This game is played by two players at a round table. They each take turns placing identical coins onto the table's surface. No two coins can overlap, and the entirety of the coin's surface must rest on the table. The loser of the game is the first person who is unable to put down a coin because there is no more room.

Which of the players has a winning strategy? What is it?

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(In reply to "novice" epitomized with this answer by Joni)

It's an interesting thought, and it would probably work if there were conditions placed on how the coins could be placed. But there aren't any such conditions beyond that the entire coin must lie on the table (no overlapping and no hanging off the edge) consider what happens when it's player 2's last move: there is enough space for two coins (player 2's last play and player 1's winning play) If player 2 has been careful, the two spaces are contiguous -- there is actually only one space large enough to fit two coins. Player 2 then plays her coin to the middle of that space, leaving only two spaces each only large enough for half of a coin . Player 1 cannot make his last play, and player 2 wins.

Player 1's strategy must ensure that the last two spaces are non-contiguous, separate, so that Player 2 must play one and leave the other free for him to play.

Posted by TomM on 2002-09-12 05:23:53