[bobman0330]

OK, my problem was that X alone needs to be compared to the future gain from refusing and insisting on N.

So, when X=N-1
(N-N+1)(1-K)/K >= N-1
1/K - 1 >= N - 1
1/K >= N

So, if the game has a 1% chance of ending each round, A should offer $100 and B should accept.

Note that the amount of money in the pot is irrelevant to the solution. If N turns out to be a large portion of the total, then I think we have to look at the situation from A's perspective as well.