So, the constraints are as follows:
- First 5 numbers are digits 1 ; 69
- The 6th number is a number 1 - 26
- Order does not matter
- Numbers do not repeat
- Tickets cost $2 each
My reasoning is as follows: for the first 5 digits, there are clearly $$69 \times 68 \times 67 \times 66 \times 65$$ possibilities.
For the 6th digit, there are 26 possible choices, but we must account for the fact that the previous digits could've been within the range of $[1, 26]$ (or we end up overcounting). This is where I'm a little less certain: clearly I can't just multiply the previous number by 26. The "worst" case is where all 5 previous numbers were also in the range of $[1, 26]$; the "best" case is where all of them were in the range $[27, 69]$. My thought process was to just multiply my previous number by $21$ (because $26 - 5 = 21$), but I think that this may be undercounting because there are $$43 \times 42 \times 41 \times 40 \times 39$$ possible numbers that they could take that are greater than 26.
It's easy to account for the $2 tickets because if I want to calculate how much money it would take to buy enough tickets to guarantee a win, I'd just have to multiply my previous result by 2.
So, my proposed solution is that it would cost $$69 \times 68 \times 67 \times 66 \times 65 \times 21 \times 2$$ dollars to buy enough Lottery tickets to guarantee that you would win.
Can someone help me with the number of possible values for the final digits? Is $21$ correct, or are there more?