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
He bets 1 dollar each time (timid strategy).
Solution:
\(q=.6(Lose)\)
\(p=.4(Win)\)
\(q/p=.6/.4=1.5\)
\(M = 8\)
\(s=1\)
\(P(win8)=(1-(q/p)^s)/(1-(q/p)^M=(1-1.5)/(1-1.5^8)\)
p_win<-(1.5^1-1)/(1.5^8-1)
p_win## [1] 0.02030135He bets, each time, as much as possible but not more than necessary tobring his fortune up to 8 dollars (bold strategy).
Solution:
Here Smith bets the entire amount (up to 8). He must win each time or go broke and lose. If he wins, he will follow the following sequence:
1,2,4,8.
He starts with 1 dollar and must win 3 times in a row with probability 0.4. So, the probability of 3 successes in a row is p^4
prob<-0.4^3
prob## [1] 0.064Which strategy gives Smith the better chance of getting out of jail?
Solution
Strategy B (Bold strategy) gives Smith a better chance out of getting out of jail, 0.064> 0.020.