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An Optimal Craps Wager

1 May 2011

By Donald Catlin

The following is part of an email I received a while ago from one of my readers.

My question has to do with "dice control." Assuming that you could come up with one less seven every 36 rolls (in other words only 5 sevens in 36 rolls), what would be the optimum profit margin to attain before raising your Pass Line bet?

Although I believe in dice control, having witnessed it in action, there are different dice sets one uses depending upon what one wishes to accomplish. For example, on the come out roll I would probably use either the All-Sevens or the YO set. If either the six or eight is established as a point, I would use the 3-V set. If a four or ten is established I would use the 2-V set. And so on. (See Frank Scoblete's book Casino Craps, Shoot to Win, Triumph Books, 2010 for a description of these sets.) Each of these would probably produce different frequencies for the seven, so analyzing the Pass Line is a real quagmire. Of course, we could pick one set and stick with it, but this would not be optimal play. Therefore, rather than answer your question I am going to answer a related, albeit, simpler question: When should you raise a place 8 bet? Even here there are some questionable assumptions one must make and I'll point them out below.

Assuming a fixed dice set (such as the 3-V) and assuming that the probability of the seven could be reduced to 5/36, the probability of the remaining outcomes would be 31/36. Since there are 30 such outcomes, the probability of each is 31/36 x 1/30. Or is it? This assumes that each outcome is equally likely and I doubt if this is true. On the other hand, with no additional information (such as empirical data), I don't see any viable alternative. Since there are 5 outcomes that lead to an eight, we are assuming that the probability of an eight is 31/36 x 5/30 or 31/216. If we write 5/36 as 30/216 we can imagine that in 216 rolls the seven will occur on average 30 times and the eight 31 times. Ignoring all other outcomes, the probability of a seven occurring before an eight is 30/61 and the probability of the eight occurring before the seven is 31/61. The expected return for a 6 unit wager is therefore

7 x 31/61 - 6 x 30/61 = 37/61

Since this is a 6-unit wager, the expected return per unit wagered is the above figure divided by 6 or 37/366. This last figure is approximately 0.1.

Recalling my article last month, if one is wagering using Kelly Betting, then the wager should be 0.1 times the stake -- a stake of 60 units for a 6-unit bet.

When should you raise your stake? Since you have to bet in multiples of 6, you can round 10% of your current stake to the nearest multiple of 6. Thus if your stake is in excess of 90 units, you should bet 12 units. If your stake grows to in excess of 150 units, you should bet 18 units. If you stake falls below 150 units, you should drop back to a wager of 12 units. And so on. See you next month.


Don Catlin can be reached at 711cat@comcast.net

Donald Catlin
Don Catlin is a retired professor of mathematics and statistics from the University of Massachusetts. His original research area was in Stochastic Estimation applied to submarine navigation problems but has spent the last several years doing gaming analysis for gaming developers and writing about gaming. He is the author of The Lottery Book, The Truth Behind the Numbers published by Bonus books.

Books by Donald Catlin:

Lottery Book: The Truth Behind the Numbers