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Two men sat at a square table, each with a very large number of stakes. They had entered a contest of wits, or so they called it. The winner of the contest would be the last person to place a stake onto the table without disturbing another.

They didn't seem to mind that the stakes weren't all identical, but they reasoned that all of them were close to being four inches long, with all sides flat.

The second man sat smug in his chair, a clever strategy worked out in his head. However, when the first man had made his move, the second man's smile vanished.

What is the best strategy to win this game?

What did the first man do to disrupt the second's strategy?


marked as duplicate by JonTheMon, Hugh Meyers, manshu, StephenTG, Aggie Kidd Mar 9 '16 at 14:58

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  • $\begingroup$ Are you sure this is a logic puzzle? $\endgroup$ – Fimpellizieri Mar 9 '16 at 0:19
  • $\begingroup$ I think it's a quite good generalization of the typical one with a perfectly round table and both player's have the exact same sized circular stake/coin/glass. Made me think - this works with a square table and different sized and shaped pieces so long as the first can always match the second in a perfectly mirrored image. Stakes can be reversed (i.e. upside down) so this works! :) ) $\endgroup$ – Paul Evans Mar 9 '16 at 0:51
  • $\begingroup$ Corollary: one has to assume that there is at least one perfectly symmetrically shaped stake with respect to the table's shape (or one larger than the table!) for the first move to succeed. Also the table could be the shape of any regular polygon (including stars) or, of course, a circle. $\endgroup$ – Paul Evans Mar 9 '16 at 2:27

The first man:

Placed a stake (maybe a very large one - but doesn't matter) in the very center of the table.

Why the second man's smile vanished:

Because they both had a "very large number of stakes" so no matter where the second man puts whatever stake - the first man could simply place the same stake on the diagonal of the middle to that second (or 4th or 6th) mirroring the second - he always has a safe place to put a stake. Until the second falters - then the first man wins.

  • 1
    $\begingroup$ Why can't the second man balance his stake on top of the first one without disturbing it? It has square sides after all. $\endgroup$ – Gordon K Mar 9 '16 at 10:19
  • $\begingroup$ That's disturbing the first piece. $\endgroup$ – Paul Evans Mar 9 '16 at 10:23

It seems obvious:

The first man placed his stake balanced precariously on one edge (or on the point of the stake), so that pretty much any touching of the table or placing of another stake would cause it to fall.

The "best strategy":

Go first and place a stake balanced very carefully so that your opponent is most likely to upset it no matter what they do.

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    $\begingroup$ This was my thinking. Put it right on the edge about to drop. $\endgroup$ – Z. Dailey Mar 9 '16 at 2:19

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