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During the drawing lottery falls six balls with numbers from $1$ to $36$. The player buys the ticket and writes in it the numbers of six balls, which in his opinion will fall out during the drawing lottery. The player wants to buy several lottery tickets in order to be guaranteed to guess at least two numbers in at least one ticket. Will there be enough to buy $12$ lottery tickets?

My work. The maximum number of numbers pairs in $12$ tickets is equal to $12 \binom{6}{2}=12 \cdot 15$. During the drawing lottery falls $ \binom{6}{2}=15$ numbers pairs. The total number of numbers pairs is equal to $\binom{36}{2}=18 \cdot 35$. I have no ideas of solving the problem.

RobPratt
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Witold
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    This is apparently from a contest. Can you give a link to the contest site to make sure that the contest is over, please? You see, we have a strict policy not to discuss questions from on-going contests- Many regulars have some connections with past and present contests, and are keen to protect the integrity of contests organized by others also. – Jyrki Lahtonen Jun 27 '19 at 12:08
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    http://tym.in.ua/2019/05/30/tym-2019-problems/ – Witold Jun 27 '19 at 12:14
  • Thanks ${}{}{}$ – Jyrki Lahtonen Jun 27 '19 at 12:22

3 Answers3

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Best known is 47 tickets.
The lower bound would be 42 tickets, but no-one has found a solution with fewer than 47 tickets.

A table of similar results is at La Jolla Covering Repository.

Ed Pegg
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    https://en.wikipedia.org/wiki/Block_design – Ed Pegg Jun 27 '19 at 19:16
  • In my opinion, you wrote a solution of another problem. In the table from your link is written that $C(4,3,2)=3$. But in this problem $C(4,3,2)=1$. – Witold Jun 28 '19 at 21:00
  • Solution of the problem is incorrect. – Witold Jun 29 '19 at 12:10
  • @Witold Please could you explain why you think this is incorrect. In particular, what is the relevance of the $C(4,3,2)$ covering to the original problem? Also, I think you can leave out one ticket from the $47$ ticket solution, as long as it contains a duplicated pair. – nickgard Jun 29 '19 at 12:38
  • First, the author did not write a solution to the problem, but only gave a link to some tables. Secondly, I searched in the internet for a solution of "lotto design". This solution of "lotto design" requires each number to enter at least one ticket. But this is not required. This is a completely another problem. In this example $C(4,3,2)$, it is easy to see that this is another problem. – Witold Jun 29 '19 at 12:55
  • "This is a completely another problem." I (among others) don't think so. – leonbloy Jun 29 '19 at 13:10
  • In another answer Ross Millikan proved that $16$ tickets are enough! But the author of this answer writes about $42$ tickets. This is a completely another problem. – Witold Jun 29 '19 at 13:24
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    I believe the difference in the two problems is whether the tickets must cover every 2-subset of 1..36 or instead just cover at least one 2-subset of every 6-subset of 1..36. – RobPratt Jun 29 '19 at 13:50
  • In this problem said " at least one ticket". – Witold Jun 29 '19 at 13:52
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In fact, 9 tickets are enough. See this paper.


 1  2  3  4  5  6 
 7  8  9 10 11 12 
13 14 15 16 17 18 
19 20 21 22 23 24 
19 20 21 25 26 27 
22 23 24 25 26 27 
28 29 30 31 32 33 
28 29 30 34 35 36 
31 32 33 34 35 36

This is a "sum of disjoint covers" obtained from three copies of C(6,6,2) and two copies of C(9,6,2), as described in Theorem 2.6.

RobPratt
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  • It written that $L (35,6,6,2) = 9$ (Lemma 4.8). Where is it written $L (36,6,6,2) = 9$? If you write a solution, then I will accept your answer as the best. – Witold Jun 29 '19 at 14:17
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    Theorem 2,3 here shows $L(n,6,6,2)$ for $35 \le n \le 54$. – RobPratt Jun 29 '19 at 14:56
  • Very nice solution! Thank you very much! For technical reasons I can reward you $150$ bounty in $19$ hours. – Witold Jun 29 '19 at 16:31
  • Strangely, the app on my phone seems to ignore the line break before each line of numbers. On my computer in a browser, the linebreaks are shown as desired. I think this is just an issue with the app, but a MathJax array would work everywhere, I think. – David K Jun 29 '19 at 18:27
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$10$ tickets are enough. Start by buying $(1,2,3,4,5,6)$ and the other five blocks of six numbers. If you have not won yet there must be one number in each block of six. Now buy $(1,2,3,7,8,9),(1,2,3,10,11,12),(4,5,6,7,8,9)$ and $(4,5,6,10,11,12)$. One of these last four must win because the number in $1-6$ and the number in $7-12$ are both in one of them.

Thanks to David K for the correction.

Ross Millikan
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  • A very nice example of $16$ tickets! Thanks! – Witold Jun 29 '19 at 13:22
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    I don’t see the need for the second batch of six tickets. If the first six don’t win, you know the draw included one number from the first ticket and one from the second. That is, of the $36$ such pairs that are possible, at least one was drawn. Each of your last four tickets covers a disjoint subset of $9$ of those pairs, so a win is guaranteed. What am I missing? – David K Jun 29 '19 at 13:26
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    @DavidK; You are not missing anything. Thanks much. – Ross Millikan Jun 29 '19 at 13:29
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    @HardikKalra: I can write the numbers I gave on $10$ tickets. Then when six numbers are drawn, at least one ticket will have two winning numbers. That seemed to be what you asked for. – Ross Millikan Jun 29 '19 at 13:38
  • Very nice solution! Thank you very much! – Witold Jun 29 '19 at 13:43