#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
knitr::opts_chunk$set(echo = TRUE)
library(tinytex)## Warning: package 'tinytex' was built under R version 4.3.3library(markovchain)## Warning: package 'markovchain' was built under R version 4.3.3## Package: markovchain
## Version: 0.9.5
## Date: 2023-09-24 09:20:02 UTC
## BugReport: https://github.com/spedygiorgio/markovchain/issues#a.timid strategy
#1.simulation solution:
p <- 0.4 # P of success
q <- 1 - p # P of fail
start <- 1
sim_len <- 1000000
w <- 0
l <- 0
bet <- 1
for (s in 1:sim_len) {
sum <- start
while (sum > 0 && sum < 8) {
if (runif(1, 0, 1.0) <= p) {
sum <- sum + bet
} else {
sum <- sum - bet
}
}
if (sum == 0) {
l <- l + 1
} else {
w <- w + 1
}
}
print(paste('wins: ', w, 'losses: ', l))## [1] "wins: 20445 losses: 979555"(w / (w+l))## [1] 0.020445#p=0.020081
#2. Math solution:
Q <- matrix(
c(0.0, 0.4, 0.0, 0.0, 0.0, 0.0, 0.0,
0.6, 0.0, 0.4, 0.0, 0.0, 0.0, 0.0,
0.0, 0.6, 0.0, 0.4, 0.0, 0.0, 0.0,
0.0, 0.0, 0.6, 0.0, 0.4, 0.0, 0.0,
0.0, 0.0, 0.0, 0.6, 0.0, 0.4, 0.0,
0.0, 0.0, 0.0, 0.0, 0.6, 0.0, 0.4,
0.0, 0.0, 0.0, 0.0, 0.0, 0.6, 0.0
),
nrow = 7,
ncol = 7,
byrow = TRUE
)
R <- matrix(
c(0.6, 0.0,
0.0, 0.0,
0.0, 0.0,
0.0, 0.0,
0.0, 0.0,
0.0, 0.0,
0.0, 0.4
),
nrow = 7,
ncol = 2,
byrow = TRUE
)
I <- diag(7)
(N <- solve(I-Q))## [,1] [,2] [,3] [,4] [,5] [,6] [,7]
## [1,] 1.6328311 1.054718 0.6693101 0.4123711 0.2410785 0.1268834 0.05075337
## [2,] 1.5820777 2.636796 1.6732752 1.0309278 0.6026963 0.3172086 0.12688343
## [3,] 1.5059477 2.509913 3.1792228 1.9587629 1.1451229 0.6026963 0.24107851
## [4,] 1.3917526 2.319588 2.9381443 3.3505155 1.9587629 1.0309278 0.41237113
## [5,] 1.2204600 2.034100 2.5765266 2.9381443 3.1792228 1.6732752 0.66931007
## [6,] 0.9635210 1.605868 2.0340999 2.3195876 2.5099128 2.6367962 1.05471848
## [7,] 0.5781126 0.963521 1.2204600 1.3917526 1.5059477 1.5820777 1.63283109(B <- N %*% R)## [,1] [,2]
## [1,] 0.9796987 0.02030135
## [2,] 0.9492466 0.05075337
## [3,] 0.9035686 0.09643140
## [4,] 0.8350515 0.16494845
## [5,] 0.7322760 0.26772403
## [6,] 0.5781126 0.42188739
## [7,] 0.3468676 0.65313243#p=0.02030135#b.bold strategy
#1. simulation solution:
p <- 0.4 # P of success
q <- 1 - p # P of fail
start <- 1 #
sim_len <- 1000000
w <- 0
l <- 0
for (i in 1:sim_len) {
sum <- start
while (sum > 0 && sum < 8) {
bet <- min(sum, 8-sum)
if (runif(1, 0, 1.0) <= p) {
sum <- sum + bet
} else {
sum <- sum - bet
}
}
if (sum == 0) {
l <- l + 1
} else {
w <- w + 1
}
}
print(paste('wins: ', w, 'losses: ', l))## [1] "wins: 64096 losses: 935904"(w / (w+l))## [1] 0.064096#P=0.063479
#2. Math solution
matx<- matrix(c(1,0,0,0,0,0,0,0,0,
0.6,0,0.4,0,0,0,0,0,0,
0.6,0,0,0,0.4,0,0,0,0,
0.6,0,0,0,0,0,0.4,0,0,
0.6,0,0,0,0,0,0,0,0.4,
0,0,0,0.6,0,0,0,0,0.4,
0,0,0.6,0,0,0,0,0,0.4,
0,0.6,0,0,0,0,0,0,0.4,
0,0,0,0,0,0,0,0,1), byrow = T, nrow = 9)
print(matx)## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
## [1,] 1.0 0.0 0.0 0.0 0.0 0 0.0 0 0.0
## [2,] 0.6 0.0 0.4 0.0 0.0 0 0.0 0 0.0
## [3,] 0.6 0.0 0.0 0.0 0.4 0 0.0 0 0.0
## [4,] 0.6 0.0 0.0 0.0 0.0 0 0.4 0 0.0
## [5,] 0.6 0.0 0.0 0.0 0.0 0 0.0 0 0.4
## [6,] 0.0 0.0 0.0 0.6 0.0 0 0.0 0 0.4
## [7,] 0.0 0.0 0.6 0.0 0.0 0 0.0 0 0.4
## [8,] 0.0 0.6 0.0 0.0 0.0 0 0.0 0 0.4
## [9,] 0.0 0.0 0.0 0.0 0.0 0 0.0 0 1.0Names <- c("0","1","2","3","4","5","6","7","8")
bold <- new("markovchain", states = Names, byrow=TRUE, transitionMatrix = matx)
bold_p <- absorptionProbabilities(bold)
print(bold_p)## 0 8
## 1 0.93600 0.06400
## 2 0.84000 0.16000
## 3 0.80160 0.19840
## 4 0.60000 0.40000
## 5 0.48096 0.51904
## 6 0.50400 0.49600
## 7 0.56160 0.43840#p=0.06400#c.The BOLD strategy gives Smith the better chance of getting out of jail