Playing it smart - Do you play craps by rote or do you understand the bets you make?

In "Diamonds Are Forever, "James Bond makes some dumb craps bets. He shoots and wins (it's a movie, not a casino). Plenty O'Toole purrs, "You handle those cubes like a monkey handles coconuts."

He'd have to, given the wagers in question. But, what if a solid citizen could actually handle those coconuts well enough to totally eliminate the chance of one side of one die appearing?

The accompanying table shows how many ways each final outcome and the total of all results could be made were this possible. The probability of any such result would be the number of ways divided by the total. For instance, excluding a two on one die, the probability of the spots adding up to eight would be 4/30.

Ways craps outcomes can be formed normally and with one face avoided on one die:

		face omitted from one die
	outcome	normal	1	2	3	4	5	6
	2	1	0	1	1	1	1	1
	3	2	1	1	2	2	2	2
	4	3	2	2	2	3	3	3	
	5	4	3	3	3	3	4	4
	6	5	4	4	4	4	4	5
	7	6	5	5	5	5	5	5
	8	5	5	4	4	4	4	4
	9	4	4	4	3	3	3	3
	10	3	3	3	3	2	2	2
	11	2	2	2	2	2	1	1
	12	1	1	1	1	1	1	0
	total	36	30	30	30	30	30	30

Here are examples of how to use this table to find edge, assuming standard payouts. You could do the arithmetic by hand, but nobody will call you a wimp for using a calculator.

Say you can keep one die from landing on a two. An 11, which pays 15-to-1, can normally be made two out of 36 possible ways (5-6 and 6-5); edge is (15/1)x(2/36) - 1x(34/36) = -11.1 percent "minus" because the house is favored. Excluding the two from one die face, the 11 can still be made two ways (5-6 and 6-5) but the base is 30, not 36. So the edge is (15/1)x(2/30) - 1x(28/30) = +6.67 percent "plus" because you have the advantage.

You can work out the edge for place bets almost as easily. Pretend you can exclude the six from one face. What's the edge on the 10? Normally, it's -6.67 percent. Excluding one six, you win two ways (5-5 or 4-6) and lose five (1-6, 2-5, 3-4, 4-3, and 5-2). So the chance of winning is 2/(2+5) and of losing is 5/(2+5). Payoff is 9-to-5. This makes edge (9/5)x(2/7) - 1x(5/7) = -20 percent. Edge for the other Place bets with one six omitted is +5.00 percent on four, +6.67 percent on five, +8.33 percent on six, -3.70 percent on eight, and -10 percent on nine.

Pass and Don't Pass are more complicated because you have to figure the come-out roll then the chance of each number being the point and winning or losing. A spread sheet program on a computer would help. Here are answers for those who trust me.

On Pass, you have the best of it by nearly 6 percent and just over 1 percent by excluding a one and a two, respectively. You're slightly under the gun if you can kill the six. Getting rid of either a three or four penalizes you by more than 3 percent. And avoiding a five brings the disadvantage well above 5 percent.

Don't Pass favors you by better than 2 percent avoiding a five and close to 1 percent without one six. You're slightly negative by excluding a three or four. But you get whacked with over 4 and 9 percent, respectively, if a two and one are among the missing.

True, this is all academic since it ain't apt to happen. But it's instructive in that the issue tests whether you really understand craps. Or, whether by merely learning the names of the bets and how to make them in the words of Oscar Wilde, you know the price of everything and the value of nothing. The punters' poet, Sumner A Ingmark, posed much the same puzzle like this:

Proficiency dictates that gamblers devote, An effort to learning what wagers denote, And not simply follow some system by rote.