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)

Answer:

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 s dollars is given as:

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

p <- 0.4
q <- 1-p
s <- 1
N <- 8
P <- (1 - (q/p)^s) / ( 1 - (q/p)^N)
P
## [1] 0.02030135

The probability that Smith will win using a timid strategy is 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)

Answer:

The probability that Smith wins three rounds in a row is: \(p * p * p = 0.4 * 0.4 * 0.4 = 0.064\)

p.bold <- p^3
p.bold
## [1] 0.064

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

Answer:

Bold strategy gives Smith the better chance of getting out of jail, because bold strategy has a probability of 0.064 which is higher than the timid strategy which has a probability of 0.0203.