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
he bets 1 dollar each time (timid strategy).
he bets, each time, as much as possible but not more than necessary to bring his fortune up to 8 dollars (bold strategy).
Which strategy gives Smith the better chance of getting out of jail?
\[{P_1[S_8<S_0] = \frac{(\frac{q}{p})^s-1}{(\frac{q}{p})^M-1} } \] \[ = { \frac{(\frac{0.6}{0.4})^1-1}{(\frac{0.6}{0.4})^8-1} } \] \[ = { \frac{(1.5) - 1}{(1.5)^8 -1} } \] \[ = 0.020301 \]
p <- 0.4
q <- 1 - p
s <- 1
M <- 8
((q/p)^1 - 1 ) / ((q/p)^M - 1 )## [1] 0.02030135Starting from $1, and betting as much as possible, Smith can double his money or lose everything. As each bet results to double or nothing, he will be up to $8 if he wins 3 times in a row, betting $1, $2, $4, and winning each time with 0.4 probability.
p^3## [1] 0.064Between the timid and bold strategy, Smith would have greater chance of getting out of jail with the bold strategy – seems like good fortune favors the bold.