  Let A be a finite subset of $\nat$. Then NIM(A;n) is the following 2-player
game: initially there are $n$ stones on the board and the players alternate
removing $a\in A$ stones. The first player who cannot move loses. This game has
been well studied.
  We investigate an extension of the game where Player I starts out with d
dollars, Player II starts out with e dollars, and when a player removes a\in A
he loses a dollars. The first player who cannot move loses; however, note this
can happen for two different reasons: (1) the number of stones is less than
min(A), (2) the player has less than $\min(A)$ dollars. This game leads to more
complex win conditions then standard NIM.
  We prove some general theorems from which we can obtain win conditions for a
large variety of finite sets A. We then apply them to the sets A={1,L}, and
A={1,L,L+1}.
