570526473

c4 193.064941311939

c5 -69.904752286696

c6 -85.770268955927

c7 276.721054209067

c9 -374.987916855954

d1 100.982005590423

d10 -51.7169119126939

d11 -59.0688708086887

d12 -55.7217141421084

d13 -51.5930580430944

d14 -64.5398179559034

d2 -173.685727106493

d3 -28.8164892769008

d4 -141.058426244729

d5 -172.520435212672

d6 -61.5407710429053

d7 114.003339618542

d9 -58.8664769640924

e1 233.179121249626

e10 35.3007579693589

e11 4.03405432942252

e12 -72.2717461326021

e13 144.393758949961

e14 17.5110796441641

e17 -108.758489911582

e18 -69.3990298810884

e19 -114.061251203356

e2 -72.6478127424059

e20 -70.3976436552606

e21 -206.192826567812

e22 -71.1614672890905

e23 -169.344108721358

e24 -72.5937280345943

e25 -205.741600763855

e26 -73.9063811523117

e27 -169.78163733803

e28 -67.941409230581

e3 154.397943691535

e4 66.3060796849447

e5 130.975552547632

e6 114.372274193839

e7 -194.072161107444

e8 16.2743819539458

e9 25.157528943044

If we give a values for X as 1, 2, 3, 4, 5, 6, 7, 8, 9 ... the Y result will be a curve, very close to the intervals curve:

Coefficient of Multiple Determination (R^2) = 0.9874443055

Since we know the intervals curve of the number till its last appearance, the goal of the next step will be to attempt to predict the next point of the intervals curve,

using the curve we already built with the above equation.

Let's see for example, we know the last 10 points of intervals curve of the number 1:

it will looks something like 1, 4, 12, 31, 1, 1, 2, 1, 2, 10

Well, now, lets build our curve using the above equation, giving a values for X = 1, 2, 3, 4, 5, ......... 100 000 (exemplary)

After getting this work done, lets compare the 10 points of the intervals curve with every set of 10 points of our curve, evaluate the Correlation function for every 2 compared sets,

and find the set of 10 points, that best matches the 10 points of the intervals curve.

The 11th point of our curve will match the next point of intervals curve in 3 - 7 % of all cases.

This is definitely a bit better than random guessing, but still not enough to break the huge house edge of the lottery.

Have fun and good luck!