patterns

ok link spread=http://www.mediafire.com/view/?37ak8m0p09bxtq6

Hello, developer, aims to reduce to 24 vtrac, and make the statistics over the 24 vtrac
  Of course, that once converted will Figures 12, but may be hybrid 4 and a pair vtrac normal
Hello, I need your help in The Following
1st step = see the use of pairs of digits repeated without vtrac
* So is 176 pairs, but in the post above half in pairs of digits with no repeated vtrac
This list must see the use of these pairs,
2nd step, see what this position, eg in the last draw, the pair vtrac without repeating digits (attention, each digit of own peers and not the digits Within the vtrac) this has to differentiate, we have two distinct pairs itself without repeating digits Within and the couple, now is about the couple, or Their worth naming, not the digit in the pair vtrac icewynd then, in the last draw the pair vtrac Gave 1 and 5th position in next
* It will not be in the same position, may be one, then you can easily dial in your planília this development, for each position in the vertical position, delays and see this, it is easy to do! f pudesemos get the system to have ambos vtrac hit numbers,
* We only need 3 vtrac to match all 6 numbers from 49/6, but it is almost impossible
hit every two to vtrac numbers, but some hits vtrac if the two numbers is That We Should take to make conbinçao hybrid, the normal pair and 4 vtrachybrid, the normal pair and 4 vtrac

Last 3 draws from Ontario 49:

14.11.2012 B      3,10,21,22,28,42 CONVERSION           Vtracs  B = 3,3,10,17,21,22

11.17.2012 B =  4,21,22,23,26,32                             VTRAC B= 1,4,7,21,22,23

11/21/2012 B= 3,4,18,24,40,46                                 Vtracs B = 3,4,15,18,21,24
* 1st draw par 03, 21 in position 1 and 5th respectively without repeating the initial and final digits
** 2nd draw 04 23 in the 1st and 4th position without repeating digit in the pair
3rd draw 04 18 in the 2nd and 3rd position without repeating digit
* So icew, must dismember a list alongside the sweepstakes are 15 positions in pairs of pairs 1,2 .... up to 5.6 = 15 positions
I have a list of 276 normal peers without the filters of repeated digits, will further reduce, oh let's enjoy the standard of normal peers without repetition
And ainsa puts outro4 vtrac in their upright positions

Hello, developer, if you want to use the concept to a new program pairs the vtrac
  You can do ok!! ie, you can create a better solft to see the statistics of pairs vtrac

Hi Dr San

Sorry I have not replyed, been very busy getting a Windows 8 compatible update online for my Lottery Statistic Analyser.

Intresting spreadsheet you sent me, with very good use of the math functions in Excel.

I will study it this coming week and see what I can do for you.

Thanks

Jamie

Hello, developer, thank you for seeing this post, you can use the concept in and make a little program, instead of macros or make a solft, you can use the 6 or hybrid vtrac
With normal pair, then the lottery 49/6 was rezuumida in 24 pairs of vtrac, but we have the problem of the position of the pair of numbers vtrac, 03.27 vtrac example, can give the two numbers, or can the number 03 position in , and 27 elsewhere, that we have to see