Flipping coins that gain and lose

You enter a game with a $100 bet that involves flipping coins. The coin has probability p that heads will come up, probability (1-p) that tails will come up. Now, if you are rewarded $10 for every heads, but lose $10 for every tails. You decide to quit if you lose or gain $20, so walk away with $80 or $120. Since you don't know how many flips it will take, what is the probability that you will end up quiting on a net gain (walking away with $120)?


Immediately, I think it's probability p, but I'm having a hard time showing why. This is close to a Negative Binomial, but that is good for trials until r successes, and doesn't count for you losing. Any thoughts?

Your ONLY states that lead to continuation are -10, Even, and +10.

Let's break it down.

Start: EVEN

First Flip:
p(+10) = p
p(-10)=q = 1-p -- The game never ends here.

Second Flip
p(+20) = p*p
P(Even) = pq+qp
P(-20) = q*q
p(game ends) = p*p + q*q = 1 - (2qp)
p(+20|Game Ends) = (p*p)/(p*p + q*q)

Third Flip: Well. the ONLY position we could have is "Even". That seems cyclical enough. What's the rest of the solution?

p(game ends on 0 flips) = 0
p(game ends on 1 flips) = 0
p(game ends on 2 flips) = p*p+q*q = A
p(game ends on 3 flips) = 0
p(game ends on 4 flips) = A^2
erc.