Heads up SNG bankroll formula by "Baller"

An RGP poster named "baller" wrote the following Heads up SNG post a while back. How sound is his thinking?

[ QUOTE ]
"I play a lot of $20SNGs and used the following to aid my bankroll mgmt.....

p = probability of winning a heads-up SNG
n = number of buy-ins in bankroll

[(1-p)/p]^n = probability of going broke based on constant p and NO RAKE

For 1% chance of ever going broke, the following are the number of buy-ins
needed:

p =52.5%, n= 45
p= 55%, n=23
p=57.5%, n=15
p=60%, n=11
p=65%, n=7

Solution:

p1 = probability of eventually going broke given 1 buy-in (regardless of path of
winning and losing)
p2= probability of eventually going broke given 2 buy-ins (regardless of path of
winning and losing)
p = probability of winning each heads-up SNG
(x, multiply........ ^, raise to the power)

p1 = (1-p) + (p x p2) --->probability of going broke with 1 buy-in
equals probability of losing the first match plus probability of winning the
first match and eventually going broke with the 2 buy-ins (after winning the
first match)...

but, p2 = p1^2 --->probability of going broke with 2 buy-ins equals p1
squared because if you start with 2 buy-ins, you eventually lose 1 buy-in with
probability p1 and then you're back to the problem where you only have one
buy-in left, and by definition you lose the last buy-in with probability
p1.....assuming losing 1 buy-in at a time are independent events

substituting in the first equation,

p1 = (1-p) + p x (p1^2)

This equation is quadratic and has two solutions, p1 =1 if p <=1/2 and
p1 = (1-p)/p if p >1/2 i.e. you go broke for sure if your probability of
winning each match equal to or less than 1/2, and you go broke with probability
(1-p)/p if you're favored to win each match with probability p. With one-buyin,
to have an even chance of not eventually going bankrupt, you need to win each
match 2/3 of the time.

If you play n times and are favored to win each match with probability p, then
the probability you'd go broke after the n matches = [(1-p/p)]^n since the n
matches are just one-buyin scenarios repeated n times (assuming independent
match probabilities, no tilt [img]/images/graemlins/smile.gif[/img]

-baller "

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