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).

\(P_j = \frac{1 - (\frac{q}{p})^j}{1 - (\frac{q}{p})^m}\)

Probability of Smith winning $8 with timid strategy is 0.0203

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

Since the prisoner bets A dollars and will lose or gain A dollars, and he bets his entire money each time until 8$. He starts with 1 dollar and must win 3 bets in a row at p=0.4

At Bet 1, If we wins, He ends up with $2.

At Bet 2, If we wins, He ends up with $4.

At Bet 3, If we wins, He ends up with $8.

Probability of Smith winning $8 with bold strategy is 0.064

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

Bold Strategy(b)