Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Background: the Belgian Lottery switched its main game this year to a draw of 6 balls out of a pool of 45, plus a bonus number which doesn't matter for the sake of this question.

Assuming someone would like to play all combinations, this means filling out $\binom{45}{6}$, or 8145060 grids, which would be a logistical nightmare and hasn't, as far as I know, attempted since 1990 (and that was with 1947792 grids).

However, the Belgian lottery also offers "multiple" betslips, in which one picks 7 to 15 numbers, so that all combination of 6 numbers out of these are played, allowing one to play up to 5005 combinations with a single betslip.

At first glance, this would seem to be a solution to the logistical issues, but to ensure each combination is played only once, there should be no overlap; that is, a combination of 6 numbers is present only on one betslip.

This would be possible using multiple betslips filled with n numbers if there exists a Steiner system S(6,n,45).

Obviously S(6,15,45) doesn't exist, as $\binom{45}{6}/\binom{15}{6}$, which would be the number of blocks (or betslips filled) isn't an integer. Similarly, we can show that S(6,14,45) and S(6,13,45) don't exist either.

Another property of Steiner systems is that the existence of S(t−1,k−1,n−1) is a necessary condition for the existence of S(t,k,n). With this property, we can also eliminate S(6,12,45) since S(5,11,44) doesn't exist as $\binom{44}{5}/\binom{11}{5}$ is not an integer. Similarly, we can rule out S(6,11,45), S(6,10,45), S(6,8,45) and (6,7,45).

However, S(6,9,45) can't be eliminated that way, as all subsystems do have an integer number of blocks. S(6,9,45) would have 96965 blocks (thus betslips), S(5,8,44) would have 19393, S(4,7,43) 3526, S(3,6,42) 574, and the known S(2,5,41) has 82. And indeed, S(6,9,45) is listed as a possible Steiner system.

So my question is, does the Steiner system S(6,9,45) actually exist? The wikipedia page states that "As of 2012, an outstanding problem in design theory is if any nontrivial (t < k < n) Steiner systems have t ≥ 6", so I assume proving that S(6,9,45) exists would be a major breakthrough. I'm not naive enough to expect that, but on the other hand, maybe it doesn't exist; and if that's the case, can it be proven?

Disclaimer: I do currently work for the Belgian Lottery, and while I ask this question out of personal interest, I may show answers to colleagues. On the other hand, considering we don't currently offer jackpots which would make interesting for anyone to attempt this, I'm confident the question is purely theoretical anyway.

share|cite|improve this question
Misread your question originally. Almost certainly the answer is no -- a S(6,9,45) solution would have incredible symmetry of a type not yet known, but that's impossible due to the Enormous Theorem. – Ed Pegg Dec 26 '12 at 22:34
That seems a more valid answer, feel free to post it as such - even though if I'm probably not going to understand it fully :-) (I'll try tough) – Joubarc Dec 27 '12 at 5:10
It's not really a proof -- the Enormous Theorem classifies all finite groups. So far, all high level Steiner systems, such as the Witt design, have had incredible symmetry. But that doesn't rule out a design lacking in symmetry. This is like a 80% proof. There is no super-symmetrical S(6,9,45) design. – Ed Pegg Dec 27 '12 at 14:17
OK, I think I see your point - and I guess I'm not too convinced one could exist without symmetry. But what about the subsystems then? If either S(5,8,44), S(4,7,43) or S(3,6,42) can be proven not to exist, that would be enough too – Joubarc Dec 27 '12 at 17:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.