TITLE: Nim game for odd number of stones
QUESTION [10 upvotes]: Consider the classical Nim game with total number of stones being odd. Then the first players wins, of course, what follows from the general description of winning positions. But is there some shorter (independent of full theory) explanation of this fact, maybe with implicit strategy or whatever?

REPLY [3 votes]: This doesn't answer the OP's question, but it occurred to me (while trying to answer the question) that there's a nice strategy-stealing argument in a somewhat artificial setting:  Suppose one pile has more than half of all the stones.  Then the first player has a winning strategy.  
It's actually (I think) a little tricky to show that the usual analysis guarantees a first-player win:  You need to show that the binary number for the dominant pile has a 1 where all the other piles' numbers have a 0.  But here's the strategy-stealing argument:
Suppose there are $n+1$ piles of size $p_0,p_1,\ldots,p_n$ with $p_0 \gt P=p_1+\cdots+p_n$.  The first player's opening move will be to take some number $1\le k\le P+1$ from pile "0."  If the second player had a winning strategy, it couldn't possibly be to take additional stones from pile 0, because the first player could have stolen that move directly.  This leaves the second player with only $P$ possible responses.  A winning strategy, if there were one, would be to assign one of these responses to each of the first player's opening moves.  (There may, in principle, be many many ways of making the assignment, but we need only imagine that the second player has decided in advance, "If the first player does this, I'll do that.")  By the pigeonhole principle, there would be two numbers, $1\le i \lt j \le P$ for which the second player's move would be the same.  There's the opening for the heist:  First take $i$ stones from pile 0, then take $j-i$ on the next move.