A game of 11 marbles wherein each player can either pick one or two marbles from the total. Starting from Player A & then Player B alternatively. This continues till all the marbles are picked. The winner is the one having odd number of marbles.
What is the strategy to be followed for Player A & B to win?. What happens for higher total number of marbles (13, 15 etc )?
(In reply to re: Trial and error
by Federico Kereki)
"Suppose you have 5 marbles, there are 3 marbles on the table, and the other player has 5 marbles, AND PICKS 1. You win if you take TWO, not one, so "do the same as the other did" doesn't work here."
In your situation, "I" am Player B in a game Player A is supposed to win. The only way I could be Player A is if I hadn't been taking what my opponent took.
My strategy is pointed at the player who should win - If he slips up, the other player should obviously take advantage of it, but I have no strategy in mind for this case.
You're right, though, my solution is flawed. I'd like to amend it:
The strategy is to take the same number of marbles the other took. If you start, you should take two.
Posted by Tamir
on 2006-10-04 08:22:33