[Aisthesis]

Just as an example (for the ultra-simple game):

All it takes for the tournament to end is for either player to win twice in a row. If the result of the first two played hands is 1 win and 1 loss for B (or A), then we're back to the beginning state.

So, if we define x as the medium stack call threshold for B, x' as the short stack call threshold for B, and x" as the big stack call threshold for B, then B's ROI is simply:
(1/4)*(2x + x' + x" + x*x" - x*x')

That's actually not too bad. It's summing the infinite series for average number of hands played that is scaring me. I'll have to try it (picking specific numbers, it always takes on a rather easy form) when I get a chance. If the values resulting from variables x, x' and x" (all between 0 and 1) will take on a coherent pattern, it might work out into something doable.

My apologies for resorting to calculus again on this one, but I can't see how to get a grip on it otherwise.

In any case if we define r as 2 times ROI (buy-in was $2), and n as average number of hands until the tournament is decided, then r/n will be EV/hand.