The Ohio 5/37 game (Buckeye5) has 7 numbers that are multiple of 5. Seven combinations can cover all the possible combinations that include these number and 3 combinations can covered all the combinations of 4's if all 5 numbers of from that group, so far that has come within one number of happing nine times in the eleven years history of the game (2604 games).
A wheel to match 4 if all 5 numbers are multiple of 5.
05-10-15-20-25
05-10-15-30-35
05-20-25-30-35
Had you played this wheel all that time, the tickets would have cost $7,812 and you would have:
match 2 = 846 for $1 x 846 = $846
match 3 = 93 for $10 x 93 = $930
match 4 = 5 for 4 x $200 = $800
Total winnings = $2,576
for a cost of $7,812-$2,576 or $5,236 or ~$470 per year. If that ever happens, you would have a 3 of 7 chance of collecting $100,000.
There are unique conditions where all the winning numbers are from a small group of numbers(7-11) that happens several time during the life of a lottery game. The trick is to find such a group and then decide how much you are willing spend to cover that conditions until it happens again. The multiple of five was just an example, but the group could be as simple as all the winning numbers had appeared twice in the previous 10 drawings and etc. The smaller the group, the easier to cover. The hard part is finding the parameters that define such a group that has won several times in the past.
I've been working on finding such a group for several lotteries, Friday I used a small group for Megamillions that cost $8 to cover ~10% and match 1+1 for $3. I was lucky because only two numbers of the group came up and I had them covered, so I'm still looking for a smaller group that I can afford to cover $100%.
Good luck to you.
RJOh