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

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.

$10$ 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.

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.