Data 605 homework 10

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).
2. he bets, each time, as much as possible but not more than necessary to bring his fortune up to 8 dollars (bold strategy).
3. Which strategy gives Smith the better chance of getting out of jail?

In this strategy, Smith bets 1 dollar each time until he either wins 8 dollars or loses all his money.

Let \(p\) be the probability that Smith wins 8 dollars before losing all his money.

There are a few scenarios to consider:

So, the probability \(p\) that Smith wins 8 dollars before losing all his money is the sum of an infinite geometric series:

\[ p = 0.4 + 0.6 \times 0.4 + (0.6)^2 \times 0.4 + \ldots \]

This is a geometric series with first term \(a = 0.4\) and common ratio \(r = 0.6 \times 0.4 = 0.24\).

Using the formula for the sum of an infinite geometric series:

\[ p = \frac{a}{1 - r} = \frac{0.4}{1 - 0.24} = \frac{0.4}{0.76} \approx 0.526 \]

In this strategy, Smith bets the maximum amount he can each time until he reaches 8 dollars or loses all his money.

Let’s denote \(p'\) as the probability that Smith wins 8 dollars before losing all his money using this bold strategy.

If Smith has \(x\) dollars, he will bet \(\min(8 - x, x)\) dollars each time. Therefore, \(p'\) can be calculated as follows:

\[ p' = P(\text{win next bet}) + P(\text{lose next bet}) \times p' \]

\[ p' = 0.4 + 0.6 \times p' \]

Solving for \(p'\):

\[ p' = \frac{0.4}{0.4 + 0.6} = \frac{0.4}{1} = 0.4 \]

Comparing the probabilities:

The gives Smith a better chance of getting out of jail because \(p\) is greater than \(p'\).