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Best of Alan Krigman
How would you set up place bets in a craps game with three five-sided dice?21 February 2011
Big bucks can be made by developing casino games that the joints adopt and players frequent. This, owing to the ongoing hefty fees the bosses pay to license proprietary offerings. However, of the numerous games being devised, few get considered seriously. And, of those that do, only a tiny fraction attain the trial stage, let alone permanent spots in the gambling repertoire.
Many proposed games are rejected because profit-minded casino bigwigs doubt they'll earn the casino more than the money generated by what's already available, for instance by pulling in new patrons rather than siphoning existing clientele off what they already enjoy, rekindling interest among solid citizens bored or dispirited with traditional gambling fare, or increasing the "hold" without discouraging players by depleting their bankrolls too quickly.
Some supposed innovations meet the subjective criteria, yet are flawed because the math doesn't wash. You'd be surprised at how many inventors are clueless about how edge evolves in casino games, and don't include asymmetries to provide house advantage. Or, worse, who try to make their games attractive by arbitrarily picking high payoffs that turn out to give away the store.
To get an idea how figures are derived so they make sense, picture a variation of craps in which totals are formed by adding the numbers from 1 through 5 on cards drawn from three special decks. The decks would comprise 15 cards, three each marked from 1 to 5. The dealer takes one card from each deck. Totals therefore range from 3 (1-1-1) to 15 (5-5-5). The accompanying table shows how many ways each possible total can be formed.
when formed from three random elements numbered 1 through 5
Total: 3 4 5 6 7 8 9 10 11 12 13 14 15 Ways: 1 3 6 10 15 18 19 18 15 10 6 3 1
For simplicity, pretend the only wagers allowed are Place bets on any numbers but nine. The bets win if the total of the three drawn cards matches the number in question; they lose if the sum is nine. The casino honchos might consider the game worth a look-see because card-based craps has already proven to be popular in California, restriction to Place bets eliminates much of the confusion that intimidates newbies at traditional dice tables , and numbers formed in relatively few ways are longshots and can therefore offer lucrative payoffs. But, what about edge?
Establishing edge in this game starts with the probabilities of winning and losing each bet. These figures are inherent in the structure of the game, and can be determined from the table of totals and ways. The 3 or 15, for instance, each has one way to win and 19 to lose. The full set of deciding draws for these numbers is accordingly 1 + 19 or 20, so the probabilities are 1/20 of success and 19/20 of failure. Probabilities for the other numbers are obtained similarly.
Payoffs for wins on the numbers must then be ascertained. High values would entice bettors. Low values are a trade-off between levels of edge sufficient to generate decent earnings for the house yet not so great as to deter players. One approach to finding suitable payoffs might start by using some high school algebra to calculate baselines for $1 bets that yield zero edge, then cutting the amount back to reach some medium both the bosses and the bettors can abide.
Here's how to get the zero-edge payoff for 3 or 15. Write a formula for edge that multiplies the "unknown" payoff (P) by 1/20, subtracts the product of the known $1 loss and 19/20, and equates the difference to zero. The result is P x (1/20) - 1 x (19/20) = 0. Solve for P (the algebra part) to get P = (19/20)/(1/20) or 19. Cut this to 18, put it back into the formula, and do the arithmetic. You'll have 18 x (1/20) - 1 x (19/20), which is -0.05, or -5 percent (minus indicating the house is favored). At 17, edge is 17 x (1/20) - 1 x (19/20), which equals -0.10 or -10 percent. A 5 percent edge would be reasonable for an 18-to-1 payoff; 10 percent would be excessive for 17-to-1.
The wicket grows stickier if the zero-edge equation with $1 bet gives a level for P where rounding down to the next whole dollar doesn't work. A seven or 11, for instance, has 15 ways to win and 19 to lose – 34 ways to reach a decision. The equation for a $1 bet with zero edge is P x (15/34) - 1 x (19/34) = 0. Solving for P yields 19/15 or $1.27. Cut the payoff down to $1 and edge would be too high at 1 x (15/34) - 1 x (19/34) or -11.8 percent. You could require that the bet be in multiples of $5, such that the zero-edge payoff was 5 x $1.27 or $6.35. Were a $5 bet to win $6, edge would be [6 x (15/34) - 5 x (19/34)]/5, or a more acceptable 2.9 percent.
Various kinds of biases may give rise to house edge. Some are as straightforward as the offset between the odds of winning and the payoff. Others are more complex and calculations go well beyond algebra and arithmetic. Regardless, games having levels and rules plucked from thin air, without mathematical analysis, are doomed. As the poet, Sumner A Ingmark, proclaimed:
Best of Alan Krigman