Why is the payoff range at sic bo so wide?

Question: In sic-bo, payouts for various three-dice totals run from a high of "1 wins 60" for a four or 17, to a low of "1 wins 6" for a nine, 10, 11, or 12. Why is the spread so wide?

Answer: The spread in payouts counterbalances the spread in the chances of making each total. "Outside" numbers are toughest and pay most. "Inside" totals are easiest and pay least. Intermediate totals fall between these limits.

For instance, there are only three ways to make four (1-1-2, 1-2-1, and 2-1-1) and three to make 17 (6-6-5, 6-5-6, and 5-6-6). I won't enumerate ways of totaling nine, 10, 11, or 12 -- but it works out to 25 each for nine and 12, and 27 each for 10 and 11.

Altogether, three dice can land in 6 x 6 x 6 or 216 ways. So if you bet on either four or 17, you have three ways to win and 216 - 3 or 213 ways to lose. Odds are therefore 213-to-3, which equals 71-to-1. Payouts are a correspondingly high 60-to-1.

Similarly, betting on either 10 or 11, you have 27 ways to win and 216 - 27 or 189 ways to lose. Odds are therefore 189-to-27 or 7-to-1. Payouts are a correspondingly low 6-to-1.

Numerically, the odds against winning always exceed the payout. In the above, 71-to-1 versus 60-to-1 on four and 17, 7-to-1 versus 6-to-1 on 10 and 11. If chances and payouts were equal, the casino would have no edge. The difference between these values is how the house gets its advantage on every bet you make.