Bold Play is Optimal When the Trials are Unfair

Java Applet Interactive red and black game

Java Applet Simulation of the red and black experiment

In this section, we will assume that p 1 / 2 so that the trials are unfair, or at least not favorable. We will show that bold play is optimal. As before, we will take the target fortune as the basic monetary unit and allow any valid fraction of this unit as a bet. Thus, the sets of fortunes and bets are

A = [0, 1], B_x = [0, min{x, 1 - x}] for x in A.

Let F denote the win probability function. for bold play. To show that bold play is optimal, we just need to verify that the condition for optimality holds:

1. pF(x + y) + qF(x - y) F(x) for x in A, y in B_x.

1. Show that the optimality condition (1) is equivalent to

D(r, s) = F[(r + s) / 2] - pF(s) - qF(r) 0 for 0 r s 1

2. Use the continuity of F to show that it suffices to prove the inequality in Exercise 1 when r and s are binary rationals.

We will now use induction on m to show that the inequality in Exercise 1 holds when r and s are binary rational with rank m or less, where m = 0, 1, ...

3. Show that the inequality in Exercise 2 holds when r and s have rank 0; that is, show that the inequality holds for

1. r = 0, s = 0,
2. r = 0, s = 1,
3. r = 1, s = 1.

Suppose now that the inequality in Exercise 1 holds when r and s have rank m or less, for a particular m. Suppose that r and s have rank m + 1 or less. We will show that the inequality holds in each of the four cases

1. r s 1 / 2
2. 1 / 2 r s
3. r (r + s) / 2 1 / 2 s
4. r 1 / 2 (r + s) / 2 s

The basic functional equation for F will be the main tool.

4. Show that in case (a),

D(r, s) = pD(2r, 2s)

5. Show that in case (b),

D(r, s) = qD(2r - 1, 2s - 1)

6. For case (c), fill in the details of the following steps:

D(r, s) = pF(r + s) - p[p + qF(2s - 1)] - q[pF(2r)]

But 1 / 2 r + s 1 so

F(r + s) = p + qF(2r + 2s - 1)

and 0 r + s - 1 / 2 1 / 2 so

F(r + s - 1 / 2) = pF(2r + 2s - 1)

Substituting gives

D(r, s) = q[F(r + s - 1 / 2) - pF(2s - 1) - pF(2r)]

If 2s - 1 2r then

D(r, s) = (q - p)F(2s - 1) + qD(2s - 1, 2r)

If 2r 2s - 1 then

D(r, s) = (q - p)F(2r) + qD(2r, 2s - 1)

7. For case (d), fill in the details of the following steps:

D(r, s) = [p + qF(r + s - 1)] - p[p + qF(2s - 1)] - q[pF(2r)]

But 0 r + s - 1 1 / 2 so

F(r + s - 1) = pF(2r + 2s - 2)

And 1 / 2 r + s - 1 1 so

F(r + s - 1 / 2) = p + qF(2r + 2s - 2)

Substituting gives

D(r, s) = p(q - p) + p[F(r + s - 1 / 2) - qF(2s - 1) - qF(2r)]

If 2s - 1 2r then

D(r, s) = p(q - p)[1 - F(2r)] + pD(2s - 1, 2r)

If 2r 2s - 1 then

D(r, s) = p(q - p)[1 - F(2s - 1)] + pD(2r, 2s - 1)

8. Use the induction hypothesis and the results of Exercises 4-7 to show that when r and s have rank m + 1 or less,

D(r, s) 0

We now know that bold play is optimal when the trials are unfair.

9. In the red and black game, set a = 16, x = 8, and p = 0.48. Define the strategy of your choice and play 100 games. Compare your relative frequency of wins with the probability of winning with bold play.

Red and Black