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
This is “The Gambler’s Ruin Problem” from pg. 487 in section 12.2 of our probability textbook. The gambler starts with a stake of size \(s\) and plays until their stake reaches the value \(M\) or the value \(0\) (absorbing states).
(a) he bets 1 dollar each time (timid strategy).
\(P_{broke} = \frac{(q/p)^z - (q/p)^M}{(1 - q/p)^M}\)
\(P_{winning} = 1 - P_{broke} = 1 - \frac{(0.6/0.4)^1 - (0.6/0.4)^8}{1 1 (0.6/0.4)^8} =\)
1 - ((1.5- (1.5^8)) / (1- (1.5^8)))## [1] 0.02030135(b) he bets, each time, as much as possible but not more than necessary to bring his fortune up to 8 dollars (bold strategy).
In a), the probability computed was the probability of making 8 moves to the right. If the gambler wagered $1 and won, they’d have $2. Then, if they wagered $2 and won, they’d have $4. Lastly, if they wagered $4 and won, they’d have $8. This is 3 moves to the right instead of 8.
\(P_{winning} = 1 - P_{broke} = 1 - \frac{(0.6/0.4)^1 - (0.6/0.4)^3}{1 1 (0.6/0.4)38} =\)
1 - ((1.5- (1.5^3)) / (1- (1.5^3)))## [1] 0.2105263(c) Which strategy gives Smith the better chance of getting out of jail?
Strategy b) gives a better chance of winning because it requires less moves to the right.