Smith is in jail and has 1 dollar; he can get out on bail if he has 8 dollars. A guard agrees to make a series of bets with him. If Smith bets A dollars, he wins A dollars with probability 0.4 and loses A dollars with probability 0.6.

Find the probability that he wins 8 dollars before losing all of his money if:


  1. He bets $1.00 each time (timid strategy).

If \(p\) is the probability that Smith wins the round, and \(q\) is the probability that he loses, the probability that Smith wins \(N\) dollars if he starts the game with \(i\) dollars is:

\[ P_i = \left\{ \begin{aligned} \frac{1 - (\frac{q}{p})^i}{1 - (\frac{q}{p})^N} &\qquad \text{if } \ q \neq p \\ \\ \frac{i}{N} &\qquad \text{if } \ q = p = 0.5 \\ \end{aligned} \right. \]


Furthermore, we also know that:

\[ P_i = \left\{ \begin{aligned} 0 &\qquad \text{if } \ i = 0 \quad \text{i.e., he has no more money to bet}\\ 1 &\qquad \text{if } \ i = N \quad \text{i.e., he wins the game}\\ \end{aligned} \right. \]


So, if Smith bets $1.00 each time using a timid strategy, we can calculate \(P_i\):

i <- 1
N <- 8
p <- 0.4
q <- 0.6
A <- 1
q_p <- q/p


p_1 <- (1 - q_p^i) / (1 - q_p^N)
p_1
## [1] 0.02030135

The probability that Smith will win using a timid strategy is about 2%.



  1. He bets, each time, as much as possible but not more than necessary to bring his fortune up to $8.00 (bold strategy).

A bold strategy means that Smith bets his entire available money each round of the game until he reaches $8.00 or goes bankrupt.

If he wins continuously, it will take him 3 rounds to beat the game, going from $1.00 to $2.00, $2.00 to $4.00, and $4.00 to $8.00.

The probability that he wins three rounds in a row is \(0.4 \times 0.4 \times 0.4 = 0.064\), or 6.4%.



  1. Which strategy gives Smith the better chance of getting out of jail?

The bold strategy gives Smith a better chance of winning the game and getting out of jail.