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 .4 and loses A dollars with probability .6. Find the probability that he wins 8 dollars before losing all of his money if

  1. he bets 1 dollar each time (timid strategy).

This is a gambler’s ruin problem where \(p \neq q\), so we will utilize the formula \(P_a=\frac{(\frac{q}{p})^s−1}{(\frac{q}{p})^{M}-1}.\)

So \(p=.4\) and \(q=.6\), therefore \(\frac{q}{p}=\frac{.6}{.4}=1.5\) \(s=1, M=8\) and we plug into the formula

\(P_a=\frac{(\frac{q}{p})^s−1}{(\frac{q}{p})^{M}-1}=\frac{(1.5)^1−1}{(1.5)^{8}-1}=0.02030134813\)

So the chances are very slim (.02) if he bets in a timid way.

  1. he bets, each time, as much as possible but not more than necessary to bring his fortune up to 8 dollars (bold strategy).

If he wins each time and bets the maximum amount he can each turn, he will bet use 3 turns betting $1, 2, 4, 8 successively. This would require 3 successive wins. So the odds this strategy would work is \(.4^3=.064\).

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

The bold strategy is a better chance to get out of jail. There is a higher probability that he will get to $8 this way, probably due to the fact that it will take less turns to win the necessary amount to receive bail.