You and a friend play a game in which there are an odd number of rocks. You can take 1, 2 or 3 rocks on your turn (alternating turns with your opponent); when all rocks have been taken, the person who has taken an odd number of rocks is the winner.

If you are the first to go, what strategy should you use in order to have the best chance of winning?

(In reply to re: strategy by SilverKnight)

There is no winning strategy for the first player with 5 initial rocks.  Considered from the opponents view, that player has an even number of rocks (zero) and has left 5 for the first player.  That's a winning strategy for the non-first player.

Looked at the other way, if the first player took 1 he'd have an odd number and be leaving the opponent 4--not a winning strategy.  If he took 2 or 3, that would leave 3 or 2-- not part of any winning strategy regardless of odd or even holdings.

Posted by Charlie on 2004-04-20 19:46:46