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).
he bets, each time, as much as possible but not more than necessary to bring his fortune up to 8 dollars (bold strategy).
Which strategy gives Smith the better chance of getting out of jail?
To answer this first question, I used the binomial distribution function: P(X) = (n choose X)(pX)((1-p)(n-X))
We know that n, the number of trials, is 8. This is because Smith is betting $1 each time in hopes of getting to 8. Since we are calculating the probability of getting $8, or P(8), X=8 succeses. We are also given the probability .4 for p.
Because of that, we can substitute like so: P(8) = (8 choose 8)(0.48)((1-0.4)(8-8)) = (1)(0.00065536)(1) = 0.00065536
There is a probability of about .07% of Smith winning $8 by playing timidly.
Because Smith is essentially betting everything he has to get to $8, we can think of the bold strategy like so:
Bet 1: bets $1 –> $2 Bet 2: bets $2 –> $4 Bet 3: bets $4 –> $8
This means that Smith would need to have 3 consecutive wins in this case. We know that the probability of a win is .4. The probability of 3 consecutive wins can be determined like so:
P(3)=(.4)^3 = .064
There is a probability of about 6.4% of Smith winning $8 by playing boldly.
In this case, it looks like the bold strategy has a higher precentage of success. It would, however, require Smith to win a consecutive amount of times, and depending on the number of total bets or trials, it might not be worth it in the end. I would still say that the bold strategy is better, simply because at the end of the day he would only be $8 short. It could be slightly more beneficial for him to go with the bold strategy!