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Last month, after explaining hold percentages, I remarked that there is no theoretical hold percentage for a game. I then went on to show the reason -- the whim of the player can affect the hold percentage. In my last paragraph, however, I claimed that if one makes an additional assumption about how a rational player might play, then one can calculate a hold percentage. Specifically, if one assumes that a player will buy into a game with stake s and agree to quit playing when he either goes broke or reaches an amount a, then we can calculate the hold percentage for such a player. In this article I'll show you how.
The Gambler's Ruin Problem
In order to carry out the aforementioned calculation, we need to develop a classical problem in probability theory known as the Gambler's Ruin Problem. This age-old problem assumes, as above, that a gambler buys into a game with s units and plays single unit wagers until he either goes broke or reaches an amount a. The assumption is that the game is a series of independent trials and that the probability of winning or losing at a single trial is p or q, respectively. We assume that p and q are not equal. The question is, what is the probability that the gambler goes broke? An equivalent question is, what is the probability that the gambler reaches his goal? We address the latter here.
Let Wx represent the probability that the gambler will reach wealth level a if his current stake is x units. Then with a bit of thought we can see that
It is easily seen that the sum of any two solutions to (1) is also a solution and that a constant multiple of any solution to (1) is again a solution. So all we need to do is find two solutions to (1), multiply each by an arbitrary constant and add them up. Well, one solution is easy; any constant is a solution to (1) (because p + q = 1). Finding a second solution is a bit harder. Fortunately for us, however, the mathematicians Laplace, De Moivre, Lagrange, and Bernoulli found the other solution many years ago. It is
Very well, our general solution to (1) is
The Hold Percentage
With the formula (8) in hand, it is now easy to calculate a hold percentage for a player who plays according to the procedure above. The player buys in for x units. If the player plays and finally wins, his profit is a - x. On the other hand, if he goes broke his profit is -x. The expected return for this player is:
Suppose that a Pass Line player at Craps buys in for $500, plays two $25 chips with no odds, and plays to either $1000 or until he is busted. Since the unit bet is $50, the player has bought in for 10 units and is playing for 20 units, that is x = 10 and a = 20. Since p = 976/1980 and q = 1004/1980, we have q/p = 1004/976. Using our trusty calculator we find that (q/p)10 = 1.326 and (q/p)20 = 1.761, whence by expression (8) we have Wx = 0.4284. From (11) it follows that the hold percentage is 14.32%. If you would like to test your understanding of these calculations, lower the unit bet to $25 and the hold percentage should rise to 27.52%.
The House Edge
We have seen that the manner of play affects the hold percentage. What about the house edge? The house advantage on a single trial is just 100(q - p). Does how we play affect this? To answer this we need to address an issue related to the Gambler's Ruin Problem, namely, how long can one expect to play? Let us define Nx to be the expected number of trials until either ruin or wealth level a occurs, given that the player's current wealth is x. If the player's current wealth is x, then the expected number of plays is just 1 plus the expected number of plays on the next trial. This latter figure is just pNx+1 + qNx-1. It follows, therefore, that
Very well, if we buy into a game with x units, we can expect to make Nx unit bets before we either go broke or reach level a, Nx being given by (14). Since our expected loss is given by x - aWx (see expression (10)), the house edge is just this last expression divided by the right-hand side of (14) expressed as a percentage. But if you remember your fifth grade fractions (to divide, invert and multiply), the result is just 100(q - p), exactly the same as for a single trial.
So, although you can change the hold percentage by altering your level of play as well as your goal, you can't change the house edge. In any case, whatever you do you can't change a negative game into a positive one. For some further discussion of these matters, specifically the effect of pressing when you're ahead, see my Easy as Pi! column in the Fall issue of The New Chance and Circumstance. See you next month.
This article is provided by the Frank Scoblete Network. Melissa A. Kaplan is the network's managing editor. If you would like to use this article on your website, please contact Casino City Press, the exclusive web syndication outlet for the Frank Scoblete Network. To contact Frank, please e-mail him at email@example.com.