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

  1. he bets 1 dollar each time (timid strategy).

The probability that the chain reaches state 8 before reaching state 0, starting from state 1 is is equal to 0.02030135.

\[P(S_8 < S_0|X_0 = i) \\ P_i = P_{i+1}*P + P_{i-1}*q \\ P_i = P_{i+1}*0.4 + P_{i-1}*0.6\]

q = 0.6
p = 0.4
n = 8

for (i in 1:8){
  bets <- (1-(q/p)^i)/(1-(q/p)^n)
  print(bets)
}
## [1] 0.02030135
## [1] 0.05075337
## [1] 0.0964314
## [1] 0.1649485
## [1] 0.267724
## [1] 0.4218874
## [1] 0.6531324
## [1] 1
  1. he bets, each time, as much as possible but not more than necessary to bring his fortune up to 8 dollars (bold strategy).

He needs to keeps a minimum winning streaks to reach 8 dollars. His winning sequence will be:

$1 – P=0.4 –> $2 – P=0.4 –> $4 – P=0.4 –> $8

win_streak = 0.4*0.4*0.4

win_streak
## [1] 0.064
  1. Which strategy gives Smith the better chance of getting out of jail?

Bold strategy (P=0.064) has better chance than timid strategy (P=0.02030)

Bold strategy will give Simth a better change of getting out of jail