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.

[jay_shark]

This is Gambler's Ruin .

Two Gamblers A and B bet on the outcomes of successive flips of a coin . For each flip , if the coin comes up heads , A collects 1 unit from B whereas if it comes up tails ,A pays 1 unit to B . They continue to do this until one runs out of money . We can make the assumption that the coin is bias with some probability p of landing heads . A question of interest is to determine the probability that A ends up with all the money if he starts with x units and B has y units .

First you should realize that there are 3 possible outcomes of this gambling game . Either A wins , B wins , or the game goes on forever with no one winning . It turns the last part has probability 0 of occurring and that either A or B must have all the money in the end .

I'll give you the formula without the derivation but if you want the derivation I'll have to charge you ...haha jk .

Here is the formula :

Px = [1-(q/p)^x]/(1-(q/p)^(x+y)) if p is not 1/2
Px = x/(x+y) if p =1/2

So as an example , if A were to start with 10 units and B were to start with 20 units then the probability A wins if p =60% is 98.26% which means A has a RoR of 100-98.26% =1.734% .

So if you are a heads up player and you have a bankroll of 10X your buy in and you play repeatedly against a player with 20X the buy in with p=0.6 , you'll wind up with all the money about 98.26% of the time .

[Bottled Rockets]

Awesome, this is EXACTLY what I might have figured out in 3 days of work. Thank you!! Formula is behaving beautifully.