No, my question isn’t “how do you cook them?”

What I’d like to know is whether laying odds on a Don’t Pass bet is a good idea.

For those who’ve never played craps, the shooter throws two dice. If a 7 or 11 comes up, the pass line wins. If a 2, 3, or 12 comes up, the pass line loses. Anything else is his “point”, and the shooter must then roll his “point” again before he rolls a 7. If he does make his point, then the pass line wins and the cycle starts over. (If he rolls a 7, then the pass line loses and the dice are given to the next shooter; that’s called ‘fading’ or ‘sevening-out’.)

If you’re betting the Don’t Pass line, you lose if the first roll is 7 or 11, you win if it’s 2 or 3, and you push if it’s 12. (Sometimes the push is for 2 instead of 12, it will be noted on the table.) And if the shooter rolls anything else, you win your bet if he rolls a 7 before the point comes up again. Since 7 is the most common combination, once the shooter establishes a point you have a positive expectancy on your bet: you get paid even money on something that will happen more than half the time.

You have the option of laying odds once a point is established. If the point is 6 or 8, you lay 6-5 odds. (For the sake of simplicity, let’s say your Don’t Pass bet is $10.) So, on a 6 or 8, you’d put up an additional $12 to win $10; on 5 or 9, you’d have to lay $15 to win $10; on 4 or 10, you’d have to lay $20 to win $10. You would get this $10 in addition to the $10 you put on the Don’t Pass line.

My question is, does laying odds make statistical sense? If your Don’t Pass bet has a positive expectancy at even money, wouldn’t laying true odds make your total bet less of a positive expectancy? Or does the ability to add to your wager and still keep the positive expectancy (even though it has been diminished) help overcome the negative expectancy on the first roll? (Remember that on the first roll, you’d lose on 8 of the 36 possible combinations while winning on only 3.)