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
the strategy by Liam)
its not a bad strategy, but the flaw is in that if your opponent also adopts the same strategy. at the end both of you would have taken an odd number, *2= even number of total taken. so there's a chance of your opponent manuvering you into losing.
e.g. there are 5 left, and its your opponent's turn. if he takes 1, there'll be 4 left.
if you take 3 he'll take the last and win.
if you take 2, he'll take 1, you get left with the last one, and he still wins.
if you take 1, he'll take 3 and still win.
but hopefully he's not that smart. or its your turn, not his.

Posted by neko
on 20040816 08:43:33 