Two boys , Brian and Gabe, decide to play a game.

They lay 20 coins on the table. Each player took turns taking 1, 2, or 3 coins. Whoever took the last coin won.

Brian thought if he went first, he'd win for sure since he had the advantage of making the first move. So Brian did go first. But Gabe won every time!

Obviously Gabe had a winning strategy, but what was it? Could Brian have outsmarted it?


1 Answer 1


Always leave a multiple of 4 coins. So if Brian takes 1, Gabe takes 3. 2-2, 3-1.

Starting with a multiple of 4, player 2 can always win with perfect play. Starting with any other number, player 1 can win with perfect play.

  • $\begingroup$ Great job! Short, Simple, and overall Excellent explaination! $\endgroup$
    – warspyking
    Commented Oct 6, 2014 at 1:55
  • $\begingroup$ Great answer! Player 1 can win in any other case situation because he can just make the coins a multiple of 4. Remember that any amount of coins have the form $4n,4n+1,4n+2,4n+3$ $\endgroup$
    – durum
    Commented Oct 6, 2014 at 10:27

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