Suppose that you have a fair coin. You start with $0. You win 1$ each time you get a head and loose $1 each time you get tails. Calculate the probability of getting $2 without getting below $0 at any time.

1/3

1/3 Because if he requires 2 heads = $2, then it means he require HH out of {HH,HT ,TH,TT} We cannot consider TT because it would turn to zero. Therefore, he has 1 chance out of 3. Hence 1/3.

Marchov chain with 2 and -1 absorbing OR geometric series 1/4^n from 1 to inf=1/3

p = 0.5(0.5+0.5p)

Each flip has a .5 possibility of being head. (.5)(.5)=.25 or 1/4