Possible Duplicate:
Probability of a random binary string containing a long run of 1s?
EDIT: Cocopuffs below has partially answered the question, but the critical base case $L=2$ to his inductive argument is missing and it's not obvious how to fill the gap.
This original problem is described in this thread at community.wizards.com.
It originated as a question about a Magic: the Gathering scenario, but I will rephrase it here in general audience terms.
Suppose you have \$0, and your friend has \$L. You start flipping coins. Every time you flip heads, you win \$2 and your friend wins \$1. Every time you flip tails, you lose all of your money. What is the probability you will eventually have (at least) as much money as your friend?
As outlined in the thread above, the probability $p(L)$ of you eventually matching your friend is given by the recurrence relation
$p(L) = \frac{1}{2^{L-1}} + \sum_{i=1}^{L-1} \frac{p(L+i)}{2^i}.$
Other than the trivial observation that $p(L) = 1$ is a solution to this recurrence, I haven't been able to make headway. How do I prove this solution is unique (and therefore the right one)?