Abraham and Blaise each have $\$10$. They repeatedly flip a fair coin. If it comes up heads, Abraham gives Blaise $\$1$. If it comes up tails, Blaise gives Abraham $\$1$. What is the expected number of flips until one of them runs out of money?
So I know that the "frog" is on lily pads labeled 0-20, where each lily pad corresponds to Abraham's money. I need to figure out the expected value is of the flips until the frog reaches 0 or 20, given that it's a fair chance to move up or down, and that we start on 10.
I notice that you are an eighth grader and you have solved this gambler's ruin problem by using frog on lily pad (frankly, I don't see how you have obtained the answer) despite the reference by Math1000 not mentioning anything about expected duration of the classical gambler's ruin problem - very impressed.
You may refer to page 17 of the reference below: http://www.johnboccio.com/research/quantum/notes/ruin.pdf
where $a = \$10$ and $z = 10 + 10 = \$20$ for $p=q=0.5$.
Hence equation (34) gives
Dz = 10 (20 - 10) = 100
Regards