This question is actually very similar to the "double your bet" strategy for baccarat, or any other close to even odds game. Casinos typically will always have a table max bet for these games to prevent some billionaire from coming in and keep doubling their bet until win back all of their loses. The number of times you can lose before you hit the limit is the breaking point. In your scenario, the breaking point is 7, calculated this way:
100 * (2^x) > 10,000
The lowest X where the above equation is true is 7. So the worst case scenario for you is if you are called and lose 7 times in a row. If you were heads up in a tournament playing this way, the chances of you losing the tournament if your opponent just called every hand are 1/128. However, given that the person will not call every hand, and instead only call with better than average hands, you're less likely to win each hand that you are played against. Presumably the person you are playing against may be able to use a strategy that gives him an edge, I would guess around 20% or so, making the position worse, perhaps making your chances of winning in the range of 1/20-1/40. (I'm too tired to do the math on this right now, and I'm not 100% sure I'd get it right even if I tried.)
In your situation, a cash game, with multiple people at the table, on the positive side, you're less likely to continuously double up the same person 7 times in a row, but on the negative side, you're also less likely to win against multiple people, especially since you'll be against an even stronger than average hand that calls you. If the table got together and devised a strategy, surely they could break you. I don't know what that strategy would be exactly, and I believe it would depend on the number of opponents at the table, but it seems like it's easy to imagine a scenario where X people call with Y or better, and you would get called often enough to lose more than you pick up in blinds when everyone folds.
As for a mathematical answer to your question, I believe the number of opponents (P) you have will need to be figured into the equation.
Update: After further thought, I believe we can solve this. If we reduce your scenario to heads up, and make the assumption that villain calls all in every hand, (I assume that isn't the best strategy for villain, but it's one strategy, and one I believe we can quantify), and continues to rebuy at 100bb every time he busts out, then I believe the following:
- Villain will always win eventually.
- The formula for calculating the expected number of hands hero will last has an e in it (Eulers number).
- This is just pure speculation, but I believe the expected number of hands hero will last is less than 1 order of magnitude of K (less than 10*K).
My thinking is this (for simplicty I'm going to assume hero has 128 times villain's stack rather than 100 times:
At 128 to 1, villain has to win 7 in a row to bust hero. Villain has 128 tries to achieve this. If Villain fails, he then has 256 tries to hit 8 in a row, which is 1 in 256. If he still fails to do that, he then has 512 tries to hit a 1 in 512 shot, etc. The probability of hitting a 1 in X shot with X tries, as X goes to infinity, is related to the number Phi. (I forgot what the actual formula is, but I know where to go to find out.)
That was slightly over simplifying, since once villain's stack is greater than hero, there is some opportunity for them to toggle back and forth a few times before one busts out the other, but I'm assumming that will have a negligible effect on the total answer.
That's where I'm at, I'll update again when I get the actual formula...