Let us assume that we play a nim-game (last person who draws a card, wins) and are in a position $a_1, a_2$ (sizes of piles). Assume that we have nim-sum (NS) $a \neq 0$.
How many moves are there?
How many winning moves are there from this position?
Question 1: Since the sizes of my piles are $a_1,a_2$, there should be a total of $a_1+a_2$ moves that I could make, correct?
Question 2: If I want to win, I need to be in a position where every time it's my turn, the NS $\neq0$. Then, I need to make sure that every move I make creates a situation where the other person always gets a NS $=0$. We know that if the other person gets a NS $=0$, then when it's my turn it will always be true that the NS $\neq 0$, and this loop will continue until I win.
But how many winning moves are there? To go from a NS $\neq 0$ to NS $=0$, one winning strategy is to always take the pile with the biggest size (we know that $a_1\neq a_2$ when NS $\neq 0$), and re-arrange it in a way such that NS $=0$, but I'm in a bind as to how many winning moves there are.
EDIT: See this link for an explanation of the game and what the nim-sum is.