Hi.
I've been trying to see how to beat these lottery wheels (http://lottery.merseyworl
What I don't understand is why 163 tickets is required to guarantee a 3 ball win.
Any SIMPLE explanations?
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Ok. So far I'm with you.
What I'm trying to do is build a program that can produce the smallest set of tickets.
"So, the more tickets we buy, the more such 3-ball combinations we cover. What we need, is a set of tickets that covers all 13983816 possible draws."
Yes, that's the current 163 tickets.
But what is the minimum number needed? There must be a mathematical proof.
>> What I'm trying to do is build a program that can produce the smallest set of tickets.
A genetic algorithm might be well suited to give you a (sub)optimal solution to this problem.
>> But what is the minimum number needed? There must be a mathematical proof.
There might be an easy way to obtain the optimal solution, but I'd say chances are against that. I've got no proff for that assertion.
You can easily find an absolute lower limit to the amount of tickets needed : 13983816 / (20 * 15180) = 46.06 -> 47
But that lower limit is not likely to be met, since it requires that all (20 * 47) = 940 3-ball combinations for all 47 tickets are distinct, and the same draw is never covered more than once. That sounds impossible (and I'm sure someone with more time could prove that for you ;) ).
I spent some time on this a few years ago. If you analyse the 163 rows against all possible draws ( A long winded way of doing it, but I had a lot of time) then you will find that there are only two numbers, both in different rows that can be considered obsolete. If they were both in the same row then there is a small possibility that the row could have been paired with another to make the wheel 162 rows, but not with them in separate rows.
Changing any of the other numbers in any rows meant that some of the 14m possible draws were not covered with a 3 match.
I've been digging around, Unfortunately it's beginning to look like the code has gone where all old code goes when you leave work and change pc's a couple of times. I have turned up one output file - here's a couple of lines from it:
1 2 3 4 5 6 |1,2,3,|15160|1,2,4,|14170
1 2 7 47 48 49 |1,2,7,|11460|1,2,47,|1064
There are 161 more lines similar to those. It shows each row from the wheel and the number of draw rows covered by each of the contained 20 triples as well as the total number of rows covered by the wheel row. I do remember that I was hoping to find some triples that weren't used much or at all as their covers had been covered by another wheel row and so possibly shorten those wheel rows with the aim of reducing the wheel below 163.
I will keep looking for the code, but I haven't had much luck yet - not even sure which language I was using.
As far as a mathematical explanation goes, I find/found it odd that even the first few rows covered quite varying numbers of the lottery perms. This is possibly because I had marked a row off as covered and not counted it again, but without the code I can't even say that for certain.
Bear in mind that due to the central limit theorem (from statistics) its less likely for a combination of ball numbers that are either all low or all high to be drawn than it is for a mix of high, middle and low numbers to be drawn. This complicates odds of picking the most likelynumber drawing permutations above.
I've found a bit of my code - it's java and I've remembered what I was trying to do. Method was as I said above to find any triples in any of the 163 (I called them 'tickets') that weren't covering as many rows as others. I was marking a row off as covered for a single hit and so the number of covers by each ticket reduced as it went through the list. I tried rotating the rows so that 1 became second , 163 became first etc. and collecting stats for all these and lastly, but I don't think I ran it, was some code to permutate the order of all the 163 tickets (biiiiig task) and run each through the stats procedure. Hopefully, there would have been, somewhere in the code, a way of identifying whether a ticket triple in a particular order actually had no useful covers and possibly find a permutation of tickets that had whole tickets that could be missed out or could be merged with another. I think the length of time needed to run those permutations combined with a change of project at work was were the reasons I didn't continue with it. There was a competition at the time to find a wheel of less than 163 tickets, at least you could get your name into the history books if you do manage to find a shorter wheel.
I didn't mean that you won't win ever, I was trying to emphasise to both aburr and Nanco that this excercise is about a covering wheel and not about trying to 'beat' the lottery.
I think getting a proof of the minimum number of tickets needed to guarantee 3 from 6 from 49 is going to be difficult.
I'm putting together some code to try and automate a wheel to cover all the 3ball combos. I'm pretty sure I'm not going to get anywhere near 163.
I've got a verifier (a program that will supply every single ticket to my wheel) and gives stats about it.
If I get anywhere I'll come back to this.
I'm going to work with 3 pairs, taking the lowest used pairs each time to make the 6 ball line. Making sure that the pairs content are unique (so, ab/bc/cd would not be valid).
So, say (ab/cd/ef) => 20 3balls which I'll then use to find the number of uses of each of the 3s. Each set of 3 will create a "weight" value (the number of matches so far).
If the weight is greater than the current weight, then skip it, but track it as a potential candidate weight.
If we exhaust all the combinations, then the new weight will be used as the minimum weight.
I hope that makes some sense. Not sure how long this will take to run or even if it will work!
> I think getting a proof of the minimum number of tickets needed to guarantee 3 from 6 from 49 is going to be difficult.
You could first prove that if someone buys N-1 tickets you can always find a drawing such that there is no 3-ball win. Then you produce a set of N tickets that guarantees a 3-ball win. Then you've proven that N is the minimum.
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by: Infinity08Posted on 2009-08-04 at 07:11:14ID: 25013991
>> What I don't understand is why 163 tickets is required to guarantee a 3 ball win.
That's just the current minimum that someone found. There might be a solution with less than 163 tickets.
>> Any SIMPLE explanations?
I'll try to keep it simple, but that also means that it will only cover the basics :
There are (49! / (6! * 43!)) = 13983816 possible ways of drawing 6 balls out of 49.
Each of these has (6! / (3! * 3!)) = 20 different 3-ball combinations.
There are (49! / (3! * 46!)) = 18424 possible 3-ball combinations.
Over all possible ways of drawing 6 balls out of 49, each of the 18424 3-ball combinations will occur ((13983816 * 20) / 18424) = 15180 times.
So, with one 3-ball combination, we cover 15180 out of 13983816 of all possible draws.
We're looking for which tickets we need, to have at least 1 match with such a 3-ball combination.
Remember that each ticket covers 20 different 3-ball combinations.
So, the more tickets we buy, the more such 3-ball combinations we cover. What we need, is a set of tickets that covers all 13983816 possible draws.
So, suppose we buy a ticket (1, 2, 3, 4, 5, 6). The 3-ball combinations we cover are :
(1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 2, 6), (1, 3, 4), (1, 3, 5), (1, 3, 6), (1, 4, 5), (1, 4, 6), (1, 5, 6),
(2, 3, 4), (2, 3, 5), (2, 3, 6), (2, 4, 5), (2, 4, 6), (2, 5, 6), (3, 4, 5), (3, 4, 6), (3, 5, 6), (4, 5, 6)
This single ticket covers all possible draws that have at least one of these 20 3-ball combinations.
The idea is to find a combination of tickets that covers all 13983816 possible draws in this way.