Solution to the Problem:
The answers are 40 tickets to be certain of winning and 10 tickets to have a 51% chance of winning.
Here is the solution:
1. To be certain of winning, you need to buy 40 tickets, having the numbers
1, 2, 3, 4, 5, 6, 8, 9, 10, 12,
15, 16, 18, 20, 24, 25, 27, 30, 32, 36,
40, 45, 48, 50, 54, 60, 64, 72, 75, 80,
90, 96, 100, 108, 120, 125, 144, 150,
180, and 216.
These are the only possible products from the three dice.
2. To have a 51% chance of winning, you need to buy 10 tickets:
6, 12, 18, 20, 24, 30, 36, 48, 60, & 72.
These are the products of 111 ordered combinations, which is a little more than 51% of 216
(the products 6, 18, 20, 48, and 72 can each be formed by nine ordered combinations -- called permuations; For example,
6 can be obtained by multiplying 1x1x6, 1x6x1, 6x1x1, 1x2x3, 1x3x2, 2x1x3, 2x3x1, 3x1x2, or 3x2x1.
The products 30, 36, and 60 can each be formed 12 ways with three dice, while the products 12 and 24 can each be formed 15 ways).