In a two-player, zero-sum game with perfect information, a losing position is one from which any legal move leads to a winning position and, a winning position is one from which a losing position may be reached in exactly one legal move, and all positions where no legal moves are possible are losing positions. Nim, viewed in the light of the above recursive definition, involves both players trying to force the other player into a losing position so that irrespective of the move they make, they return you to a winning position.
Update: After Jean-Pierre clarified his question (exactly one PASS move available for both players i.e. after one uses the PASS, then no one else can during the rest of the game)
Conclusion: The first player (P1) is guaranteed a win provided he is not in a winning position from which another winning position cannot be reached.
Strategy: If P1 is in a losing position then he will simply pass, ensuring a win eventually for P1. If P1 is in a winning position, he will play sub-optimally to give to P2 another winning position. If P2 passes now, then P1 can win the game by playing optimally. If P2 chooses to play optimally then P1 can pass and win as in the case when P1 is in a losing position.
If the above assumption fails to hold, i.e., if the winning position P1 is in, is one from which only other winning positions may be reached, then P2 will win by stealing P1's strategy as described above.
The possible moves are
(1 - n) take n coins of the pile 1
(2 - n)take n coins of the pile 2
(3) take one coin of both piles
The trick in this game is to give your player one of the following kinds of patterns:
(a) N ; N
(b) N + 1; N + 2
(c) N + 2; N + 1
Where $N$ represents a multiple of 3 (can be 0. In those positions, the other player cannot win in one round, and we can always answer them in such a way we can give them another piles of those kinds.
We will represent B as the player with the winning strategy, and gives to player A one of the referred positions, wlog (a): if A plays (3), B takes one coin, if A plays (1 - n) or (2- n), the:
if n=1 mod 3, B plays n+1 on the other stack
if n=2 mod 3, B plays n-1 on the other stack
if n=0 mod 3, B plays n on the other stack
In that fashion, B always leaves the game in one of the winning kinds. For (b) and (c), the strategy is similar.
Summing up:
Suppose $N$ is 1 module 3, then A has to use (3) and plays as I explain after, having the winning strategy.
Suppose $N$ is 2 module 3, then A has to use (1 - 1) or (2 - 1), i.e. has to take one coin from a pile at choice. Then plays as I explain after, having the winning strategy.
If $3|N$, then B can always answer
Best Answer
This is a well known result. Surely you can google up something, or failing that, go to a math library and read a solution from "Winning Ways I-II" by Berlekamp, Conway and Guy.
But I'm glad to hear that you want to work it out yourself. Here are a couple of hints. Call a position in the game of Nim balanced, if the sequence of numbers of remaining coins in the piles satisfies the condition "even number of 1s in any position", and call it unbalanced otherwise.
1) Show that if a player is given a balanced position, then any move s/he will make leaves an unbalanced position.
2) Show that if we are given an unbalanced position, there exists a move (at least one, possibly more) that will leave the other player with a balanced position.
3) Note that the desired all-zeros end position is balanced. Apply a little bit of recursive thinking (or induction, if you like).