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).

We utilize the Gambler’s Ruin equation: The gambler starts with a stake of size s and plays until their stake reaches the value M = 0
\[ P_{broke} = \frac{(\frac{q}{p})^z-(\frac{q}{p})^M}{(1-\frac{q}{p})^M} \\ P_{win} = 1 - P_{broke} \] M=8, z=1, p=0.4, and q=0.6, since the probability of Smith winning A dollars is 0.4, losing A dollars is 0.6, Smith starts with a stake of 1 (1 dollar), and it’ll be played until Smith reaches a capital of 8

\[ = 1 - \frac{(\frac{0.6}{0.4})^1-(\frac{q}{p})^8}{(1-\frac{0.6}{0.4})^8} \\ \]

1 - ((1.5-(1.5^8)) / (1-(1.5^8)))
## [1] 0.02030135
  1. he bets, each time, as much as possible but not more than necessary to bring his fortune up to 8 dollars (bold strategy).

Here, as a bold move, Smith bets the whole amount, upto $8. He starts with 1 dollar and must win 3 times in a row with probability 0.4

He places $1 and wins to get $2. He then places the $2 and wins, to get $4. Finally, if he places the $4 and wins, they make $8.

The probability of 3 successes in a row is

0.4 ^ 3
## [1] 0.064
  1. Which strategy gives Smith the better chance of getting out of jail?

The bold strategy has a better chance of winning because it requires only 3 moves and has higher probability.