The Probability of Winning with Bold Play

Java Applet Interactive red and black game

Java Applet Simulation of the red and black experiment

Bold Play

Recall that with bold play, the gambler on each trial bets either his entire fortune or the amount needed to reach the target fortune, whichever is smaller. We are interested in the probability that the player reaches the target. The first interesting fact is that only the ratio of the initial fortune to the target fortune matters, quite in contrast to timid play.

1. Suppose that the gambler plays boldly with initial fortune x and target fortune a. Argue that for any c > 0, the process

cX_i, i = 0, 1, 2, ...

is the fortune process for bold play with initial fortune cx and target fortune ca.

The Probability of Winning

Because of the result in Exercise 1, it is convenient to use the target fortune as the monetary unit and to allow irrational, as well as rational, initial fortunes. Thus, we will denote the probability that the bold gambler reaches a = 1 starting from x in [0, 1] by F(x). By Exercise 1, the probability that the bold gambler reaches some other value of a, starting from x in [0, a], is F(x / a).

2. By conditioning on the outcome of the first trial, show that F satisfies the functional equation

F(x) = pF(2x) for x in [0, 1 / 2]
F(x) = p + qF(2x - 1) for x in [1 / 2, 1]

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

Binary Expansions

One of the keys to our analysis is to represent the initial fortune in binary form. The binary expansion of x in [0, 1) is

where u_i is in {0, 1} for each i. This representation is unique except when x is a binary rational of the form

x = k / 2ⁿ where n = 1, 2, ... and k = 1, 2, ... 2ⁿ - 1

The smallest possible value of n in this representation (after dividing out common powers of 2), is called the rank of x. For a binary rational x of rank n, we will use the standard terminating representation where

u_n = 1 and u_i = 0 for i > n

Rank can be extended to all numbers in [0, 1) by defining the rank of 0 to be 0 (0 is also considered a binary rational) and by defining the rank of a binary irrational to be infinity.

Thus, we will define the following functions of x in [0, 1):

• u_i(x) = i'th coordinate in the binary expansion of x
• z_i(x) = k - [u₀(x) + u₁(x) + ··· + u_k(x)] = the number of 0's in the first k binary digits.
• n(x) = the rank of x.

3. Show that

u_i(2x) = u_{i + 1}(x) if x in (0, 1 / 2)
u_i(2x - 1) = u_{i + 1}(x) if x in [1 / 2, 0)

4. Suppose that gambler starts with initial fortune x in (0, 1). Show that the gambler eventually reaches the target 1 if and only if there exists a positive integer k such that

I_j = 1 - u_j(x) for j = 1, 2, ..., k - 1 and I_k = u_k(x)

We will now introduce an interesting random variable that will play a fundamental role in our analysis. Let

Note that W is a well defined random variable and that W takes values in [0, 1].

5. Suppose that the gambler starts with initial fortune x in (0, 1). Use the result of Exercise 4 to show that the gambler reaches the target 1 if and only if W < x.

6. Show that W has a continuous distribution. That is, show that P(W = x) = 0 for any x in [0, 1].

From the results of Exercises 5 and 6, it follows that F is simply the distribution function of W. In particular, F is an increasing function, and since W is continuous, F is a continuous function.

For Exercises 7–10 below, let

x = k / 2^m where m in {1, 2, ...}, k in {0, 1, ... 2^m - 1} and y = (k + 1) / 2^m

7. Show that either x or y has rank m.

8. Argue that the only sequence of outcomes that results in ruin for the gambler when the initial fortune is x, but success when the initial fortune is y is the sequence

I_j = u_j(x) - 1 for j = 1, 2, ..., m

9. Use the result of Exercise 8 to show that

F(y) = F(x) + p^zm(x) q^{m - zm(x)}.

10. Show that

11. Show that

F(1 / 8) = p³, F(2 / 8) = p², F(3 / 8) = p² + p²q, F(4 / 8) = p
F(5 / 8) = p + p²q, F(6 / 8) = p + pq, F(7 / 8) = p + pq + pq²

12. Use the result of Exercise 9 to show that F is strictly increasing on [0, 1]. This means that the distribution of W has support [0, 1]; that is, there are no subintervals of [0, 1] that have positive length, but 0 probability.

13. Use induction on the rank to show that any two solutions of the functional equation in Exercise 2 must agree at the binary rationals. Therefore, any solution of the functional equation in Exercise 2 must satisfy the results in Exercises 9 and 10.

14. Use the result of Exercise 13 to show that F is the unique continuous solution to the functional equation in Exercise 2.

15. Suppose that p = 1 / 2. Show that F(x) = x satisfies the functional equation in Exercise 2.

Thus, for fair trials, the probability that the bold gambler reaches a starting from x is x/a, just as it is for the timid gambler.

From Exercise 15, note that when p = 1 / 2, W has the uniform distribution on [0, 1]. When p is not 1 / 2, the distribution of W is quite strange. To state the result succinctly, we will indicate the dependence of the of the probability measure P on the parameter p. Now, define

that is, the set of x in (0, 1) for which the relative frequency of 0's in the binary expansion is p.

16. Use the strong law of large numbers to show that

From Exercise 16, it follows that when p is not 1/2, W does not have a density, even though it is a continuous random variable. Our proof is by contradiction; if W did have a density f then

In turn, this means that when p is not 1/2, F has derivative 0 at almost every point in [0, 1]. The following picture shows the graphs of F when p = 0.2 and 0.4.

17. In the red and black experiment, select Bold Play. Vary x, a, and p with the scroll bars and note how the distribution of final fortune changes. In particular, note that the win distribution depends only on x / a. Now with a = 64, x = 24, and p = 0.48, run the experiment 1000 times with an update frequency of 100 and note the apparent convergence of the relative frequency function to the density function.

Red and Black