Guaranteed to guess two numbers in one from twelve lottery tickets, if falls $6$ balls with numbers from $1$ to $36$

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

$10$ 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.