1.) Because Smith is stuck when he is out of cash and out on bail when he has 8 dollars, I made both absorbing states, as he will no longer be playing then. The matrix TimidM is the transtion matrix for the states needed to make bail.

r0<-c(1,0,0,0,0,0,0,0,0)
r1<-c(.6,0,.4,0,0,0,0,0,0)
r2<-c(0,.6,0,.4,0,0,0,0,0)
r3<-c(0,0,.6,0,.4,0,0,0,0)
r4<-c(0,0,0,.6,0,.4,0,0,0)
r5<-c(0,0,0,0,.6,0,.4,0,0)
r6<-c(0,0,0,0,0,.6,0,.4,0)
r7<-c(0,0,0,0,0,0,.6,0,.4)
r8<-c(0,0,0,0,0,0,0,0,1)
Timiddf<-rbind(r0,r1,r2,r3,r4,r5,r6,r7,r8)
TimidM<-as.matrix(Timiddf)
row.names(TimidM)<-c("$0","$1","$2","$3","$4","$5","$6","$7","$8")

TimidMar<-new("markovchain",states=c("$0","$1","$2","$3","$4","$5","$6","$7","$8"), transitionMatrix=TimidM)
steadyStates(TimidMar) #double checking that everything is pushed to the two end states
##      $0 $1 $2 $3 $4 $5 $6 $7 $8
## [1,]  0  0  0  0  0  0  0  0  1
## [2,]  1  0  0  0  0  0  0  0  0
#looking for n->infinity, 10000 is enough to reach a stead state

TimidMar^10000
## Unnamed Markov chain^10000 
##  A  9 - dimensional discrete Markov Chain defined by the following states: 
##  $0, $1, $2, $3, $4, $5, $6, $7, $8 
##  The transition matrix  (by rows)  is defined as follows: 
##           $0 $1 $2 $3 $4 $5 $6 $7         $8
## $0 1.0000000  0  0  0  0  0  0  0 0.00000000
## $1 0.9796987  0  0  0  0  0  0  0 0.02030135
## $2 0.9492466  0  0  0  0  0  0  0 0.05075337
## $3 0.9035686  0  0  0  0  0  0  0 0.09643140
## $4 0.8350515  0  0  0  0  0  0  0 0.16494845
## $5 0.7322760  0  0  0  0  0  0  0 0.26772403
## $6 0.5781126  0  0  0  0  0  0  0 0.42188739
## $7 0.3468676  0  0  0  0  0  0  0 0.65313243
## $8 0.0000000  0  0  0  0  0  0  0 1.00000000

Smith has roughly a 2% chance of actually making bail.

2.) For the bold strategy, we do not actually need any Markov Chains. There are three bets ($1->$2, $2->$4, and $4->$8) that Smith needs to get to make bail, otherwise, he is stuck. Given that these are three independent trials that all need to be successes, the probability is simply \((.4)^3\).

.4^3
## [1] 0.064

Smith has a 6.4% chance of making bail through this strategy.

3.) Smith has lousy odds either way. But, the go for broke strategy yields hits 3 times more often than the timid one. His best bet is to limit the number of bets.