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 he bets 1 dollar each time (timid strategy).
This is similar to the Gambler’s Ruin problem.
If \(p \neq q\), the probability of each state is \(P_{i}=\frac{1-(\frac{q}{p})^i}{1-(\frac{q}{p})^N}\).
If \(p=q=\frac{1}{2}\), then \(P_{i}=\frac{i}{N}\).
\(P_{1}=\frac{1-(\frac{0.6}{0.4})^1}{1-(\frac{0.6}{0.4})^8} \approx 0.0203\).
i <- 1
N <- 8
p <- 0.4
q <- 0.6
(1-(q/p)^i)/(1-(q/p)^N)## [1] 0.02030135#m <- matrix(c())Find 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 (bold strategy).
If Smith bets as much as possible but not more than necessary to bring his fortune up to 8 dollars, he needs to make 3 consecutive wins and no loses.
1st bet: He bets $1 and wins. He will have $2.
2nd bet: He bets $2 and wins. He will have $4.
3rd bet: He bets $4 and wins. He will have $8.
\(P=(0.4)^{3}=0.064\)
p^3## [1] 0.064Which strategy gives Smith the better chance of getting out of jail?
If Smith uses the timid strategy, the chance of him getting out of jail is 0.0203. If Smith uses the bold strategy, the chance of him getting out of jail is 0.064. Smith will have a higher chance of getting out of jail using the bold strategy. However, it would be risky because he must make $8 with 3 consecutive wins.