NLHE 5% edge --> chance of dropping X buyins?

[Bottled Rockets]

Q: Model that you win 55% of the time the money goes in & that you keep playing this game forever. What is the chance that you will ever drop say 7 buyins below where your bankroll starts right now? (It might happen in the first 7 hands in a row, and it might take 7000 hands.)

A: I thought I solved it, but there's some strange behavior in my solution. Here's what I have,
=(45%*(1+(45%*55%*(1/(1-2*(45%*55%)))))^7
Logic: I derived an infinite series for all the hand sequences that ever drop a single buyin, and figured that all the various ways to drop 7 buyins correspond exactly to any 7 ways in a row to drop 1 buyin. My derivation might be flawed; it was first principles.
The strange behavior: As I make the edge smaller & smaller, like even 50.01% vs 49.99%, the chance to EVER drop a single buyin never exceeds 75%.

A HA! I see what I'm doing wrong. My method works only for dropping a buyin BEFORE swinging even a single buyin up. So, I need to embed the above formula in another infinite series accounting for all the ways to swing up Y buyins and then down X+Y buyins...wow.

There's gotta be prior art on this. I'd love to see it.