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?

you always take 2 stones until there are somewhere 0-6 stones left. if its your turn and there is 1 left, you of course win (2n+1).

if there are 2 left, again you take only 1, the opponent takes the last one and you still have 2n+1

3 left - you take all 3, 2n+3

4 left - you take 3 again, 2n+3

5 left - you take 1, the opponent is left with 4, and after him taking any amount, we are left with 1,2,3 again - and we know how to win.

6 left - we take 2, same story

7 left - we take 3

Posted by Archie on 2004-06-10 15:34:57