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02232011, 06:59 AM

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Number selection method for Pick5 and Pick6 lotteries
Hi everyone
Here’s a system I’ve been thinking about for number selection in Pick 5 or Pick 6 lotteries. As an example, I will use the results for Euro Millions (5/50), main numbers, without taking into account the stars. The method uses the 20 last draws results. I’ve studied the results and noticed that in a 5/50 lottery, on average:
 there are 13 of the last 20 results with 0 winning numbers from the next drawing (in 17% of the drawings)
 there are 12 of the last 20 results with 0 winning numbers from the next drawing (in 18% of the drawings).
Since 13 is a lucky number, let’s use 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 20 drawings before, 13 drawings will NOT have 12, 13, 36, 41 or 46. To put things in perspective: this happens approximately 9 times per year! That's not too bad.
So how do we use this information for 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 these results will have 0 winning numbers. Now comes the hardest part: we have to guess, which 13 of these 20 results will have 0 winning numbers (there are actually 77.520 possibilities to select from, but this is still a vast improvement of the 5/50 lotteries odds which are 2.118.760 combinations). Let’s mark the results 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, if we selected the 13 drawings with 0 winning numbers correctly, we now know which numbers 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 can be 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 which we did not select to have 0 winning numbers, will have at least 1 winning number (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 nonwinning numbers 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 which have only one number left:
14/08/2009 5
21/08/2009 4
18/12/2009 14
As we are sure that each line contains at least 1 winning number, we are now sure that 4, 5 and 14 will be winners in the next drawing. That’s pretty cool huh?
Let’s now combine these numbers with a 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 four winning numbers from the next drawing. Sounds good! The other line will still have 3 winning numbers.
As for the fifth winning number, it can be any of the numbers from the group we previously determined as potential winners: 1, 2, 3, 4, 5, 14, 17, 18, 21, 27, 28, 34, 45, 48. 14 numbers to choose from, minus the 4 we already determined, 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 a guaranteed 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 from the jackpot (5 correct numbers), we also have several lines with 4 winners and 3 winners. None of the lines we played have 0, 1 or 2 winners (d’uh!).
Hey, what’s the catch? You need more than a little luck with this method. First of all, from the previous 20 drawings, 13 lines need to have 0 numbers from the following drawing (which happens 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 task at hand, is to be lucky when selecting the 13 lines which you think will have 0 winners. There are actually 77.520 possibilities for doing so as I mentioned before. 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. If nothing else, the system provides you with a fun way to make a number selection, 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.

02232011, 10:29 PM


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Location: South Africa
Posts: 844


Hi GameBelgium
Please post your analysis for pick 6, it is very interesting for the pick 5 so I'm going to check it out against our pick 5
BlouBul

02232011, 11:51 PM

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Join Date: Nov 2007
Location: Greece
Posts: 59


Hi GameBelgium
How you are going to elliminate the 13 draws?
what will be your method?
are you going to use a macro?
El_Patron

03102011, 05:48 AM

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Join Date: Oct 2007
Posts: 88


For the 6/42 lotto, I had the following results for the amount of lines with 0 winners in the last 20 draws:
0 0,00%
1 0,00%
2 0,46%
3 3,21%
4 4,59%
5 10,09%
6 13,76%
7 27,06%
8 16,06%
9 10,09%
10 8,26%
11 3,21%
12 2,29%
13 0,46%
14 0,46%
15 0,00%
16 0,00%
For the 5/50 lotteries, the following results:
1 0,00%
2 0,00%
3 0,00%
4 0,00%
5 0,00%
6 0,56%
7 3,33%
8 4,44%
9 8,89%
10 12,78%
11 15,56%
12 18,89%
13 17,22%
14 6,11%
15 7,22%
16 3,33%
17 1,67%
18 0,00%
19 0,00%
20 0,00%
How would I select the 13 lines with 0 winners? I would pick the 7 lines with at least 1 winner. Divide the 20 lines in 2x10 lines, pick 3 lines from the top 10 lines and pick 4 lines from the bottom 10 lines. Or vice versa. As I said: 77520 possibilities, you will still need a whole lot of luck!

03112011, 05:20 AM


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Join Date: Aug 2009
Location: UK
Posts: 285


The method you describe for Euromillions makes use of the theoretical fact that any 5/50 result has a 57.66% probability that any other result (past or future or a sequence written on a piece of paper or randomly generated) will not contain a number from that result. This would be an average over many hundreds of draws, if that many had been drawn.
Your analysis of 13/20 works out at 65%.
Looking back over only the most recent 20 draws you will find that the statistic you have discovered will vary over time, but as you say you can average it out.
I have been using a similar method for the UK lotto for some time now. You have correctly identified the issues involved, namely choosing the selection of draws you think will have at least a match of 1 and identifying those you think will not, and as you rightly say the odds against choosing the right combination of candidate past draws are high.
Whilst the method you describe for cross checking the lines you rule out (and the numbers eliminated) against the lines you rule in (hence narrow down the actual balls to repeat) is sound it is only possible for small samples such as the 20 draws you suggest.
The statistic is true for the whole of lottery history and if you were to extend the method over more draws then you would end up eliminating all 50 balls from a future draw by that method.
The other reality is that had you decided to look only at 20 alternate past draws (look back 40 but skip ones between to end up with 20) instead of consecutive ones, the same law would apply  but you would have an entirely different draw selection pool to draw from, and get different results.
You have however given me an idea of how I might modify my current method which has limited success at the moment.
Thanks for sharing this and as you say despite its shortfalls, it is fun and does make you feel in control of your destiny rather than relying on pure chance.

03112011, 07:35 AM

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Join Date: Oct 2007
Posts: 88


Quote:
You have however given me an idea of how I might modify my current method which has limited success at the moment.

Frank, that was what I wanted to mail you about, as I read in another post you used a similar system. But as you said, it's even better to discuss this here on the forum, so the other active members can think along, and maybe provide some input.
You're also right when saying that the same law of averages would apply to any 20 drawings, not only the 20 most recent ones. Would you consider it a good idea to use the same set of 20 drawings all the time? And maybe keep a record of how many lines produced how many winners, and use this as statistical information to try to determine the lines with no winners / more winners more accurately?
In the meantime, I did some further research for the 6/42 lottery, by looking at the results with 7 lines containing no winners, and turning these results in a 7/20 lottery. This generated results like the one below:
2 5 7 10 11 15 17
4 5 8 11 12 14 18
2 4 5 8 10 13 20
5 7 8 9 10 13 18
2 7 10 12 15 16 20
3 4 5 10 12 16 19
2 3 6 11 13 14 17
3 4 5 6 8 13 16
2 3 4 5 6 9 16
6 10 11 14 15 19 20
1 3 4 9 10 12 18
11 13 14 15 16 18 19
2 4 6 9 10 14 17
1 2 3 8 11 12 19
7 11 13 16 18 19 20
2 3 4 11 12 15 18
2 7 8 10 13 16 18
1 3 5 6 9 13 20
6 9 10 12 13 17 19
3 4 7 10 12 14 18
I tried to use the same selection method by looking at the results, but in almost all of the results, there were only 1 (33%) or 0 (62%) lines containing 0 winners, making it very hard to use this information to generate a p(l)ayable amount of combinations.
When looking at the most recent result of 7 lines containing no winner:
6 9 10 12 14 16 18
you will see that none of the 20 lines above has 0 winners...
So the search for a system to accurately select 7 lines from a pool of 20 continues. Every input of course, is more than welcome!

03112011, 07:45 AM

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Join Date: Oct 2007
Posts: 88


Quote:
The statistic is true for the whole of lottery history and if you were to extend the method over more draws then you would end up eliminating all 50 balls from a future draw by that method.

I can't agree with you on this one: it would eliminate 45 numbers, and leave the 5 winning numbers. You can try this out by starting from the winning numbers and backchecking against the past results. Usually, within a 100 drawings, you will have succesfully eliminated the 45 losing numbers, leaving you with the 5 winning numbers. I've also tried this without knowing the winning numbers, but the 'system' generates many thousands and thousands of combinations.

03122011, 10:55 AM


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Join Date: Aug 2009
Location: UK
Posts: 285


Well you've touched on a few issues there, and we've moved from 5/50 via 6/69 to 6/42 (turned into 7/20  you lost me there) and theres a danger of losing sight of the original question. My analysis models stick to 6/49 and Euromillions so I'm not going to drift away from my own experience with those, other than give statistical theoretical odds for no repeats in other lottery types.
In order to compile a thought out reply I'll need to access my spreadsheets which will require some time to compile. I'll be back after the weekend, meanwhile I can tell you that the probability of not matching at least one ball of a 6/42 result with any other result is 37.13% which may explain your difficulties with that.

03122011, 08:58 PM


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Location: ONtario
Posts: 184


GamesBelgium:
Tried to reason out your instruction for ON49. Only the number 1 showed up for the next winning line. Could you show us an example of using 20 line of ON49?
Madam

03142011, 04:56 AM

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Join Date: Oct 2007
Posts: 88


Madam, sure I could. Where can I find the results for the ON49 lotteries? I can work out an example with actual results then.
Frank:
 for the 6/42 lotteries, I tried to focus on 7 lines with 0 winning numbers (which leads to 7/20)
 for the 5/50 lotteries, I tried to focus on 13 lines with 0 winning numbers, and thus 7 lines (7/20) with more than 1 winning number
I did the exercise for the 6/42 Belgian lottery, isolating all past drawings which had 7 lines with 0 winning numbers. And from there, I just converted those lines into a 7/20 lottery result. Let's say that for a drawing, lines 1, 4, 5, 9, 12, 14 and 19 had 0 winning numbers, then I would convert that information into a 7/20 lottery result:
01 04 05 09 12 14 19
But as I said before, the information this yielded was not useful for more accurate selection of which 7 lines with 0 winners were most likely to pop up next.

03142011, 07:48 AM


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Join Date: Aug 2009
Location: UK
Posts: 285


Hi GameBelgium,
Where to start, thats the problem. Okay  the easiest question first, this will lead on to the rest. You basically asked whether it would be a good idea to use the same set of drawings all the time.
Well firstly your method looks at draws with a one ball or better direct repeat from the most recent draw to the next drawing with no skips in between. Secondly regardless of how you selected the 20 candidate lines you make the assumption that the 20 lines are a related set. I'll come to that.
Lets talk about the Euromillions since you gave examples of that, and I do have a speadsheet model of that.
What I am able to do is set my spreadsheet to compare past draws which are n apart (n is variable from 1 to anything). I call n a skip value, so if I set skip to 1 it goes down the whole of Euromillions past draws checking to see if any draw had any repeats in the following (consecutive) draw. It shows what balls repeated and counts up over lottery history how many draws had one ballrepeat, 2 balls repeat, no balls repeat etc. So I get instant readouts for any values of skip.
Looking at consecutive draw repeats then skip =1, it would for example compare amongst all the others draw 368 with 367 and note that 45 repeated, draw 172 with draw 171 and note that 26 and 49 repeated, 155 with 154 and find no repeats.
Summarising I can tell you that 45.53% of draws had at least one ball directly repeat in the next draw.
Varying the value of skip to 2, I get listings of comparisons of (for example) 368 with 366, 351 with 349 etc. etc. Summarising this, 42.39% of draws repeated to the next but one draw. For high skip values, I'm comparing draws hundreds apart.
If I make a selection of skips and tabulate them you'll see what I mean.
Skip...repeated
1 45.53%
2 42.39%
3 40.60%
4 43.72%
5 44.93%
6 41.48%
7 42.42%
21 43.55%
22 47.41%
47 41.18%
65 44.59%
76 45.58%
100 45.56%
AVG 43.76%
The theoretical value is 10.5766 = 0.4234 or 42.34 %. The values only differ from each other because of the relatively small sample size of 370 draws and on the UK lotto model over 1588 draws the results are much closer to the theoretical value for a 6/49. I have also replaced actual results with randomly generated ones and done the same analysis, with the same results.
All of which tells me that it does not matter theoretically where you get your 20 candidate draws from, the statistics are exactly the same for the probability of a repeat to the next or any other draw, be they consecutive or scattered throughout history. So to answer your question, it really doesn't matter.
I think what does matter is how you select the ones you think will have a repeat in the next draw. I have my own method which basically looks at the track record of a particular skip. A skip of 4 for example has worked for 5 times in a row up to today. If I thought this trend would continue then draw 367 would be one of my candidate draws for 371. On the other hand I might think that 6 times in a row was unlikely and discard it.
to.. from.. a repeat ball ?
370 366 ...yes
369 365 ...yes
368 364 ...yes
367 363 ...yes
366 362...yes
A skip of 122 also has been a winner for 5 draws in a row
to.. from. a repeat ball ?
370 248...yes
369 247...yes
368 246...yes
367 245...yes
366 244...yes
so I may decide to use 249 as a candidate draw for 371 if I guess it will continue. Note we are back in 50/50 guess territory. Its the old coin toss question  how many times in a row can you throw heads ?
Anyway I only tell you this as an illustration of how our methods differ.
Back to your system of 20 past draws which you treat as a related set. You take 20 random results, In your example they were a consecutive chunk from 2009, but I suggest that you could have taken them from anywhere with any skip or varying skip between. You flag up 13 draws you think will NOT repeat to the next draw. You then infer that the other 7 draws (by cross referencing) can have a repeat ball and by comparison with the 13 draw list deduce from that a shortlist of remaining possible repeat balls ? I do not accept that logic, since there is no connection between the 7 draws you ruled as 'in' and the 13 draws you ruled as 'out'. You could just as easily have chosen the 13 draws you ruled as 'out' with a completely different set of 7 draws you ruled as 'in'. The 42.34% statistic is true for all, not just your chunk of 20. Take any 20 draws from literally anywhere,The 42.34% statistic is true, take a different 20 draws, the 42.34% statistic is still true, but gives entirely different results. So I don't think the logic you use can work as a predictive tool.
I'm not saying I can do any better, we are up against a chaotic system. I'm just pointing out what we're up against. Once I've chosen my candidate lines by the system above, usually between 17 and 24 of them  but all ruled as possible repeaters  I then have to decide which ball(s) on those lines will repeat ? That's another story, and for a while I thought your idea would help. On reflection and as explained above, I can't see how it can be valid.
Yes it is important to try it and record how well it would have done, I have done this for the past year on my UK lotto system, with variable  mainly disappointing results. But you can then tweak your methods and see if a change made has a permanent improvement or otherwise. I don't want to discourage you from doing such a trial, I for one would like to see the results after a few months, so by all means prove me wrong.

03142011, 09:40 AM

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Join Date: Oct 2007
Posts: 88


Good statistics Frank, thanks for the input.
So basically, we're back to the same question: how do we select 7 lines from a pool of 20 and still end up with an affordable amount of lines to play.
Since you mention the repeats: if we assume that one number will repeat from the last draw (which will happen in 42% of the drawings), it's possible to eliminate 3 to 5 lines, to further reduce the odds of picking the right lines with 0 winners.
Let's say that the last drawing was the following:
07 15 18 22 25 31
The entire pool of 20 lines to choose from:
1) 2 16 17 22 23 27
2) 17 19 21 30 35 40
3) 3 7 14 15 33 34
4) 9 10 13 24 38 42
5) 3 6 17 27 32 37
6) 17 22 28 32 33 40
7) 16 17 18 30 34 35
8) 2 3 17 30 33 36
9) 7 17 28 29 40 42
10) 5 7 17 18 33 38
11) 6 15 16 18 20 28
12) 2 4 8 21 23 37
13) 2 6 15 18 20 30
14) 8 21 25 29 33 34
15) 4 21 22 27 35 40
16) 4 7 12 28 38 42
17) 1 2 24 25 41 42
18) 2 6 13 19 27 31
19) 2 3 5 9 16 40
20) 7 15 18 22 25 31
Let's assume that 15 will repeat in the next drawing. That would imply that all lines containing 15 will not be a line with 0 winners. That information would allow us to eliminate lines 20 (always when there's a repeating number), 13, 11 and 3. Leaving us with only 16 lines to choose the 7 lines with 0 winners from.
For Euro Millions, the situation is opposite. If we assume a number will repeat there, it will allow us to determine 1 to 3 lines which will have at least one winning number. When 3 lines, we only need to choose 4 more correctly to determine the 7 lines with at least one number, leaving us 13 lines with 0 winning numbers.

03152011, 06:28 AM


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Join Date: Aug 2009
Location: UK
Posts: 285


If its just a random 20 draws without making any assumptions before selecting them, then once you have your 20 and then make assumptions about what will repeat from them, then that will remove lines from your set as you say. But how many other sets of 20 could you have chosen ?
The problem is the system is very highly geared to choosing the right 20 in the first place and I'm sure you've considered how many combinations of 20 draws from 371 in Euromillions there are, they are astronomical.
If you use blocks of consecutive draws to choose your 20 from theres a lot less of them, but then you are up against the fact that the 42% to 53% split (will repeat/will not repeat) is not guaranteed within a consecutive block, due to statistical variations  those repeat statistics are only an average over hundreds of draws, not 20. You could well pick a block where 75% of them don't contain a repeating number.
So you need really sound reasons for choosing each and every candidate in your 20. Having done that then your reasoning within the chosen 20 seems sound.

03152011, 06:56 AM

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Join Date: Oct 2007
Posts: 88


Thanks again for thinking along Frank. Basically, what I do, is that I always take the last 20 drawings of the 6/42 Belgian lottery, and use those drawings to pick 7 lines from, of which I hope that they will have 0 winners. Of course, it's not always the case, that exactly 7 lines from the 20 last drawings have 0 winners, but it does happen regularly (approx. 20% of the drawings).
To give you an idea, I kept a record of how many lines of the last 20 drawings had 0 winning numbers for an extended period of time, with the following results (from december 2010 to now, but my list goes back all the way to the mid 1990's):
01/12/2010: 8
04/12/2010: 8
08/12/2010: 7
11/12/2010: 7
15/12/2010: 7
18/12/2010: 8
22/12/2010: 4
25/12/2010: 8
29/12/2010: 6
01/01/2011: 7
05/01/2011: 6
08/01/2011: 5
12/01/2011: 14
15/01/2011: 5
19/01/2011: 5
22/01/2011: 5
26/01/2011: 4
29/01/2011: 2
02/02/2011: 6
05/02/2011: 7
09/02/2011: 9
12/02/2011: 8
16/02/2011: 6
19/02/2011: 8
23/02/2011: 9
26/02/2011: 11
02/03/2011: 7
05/03/2011: 2
09/03/2011: 7
12/03/2011: 7
So, for every drawing where there's a 7, I would have a 1 in 77520 chance of winning the jackpot, which is not too bad. In the 30 drawings listes above, it happened 8 times that 7 lines of the last 20 drawings had 0 winners. That's a little above average (26,67%).

03152011, 03:40 PM


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Join Date: Aug 2009
Location: UK
Posts: 285


Unfortunately the presence of 7 lines with no winning numbers in your 20 does not signify that the 20 you are using will predict the next winning result. So It doesn't really give you a 1 in 77520 chance of winning the jackpot, its only a statistic not an explicit relationship !
You could make exacty the same statement about the next block of 20 down or a block from last year.
The only way you can prove this to yourself is to try it and see. Thats all I can suggest.
The acid test for your system if it creates a pool of numbers which might contain the next draw result is to record those numbers before the draw. Does it capture the result within the list ? How many ? keep a record.

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