Jail Gamble Game

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).

We can calculate this with the following expression:

\[ \frac{1 - \frac{P_W}{P_L}}{1 - (\frac{P_W}{P_L})^n} \]

pw = .4
pl = .6
n = 8

(1-(pl/pw))/(1-(pl/pw)^n)
## [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).

As 8 is equal to 2^3, we know that our prisoner can only win in this strategy with 3 straight wins. He starts with $1, then provided he wins-out will have $2, $4, then $8. If he loses any game, he will be out.

We know his probability of winning is \[ (P_w)^3\]

pw^3
## [1] 0.064

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

As .064>.0203, we know that the bold strategy yields the better probability.