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:

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

Overall, there are two possible outcomes: he makes it to 8 dollars, or he looses it all.

For the timid strategy, Smith has many different ways in which me might reach 8 or 0 dollars, starting from 1. It might (probably will) take more than 7 rounds. It’s worth noting that because there are a potential infinite amount of bets, the game will have to end eventually, either at 0 or 8 dollars.

We can think of this like a random walk between two players, the gaurd starting with \(\$7\) and Smith starting with \(\$1\); we should try to find the probability of one of them looisng all their money first; same thing as Smith winning all \(\$8\) or the guard taking it all. Luckily, there’s a formula for this: https://en.wikipedia.org/wiki/Gambler%27s_ruin#Unfair_coin_flipping.

\(Pr(Smith Looses) = 1-p/(1-p)^{n_2} /1-p/(1-p)^{n_1+n_2}\) Where p = 0.4 is the probability that smith wins a bet, \(n_2\) = 7 is the amount of money the gaurd has(that Smith needs to win), and \(n_1\) = 1 is the money Smith starts with. Plug all that in and we get:

## [1] 0.9796987

So about 0.98 chance of Smith loosing all his money first, and a \(1 - 0.98 = 0.02\) chance of Smith winning all the gaurd’s money first, and reaching $8.

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

The bold strategy is much simpler in this specific instance. If sucessful, he will have bet 1, then 2, then 4; this is the only path he can follow to win. If he looses at any one of the three steps, he looses all his money; this would be a more complicated question if he needed to win somthing other than \(\$2^x\).

We can look at this like a binomial, where he needs 3 out of 3 sucesses (2 -> 4 -> 8) or he looses. The probability of winning a round is 0.4, so the probability of winning 3 in a row is \(0.4^3 = 0.064\).

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

0.064 (b) > 0.02 (a): the bold strategy is better.