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.
According to formula of Gambler’s Ruin: \[ P = \frac{1 - (\frac{q}{p})^s}{1 - (\frac{q}{p})^M} \\ P = \frac{1 -(\frac{0.6}{0.4})^1}{1 - (\frac{0.6}{0.4})^8} \\ P = 0.02 \]
p <- 0.4
q <- 0.6
M <- 8
z <- 1
# Calculate the probability of reaching M before reaching 0
p_z1 <- 1 - ((q/p)^M - (q/p)^z)/((q/p)^M - 1)
p_z1
## [1] 0.02030135
#install.packages("gamblers.ruin.gameplay")
library(gamblers.ruin.gameplay)
#grp.gameplay(ini.stake, p, win.amt)
#init.stake: the initial capital (money) with which the gambler enters the game.
#p: the probability with which the gambler wins each round of the game, 0<p<1.
#win.amt: the amount of money which the gambler desires to win from the game.
bets <- grp.gameplay(1,0.4,8)
bets
The probability that Smith reaches \(8 dollars\) without ever having reached \(0\) while using the timid strategy is \(0.02\) or \(2\)%.
# create transition probability matrix
P <- matrix(rep(0,25), nrow=5, ncol=5)
P[1,1] <- 1
P[5,5] <- 1
for (i in 2:4) {P[i,i-1] <- 0.6
P[i,i+1] <- 0.4}
P
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1.0 0.0 0.0 0.0 0.0
## [2,] 0.6 0.0 0.4 0.0 0.0
## [3,] 0.0 0.6 0.0 0.4 0.0
## [4,] 0.0 0.0 0.6 0.0 0.4
## [5,] 0.0 0.0 0.0 0.0 1.0
library(expm)
# probability vector, (0,1,0,0,0), represents the starting distribution
# final distribution of the probability vector after 4 steps
c(0,1,0,0,0)%*%(P%^%4)
## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.744 0.1152 0 0.0768 0.064
Smith has a better chance of getting out of jail with the bold strategy, \(0.064\) than with the timid strategy, \(0.02\).