Lottery Math (different combinations)

I can't answer the minimum, but I can suggest a computer process that will give you an upper bound, which I suspect will be close. There are ${60 \choose 6}=50,063,860$ possible draws. Each ticket covers ${15 \choose 6}=5005$ of them. Start by dividing the numbers into batches of $15$ with as even overlap as possible. Your first four should be $1-15, 16-30, 31-45, 46-60$ Then split each group of $15$ into $4,4,4,3$ and make new tickets with that many from each group, so things like $1-4,16-19,31-34,46-48$ and so on. Keep an array of length $50,063,860$ showing all the combinations you have accounted for by making the corresponding entry $1$. After you get tired of specifying "smooth" combinations, start picking random tickets-choose $15$ random numbers and count how many new combinations each ticket accounts for. Pick, say, $100$ tickets and keep the one that covers the most new combinations. Try it for a few different starting combinations (maybe even start doing random tickets initially). Toward the end you may have to shift to starting with a set of six that is not yet covered and see what you can add best. When you have covered them all, you have a set of tickets which you have proven will cover all the possibilities. It will not prove that your group is minimum, but it may not be too bad, especially if you try a number of times.