There are three empty bowls X, Y and Z on a table. Three players A, B and C take turns playing a game. A places a piece into bowl Y or Z, B places a piece into bowl Z or X, and C places a piece into bowl X or Y. The first player to place the 1999th piece into a bowl loses. Show that irrespective of who plays first and second (thereafter the order of play is determined) A and B can always conspire to make C lose.
Solution
A always plays into bowl X, and B always plays into bowl Y (if they can). For the first 999 moves each, A, B and C will certainly play into X and Y. If C can still play, then at least one of A and B can still play into X and Y. So for the next 500 moves C and at least one of A, B will play into X and Y. During that time at most 500 pieces go into Z, so A and B are still free to play into Z. After 999 + 500 moves there are at least 3·999 + 2·500 = 3997 pieces in X and Y. After the next piece is played into X and Y, they are both full and C cannot play. But both A and B can still play into Z. If both of A and B play into X/Y beyond the 999th move, then C loses quicker.
© John Scholes
jscholes@kalva.demon.co.uk
4 March 2004
Last corrected/updated 4 Mar 04