Mega-Million MatrixPrev TopicNext Topic

• Sedertree

  

  East Jordan, MI
  United States
  Member #2
  June 1, 2001
  675 Posts
  Offline
  Jun 26, 2002, 7:02 am

  ▲▼

  Lets see how it does on Friday.
  
  01-01-28-28-27 | 01
  03-03-30-30-29 | 03
  05-05-32-32-31 | 05
  07-07-34-34-33 | 07
  09-09-36-36-35 | 09
  11-11-38-38-37 | 11
  13-13-40-40-39 | 13
  15-15-42-42-41 | 15
  17-17-44-44-43 | 17
  19-19-46-46-45 | 19
  21-21-48-48-47 | 21
  23-23-50-50-49 | 23
  25-25-52-52-51 | 25
  
  Good Luck......../George
• RJOh

  

  100

  mid-Ohio
  United States
  Member #9
  March 24, 2001
  20,272 Posts
  Offline
  Jun 26, 2002, 11:07 am

  ▲▼

  Since numbers appear twice in a line, how do you use the matrix to come up with playable combinations?
  
  RJOh

   * you don't need to buy every combination, just the winning ones * 

         
  
• Sedertree

  

  East Jordan, MI
  United States
  Member #2
  June 1, 2001
  675 Posts
  Offline
  Jun 27, 2002, 5:55 am

  ▲▼

  Quick response, here goes:
  
  01-01-28-28-27 | 01
  03-03-30-30-29 | 03
  05-05-32-32-31 | 05
  07-07-34-34-33 | 07
  09-09-36-36-35 | 09
  11-11-38-38-37 | 11
  13-13-40-40-39 | 13
  15-15-42-42-41 | 15
  17-17-44-44-43 | 17
  19-19-46-46-45 | 19
  21-21-48-48-47 | 21
  23-23-50-50-49 | 23
  25-25-52-52-51 | 25
  
  
  It's obvious that the same digits cannot be used together to make up potential winning numbers. So If you choose digit (number) 01 for the first position then you will eliminate the 01 that is in the second position. The gold ball, 01, will not be eliminated of course.
  As you wheel from left to right keep this in mind. When you begin wheeling from the 03 as the first digit (number), then you would not include 01 as the second digit since 01 and 03 were placed together in the first set.
  At first glance you would say that there are too many numbers (combinations) to cover. This is true.
  
  However there are not as many combinations as you would expect. This is something I'm working out on paper, not 60/40 Matrix, that I want to see how it does in a live test.
  
  ........./George
• RJOh

  

  100

  mid-Ohio
  United States
  Member #9
  March 24, 2001
  20,272 Posts
  Offline
  Jun 27, 2002, 9:31 am

  ▲▼

  Thank you for your explanation.
  If I understand you correctly, the matrix is a system to wheel numbers by position. I noticed that 1 as the lowest number can never occur in the second position and 52 as the highest number can never occur at all since it is listed in the third and forth position only. Do the numbers change their position ouside the matrix?
  
  Thank you
  RJOh

   * you don't need to buy every combination, just the winning ones * 

         
  
• Sedertree

  

  East Jordan, MI
  United States
  Member #2
  June 1, 2001
  675 Posts
  Offline
  Jun 27, 2002, 2:59 pm

  ▲▼

  RJOh,
  
  01-01-28-28-27 | 01
  03-03-30-30-29 | 03
  05-05-32-32-31 | 05
  07-07-34-34-33 | 07
  09-09-36-36-35 | 09
  11-11-38-38-37 | 11
  13-13-40-40-39 | 13
  15-15-42-42-41 | 15
  17-17-44-44-43 | 17
  19-19-46-46-45 | 19
  21-21-48-48-47 | 21
  23-23-50-50-49 | 23
  25-25-52-52-51 | 25
  
  >Do the numbers change their position outside the matrix?
  
  No, You were correct in the assumption that 51 is the highest digit that can appear in the last position (5). In position 3 and 4 the 52 would be eliminated.
  
  I know it would have been less confusing to just leave out the 52 etc. from those columns. I left them in for continuity.
  
  The first/lowest set that would appear in this wheeling system would be:
  
  01-03-28-30-31 | 01
  
  The last/highest possible combination in this wheel would be:
  
  23-25-48-50-51 | 25
  
  It may get confusing to wheel these columns. I admit that I didn't try wheeling them by hand, simply because I don't know how many, possibly thousands, I'll end up with.
  
  This is just one strategy that I'm trying to workout. With hundreds of thousands of possible combinations it would be impossible for a software, or algorithm, to predict an accurate and/or reasonable number to play for the independent player. The good news is that by using various scientific methods I can create a balanced matrix that will at least get you into the correct ballpark.
  
  I am getting a copy of Visual Basic tommorrow. I'll give it a go to create a simple utility to do these calculations for me. Code has probably changed since the days of QBASIC or BASICA. I don't intend to bother my programmers with this outside project right now. I need to know if I'm on the right track first or just a dead end.
  
  Hope this helps,
  
  ............/George
• RJOh

  

  100

  mid-Ohio
  United States
  Member #9
  March 24, 2001
  20,272 Posts
  Offline
  Jul 7, 2002, 10:21 am

  ▲▼

  Sorry to read about your lost, hope you are up and running real soon. That matrix idea sounds like the parameters that I used in a program I'm working on. I establish parameters for each column or positions based on the history of the drawings. Good luck to you and I hope to hear more as you develope your ideas.
  
  RJOh

   * you don't need to buy every combination, just the winning ones *