You want to buy a given number $n$ of lottery tickets, in each of which you have to guess exactly 6 numbers from 1 to 100. If you get 0 or 1 numbers right, you don't get any money. If you get $i>1$ numbers right, you'll get $x_i$ dollars (i.e. if your guess was 1,2,3,4,5,6 but the numbers were 1,2,3,4,5,7 you'll win $x_5$).
Is there an optimal strategy for choosing the numbers on the $n$ lottery tickets?
For $n=2$ (actually this guess generalizes for all $n \leq 16$), my guess is that betting on, say, $(1, 2, 3, 4, 5, 6)$ and $(1, 2, 3, 4, 5, 7)$ will result on a lower expected value of money won than, say, $(1, 2, 3, 4, 5, 6)$ and $(7,8,9,10,11,12)$ because the values "are more different" which will result in more different $k$-uples available. I wonder how I will be able to continue using this strategy when $n$ gets over $16$ (we will have to repeat numbers).
The chance that any 1 number is correct is $\frac{6}{100}$. It should be clear that when we pick our first ticket, all options are equivalent, so we can set it to (1, 2, 3, 4, 5, 6). Then, we can consider whether to choose overlapping numbers, or new numbers for the next ticket. Let us look at the extremes. We could choose the exact same numbers again for the second ticket. In this case, we double the expected value, as whichever payout we would get from the first ticket, it is doubled by the second ticket. Alternatively, we could choose all different numbers. In this case, we also double the expected value, as the chance that some amount of our numbers are correct is doubled by doubling our guesses. I haven't worked it out explicitly, but it should be clear that this holds true for all other choices of how many numbers should overlap.
Intuitively, choosing overlapping numbers increases the reward for winning, and choosing non-overlapping numbers increases the chance of winning, which have the same impact on the expected value.