You have 31 positive payout guesses (1 $5, 2 $4, 4 $3, 8 $2 and 16 $1) leaving 69 other numbers with zero or negative payouts. You don't want to have gaps larg

You have 31 positive payout guesses (1 $5, 2 $4, 4 $3, 8 $2 and 16 $1) leaving 69 other numbers with zero or negative payouts. You don't want to have gaps larger than three between your positive guesses, but there are 32 gaps for a total of 96 possibilities, or an excess of 27 over the numbers you need to cover. It seems like a lot of possibilities and I think you can get away with a minimum gap size of one, but let's assume you do 5 3-gaps at 1, 25, 50, 75, and 100 and 2-gaps everywhere else. So start with 51, then 26 and 76. Then go up or down 12, then 6, then 3. If you have a gap of two you flip a coin, if a gap of three you pick the middle one. Or if you have them write down the number and you think it has double-digits you could put your 4-gaps below 20. Start with 53 and go up or down 24, 12, 6, and 3 (unless it is below 20, then it is multiples of four.) 59 would pay you a dollar. Your starting guess could be anywhere from 37 to 64 without paying out more than a dollar, but if you start with an extreme, then low odd numbers and high even numbers will have a negative payout. However, I think you can still randomize sufficiently starting with 38 and 63, e.g. 63-31-15-7-3-1.