Start with two piles of counters, not the same length. Two players take it in turns, and can either take as many as they like from one pile, or the same number from both. The winner is simply the player who takes the last counter. You'll notice that this is a special case of Nim, in that it has specifically two piles.
This isn't intended to be a championship game, it's just a mathematical curiosity.
At first this game seems rather confusing, but in truth it's remarkably simple. I'll explain some techniques then look at the math behind them. First, you have to understand the significance of the position (2, 1) and it's reflection (1, 2). If this is the position at the end of your go, you are in an unbeatable winning position. Whatever they do next will set you up for any easy win. Now, how to get this position, and how to avoid it:
- After your go, each pile must be at least three counters long. If one pile is empty, the other player can just take the other pile whole and win. If one pile has one or two, the other player can easily cut down the other pile and give themselves the (1, 2) position. Therefore, keeping each pile at least three long is crucial.
- Keep one pile at least two longer than the other. If the piles are (n, n+1), it's easy for the other player to remove n - 1 from both piles and leave themselves with (1, 2) and thus a win. Therefore you should be avoiding this.
That's a very straight-forward overview of how to play. Avoid those two situations and you give yourself the best chance. If you find yourself in a situation where one of these is inevitable, and therefore you are doomed to lose, take the fewest counters you can. This gives your opponent the most opportunities to make a mistake.
A game theory perspective.
As this is based on game theory, I suggest you read up on safe and unsafe positions if you are unfamiliar with these notions. I've already mentioned one simple safe position: the (1, 2). There are infinitely more, and the nth safe position is ( [φn], [φ²n] ), where [ ] represents the floor function and φ the Golden Ratio. Therefore the complimentary lengths of the rows in safe positions form a Beatty sequence.