Honduras Member #20982 August 29, 2005 4715 Posts Offline

Posted: November 23, 2009, 2:30 am - IP Logged

Does anybody knows how to do it? but i am not talking about individual numbers but the entire combination...The reason why i ask is because if you could make it into a algebraic graph or function graph you could theoretically work out an equation that will intercept the graph at a particular time and know what the numbers where going to be and win....You only need to win the lotto ONCE, not twice or three times...

Also if you could make each individual number of the combination into a graph i think it will be easier to write an EQUATION OR FORMULA graph that will follow the lotto graph and meet/intercept it at multiple points and this way win...

i can not think of a way to do it...

The Forex trades: 1.6 Trillion dollars EVERY day, that´s more than the GDP of the Carribbean Central America, COMBINED. Enough to feed every crook out there for centuries...To all Geniuses & Powers Countries of the World the Planet needs breakthroughs in all Medicine, Veterinary, Biology related fields, Psychology, Population Psychology/Sociology..They need to genetically ingeneer new plants species/types to give more variety of plants and thus have more resources for combating diseases¨

mid-Ohio United States Member #9 March 24, 2001 18220 Posts Offline

Posted: November 25, 2009, 2:55 pm - IP Logged

There's software designed to make graphs and pie charts, all you have to do is enter the data in a form it understands.

* you don't need more tickets, just the right ticket * * your best chance at winning a lottery jackpot is to buy a ticket * "I will magically reveal the winning numbers after the drawing"

Honduras Member #20982 August 29, 2005 4715 Posts Offline

Posted: November 25, 2009, 9:33 pm - IP Logged

Quote: Originally posted by RJOh on November 25, 2009

There's software designed to make graphs and pie charts, all you have to do is enter the data in a form it understands.

yeah i know but i dont know how to feed it in a lotto combination form, it normally asks for 1 or 2 number input but the lotto is a string of numbers and i was saying it so you all understand and you all make the formula i dont want the formula is for you all...it will be nice if you all share...Something i was thinking and you just gave me this idea is: what if we make the lotto a 3 dimensional graph and write a 3 dimensional equation/formula graph to intercept it at a particular time...The only problem is that you might only get 3 numbers for the 3 axis yxz [remember this from my calculus book], or i got another idea what if we make the first 2 numbers a coordinates of the graph and the other 2 numbers the coordianates of the grap but if the equation/formula graph intercepts it at one point it will only be 2 numbers...That's the problem...I am talking graphically...maybe it could intercept it at more than just the 1 point and intercpet it at the 2 points...Also i was thinking if you make it a pick6 then it will be 3 coordinates the equation graph might be able to intercept it at 2 points but at 3 is very rare but i can try...but seen other graphs is probably only going to intercept it at 2 numbers....

The Forex trades: 1.6 Trillion dollars EVERY day, that´s more than the GDP of the Carribbean Central America, COMBINED. Enough to feed every crook out there for centuries...To all Geniuses & Powers Countries of the World the Planet needs breakthroughs in all Medicine, Veterinary, Biology related fields, Psychology, Population Psychology/Sociology..They need to genetically ingeneer new plants species/types to give more variety of plants and thus have more resources for combating diseases¨

Honduras Member #20982 August 29, 2005 4715 Posts Offline

Posted: November 25, 2009, 9:48 pm - IP Logged

if i can make it a trigonometric graph or like the unit circle it will be easier to write a formula/equation for it...and i just thought of something that i think is unheard of...You know how the unit circle in trigonometry forms a function graph? trigonometric function graphs are normally in 1 dimension...what if the trigonometric function graph had 3 dimensions or what if the trigonometric unit circle had a 3d dimension normaly is written in 2 dimensions what if we add z to it instead of the usual y and x...You know something you will not find this in college texbooks if this type of math exist if it doesn't exists it has been right under your noses and they had not been able to see it...I for sure have not seen such a textbook....

I am talking about the unit circle having a 3rd dimension like a sphere....What it trigonometry itself had was in 3-d...

The Forex trades: 1.6 Trillion dollars EVERY day, that´s more than the GDP of the Carribbean Central America, COMBINED. Enough to feed every crook out there for centuries...To all Geniuses & Powers Countries of the World the Planet needs breakthroughs in all Medicine, Veterinary, Biology related fields, Psychology, Population Psychology/Sociology..They need to genetically ingeneer new plants species/types to give more variety of plants and thus have more resources for combating diseases¨

Honduras Member #20982 August 29, 2005 4715 Posts Offline

Posted: December 27, 2009, 2:06 am - IP Logged

I realized something after looking at a calculus book...You could make the combination represent a line/graph and each subsequent drawing combination be the continuation of that line/graph so that each lottery will have 1 particular line/graph in a 3 dimensional field...Not what i was talking about but is good to know...

The Forex trades: 1.6 Trillion dollars EVERY day, that´s more than the GDP of the Carribbean Central America, COMBINED. Enough to feed every crook out there for centuries...To all Geniuses & Powers Countries of the World the Planet needs breakthroughs in all Medicine, Veterinary, Biology related fields, Psychology, Population Psychology/Sociology..They need to genetically ingeneer new plants species/types to give more variety of plants and thus have more resources for combating diseases¨

Miami United States Member #62793 July 9, 2008 669 Posts Offline

Posted: December 27, 2009, 2:11 am - IP Logged

Quote: Originally posted by pumpi76 on November 23, 2009

Does anybody knows how to do it? but i am not talking about individual numbers but the entire combination...The reason why i ask is because if you could make it into a algebraic graph or function graph you could theoretically work out an equation that will intercept the graph at a particular time and know what the numbers where going to be and win....You only need to win the lotto ONCE, not twice or three times...

Also if you could make each individual number of the combination into a graph i think it will be easier to write an EQUATION OR FORMULA graph that will follow the lotto graph and meet/intercept it at multiple points and this way win...

i can not think of a way to do it...

Uh oh.....

"...a chance to push everything aside, the circumstances that've controlled our lives, and do it our way now. Good, bad or otherwise. You'll maybe get lost in it, tied up in it a little bit, but if you work your way through that the real you shows up, I think. Maybe what's at your core deep down, maybe that comes out. Maybe that's what it's about." Mike Pace

United States Member #83701 December 13, 2009 225 Posts Offline

Posted: December 29, 2009, 6:18 pm - IP Logged

Quote: Originally posted by pumpi76 on November 23, 2009

Does anybody knows how to do it? but i am not talking about individual numbers but the entire combination...The reason why i ask is because if you could make it into a algebraic graph or function graph you could theoretically work out an equation that will intercept the graph at a particular time and know what the numbers where going to be and win....You only need to win the lotto ONCE, not twice or three times...

Also if you could make each individual number of the combination into a graph i think it will be easier to write an EQUATION OR FORMULA graph that will follow the lotto graph and meet/intercept it at multiple points and this way win...

i can not think of a way to do it...

I've been giving some thought to this even though I don't believe an effective system could be developed due to the fundamental assurance of a random draw. As numbers, each ball isn't completely independent of each other, for one drawing a number for pick1 excludes that from the possibilities for pick2 then there's the problem that the databases do not list the numbers in the order drawn just in their numeric order, ie.: what we have recorded as P1 isn't P1 but is just the lowest value of the picks. I suppose you could use each pick as a position on a axis to get a single number ie.: Number = P1*35^4 + P2*35^3 + P3*35^2 + P4*35^1 + B but this would make the bonus ball the least significant value and it would be sparse as many values are just not possible since each pick excludes possibilities for the next plus the picks are not in draw order so each pick is also bounded by the value of the previous pick. Whether you define P1 or the B ball the most significant digit will determine what types of function fitting will predict what best. One possibility given the arbitrary draw order might be Number = (P1-1) + (P2-P1-1)*(35-3) + (P3-P2-1)*(35-3)*(35-2-P1) + (P4-P3-1)*(35-3)*(35-2-P1)*(35-1-P2) + (B-1)*(35-3)*(35-2-P1)*(35-1-P2)*(35-0-P3). This would have the effect of making the bonus ball the most significant value and the space continuos with the possible sequences but reversing the calculation would be difficult, you would have to to the following: P1 = Number % 35 + 1; Number = INT( Number/35 ) ; P2 = Number % (35-2-P1) + P1 + 1 ; Number = INT( Number/(35-2-P1)) ; P3 = Number % (35-1-P2) + P2 + 1 ; Number = INT( Number/(35-1-P2)) ; P4 = Number % (35-0-P3) + P3 + 1 ; B = Int(Number/(35-0-P3)). This would have the problem that the number system actually changes with each sequence and hence it would be difficult for the bonus ball to be predicted reliably, it would probably be best to simply have the bonus ball be handled separately or have the bonus ball as the least significant digit with P1 being the second least ie.: Number = B + P1*(35) + P2*(35)*(35) + P3*(35)*(35)*(35-2-P1)... etc., but it does solve the continuity problem.

So far, this is the only way I can think of to incorporate the two known constraints on the randomness of the numbers, that each consecutive pick excludes numbers already drawn and the sequence is recorded in numeric order not draw order for each sequence of however many picks it is.

You'll note that it's essentially taking each pick as a plane or axis but because of the constraints, the scale of each axis is being packed. Perhaps the approach should be to stretch out the scale by normalizing to a constant base rather than use a variable base... That would give us something like Number= B + P1*35 + ((P2-P1)/(35-2-P1)*35)*35^2 + ((P3-P2)/(35-1-P2)*35)*35^3 +((P4-P3)/(35-0-P3)*35)*35^4. The problem with this is that it's like taking reality and stretching it across a frame but not stretching it evenly everywhere.

If you don't mind having a sparse mapping and simply ignoring continuity, you could just use a cryptographic hash such as ROT13 or 32bit CRC or MD5 but then any kind of function fitting or decomposition in order to make predictions could result in number sequences that just aren't possible and can't be reversed.

Mind you, once you have a way of mapping the sequences into a numeric value, you could then do autocorrelation, fourier transform decomposition, fractal decomposition, wavelet processing and whatever else you may want to try. Simple polynomial fitting and line fitting probably won't be a good idea but if you took the differential and then took the differential of a differential and continued doing so till you got a constant value then that would mean you could line fit, problem is that probably won't happen, you'll probably never get to a constant value.

United States Member #83701 December 13, 2009 225 Posts Offline

Posted: December 29, 2009, 6:56 pm - IP Logged

Quote: Originally posted by jwhou on December 29, 2009

I've been giving some thought to this even though I don't believe an effective system could be developed due to the fundamental assurance of a random draw. As numbers, each ball isn't completely independent of each other, for one drawing a number for pick1 excludes that from the possibilities for pick2 then there's the problem that the databases do not list the numbers in the order drawn just in their numeric order, ie.: what we have recorded as P1 isn't P1 but is just the lowest value of the picks. I suppose you could use each pick as a position on a axis to get a single number ie.: Number = P1*35^4 + P2*35^3 + P3*35^2 + P4*35^1 + B but this would make the bonus ball the least significant value and it would be sparse as many values are just not possible since each pick excludes possibilities for the next plus the picks are not in draw order so each pick is also bounded by the value of the previous pick. Whether you define P1 or the B ball the most significant digit will determine what types of function fitting will predict what best. One possibility given the arbitrary draw order might be Number = (P1-1) + (P2-P1-1)*(35-3) + (P3-P2-1)*(35-3)*(35-2-P1) + (P4-P3-1)*(35-3)*(35-2-P1)*(35-1-P2) + (B-1)*(35-3)*(35-2-P1)*(35-1-P2)*(35-0-P3). This would have the effect of making the bonus ball the most significant value and the space continuos with the possible sequences but reversing the calculation would be difficult, you would have to to the following: P1 = Number % 35 + 1; Number = INT( Number/35 ) ; P2 = Number % (35-2-P1) + P1 + 1 ; Number = INT( Number/(35-2-P1)) ; P3 = Number % (35-1-P2) + P2 + 1 ; Number = INT( Number/(35-1-P2)) ; P4 = Number % (35-0-P3) + P3 + 1 ; B = Int(Number/(35-0-P3)). This would have the problem that the number system actually changes with each sequence and hence it would be difficult for the bonus ball to be predicted reliably, it would probably be best to simply have the bonus ball be handled separately or have the bonus ball as the least significant digit with P1 being the second least ie.: Number = B + P1*(35) + P2*(35)*(35) + P3*(35)*(35)*(35-2-P1)... etc., but it does solve the continuity problem.

So far, this is the only way I can think of to incorporate the two known constraints on the randomness of the numbers, that each consecutive pick excludes numbers already drawn and the sequence is recorded in numeric order not draw order for each sequence of however many picks it is.

You'll note that it's essentially taking each pick as a plane or axis but because of the constraints, the scale of each axis is being packed. Perhaps the approach should be to stretch out the scale by normalizing to a constant base rather than use a variable base... That would give us something like Number= B + P1*35 + ((P2-P1)/(35-2-P1)*35)*35^2 + ((P3-P2)/(35-1-P2)*35)*35^3 +((P4-P3)/(35-0-P3)*35)*35^4. The problem with this is that it's like taking reality and stretching it across a frame but not stretching it evenly everywhere.

If you don't mind having a sparse mapping and simply ignoring continuity, you could just use a cryptographic hash such as ROT13 or 32bit CRC or MD5 but then any kind of function fitting or decomposition in order to make predictions could result in number sequences that just aren't possible and can't be reversed.

Mind you, once you have a way of mapping the sequences into a numeric value, you could then do autocorrelation, fourier transform decomposition, fractal decomposition, wavelet processing and whatever else you may want to try. Simple polynomial fitting and line fitting probably won't be a good idea but if you took the differential and then took the differential of a differential and continued doing so till you got a constant value then that would mean you could line fit, problem is that probably won't happen, you'll probably never get to a constant value.

Ooops forgot the -1 to account for the numbers starting with 1 in the last possible transformation, ie.:

Number = (B-1) + ((P1-1)/(35-3)*35 +((P2-1-P1)/(35-2-P1)*35)*35^2 + ((P3-1-P2)/(35-1-P2)*35)*35^3 + ((P4-1-P3)/(35-0-P3)*35)*35^4

To reverse you would do:

B = Number % 35 + 1

Number = INT( Number/35)

P1 = round((Number % 35)/35 * (35-3) + 1)

Number = INT( Number/35 )

P2 = round((Number % 35)/35 * (35-2-P1) + 1)

Number = INT( Number/35 )

P3 = round((Number % 35)/35 * (35-1-P2) + 1)

Number = INT( Number/35 )

P4 = round((Number % 35 )/35 * (35-P3) + 1)

Despite the variable density of the mapping, this is probably the best approach, it has the Bonus number as the least significant and P1 as the second least significant so it's essentially having the values with the least constraints being the most variable components of the value, however it does stretch the volatile portions of the subsequent numbers to fill the entire space of the other planes so I suspect that they do would be equally random so I doubt a continuous function could serve to predict the values reliably.