The Expected Number of Trials with Bold Play

Java Applet Interactive red and black game

Java Applet Simulation of the red and black experiment

We will now study the expected number of trials needed with bold play. Several results in our study of the probability of winning with bold play will be important, so you need to review this section.

First, when the gambler plays boldly, only the ratio of the initial fortune to the target fortune matters. Thus, we will let a = 1 and define

G(x) = E(N | X₀ = x) for x in [0, 1].

For any other value of a, and any x in [0, a], the expected number of trials is just G(x / a).

1. By conditioning on the outcome of the first trial, show that G satisfies the functional equation

G(x) = 1 + pG(2x) for x in (0, 1 / 2]
G(x) = 1 + qG(2x - 1) for x in [1 / 2, 1)

and show that G satisfies the boundary conditions G(0) = 0, G(1) = 0.

Note, interestingly, that the functional equation is not satisfied at x = 0 or x = 1. As before, the binary representation of an initial fortune x in (0, 1) will be fundamental in our analysis.

2. Suppose that the initial fortune of the gambler is a binary rational x in (0, 1). Show that

N = min{k = 1, 2, ...: I_k = u_k(x) or k = n(x)}

Hence, the possible values of N are 1, 2, ..., n(x).

3. Suppose that initial fortune of the gambler is a binary irrational x in (0, 1). Show that

N = min{k = 1, 2, ...: I_k = u_k(x)}

Hence, the possible values of N are 1, 2, ....

We can give an explicit formula for the expected number of trials G(x) in terms of the binary representation of x.

4. Suppose that x in (0, 1) is a binary rational. Show that

5. Use the result of Exercise 4 to show that

G(1 / 8) = 1 + p + p², G(2 / 8) = 1 + p, G(3 / 8) = 1 + p + pq, G(4 / 8) = 1
G(5 / 8) = 1 + q + pq, G(6 / 8) = 1 + q, G(7 / 8) = 1 + q + q²

6. Suppose that x in (0, 1) is a binary irrational. Show that

7. Suppose that p = 1 / 2. Show that

G(x) = 2 - 1 / 2^{n(x) - 1} if x is a binary rational
G(x) = 2 if x is a binary irrational

8. In the red and black experiment, select Bold Play. Vary x, a, and p with the scroll bars and note how the expected number of trials changes. In particular, note that the mean depends only on the ratio x / a. Now set a = 64, x = 31, and p = 0.5. Run the experiment 1000 times with an update frequency of 100 and note the apparent convergence of the sample mean to the distribution mean.

9. For fixed x, show that G is continuous as a function of p.

As a function of the initial fortune x and for fixed p, the function G is very irregular. In fact, G is discontinuous at the binary rationals in [0, 1] and continuous at every other point. The following exercises are the steps in the proof.

10. Show that for b > 0, there exists M such that for any x

11. Suppose that x in (0, 1) is a binary irrational. For the M in Exercise 10, there is a binary interval of rank M containing x:

k / 2^M < x < (k + 1) / 2^M

Show that if y is in this interval, then x and y have the same binary digits, up to order M - 1 and hence

|G(y) - G(x)| < b

12 Suppose that x is a binary rational of rank n. For m = 1, 2, ... define y_m by

u_i(y_m) = u_i(x) for i = 1, 2, ..., n
u_i(y_m) = 1 for i = n + m
u_i(y_m) = 0 otherwise

Show that y_m converges to x as m increases but that

G(x) < G(y₁) < G(y₂) < ···

Red and Black