1.1 Part (a)

he bets 1 dollar each time (timid strategy).

Anwer：

\(Let\;P(win)=q=0.4\;and\;P(lose)=q=1-p=0.6\)

\(P(win\;8\;dollars|bets\;1\;dollar\;each\;time)=\)

according to the formula of ‘Gambler’s Ruin’:

\(P_{i}(n)=\frac{1-(\frac{q}{p})^t}{1-(\frac{q}{p})^n}\;if\;p\neq q\)

\(\because\) 1 dollar each time and wins 8 dollars

\(\therefore\;i=1\;and\;n=8\)

\(\therefore\;P_{1}(8)=\frac{1-(\frac{0.6}{0.4})^i}{1-(\frac{0.6}{0.4})^n}=0.02030135\)

p = 0.4
q = 1-p

(1-(q/p)^1)/(1-(q/p)^8)

## [1] 0.02030135

1.2 Part (b)

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

Answer:

Using bold strategy, he fails to get out on bail whenever he loss any bets, therefore there is only one scenario where he can win 8 dollars:

1. Win the first time, get 2 dollars,

2. Win the second time, get 4 dollars,

3. Win the third time, get $8 dollars.

The probability of wining 3 times consectively is \(P(3 wins consectively)=0.4^3=0.064\)

0.4^3

## [1] 0.064

1.3 Part (c)

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

Answer:

becuase 0.064 > 0.0203, therefore using bold strategy giving Smith better chance of getting out of jail.