HW 10 RBalaban

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

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

Smith wins with probability \(p\), and loses with probability \(q\). The probability he wins \(N\) dollars if he starts with \(i\) dollars is

\[ P_i = \frac {1 - (\frac{q}{p})^i}{1-(\frac{q}{p})^n} \] We know that \(P_i = 0\) when \(i=0\) (no money left) and \(P_i = 1\) when \(i=N\) (he won).

i <- 1 #Starting money
p = 0.4 #P win
q = 1-p #P lose
n = 8 #final amount needed to win

P_timid <- (1-(q/p)^i)/(1-(q/p)^n)
cat("The chance of winning with the timid approach is: ", P_timid*100, "%.")

## The chance of winning with the timid approach is:  2.030135 %.

• 2. he bets, each time, as much as possible but not more than necessary to bring his fortune up to 8 dollars (bold strategy).

If he’s betting the maximum each time, he’s going to bet 1$, then 2$, 4$. If he wins, he’ll have 8$. If he loses even a single bet, he loses all his monday, therefore he has to win 3 bets in a row, each being independent and having \(P(win) = 0.4\)

P_bold = p^3
cat("The chance of winning with the bold approach is: ", P_bold*100, "%.")

## The chance of winning with the bold approach is:  6.4 %.

• 3. Which strategy gives Smith the better chance of getting out of jail?

While both strategies seem awful, he seems to have better chances with the bold approach.