Calculating a Hold Percentage

1 October 2000

By Donald Catlin

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 W_x 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

Wx = pWx+1 + qWx-1

(1)

This equation just says that the probability of success with x units is the same as either the probability of success after winning the current trial with probability p or losing it with probability q. There are also a couple of boundary conditions

W0 = 0 and Wa = 1

(2)

Do you see why?

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

Wx = (q/p)x

(3)

It is not hard to see that this works. If we substitute (3) into the right-hand side of (1) and factor out (q/p)^x-1 we obtain an expression of the form

right side = (q/p)x-1[p(q/p)2 + q]

(4)

The expression in the brackets in (4) reduces to q/p (using the fact that q+ p =1). In other words, the right-hand side reduces to (q/p)^x, which according to (3) is just W_x.

Very well, our general solution to (1) is

Wx= c + d(q/p)x

(5)

where the constants c and d are yet to be determined. These are determined by imposing the constraints in (2):

c + d = W0 = 0 c + d(q/p)a = Wa = 1

(6)

Solving the system in (6) for c and d we obtain

d = 1/[(q/p)a - 1] c = -1/[(q/p)a - 1]

(7)

Substituting the expressions in (7) into the general solution (5), it follows that

Wx = [(q/p)x - 1] / [(q/p)a - 1]

(8)

The expression (8) is our probability of success. Obviously 1 - W_x represents the probability of ruin.

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:

expected return = Wx (a - x) + (1 - Wx )(- x)

(9)

or

expected return = aWx - x

(10)

From the casino's point of view, the negative of the expression in (10) is the amount of the player's buy-in that, on average, they expect to retain. The hold percentage in this case, therefore, is simply given by

hold percentage = 100[x - aWx] / x

(11)

Notice that if a = 2x (double your money) then the hold percentage is simply 1 - 2W_x. Of course, different players will use different values for x and a and not all players make unit bets throughout their play and this, as we have seen, means that there is no theoretical hold percentage for an actual game; it is an empirical figure.

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 W_x = 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 N_x 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 pN_x+1 + qN_x-1. It follows, therefore, that

Nx = pNx+1 + qNx-1 + 1

(12)

The boundary conditions are

N0 = 0 and Na = 0

(13)

since if either level a or ruin occur there are no more trials. Since we have already solved the Gambler's Ruin Problem, the present problem is easily solved via the formula

Nx = [x - aWx] / (q - p)

(14)

It is easy to see that the formula (14) satisfies the conditions in (13). To see that (14) holds as well, simply substitute the formula on the right side of (14) into the right side of (12):

[p[x+1 - aWx+1] / (q - p) + q[x-1 - aWx-1] / (q - p) + 1

(15)

which reduces to

[x + (p - q) - a( pWx+1 + qWx-1)] / (q - p) + 1

(16)

Recalling equality (1) we see that the third term in the brackets is just W_x and noting that the term (p - q) / (q - p) is just -1, we have that (16) reduces to

[x - aWx] / (q - p) - 1 + 1 = [x - aWx] / (q - p)

(17)

which is exactly the right-hand side of (14). This proves that the right-hand side of (14) is indeed the correct expression for N_x.

Very well, if we buy into a game with x units, we can expect to make N_x unit bets before we either go broke or reach level a, N_x being given by (14). Since our expected loss is given by x - aW_x (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.