## The gamblers ruin in a biased game

Most casinos have a slight edge on the gamblers playing there. This means that on average your holdings will decrease (the casino's edge) at rate m, as well as change due to the random fluctuations of the game. To capture this idea, we replace Eq. (7.11) by dX = X (t + dt)-X (t) = - mdt + dW (7.20)

Show that dXis normally distributed with mean -mdt and variance dt + o(dt) by evaluating E{dX} and E{dX2} using Eq. (7.20) and the results of Exercise 7.1.

As before, we compute w(x), the probability that X(t) hits C before 0, but now we recognize that the average must be over dX rather than dW, since the holdings change from x to x + dX due to deterministic (mdt) and stochastic (dW) factors. The analog of Eq. (7.13) is then u(x) = EdX{u(x + dX)} = Edr{m(x - mdt + dW)} (7.21)

We now Taylor expand and combine higher powers of dt and dW into a term that is o(dt)

u(x) = EdX|w(x) + (—mdt + dW )ux + 2 (-mdt + dW )2mii + o(dt)

We expand the squared term, recognizing that O(dW2) will be order dt, take the average over dX, divide by dt and let dt! 0 (you should write out all of these steps if any one of them is not clear to you) to obtain

which we need to solve with the same boundary conditions as before w(0) = 0, w(C) = 1. There are at least two ways of solving Eq. (7.23). I will demonstrate one; the other uses the same method that we used in Chapter 2 to deal with the von Bertalanffy equation for growth.

Let us set w = wx, so that Eq. (7.23) can be rewritten as wx = 2mw, for which we immediately recognize the solution w(x) = k1e2mx, where k1 is a constant. Since w(x) is the derivative of w(x) we integrate again to obtain

where k2 and k3 are constants and, to be certain that we are on the same page, try the next exercise.

What is the relationship between kj and k2?

When we apply the boundary condition that w(0) = 0, we conclude that k3 = —k2, and when we apply the boundary condition w(C) = 1, we conclude that k2 = 1/(e2mC — 1). We thus have the solution for the probability of reaching the limit of the casino in a biased game:

and now things are very bleak: the chance that you win is, for almost all situations, vanishingly small (Figure 7.6).

Once again, we can ask about how long you can stay in the game and, possibly, about connections between the biased and fair gambles. I leave both of these as exercises.

Figure 7.6. When the game is biased, the chance of reaching the limit of the casino before going broke is vanishingly small. Here I show u(x) given by Eq. (7.25) for m = 0.1 and C = 100. Note that if you start with even 90% of the casino limit, the situation is not very good. Most of us would start with x < C and should thus just enjoy the game (or develop a system to reduce the value of m, or even change its sign.)

## Post a comment