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
The probability that he wins $8 before losing all his money if he starts with $1 is 2.03%
transition_matrix <- matrix(c(1,0,0,0,0,0,0,0,0,
0.6,0,0.4,0,0,0,0,0,0,
0,0.6,0,0.4,0,0,0,0,0,
0,0,0.6,0,0.4,0,0,0,0,
0,0,0,0.6,0,0.4,0,0,0,
0,0,0,0,0.6,0,0.4,0,0,
0,0,0,0,0,0.6,0,0.4,0,
0,0,0,0,0,0,0.6,0,0.4,
0,0,0,0,0,0,0,0,1),nrow=9,byrow=TRUE)
colnames(transition_matrix) <- c(0,1,2,3,4,5,6,7,8)
rownames(transition_matrix) <- c(0,1,2,3,4,5,6,7,8)
Q <- matrix(c(0,0.4,0,0,0,0,0,
0.6,0,0.4,0,0,0,0,
0,0.6,0,0.4,0,0,0,
0,0,0.6,0,0.4,0,0,
0,0,0,0.6,0,0.4,0,
0,0,0,0,0.6,0,0.4,
0,0,0,0,0,0.6,0),nrow=7,byrow=TRUE)
I <- diag(7)
N <- solve(I-Q)
r <- matrix(c(0.6,0,0,0,0,0,0,0,0,0,0,0,0,0.4),nrow=7,byrow=TRUE)
B = N %*% r
B[1,2]
## [1] 0.02030135
We could also solve this using the Gambler’s ruin probability function:
q = 0.6
p = 0.4
n = 1
N = 8
((1-(q/p)^n)/(1-(q/p)^N))
## [1] 0.02030135
In this case, Smith has to always win the bet otherwise, he will lose all of this money. This route to winning only includes: * Bet #1: bet $1, win and then have $2
* Bet #2: bet $2, win and then have $4
* Bet #3: bet $4, win and then have $8
The probability of this is 6.4%
p = 0.4
win_prob = p^3
win_prob
## [1] 0.064