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Some basic math formulas applied to pick3/pick4

Topic closed. 13 replies. Last post 3 years ago by lottologix.

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pavizlo$'s avatar - binary
OKC, OK
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Posted: May 4, 2013, 2:50 am - IP Logged

Some basics math formulas applied to Pick3/Pick4

First, let use some abbreviations:

Let call the 3 digits of the current (today’s) drawing as Xc, Yc, Zc ( the index c means current). Xc is the first digit, Yc the second digit, and Zc the third digit.

Let call the 3 digits that will be drawn next (tomorrow’s) as Xn, Yn, Zn (the index n means next). Xn is the first digit, Yn the second digit, and Zn the third digit

Let call “M” the mirror.

M(Xc) means “mirror of the first digit of the current drawing.

M(Yc) is the mirror of the second digit of the current drawing.

M(Zc) is the mirror of the 3rd digit of the current drawing.

Using the above definitions the following equations are true and can be verified using any state pick3 drawing results:

Xn = M(Xc) + K1.   K1 is a variable digit that can be any of these 0,1,2,3,4,5,6,7,8,9.

Yn = M(Yc) + K2.   K2 is a variable digit that can be any of these 0,1,2,3,4,5,6,7,8,9.

Zn = M(Zc) + K3.   K3 is a variable digit that can be any of 0,1,2,3,4,5,6,7,8,9.

This method gives a straight hit. The most tedious problem is choosing the digits K1, K2, and K3.

Here an example to illustrate the above formulas using Oklahoma pick3

04/30/13

               
   

5

   
   

6

   
   

9

   

Winners

04/29/13

               
   

8

   
   

7

   
   

6

   

Winners

Xc= 8;  M(Xc)=M(8)= 3

Yc=7;   M( Yc) =M(7)=2

Zc=6;  M(Zc)=M(6)= 1

Xn=5

Yn=6

Zn=9

Now, let apply the formulas

Xn= M(Xc)+K1; 5=3+K1 here we can see our K1=2

Yn= M(Yc)+K2; 6=2+k2 here our K2=4

Zn=M(Zc)+K3;  9=1+K3 here we can see that K3=8

These formulas work for any state pick3/pick4, the main problem is the choice of our variable digits K1, K2, and K3.

The same concept can be applied to pick4 by adding the the 4th equation.

GOOD LUCK EVERYONE

    helpmewin's avatar - dandy
    u$a
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    Posted: May 4, 2013, 8:13 am - IP Logged

    Thanks for sharing Smiley

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      Georgia
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      Posted: May 4, 2013, 8:21 am - IP Logged

      Thank you for sharing!  May you have a blessed and prospers day.  Sun Smiley


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        Posted: May 4, 2013, 12:42 pm - IP Logged

        Sorry, I accidentally gave you one star. I meant to give you 5.

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          NASHVILLE, TENN
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          Posted: May 4, 2013, 8:37 pm - IP Logged

          Would using your equations work equally as well if we were to discard the Xn, Yc, and so on and just go with the K1, K2, & K3 numbers?

          Just pick the correct K number and make your wager.

            pavizlo$'s avatar - binary
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            Posted: May 4, 2013, 9:59 pm - IP Logged

            Would using your equations work equally as well if we were to discard the Xn, Yc, and so on and just go with the K1, K2, & K3 numbers?

            Just pick the correct K number and make your wager.

            Thanks, Gasmeterguy, at least you said something positive on discarding the xn, yn, zn and just pick the correct k.

            GOOD LUCK EVERYONE

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              Kentucky
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              Posted: May 4, 2013, 10:12 pm - IP Logged

              "the main problem is the choice of our variable digits K1, K2, and K3."

              And you said K1, K2, and K3 are "a variable digit that can be any of these 0,1,2,3,4,5,6,7,8,9." It looks like you're asking for help choosing one of ten digits in all three digit positions.

              I'm not trying to destroy your idea and many of us want to help, but the odds of finding the variable digit are the same as matching any three (or four) digit number.

                pavizlo$'s avatar - binary
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                Posted: May 4, 2013, 10:30 pm - IP Logged

                "the main problem is the choice of our variable digits K1, K2, and K3."

                And you said K1, K2, and K3 are "a variable digit that can be any of these 0,1,2,3,4,5,6,7,8,9." It looks like you're asking for help choosing one of ten digits in all three digit positions.

                I'm not trying to destroy your idea and many of us want to help, but the odds of finding the variable digit are the same as matching any three (or four) digit number.

                Thanks Stack47. By studying some drawing history, i noticed that most of the K values were in the range of 0 to 4. Once again this range will depend for each state.

                GOOD LUCK EVERYONE

                  AlgorithmGuru's avatar - avatar
                  Pittsburgh, PA
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                  Posted: May 29, 2013, 6:42 am - IP Logged

                  Some basics math formulas applied to Pick3/Pick4

                  First, let use some abbreviations:

                  Let call the 3 digits of the current (today’s) drawing as Xc, Yc, Zc ( the index c means current). Xc is the first digit, Yc the second digit, and Zc the third digit.

                  Let call the 3 digits that will be drawn next (tomorrow’s) as Xn, Yn, Zn (the index n means next). Xn is the first digit, Yn the second digit, and Zn the third digit

                  Let call “M” the mirror.

                  M(Xc) means “mirror of the first digit of the current drawing.

                  M(Yc) is the mirror of the second digit of the current drawing.

                  M(Zc) is the mirror of the 3rd digit of the current drawing.

                  Using the above definitions the following equations are true and can be verified using any state pick3 drawing results:

                  Xn = M(Xc) + K1.   K1 is a variable digit that can be any of these 0,1,2,3,4,5,6,7,8,9.

                  Yn = M(Yc) + K2.   K2 is a variable digit that can be any of these 0,1,2,3,4,5,6,7,8,9.

                  Zn = M(Zc) + K3.   K3 is a variable digit that can be any of 0,1,2,3,4,5,6,7,8,9.

                  This method gives a straight hit. The most tedious problem is choosing the digits K1, K2, and K3.

                  Here an example to illustrate the above formulas using Oklahoma pick3

                  04/30/13

                                 
                     

                  5

                     
                     

                  6

                     
                     

                  9

                     

                  Winners

                  04/29/13

                                 
                     

                  8

                     
                     

                  7

                     
                     

                  6

                     

                  Winners

                  Xc= 8;  M(Xc)=M(8)= 3

                  Yc=7;   M( Yc) =M(7)=2

                  Zc=6;  M(Zc)=M(6)= 1

                  Xn=5

                  Yn=6

                  Zn=9

                  Now, let apply the formulas

                  Xn= M(Xc)+K1; 5=3+K1 here we can see our K1=2

                  Yn= M(Yc)+K2; 6=2+k2 here our K2=4

                  Zn=M(Zc)+K3;  9=1+K3 here we can see that K3=8

                  These formulas work for any state pick3/pick4, the main problem is the choice of our variable digits K1, K2, and K3.

                  The same concept can be applied to pick4 by adding the the 4th equation.

                  I understand the idea behind this and if the "trend" you said later about the k digit being USUALLY 0-4 could be beneficial.  However I'm curious if you have tried this theory by moving the mirror so to speak.  Technically a mirror is assigned arbitrarily.  Perhaps there is a "universal" idea of what a mirror is and how its constituted, in reality a mirror is just a form of a cypher and moving the digits from one place to another since the numbers still run consecutively.  In which case you could move the numbers or attempt a different form of the mirror (for instance 0-9 and 9-0 respectively).  I would then attempt to confirm through whatever method you used to confirm your original theory based on the new mirror.  However I would have to say that in doing so if you can prove that both mirrors work then I would say the theory doesn't hold enough weight, so you'd in fact be discrediting the idea.  Also you didn't explain what to do in cases when the drawn number is less than the mirrored number for instance if the mirror of a draw is 7 and the next number drawn is a 5 what do you add to 7 to get to 5?  Does the addition wrap around so to speak?  Or by this theory is the mirror number always less than the draw?  In which case I would say what happens in the case of 000 being drawn (or if you use the 0's as tens, then what about 111?)  Especially 111 because the only non wrap around addition that would work on that is in the case of 000 being the mirror in which case a triple digit (in any mirror) would always preceed a draw of 111 (unless the wrap is included) - which can easily prove this theory false.  In other words a wrap is just (x + y) modulo 10.  ->  However that's not how the formula is written just looking for clarification.  The reason I say a lot of this is because when I used to play around with math in high school I was always trying to find shortcuts to long handing lots of calculations in algebra by finding how the numbers were interrelated.  I discovered two things, in SOME cases there are definitely relationships between numbers (as in functions, formulas etc.) that can translate quickly from a set of numbers to the appropriate answer.  I also discovered that in MOST cases (if not all) you can find a relationship between numbers that is arbitrary.  What struct me as ironic is that k can be any digit between 0-9.  I had to read the idea a few times to understand why this was.  It looks to me as if you can assign a variety of formulas to lottery numbers using the digits 0-9 that if the answer falls within the constraints of being within 0-9 will always hold true and verifiably and accurately.  The problem is that its just a redistribution of the same problem with extraneous calculations that bring one no closer to a solution within a given budget, or profitably at that.   HOWEVER I do like the idea if the constraints can be controlled to cull the numbers so to speak.  For instance if you could define K:  Such that K is an integer value between 0 and 4 (or hell even 5 for that matter) Then I would say you've effectively eliminated half of the possibilities and that is something indeed extraordinary.  But being that k is a value between 0-9 (and potentially the formula is:

                  Xn / Yn / Zn = (M(Xc) + K1)mod10 / (M(Yc) + K2)mod10 / (M(Zc) + K3)mod10

                  all that is being said is that if you add some integer between 0-9 to another integer in the range 0-9 and calculate the modulo of it by 10 then you will come up with a number in the set 0-9 inclusive.  Which is like saying that all the numbers drawn will be in the range of 0-9.  Which by definition of the lottery (daily number) we know to be true.  I hope this doesn't come off as an attack.  I really love hearing the mathematical ideas of lottery lovers and I say all this for two reasons.  The main one being its always possible that I'm not reading the idea correctly or that something integral to the idea was left out that I haven't included in my understanding and I would love to hear it, if that is the case.  The second reason is, that if I was pursuing an idea, a mathematical breakdown or process and it was incorrect for some provable reason, I would hope someone would say so.  At least then I can redirect my energies to other methods that bear more fruit.  So if I have missed something in my analysis of this particular method please let me know.  If there is more information that you are willing to share regarding this method then please let me know.  If you found my analsys helpful and accurate, then my apologies on finding the fault, however I  hope to hear new ideas that you may have.   It is a great idea in one respect, finding a way to alter then number in such a way that a smaller subset of numbers can be used to find the next draw is an interesting idea that I'll mull over.  Thanks for the post. 

                  Quick afterthought... Something else that would be extraordinary would be if there wasn't a K1, K2, and K3 but merely K.  If you could add a single integer to a formula that maniuplated each number then in effect your odds would drop to one in ten.  You see by adding K1, K2, K3 you do have the same odds, however since the numbers are separate then the odds are one in ten X one in ten X one in ten or effectively 10*10*10 = 1000 which is the odds of picking any 3 digits as another poster said.  However if K's odds are applied universally across all three numbers then you have reduced the odds to a mere 1/10.  Good luck to all.

                  If it's not consistent, it's irrelevant :)

                    pavizlo$'s avatar - binary
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                    Posted: May 29, 2013, 11:40 pm - IP Logged

                    Great positive analysis Guru i like your idea,this can be used to redefine the above theory. 

                    Thank you.

                    GOOD LUCK EVERYONE


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                      Posted: June 8, 2013, 12:24 pm - IP Logged

                      Thanks for the idea. I don't see the reason for using the mirror instead of just using Xn and Xc, its just a static exchange to me. Perhaps you can clarify.

                      Anyway, FYI a quick check of CA pick 3 in over 1500 draws (just using Xc-Xn etc seems to produce k1=k2=k3 delta only 0.63% of the time.

                      The other thing is  if we average k1,k2,k3 for these, they are all =0, meaning that the +/- k's all average out for those draws.

                      Indicating that it's almost perfectly random. Crying

                        olplugger's avatar - moon2
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                        Posted: June 9, 2013, 11:02 am - IP Logged

                        Thanks so much for sharing!  Thumbs Up

                        "Whatever the human mind can conceive and believe it can achieve."

                        Napoleon Hill, Author of "Think & Grow Rich"

                                                                                                                                                               

                          chasb123's avatar - Lottery-062.jpg
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                          Posted: July 5, 2013, 1:27 pm - IP Logged

                          Thanks for the insight, you brought me right back where I first started. .. WOW

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                            Posted: July 11, 2013, 10:11 am - IP Logged

                            In a way I agree but in a way I also disagree. The problem I see is that you are using individual numbers.  I have noticed that I tend to get more accurate numbers if I use boxed. Whole numbers. Picking digits tends to be allover the place. I am just. Not sure how I would make boxed mirrors to test it. I am just not that familiar with using mirror numbers because until this post I have seen no value in it. Ya lear
                            n something new every day:)