I wish to share with you a lotto 6/49 prediction methodology, that may increase a bit the chance to guess the profitable numbers on the subsequent draw. It’s based mostly on the intervals of the numbers, e.g. the variety of attracts between two appearances of the identical quantity.

Suppose the number one seems after 7 attracts, we write 7 as first variety of the sequence, then identical no 1 comes after 8 attracts, we write 8 and so forth.

By this fashion we will construct the sequence of intervals for the number one, it appears to be like one thing like: 7, 8, 30, 3, 10, 7, 5, 2 …

The purpose is to acquire a mathematical equation, so we will construct the intervals curve, utilizing a sequence of numbers, that we already know.

For instance, utilizing the sequence 1, 2, 3, 4, 5 ….

I spent plenty of time, analysing the databases of essentially the most 6/49 lotteries, trying to find appropriate equation, to breed all intervals curves for the 49 numbers.

Under 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 next:

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 consequence shall be a curve, very near the intervals curve:

Coefficient of A number of Willpower (R^2) = 0.9874443055

Since we all know the intervals curve of the quantity until its final look, the purpose of the subsequent step shall be to aim to foretell the subsequent level of the intervals curve,

utilizing the curve we already constructed with the above equation.

Let’s examine for instance, we all know the final 10 factors of intervals curve of the number one:

it can appears to be like one thing like 1, 4, 12, 31, 1, 1, 2, 1, 2, 10

Nicely, now, lets construct our curve utilizing the above equation, giving a values for X = 1, 2, 3, 4, 5, ……… 100 000 (exemplary)

After getting this work achieved, lets evaluate the ten factors of the intervals curve with each set of 10 factors of our curve, consider the Correlation perform for each 2 in contrast units, and discover the set of 10 factors, that greatest matches the ten factors of the intervals curve.

The eleventh level of our curve will match the subsequent level of intervals curve in 3 – 7 % of all circumstances. That is positively a bit higher than random guessing, however nonetheless not sufficient to interrupt the massive home fringe of the lottery.

Have enjoyable and good luck!