The second player can always win the Game of van der Waerden. The first round, player 2 picks whichever of 1,2,8, and 9 is closest to player 1's first move with the same parity. The second move, pick one of those four with the opposite parity, from the other end of the range if possible (e.g., if the your first pick was 2, take 9 if possible, otherwise 1). On player 2's third move, with one exception it is always possible to leave a position such that at least 2 of the 3 remaining numbers will not create an arithmetic sequence; on the next turn, 1 will be left, forcing the win. The exception is when player 1 has 1,2,5 and player 2 has 8,9 (or similarly, player 1 has 5,8,9 and player 2 has 1,2). In this case, take 4 (6); 3 (7) will then be playable on the next turn, since player 1 can't take it. I would be quite surprised if player 2 is not the winner of the game of Ramsey, as well. Besides having an extra edge all the time, the first move involves no real choice - all edges are equivalent - so the usual advantage of going first is not there. (Of course, player 2's first move has only 2 distinct options.) And, player 1 winds up with 1 more edge. Franklin T. Adams-Watters David asks: << Can anybody tell me who is the winner of the following two games. Game of Ramsey. Two players alternately pick edges from a complete graph on 6 vertices until one of them has collected three edges that form a triangle at which point his Opponent is the winner. Game of van der Waerden. Two players alternately pick numbers from the set {1,2, . .,9} until one of them has collected three numbers which form an arithmetic progression at which point his Opponent is the winner. The point is, of course, that these games cannot end in a draw.
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