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A game of Nim (Posted on 2002-04-23) Difficulty: 3 of 5
This game is a variation of the old Eastern game "Nim".

Starting with a deck of cards, two players take turns removing cards from the deck. During his turn a player may take one, two or three cards. The winner is the person who takes the last card (or cards) from the deck.

There is a strategy that the first player can follow in order to assure his victory, but this strategy will not work for all the deck configurations.

What is this strategy, which of the following decks will it work for, and why?

  • 32 cards (7s through Aces)
  • 36 cards (6s through Aces)
  • 52 cards (2s through Aces)
  • 54 cards (2s through Aces + 2 Jokers)

See The Solution Submitted by levik    
Rating: 3.8000 (10 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts One way to do it? Comment 3 of 3 |
There is a stategy what works for most games like this: (t-1)/(c+1)where t = the total number of cards and c = the number of cards you can take on a turn.
As long as there is a remainder after this division takes place, you can go first and win the game. Just start by taking the remainder, and after that take 1 when they take 3, 2 when they take 2, and 3 when they take 1.
If there is a remainder, then try to trick them into going first. Otherwise, try to reduce the number to a multiple of 4, like levik said.
  Posted by Captain Paradox on 2004-01-15 08:23:53
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