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Number selection method for Pick5 / Pick 6 lotteries

Topic closed. 2 replies. Last post 6 years ago by ACPutz.

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Gent
Belgium
Member #73987
April 30, 2009
33 Posts
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Posted: February 23, 2011, 7:50 am - IP Logged

Hi everyone

 

Here’s a system I’ve been thinkingabout for number selection in Pick 5 or Pick 6 lotteries. As an example, I willuse the results for Euro Millions (5/50), main numbers, without taking intoaccount the stars. The method uses the 20 last draws results. I’ve studied the resultsand noticed that in a 5/50 lottery, on average:

- there are 13 of the last 20results with 0 winning numbers from the next drawing (in 17% of the drawings)

- there are 12 of the last 20results with 0 winning numbers from the next drawing (in 18% of the drawings).


Since 13 is a lucky number, let’suse this information for number selection.

First of all, what does this mean?Let’s say that the following numbers are drawn in Euro Millions:

12          13          36          41          46

You have a 17% chance that in the 20drawings before, 13 drawings will NOT have 12, 13, 36, 41 or 46. To put thingsin perspective: this happens approximately 9 times per year! That's not too bad.

 

So how do we use this informationfor number selection? Let’s take 20 (not so) random drawings for Euro Millions (these are actual results!):

7/08/2009          10          20          22          24          31

14/08/2009        5            8            24          30          49

21/08/2009        4            7            16          31          42

28/08/2009        8            36          37          41          49

4/09/2009          6            9            20          38          39

11/09/2009        12          15          35          42          43

18/09/2009        6            16          30          38          41

25/09/2009        6            17          18          21          34

2/10/2009          22          23          24          29          44

9/10/2009          7            11          29          46          50

16/10/2009        12          23          30          31          47

23/10/2009        6            18          20          29          31

30/10/2009        9            33          35          38          40

6/11/2009          11          19          34          43          45

13/11/2009        13          15          25          26          32

20/11/2009        5            9            28          43          47

27/11/2009        5            8            13          15          33

4/12/2009          18          19          25          30          44

11/12/2009        20          41          43          44          46

18/12/2009        14          30          32          35          49

 

We will presume that 13 of theseresults will have 0 winning numbers. Now comes the hardest part: we have toguess, which 13 of these 20 results will have 0 winning numbers (there areactually 77.520 possibilities to select from, but this is still a vastimprovement of the 5/50 lotteries odds which are 2.118.760 combinations). Let’s mark theresults we selected with a 0 at the end of the drawing:

 

7/08/2009          10          20          22          24          31          0

14/08/2009        5            8            24          30          49

21/08/2009        4            7            16          31          42

28/08/2009        8            36          37          41          49          0

4/09/2009          6            9            20          38          39          0

11/09/2009        12          15          35          42          43          0

18/09/2009        6            16          30          38          41          0

25/09/2009        6            17          18          21          34

2/10/2009          22          23          24          29          44          0

9/10/2009          7            11          29          46          50          0

16/10/2009        12          23          30          31          47          0

23/10/2009        6            18          20          29          31          0

30/10/2009        9            33          35          38          40          0

6/11/2009          11          19          34          43          45

13/11/2009        13          15          25          26          32          0

20/11/2009        5            9            28          43          47

27/11/2009        5            8            13          15          33

4/12/2009          18          19          25          30          44          0

11/12/2009        20          41          43          44          46          0

18/12/2009        14          30          32          35          49

 

What do we learn from this? Well, ifwe selected the 13 drawings with 0 winning numbers correctly, we now know whichnumbers will not appear in the following drawing:

6, 7, 8, 9, 10, 11, 12, 13, 15, 16,19, 20, 22, 23, 24, 25, 26, 29, 30, 31, 32, 33, 35, 36, 37, 38, 39, 40, 41, 42,43, 44, 46, 47, 49, 50.

So we also know, which numbers canbe winners in the following drawing:

1, 2, 3, 4, 5, 14, 17, 18, 21, 27,28, 34, 45, 48

 

Let’s move on. The 7 drawings whichwe did not select to have 0 winning numbers, will have at least 1 winningnumber (in most cases even, exactly 1 winning number):

14/08/2009        5            8            24          30          49

21/08/2009        4            7            16          31          42

25/09/2009        6            17          18          21          34

6/11/2009          11          19          34          43          45

20/11/2009        5            9            28          43          47

27/11/2009        5            8            13          15          33

18/12/2009        14          30          32          35          49

 

We can remove the non-winningnumbers from these lines, leaving the following numbers as potential winners:

14/08/2009        5           

21/08/2009        4           

25/09/2009        17          18          21          34

6/11/2009          34          45

20/11/2009        5            28

27/11/2009        5           

18/12/2009        14

 

Let’s take a look at the lines whichhave only one number left:

14/08/2009        5           

21/08/2009        4           

18/12/2009        14

As we are sure that each line containsat least 1 winning number, we are now sure that 4, 5 and 14 will be winners inthe next drawing. That’s pretty cool huh?

 

Let’s now combine these numbers witha line on which 4, 5 or 46 do not appear:

6/11/2009          34          45

which will give the following combinations:

4            5            14          34

4            5            14          45

One of these two lines has now fourwinning numbers from the next drawing. Sounds good! The other line will still have 3 winning numbers.

 

As for the fifth winning number, itcan be any of the numbers from the group we previously determined as potentialwinners: 1, 2, 3, 4, 5, 14, 17, 18, 21, 27, 28, 34, 45, 48. 14 numbers  to choose from, minus the 4 we alreadydetermined, leaves 10 numbers to choose from. A 1 in 10 chance to hit the 5/5.Or if you want to play every combination: make 10 x 2 = 20 combinations for aguaranteed 5/5.

1            4            5            14          34

2            4            5            14          34

3            4            5            14          34

4            5            14          17          34

4            5            14          18          34

4            5            14          21          34

4            5            14          27          34

4            5            14          28          34

4            5            14          34          45

4            5            14          34          48

1            4            5            14          45

2            4            5            14          45

3            4            5            14          45

4            5            14          17          45

4            5            14          18          45

4            5            14          21          45

4            5            14          27          45

4            5            14          28          45

4            5            14          34          45

4            5            14          45          48

 

Let’s now have a look at the actualdraw result:

25/12/2009        4            5            14          17          34

 

That’s right, it worked! Aside fromthe jackpot (5 correct numbers), we also have  several lines with 4 winners and 3 winners. None of the lines we playedhave 0, 1 or 2 winners (d’uh!).

 

Hey, what’s the catch? You need morethan a little luck with this method. First of all, from the previous 20drawings, 13 lines need to have 0 numbers from the following drawing (whichhappens in 17% of the drawings, which is quite ok). For this example, I selected a pool of 20 drawings from which I knew there were 13 lines with 0 winning numbers from the next drawing. And then, the biggest taskat hand, is to be lucky when selecting the 13 lines which you think will have 0winners. There are actually 77.520 possibilities for doing so as I mentionedbefore. I selected the 13 lines with 0 winning numbers correctly, because I knew the result from the winning drawing in advance. Other than that, no catch, this works exactly as I described. Ifnothing else, the system provides you with a fun way to make a numberselection, and still end up with an affordable amount of combinations to play.You will however, need some more luck when adding the stars to play with EuroMillions, but I for one would be more than happy to win the amount of money for5 winning numbers, without the stars.

 

I also made the same analysis for a Pick6/42 lottery. If anyone is interested, I will post the results of that analysis in this post too.

    LANTERN's avatar - kilroy 28_173_reasonably_small.jpg
    Tx
    United States
    Member #4570
    May 4, 2004
    5180 Posts
    Offline
    Posted: February 25, 2011, 11:52 am - IP Logged

    Hi everyone

     

    Here’s a system I’ve been thinkingabout for number selection in Pick 5 or Pick 6 lotteries. As an example, I willuse the results for Euro Millions (5/50), main numbers, without taking intoaccount the stars. The method uses the 20 last draws results. I’ve studied the resultsand noticed that in a 5/50 lottery, on average:

    - there are 13 of the last 20results with 0 winning numbers from the next drawing (in 17% of the drawings)

    - there are 12 of the last 20results with 0 winning numbers from the next drawing (in 18% of the drawings).


    Since 13 is a lucky number, let’suse this information for number selection.

    First of all, what does this mean?Let’s say that the following numbers are drawn in Euro Millions:

    12          13          36          41          46

    You have a 17% chance that in the 20drawings before, 13 drawings will NOT have 12, 13, 36, 41 or 46. To put thingsin perspective: this happens approximately 9 times per year! That's not too bad.

     

    So how do we use this informationfor number selection? Let’s take 20 (not so) random drawings for Euro Millions (these are actual results!):

    7/08/2009          10          20          22          24          31

    14/08/2009        5            8            24          30          49

    21/08/2009        4            7            16          31          42

    28/08/2009        8            36          37          41          49

    4/09/2009          6            9            20          38          39

    11/09/2009        12          15          35          42          43

    18/09/2009        6            16          30          38          41

    25/09/2009        6            17          18          21          34

    2/10/2009          22          23          24          29          44

    9/10/2009          7            11          29          46          50

    16/10/2009        12          23          30          31          47

    23/10/2009        6            18          20          29          31

    30/10/2009        9            33          35          38          40

    6/11/2009          11          19          34          43          45

    13/11/2009        13          15          25          26          32

    20/11/2009        5            9            28          43          47

    27/11/2009        5            8            13          15          33

    4/12/2009          18          19          25          30          44

    11/12/2009        20          41          43          44          46

    18/12/2009        14          30          32          35          49

     

    We will presume that 13 of theseresults will have 0 winning numbers. Now comes the hardest part: we have toguess, which 13 of these 20 results will have 0 winning numbers (there areactually 77.520 possibilities to select from, but this is still a vastimprovement of the 5/50 lotteries odds which are 2.118.760 combinations). Let’s mark theresults we selected with a 0 at the end of the drawing:

     

    7/08/2009          10          20          22          24          31          0

    14/08/2009        5            8            24          30          49

    21/08/2009        4            7            16          31          42

    28/08/2009        8            36          37          41          49          0

    4/09/2009          6            9            20          38          39          0

    11/09/2009        12          15          35          42          43          0

    18/09/2009        6            16          30          38          41          0

    25/09/2009        6            17          18          21          34

    2/10/2009          22          23          24          29          44          0

    9/10/2009          7            11          29          46          50          0

    16/10/2009        12          23          30          31          47          0

    23/10/2009        6            18          20          29          31          0

    30/10/2009        9            33          35          38          40          0

    6/11/2009          11          19          34          43          45

    13/11/2009        13          15          25          26          32          0

    20/11/2009        5            9            28          43          47

    27/11/2009        5            8            13          15          33

    4/12/2009          18          19          25          30          44          0

    11/12/2009        20          41          43          44          46          0

    18/12/2009        14          30          32          35          49

     

    What do we learn from this? Well, ifwe selected the 13 drawings with 0 winning numbers correctly, we now know whichnumbers will not appear in the following drawing:

    6, 7, 8, 9, 10, 11, 12, 13, 15, 16,19, 20, 22, 23, 24, 25, 26, 29, 30, 31, 32, 33, 35, 36, 37, 38, 39, 40, 41, 42,43, 44, 46, 47, 49, 50.

    So we also know, which numbers canbe winners in the following drawing:

    1, 2, 3, 4, 5, 14, 17, 18, 21, 27,28, 34, 45, 48

     

    Let’s move on. The 7 drawings whichwe did not select to have 0 winning numbers, will have at least 1 winningnumber (in most cases even, exactly 1 winning number):

    14/08/2009        5            8            24          30          49

    21/08/2009        4            7            16          31          42

    25/09/2009        6            17          18          21          34

    6/11/2009          11          19          34          43          45

    20/11/2009        5            9            28          43          47

    27/11/2009        5            8            13          15          33

    18/12/2009        14          30          32          35          49

     

    We can remove the non-winningnumbers from these lines, leaving the following numbers as potential winners:

    14/08/2009        5           

    21/08/2009        4           

    25/09/2009        17          18          21          34

    6/11/2009          34          45

    20/11/2009        5            28

    27/11/2009        5           

    18/12/2009        14

     

    Let’s take a look at the lines whichhave only one number left:

    14/08/2009        5           

    21/08/2009        4           

    18/12/2009        14

    As we are sure that each line containsat least 1 winning number, we are now sure that 4, 5 and 14 will be winners inthe next drawing. That’s pretty cool huh?

     

    Let’s now combine these numbers witha line on which 4, 5 or 46 do not appear:

    6/11/2009          34          45

    which will give the following combinations:

    4            5            14          34

    4            5            14          45

    One of these two lines has now fourwinning numbers from the next drawing. Sounds good! The other line will still have 3 winning numbers.

     

    As for the fifth winning number, itcan be any of the numbers from the group we previously determined as potentialwinners: 1, 2, 3, 4, 5, 14, 17, 18, 21, 27, 28, 34, 45, 48. 14 numbers  to choose from, minus the 4 we alreadydetermined, leaves 10 numbers to choose from. A 1 in 10 chance to hit the 5/5.Or if you want to play every combination: make 10 x 2 = 20 combinations for aguaranteed 5/5.

    1            4            5            14          34

    2            4            5            14          34

    3            4            5            14          34

    4            5            14          17          34

    4            5            14          18          34

    4            5            14          21          34

    4            5            14          27          34

    4            5            14          28          34

    4            5            14          34          45

    4            5            14          34          48

    1            4            5            14          45

    2            4            5            14          45

    3            4            5            14          45

    4            5            14          17          45

    4            5            14          18          45

    4            5            14          21          45

    4            5            14          27          45

    4            5            14          28          45

    4            5            14          34          45

    4            5            14          45          48

     

    Let’s now have a look at the actualdraw result:

    25/12/2009        4            5            14          17          34

     

    That’s right, it worked! Aside fromthe jackpot (5 correct numbers), we also have  several lines with 4 winners and 3 winners. None of the lines we playedhave 0, 1 or 2 winners (d’uh!).

     

    Hey, what’s the catch? You need morethan a little luck with this method. First of all, from the previous 20drawings, 13 lines need to have 0 numbers from the following drawing (whichhappens in 17% of the drawings, which is quite ok). For this example, I selected a pool of 20 drawings from which I knew there were 13 lines with 0 winning numbers from the next drawing. And then, the biggest taskat hand, is to be lucky when selecting the 13 lines which you think will have 0winners. There are actually 77.520 possibilities for doing so as I mentionedbefore. I selected the 13 lines with 0 winning numbers correctly, because I knew the result from the winning drawing in advance. Other than that, no catch, this works exactly as I described. Ifnothing else, the system provides you with a fun way to make a numberselection, and still end up with an affordable amount of combinations to play.You will however, need some more luck when adding the stars to play with EuroMillions, but I for one would be more than happy to win the amount of money for5 winning numbers, without the stars.

     

    I also made the same analysis for a Pick6/42 lottery. If anyone is interested, I will post the results of that analysis in this post too.

    Thanks!

    Do post the 6/42 lottery results also! Please!

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      August 26, 2009
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      Posted: February 25, 2011, 12:11 pm - IP Logged

      Reallly good stuff.  Thanks Game.