Question 1

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 use the Gambler’s Ruin formula:

\[\boldsymbol{P}=\frac{1-\left(\frac{q}{p}\right)^s}{1-\left(\frac{q}{p}\right)^M}\]

Where \(P=.6\), \(q=.4\), \(M=8\) and \(s=1\).

# set variables
P <- 0.4
q <- 0.6
M <- 8
s <- 1

# compute
(P_1_dollar <- (1 - (q/P)^s) / (1 - (q/P)^M))
## [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).

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. We can use the binomial distribution to calculate:

# set variables
x <- 3
s <- 3
p <- 0.4

# compute
dbinom(x, s, p)
## [1] 0.064

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

Bold strategy gives him a better chance of getting out of jail 6.4% vs ~2%.