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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).

Gambler’s ruin equation:

\[\frac{1-(\frac{q}{p})^s}{1-(\frac{q}{p})^m}\]

p = .4
q = .6
s = 1
m = 8
(1 - (q/p)^s)/(1 - (q/p)^m)
## [1] 0.02030135

Only a 2.03% he makes it out.

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

After one round - .4 he has $2. After two rounds - .16 he has $4. After three rounds:

.16 * .4
## [1] 0.064

There are alternate ways to do this via transition matrices or simulation, but a back-of-the-envelope works for this relatively simple scenario

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

Clearly betting as much as possible (but not more than necessary) leads to a beter outcome than betting as little as possible. This matches our intuition given the probabilities.