Please refer to the Assignment 10 Document.

1 Problem Set 1

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

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

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

(c.) Which strategy gives Smith the better chance of getting out of jail?

1.1 PS1 Answer

1.1.1 Part (a)

Using the Gambler’s Ruin equation, \(P_{i} = \left\{\begin{matrix} \frac{1-(\frac{q}{p})^{i}}{1-(\frac{q}{p})^{N}} & if\;p\neq q\\ \frac{i}{N} & if\;p = q =0.5 \end{matrix}\right.\)

Given that \(i=1, N=8, p=0.4, q=0.6,\) the probability of Smith wins 8 dollars before losing all of his money if he bets 1 dollar each time is 0.0203.

## [1] 0.02030135

1.1.2 Part (b)

Using bold strategy, Smith is not allowed to lose any bets as that will lose all his money and that contradicts with the given condition.

To bring his fortune up to 8 dolloars, he has to win 3 bets in a row.

  • Bet#1: bet 1 dollar, win the bet and gain 1 dollar, end up with 2 dollars.

  • Bet#2: bet 2 dollars, win the bet and gain 2 dollars, end up with 4 dollars.

  • Bet#3: bet 4 dollars, win the bet and gain 4 dollars, end up with 8 dollars.

Therefore, the probability of using bold strategy \(= (0.4)^{3} = 0.064\)

## [1] 0.064

1.1.3 Part (c)

The bold strategy gives Smith the better chance of getting out of jail as bold strategy has a higher probability of Smith winning 8 dollars before losing all of his money.