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# two diagonals

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

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bgonÃ§alves
Brasil
Member #92564
June 9, 2010
2202 Posts
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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,

Economy class
Belgium
Member #123700
February 27, 2012
4035 Posts
Offline
 Posted: August 29, 2012, 5:53 pm - IP Logged

Schick mir dann die Kopie!

bgonÃ§alves
Brasil
Member #92564
June 9, 2010
2202 Posts
Offline
 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:

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