[well]

Here's my simplification...

I'll disregard the blinds for now, and I don't think it matters that much.
Furthermore, I will not count comebacks, i.e. once you have less than your opponent, you'll be ruined.

Suppose the opponent will call with any hand that has an equity of X over a random hand.

Then there are three numbers to our interest:

- the probability that he will fold the hand (we will look at this as a tie, t)
- the probability that he will call and win, w
- the probability that he will call and lose, l

And what we want to know, is for what N there is a probability P<.5 that we will lose (at least) N times before winning one.

The probability that the next non-tie outcome will be a win, is w+tw+t^2w+... = w/(1-t) = W.
For a loss, this is l/(1-t) = L.

So, with P=L^N, this yields N> log .5 / log L

With X=.5, N>1.12
X=.51, N>1.15
X=.55, N>1.33
X=.6, N>2.07

For X=.85 (i.e. he calls with Aces and with Aces only), N>27.56.

The higher X is, the less hands your opponent will play, the more relevant the blinds become.
So for the only Aces case, I don't think that is too good an approximation.

Regards,

Well.