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:

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

Utilizing Gambler’ Ruin Equation,

\(P = \frac{1-(\frac{q}{p})^s}{1-(\frac{q}{p})^M}\)

where M = 8; s = 1; p = 0.4; q = 0.6

\(P = \frac{1-(\frac{0.6}{0.4})^1}{1-(\frac{0.6}{0.4})^8}\)

prob <- (1 - (0.6/0.4)^1) / (1 - (0.6/0.4)^8)
prob
## [1] 0.02030135

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

If Smith wins each time, he will follow below sequence. Smith starts with $1, so he can bet a maximum of $1. If we wins, his new stake is k=$2. Smith starts with $2, so he can bet a maximum of $2. If we wins, his new stake is k=$4. Smith starts with $4, so he can bet a maximum of $4. If we wins, his new stake is k=$8.

Now we know that if Smith win 3 times in a row, he can win $8 that will enough for him to bail. Since each state has a probability of 0.4 for wining, we need to compute prob. of 3 success in a row,

p = 0.4
p^3
## [1] 0.064

(c) Which strategy gives Smith the better chance of getting out of jail?

Timid strategy gives Smith a 2.03% chance of wining while Bold strategy give Smith a 6.4% of chance at winning. Based on the answers, Bold Strategy gives Smith the better chance of getting out of jail.