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two diagonals

Topic closed. 2 replies. Last post 5 years ago by dr san.

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bgonçalves
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June 9, 2010
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Posted: August 29, 2012, 5:14 pm - IP Logged

Hello, who can help me create a macro to see it = 'Example is a lottery of 50/5, but can be used for various lotteries 39/5 49/6 60/5 etc. = After several tests and applying triangulation filters, which were more faithful to each other were what I call the diagonal 1-48, 6-49 and diagonal endings which then I shall explain who is who. However it must be said that the terminations of the filter functions as a filter applied within the diagonals 1-48 because each diagonal 1-48 has only two terminations.
Let's face it: In this picture we have 13 diagonals that make up the diagonals of 1-48, 1-48 baptized because of it starts at 1 and ends at 48. If fixed, these each diagonal contains only two terminations. Diag.1 - 01 Diag.2 - 02:07 Diag.3 - 03:08:13 Diag.4 - 04.09.14.19 Diag.5 - 05.10.15.20.25 Diag.6 - 06.11.16.21.26.31 Diag.7 - 12.17.22.27.32.37 Diag.8 - 18.23.28.33.38.43 Diag.9 - 24.29.34.39.44 Diag.10 - 30.35.40.45 Diag.11 - 36.41.46 Diag.12 - 42.47 Diag.13 - 48


Regarding Diagonals of 6-49 have 14 diagonals that comprise it, it also baptized because of 6-49 starts at 6 and ends at 49. If we analyze these endings also diagonal, we can notice that in each one there is a repeat termination. And each diagonal consists of 6-49: Diag.1 - 06 Diag.2 - 05:12 Diag.3 - 04:11:18 Diag.4 - 03.10.17.24 Diag.5 - 02.09.16.29.30 Diag.6 - 01.08.15.22.29.36 Diag.7 - 07.14.21.28.35.42 Diag.8 - 13.20.27.34.41.48 Diag.9 - 19.26.33.40.47 Diag.10 - 25.32.39.46 Diag.11 - 31.38.45 Diag.12 - 37.44 Diag.13 - 43.50 Diag.14 - 49


Now, given that only come out in EuroMillions 5 balls in the framework above, it suffices to choose only 5 diagonals also playing in both the 1-48 and 6-49, and 5 endings, not obvious? When doing this application will almost always the numbers reduced to a amount of 5 to 11 numbers endings. But also, if the choices are "poorly done" we have a reduction of less than 5 numbers, which will allow for the addition of a few more choices.
However, the problem lies in that diagonal endings and choose! On this issue, you must be a sensitivity and a little luck (just a little!) ...
For example ... and even just an example! If we choose Diag.1-48 in the following 5 diagonals: Diag.3 - 03:08:13 Diag.5 - 05.10.15.20.25 Diag.6 - 06.11.16.21.26.31 Diag.8 - 18.23.28.33.38.43 Diag.11 - 36.41.46
And in Diag. 6-49 as follows: Diag.4 - 03.10.17.24 Diag.6 - 01.08.15.22.29.36 Diag.8 - 13.20.27.34.41.48 Diag.9 - 19.26.33.40.47 Diag.11 - 31.38.45
As you can see, combining the filters together, we were only a total of 12 numbers, which are the ones that put in bold (and 03.08.10.13.15.20.26.31.33.36.38 41)
Now, by applying the filter of endings, if we choose the endings 0,1,3,6 and 8 are left with a final result of 11 numbers:
After we get to the end result, we must first of all look at our pocket ... If we can play a key opportunity with 11 numbers and 2 stars (€ 924.00), which is a multiple bet that allows the greatest amount of numbers, however is not the most expensive bet ... However, in my opinion is a waste of money to play with multiple.
As for the stars, you ask, "diagonals there for them?" Yes clear that the method of intersection of diagonals also applies,

    SergeM's avatar - slow icon.png
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    Posted: August 29, 2012, 5:53 pm - IP Logged

    Schick mir dann die Kopie!

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      bgonçalves
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      Posted: August 29, 2012, 7:01 pm - IP Logged

      Schick mir dann die Kopie!

      hello SERGEM

      However it must be said that the terminations of the filter functions as a filter applied within the diagonals 1-48 because each diagonal 1-48 has only two terminations.
      Let's face it: In this picture we have 13 diagonals that make up the diagonals of 1-48, 1-48 baptized because of it starts at 1 and ends at 48. If fixed, these each diagonal contains only two terminations. Diag.1 - 01 Diag.2 - 02:07 Diag.3 - 03:08:13 Diag.4 - 04.09.14.19 Diag.5 - 05.10.15.20.25 Diag.6 - 06.11.16.21.26.31 Diag.7 - 12.17.22.27.32.37 Diag.8 - 18.23.28.33.38.43 Diag.9 - 24.29.34.39.44 Diag.10 - 30.35.40.45 Diag.11 - 36.41.46 Diag.12 - 42.47 Diag.13 - 48


      Regarding Diagonals of 6-49 have 14 diagonals that comprise it, it also baptized because of 6-49 starts at 6 and ends at 49. If we analyze these endings also diagonal, we can notice that in each one there is a repeat termination.
      And each diagonal consists of 6-49: Diag.1 - 06 Diag.2 - 05:12 Diag.3 - 04:11:18 Diag.4 - 03.10.17.24 Diag.5 - 02.09.16.29.30 Diag.6 - 01.08.15.22.29.36 Diag.7 - 07.14.21.28.35.42 Diag.8 - 13.20.27.34.41.48 Diag.9 - 19.26.33.40.47 Diag.10 - 25.32.39.46 Diag.11 - 31.38.45 Diag.12 - 37.44 Diag.13 - 43.50 Diag.14 - 49


      Now, given that only come out in EuroMillions 5 balls in the framework above, it suffices to choose only 5 diagonals also playing in both the 1-48 and 6-49, and 5 endings, not obvious? When doing this application will almost always the numbers reduced to a amount of 5 to 11 numbers endings. But also, if the choices are "poorly done" we have a reduction of less than 5 numbers, which will allow for the addition of a few more choices.
      However, the problem lies in that diagonal endings and choose! On this issue, you must be a sensitivity and a little luck (just a little!) ...
      For example ... and even just an example! If we choose Diag.1-48 in the following 5 diagonals: Diag.3 - 03:08:13 Diag.5 - 05.10.15.20.25 Diag.6 - 06.11.16.21.26.31 Diag.8 - 18.23.28.33.38.43 Diag.11 - 36.41.46
      And in Diag. 6-49 as follows: Diag.4 - 03.10.17.24 Diag.6 - 01.08.15.22.29.36 Diag.8 - 13.20.27.34.41.48 Diag.9 - 19.26.33.40.47 Diag.11 - 31.38.45
      As you can see, combining the filters together, we were only a total of 12 numbers, which are the ones that put in bold (and 03.08.10.13.15.20.26.31.33.36.38 41)
      Now, by applying the filter of endings, if we choose the endings 0,1,3,6 and 8 are left with a final result of 11 numbers: