You have two expert players playing heads up:

Let's say Durrr vs. Ivey, or Antonius vs. Urindanger, or whatever...

Is there a way of mathematically determining the longest possible match before edge/variance would make the match (almost) certainly over? (This is assuming the edge one player would have over the other is very small)

Depends almost entirely on how fast the blinds go up and at what rate.

the formula is going to be something like: when the average chip stack divided by blinds and antes = 12 or less the tournament is now luckfest crapshoot.

Sorry...I meant to imply a cash game, or at least a tourney where the blinds never go up.

No. However, there are ways to determine how likely it is that the match will last less than x hands. The amount of hands played before an elimination depends heavily on the profile of the players. A match between two loose agressive players is likely to last much shorter than a match between two tight passive players.

I was thinking the same thing. For instance, if Hellmuth could play against himself, well....how long before the sun collapses?

I guess it's impossible to answer this is any real meaningful way, still, care to give your best estimate?

I doubt the match would last very long. You underestimate the amount of luck in poker.

There is a way to approach this question. The match is over on an all-in hand with the smaller stack losing. With equal and good players, it's reasonable to assume any all-in hand is a 50% shot. This might not be true if one player is close to being blinded out, but that doesn't affect the probability much, because once one player gets down to a few big blinds, he's got very little chance anyway.

The style of the players will determine how often there is an all-in call. Some players might have one every five minutes, others every five hours. So guess whatever you want here.

Once we know that, all we need to know is the probability of flipping a fair coin N times before getting a head. Every all-in call has a 50% chance of ending the match. I'm ignoring the small probability of an exactly equal all-in call, for example if there is one on the first hand, where it has a 100% probability of ending the match.

The chance is 50% that the match ends on the first all-in call, 25% on the second, 12.5% on the third, and so on. If we had 1,024 matches, 1 would go beyond 10 all-in calls, 1 would end on the 10th all-in call, 2 on the 9th, 4 on the 8th, 8 on the 7th, 16 on the 6th, 32 on the 5th, 64 on the 4th, 128 on the 34rd, 256 on the 2nd and 512 on the first.

Should all be raked gone in like 2 hours

Is it just me, or would the answer lie in the bb/100 edge one got over the other?

Lets say ivey beat durrr for 2 big bets/100 hands. They each have 500 big bets in their stack. 500BB/2 * 100. The match would then, given an accurate win rate, last for 25000 hands on average.

Edit: If hellmuth is playing hellmuth where noone has an edge over the other it would all come down to how often variance would reach all the way down to the felt in a 1000bb stack. I cant do that math.

Am I really the only one to think this is a schoolbook example of a risk-of-ruin problem?

It is related to that problem. Although a better question is to compute the average duration of the sng.

The average duration of the sng can be computed if we assume that both players are approximately equal in skill; one may be slightly superior to the other.

Given that both players start off with n chips and flip a coin to compete for each chip, then the average duration of the game is to flip n^2 many times.

http://www.mathpages.com/home/kmath084/kmath084.htm

For simplicity, if we assume that the game has a standard deviation of 100 bb/100 hands then the game is approximately equivalent to having 10 chips apiece and the average length of that game is 10^2=100 flips.

Since the standard deviation is measured per 100 hands, the average length of the sng is about 100*100 = 10,000 hands.