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Number selection method for Pick5 / Pick 6 lotteriesPrev TopicNext Topic
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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.
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Quote: Originally posted by GameBelgium on Feb 23, 2011
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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Reallly good stuff. Thanks Game.
