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).
  2. he bets, each time, as much as possible but not more than necessary to bring his fortune up to 8 dollars (bold strategy).
  3. Which strategy gives Smith the better chance of getting out of jail?

Solution:

This follows the Gambler’s Ruin Problem where p = 0.4 and q = 0.6.

(a)

For the timid strategy, we look at a chain of 8 different probabilities since we are looking to find the probability that Smith wins 8 dollars before losing all of his money. In this case the probability would be the first part in the chain.

p <- 0.4
q <- 0.6
k <- q / p 
Timid_strategy <- (1 - (k ^ 1)) / (1 - (k ^ 8))
round(Timid_strategy, 4)
## [1] 0.0203

The probability that he wins 8 dollars before losing all of his money if he bets 1 dollar each time, otherwise known as the timid strategy, is 0.0203.

(b)

For the bold strategy, each time Smith bets he bets as much as possible. If we assume that each time he bets he wins, then since he starts with 1 dollar his winnings would follow the pattern 1, 2, 4, and 8 since him winning would give him 2 dollars, then 4 dollars, then 8 dollars. So by looking at this we see that he needs to win 3 times in a row.

(0.4) ^ 3
## [1] 0.064

The probability that he wins 8 dollars before losing all of his money if he bets, each time, as much as possible but not more than necessary to bring his fortune up to 8 dollars is 0.064.

(c)

The strategy that gives Smith a better chance of getting out of jail is the bold strategy. This is because the bold strategy gives him a 6.4% chance of winning as opposed to the timid strategy that gives him a 2.03% chance of winning.