6-49 Prediction method in Lotto
I would like to share with you a 6/49 lottery prediction method, which can slightly increase the probability of guessing the winning numbers in the next draw. It is based on the intervals of the numbers, for example, the number of draws between two occurrences of the same number.
Suppose the number 1 appears after 7 draws, we write 7 as the first number in the sequence, then the same number 1 appears after 8 draws, we write 8, etc.
In this way we can build the sequence of intervals for the number 1, it looks like: 7, 8, 30, 3, 10, 7, 5, 2 …
The objective is to obtain a mathematical equation, so that we can construct the interval curve, using a sequence of numbers, which we already know.
For example, using the sequence 1, 2, 3, 4, 5 ….
I spent a lot of time going through the databases of most 6/49 lotteries, looking for the right equation, to reproduce all the interval 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 parameter values are as follows:
to 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 values for X as 1, 2, 3, 4, 5, 6, 7, 8, 9 … the result of Y will be a curve, very close to the interval curve:
Multiple determination coefficient (R ^ 2) = 0.9874443055
Since we know the interval curve of the number until its last occurrence, the objective of the next step will be to try to predict the next point on the interval curve,
using the curve that we already built with the previous equation.
Let’s see for example, we know the last 10 points of the interval curve of number 1:
it will look something like 1, 4, 12, 31, 1, 1, 2, 1, 2, 10
Well now, let’s build our curve using the above equation, giving values for X = 1, 2, 3, 4, 5, … 100,000 (exemplary)
After doing this work, let’s compare the 10 points on the interval curve with each set of 10 points on our curve, evaluate the correlation function for every 2 sets compared, and find the set of 10 points that best matches the 10 points of the interval curve.
The eleventh point on our curve will coincide with the next point on the interval 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 lottery edge.
Have fun and good luck!
