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 )?
Sorry the title is incorrect, this is not a full solution.
If the game only had 3 marbles, then A would never win. If A takes one, B takes one, A takes one and loses. If A takes 2, he loses as soon as B takes one.
One winning (partial) strategy would be to finish a move, while holding an odd number of marbles such that there are 3 marbles left and it's the other player's turn. Then just take one marble on your next move and you have an even number and you win.
Edited on October 3, 2006, 4:52 pm
Edited on October 3, 2006, 4:53 pm
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Posted by Larry
on 2006-10-03 16:33:48 |