Virtual Laboratories > Red and Black > 1 [2] 3 4 5

2. Timid Play


Recall that with timid play, the gambler makes a small constant bet, say $1, on each trial until he stops. Thus, on each trial, the gambler's fortune either increases by 1 or decreases by 1, until the fortune reaches either 0 or the target a (a positive integer). Thus, the fortune process is a random walk with 0 and a as absorbing barriers. Recall that we denote this process by

Xi, i = 0, 1, 2, ...

As usual, we are interested in the probability of winning and the expected number of trials. The key idea in the analysis is that after each trial, the fortune process simply starts over again, but with a different initial value. This is an example of the Markov property and is fundamentally important in probability theory. Our analysis based on the Markov property suggests that we treat the initial fortune as a variable.

The Probability of Winning

We will denote the probability that the gambler reaches the target a, starting with an initial fortune x by

f(x) = P(XN = a | X0 = x) for x = 0, 1, ..., a.

Mathematical Exercise 1. By conditioning on the outcome of the first trial, show that f satisfies

  1. f(x) = qf(x - 1) + pf(x + 1) for x = 1, 2, ..., a - 1 (difference equation)
  2. f(0) = 0, f(a) = 1 (boundary conditions)

The difference equation in Exercise 1 is linear, homogeneous and second order.

Mathematical Exercise 2. Show that the characteristic equation of the difference equation in Exercise 1 is

pr2 - r + q = 0

and that the roots are r = 1 and r = q / p.

Mathematical Exercise 3. Show that if p is not 1/2 then the roots in Exercise 2 are distinct. Show that, in this case, the probability that the gambler reaches his target is

f(x) = [(q / p)x - 1] / [(q / p)a - 1] for x = 0, 1, ..., a.

Mathematical Exercise 4. Show that if p = 1/2, the characteristic equation has a single root 1 that has multiplicity 2. Show that, in this case, the probability that the gambler reaches his target is simply the ratio of the initial fortune to the target fortune:

f(x) = x / a for x = 0, 1, ..., a.

From Exercises 3 and 4, we have the distribution of the final fortune XN in all cases:

P(XN = 0 | X0 = x) = 1 - f(x), P(XN = a | X0 = x) = f(x).

Simulation Exercise 5. In the red and black experiment, choose Timid Play and set a = 32 and p = 0.45. Vary x form 0 to 32 with the scroll bar and note how the distribution of the final fortune changes. Now with x = 24, run the experiment 1000 times with an update frequency of 100 and note the apparent convergence of the relative frequency function to the true density.

Properties

Mathematical Exercise 6. Show that as a function of x, for fixed p and a, f(x) increases from 0 to 1 as x increases from 0 to a.

Simulation Exercise 7. In the red and black experiment choose Timid Play and set a = 64 and x = 16. Vary p from 0 to 1 with the scroll bar and note how the distribution of the final fortune changes. Now with p = 0.55, run the experiment 1000 times with an update frequency of 100 and note the apparent convergence of the relative frequency function to the true density.

Mathematical Exercise 8. Show that f(x) is continuous as a function of p, for fixed x and a. Specifically, use L'Hospital's Rule to show that the expression in Exercise 3 converges to the expression in Exercise 4, as p converges to 1/2.

Simulation Exercise 9. In the red and black experiment, choose Timid Play and set a = 64 and x = 32. Vary p from 0 to 1 with the scroll bar and note how the distribution of the final fortune changes. Now with p = 0.45, run the experiment 1000 times with an update frequency of 100 and note the apparent convergence of the relative frequency function to the true density.

Mathematical Exercise 10. Show that for fixed x and a, f(x) increases from 0 to 1 as p increases form 0 to 1.

Constant Bets

What happens if the gambler makes constant bets, but with an amount higher than 1? The answer to this question may give insight into what will happen with bold play.

Simulation Exercise 11. In the red and black game, set the target fortune to 16, the initial fortune to 8, and the win probability to 0.45. Play 10 games with each of the following strategies. Which seems to work best?

  1. Bet 1 on each trial (timid play).
  2. Bet 2 on each trial.
  3. Bet 4 on each trial.
  4. Bet 8 on each trial (bold play).

We will need to embellish our notation to indicate the dependence on the target fortune:

f(x; a) = P(XN = a | X0 = x).

Now fix p and suppose that the target fortune is 2a and the initial fortune is 2x. If the gambler plays timidly, then of course, his probability of reaching the target is f(2x; 2a). On the other hand:

Mathematical Exercise 12. Suppose that the gambler bets 2 on each trial. Argue that

Xi / 2, i = 0, 1, 2, ...

corresponds to timid play with initial fortune x and target fortune a and that therefore the probability that the gambler reaches the target is f(x; a)

Thus, we need to compare the probabilities f(2x; 2a) and f(x; a).

Mathematical Exercise 13. Show that

  1. f(2x; 2a) = f(x; a)[(q / p)x + 1] / [(q / p)a + 1]
  2. f(2x; 2a) < f(x; a) if p < 1 / 2; f(2x; 2a) > f(x; a) if p > 1 / 2.

Thus, it appears that increasing the bets is a good idea if the trials are unfair, a bad idea if the trials are favorable, and makes no difference if the trials are fair.

Mathematical Exercise 14. Generalize Exercises 12 and 13 to compare timid play with the strategy of betting $k on each trial (let the initial fortune be kx and the target fortune ka).

The Expected Number of Trials

Now let us consider the expected number of trials needed with timid play, when the initial fortune is x:

g(x) = E(N | X0 = x) for x = 0, 1, ..., a.

Mathematical Exercise 15. By conditioning on the outcome of the first trial, show that g satisfies the difference equation

  1. g(x) = qg(x - 1) + pg(x + 1) + 1 for x = 1, 2, ..., a - 1 (difference equation)
  2. g(0) = 0, g(a) = 0 (boundary conditions).

The difference equation in the last exercise is linear, second order, but non-homogeneous. The corresponding homogeneous equation is the equation satisfied by the win probability function f. Thus, only a little additional work is needed here.

Mathematical Exercise 16. Show that if p is not 1/2 then

g(x) = x / (q - p) - [a / (q - p)][(q / p)x - 1] / [(q / p)a - 1] for x = 0, 1, ..., a.

Mathematical Exercise 17. Show that if p = 1/2 then

g(x) = x (a - x) for x = 0, 1, ..., a.

For many parameter settings, the expected number of trials is surprisingly large. For example, suppose that p = 1/2 and the target fortune is 100. If the gambler's initial fortune is 1, then the expected number of trials is 99, even though half of the time, the gambler will be ruined on the first trial. If the initial fortune is 50, the expected number of trials is 2500.

Simulation Exercise 18. In the red and black experiment, select Timid Play. Vary the initial fortune, the target fortune and the win probability and notice how the expected number of trials changes. Now with x = 16, a = 32, and p = 0.5, run the experiment 1000 times with an update frequency of 100. Note the apparent convergence of the sample mean number of trials to the expect value.

Simulation Exercise 19. In the red and black experiment, select Timid Play. Set the target fortune to 128, the initial fortune to 64 and the trial win probability to 0.5. Run the experiment 100 times and note the large size and large variation of the number of trials.