6 – forty nine Prediction Method in Lotto

I would like to share with you a lotto 6/forty nine prediction system, that can increase a little bit the likelihood to guess the profitable numbers on the next draw. It is dependent on the intervals of the numbers, e.g. the variety of draws concerning two appearances of the same variety.

Suppose the variety 1 appears after seven draws, we write seven as very first variety of the sequence, then same variety 1 will come after 8 draws, we write 8 and so forth.
By this way we can construct the sequence of intervals for the variety 1, it looks something like: seven, 8, thirty, three, 10, seven, five, two …

The purpose is to get hold of a mathematical equation, so we can construct the intervals curve, making use of a sequence of numbers, that we currently know.
For example, making use of the sequence 1, two, three, four, five ….

I expended a lot of time, analysing the databases of the most 6/forty nine lotteries, hunting for suitable equation, to reproduce all intervals curves for the forty nine numbers.

Beneath 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 seven.29968401551873
a3 -sixteen.685835427847
a4 four.03362006820856
a5 11.7878901996141
a6 -.929875140722455
b1 -two.18308812702256
b10 two.19257627739827
b11 .646184039028009
b12 three.12875081362303
b13 -two.63990819911078
b14 -1.23445265954403
b17 1.68432237929677
b18 1.80681539787069
b19 -1.00807239478445E-02
b2 five.6223457630153
b20 1.8198683870071
b21 three.42192985805353
b22 1.86269712706211
b23 .540543148349822
b24 1.86248490223944
b25 1.22919827028682
b26 1.88811410276383
b27 .542228454843728
b28 1.85312655171971
b3 -two.93793519786925
b4 three.05516005231002
b5 four.15565748199625
b6 1.99914999103218
b7 1.42403484496882
b8 1.12315432067913
b9 .58752842233569
c1 nine.58219972192211
c10 444.826536028089
c11 -256.60103094578
c12 -590.091497107117
c13 173.815562399882
c14 -605.344333544192
c2 a hundred and sixty.540667471316
c3 235.597570526473
c4 193.064941311939
c5 -sixty nine.904752286696
c6 -eighty five.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 -sixty one.5407710429053
d7 114.003339618542
d9 -fifty eight.8664769640924
e1 233.179121249626
e10 35.3007579693589
e11 four.03405432942252
e12 -72.2717461326021
e13 a hundred and forty four.393758949961
e14 17.5110796441641
e17 -108.758489911582
e18 -sixty nine.3990298810884
e19 -114.061251203356
e2 -72.6478127424059
e20 -70.3976436552606
e21 -206.192826567812
e22 -seventy one.1614672890905
e23 -169.344108721358
e24 -72.5937280345943
e25 -205.741600763855
e26 -73.9063811523117
e27 -169.78163733803
e28 -sixty seven.941409230581
e3 154.397943691535
e4 66.3060796849447
e5 130.975552547632
e6 114.372274193839
e7 -194.072161107444
e8 sixteen.2743819539458
e9 twenty five.157528943044

If we give a values for X as 1, two, three, four, five, 6, seven, 8, nine … the Y consequence will be a curve, really near to the intervals curve:

Coefficient of Various Dedication (R^two) = .9874443055

Due to the fact we know the intervals curve of the variety till its previous visual appeal, the purpose of the next stage will be to try to forecast the next point of the intervals curve,
making use of the curve we currently developed with the higher than equation.

Let us see for example, we know the previous 10 points of intervals curve of the variety 1:
it will looks something like 1, four, twelve, 31, 1, 1, two, 1, two, 10

Perfectly, now, lets construct our curve making use of the higher than equation, offering a values for X = 1, two, three, four, five, ……… 100 000 (exemplary)

Immediately after finding this work accomplished, lets compare the 10 points of the intervals curve with each and every set of 10 points of our curve, consider the Correlation operate for each and every two when compared sets, and discover the set of 10 points, that very best matches the 10 points of the intervals curve.

The 11th point of our curve will match the next point of intervals curve in three – seven % of all scenarios. This is surely a little bit superior than random guessing, but even now not sufficient to split the enormous residence edge of the lottery.

Have exciting and good luck!

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