[AaronBrown]

Well, I can give you a start to this question.

Begin with a simpler question. You have chips for n hands, paying the antes and blinds. Your goal is to win one hand, but you lose with game over if you lose one hand. The solution is to wait for a hand such that your chance of losing is less than 3^(k-1)/2^(2*k-1), where k is the number of hands have left if you fold this one.

For example, if you can afford one more hand after this, k=1 and you would call the bet if your chance of losing was less than 3^0/2^1 = 1/2. If you could afford 4 more hands, you would call the bet if your chance of losing was less that 3^3/2^7 = 27/128. If this is the last hand you can afford, the formula does not work, you have to call on anything.

Going back to your game, you have a choice on the first hand: either call or put yourself in a position that you'll have to call and win two hand to win the tournament. There is some probability of winning such that you would call the first bet.

If you don't call the first bet, you know you have to win twice. If you lose another half of your initial stake, you'll have to win three times.

I think you can get a pretty good approximation by treating the number of folds before you increase the number of wins you need as the limit in the formula above. For example, suppose you start with enough chips for 10 folds. You fold the first hand and have chips for 9 folds. If you get down to 4 folds, you'll have to win three times instead of twice. That's such a disadvantage, that it makes sense to play similar to what you would do if you needed a win in the next five hands. That's the 27/128 chance of losing from above. It's not exact, but my intuition says it's pretty close.

I'm sure there's a way to figure the optimal calling win percentage in each of these cases. You're right that it depends on the number of rounds you can afford to fold.