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 0.4 and loses A dollars with probability 0.6. Find the probability that he wins 8 dollars before losing all of his money if
He bets 1 dollar each time
Using the equation in chp12 page 489,
\[ q_z = \frac{(q/p)^z - 1}{(q/p)^m -1} \] We have,
\[ \begin{align} z = 1 \ && p = 0.4 \\ m = 8 && q = 0.6 \end{align} \]
Plugging them in the equation,
\[ \Rightarrow \frac{(04./0.6)^1 - 1}{(0.4/0,6)^8 -1} = \frac{1.5 - 1}{1.5^8 -1} \approx 0.020301 \Leftrightarrow 2.03\% \]
he bets, each time, as much as possible but not more than necessary to bring his fortune up to 8 dollars
For this strategy, Smith is going to bet the as much as possible to reach $8. He starts with $1 and he needs 3 consecutive wins. His bets would be the 1, 2, 4 and his balance would be 2,4,8 which would get him bail on the third win.
So, Smith to win 3 independent trials in a row since if he loses once he would lose all this money and the game ends.
\[ p = 0.4 \\ P(\text{Smith wins with the bold strategy}) = p * p * p = 0.4^3 = 0.064 \Leftrightarrow 6.4\% \]
Which strategy gives Smith the better chance of getting out of jail?
The bold strategy gives Smith a 6.40% change of getting out of jail compared to the timid strategy of having a 2.03% chance of getting out of jail. The bold strategy is also the riskier strategy because Smith can not afford to lose a single bet while the timid strategy has room to lose bets and recover those loses.