This solution is by Clayton W. Dodge of the University of Maine at Orono:

Number the boxes 1-9 from top left to lower right, so that the first column is 1-4-7. Now:

1. The middle box must be empty. If it were marked, say with an X, then the board’s second X would have to be adjacent to it somewhere, and an O must complete that row, column, or diagonal. But now wherever we place the remaining O, the O player can immediately force a win by making a double threat if she has the move. X would be guilty of malpractice for allowing this disaster to come about, and we know she’s an expert player. So it must be X’s turn to move. But now we’ve deduced whose turn it is, and we’ve been told this is impossible. So this whole scenario cannot be the case — the center box must be empty.

2. No two similar marks can appear symmetrically around box 5, since then (if that player has the move) it leads to an immediate win or (if that player doesn’t have the move) we’re violating observation 1 above.

3. The same mark, say X, can’t appear in two adjacent corners, because then an O must appear between them, as well as in another box that will allow O to make two simultaneous threats if she has the move.

4. The same mark, say X, can’t appear in two middle-of-an-edge squares. If Xs appeared in, say, 2 and 8 we’d be violating observation 2. If they appeared in boxes 2 and 4, then one O must go in box 1 (to prevent the double threat) and another in box 6 or 8 to prevent X from winning with box 5. But then O can win with box 9.

5. The same mark, say X, can’t occupy two adjacent squares (say 1 and 2), since then an O must go in box 3 and another box that would allow O to make a double threat if she has the move.

6. Putting all this together, each player has marked one corner and one box opposite to it in the middle of an edge. There are four possibilities:

• If X has played to 1 and 6 and O to 2 and 9, then X can play to 4 and force a win.
• If X has played to 1 and 6 and O to 2 and 7, then X can play to 5 and force a win.
• If X has played to 1 and 6 and O to 4 and either 3 or 9, then expert play will always lead to a draw.

So the envelope must bear one of these two diagrams (or its symmetrical counterpart):

(From Pi Mu Epsilon Journal 5:10 [Spring 1974], 534.)

White wins with the craziest move on the board, 1. Qd4!, offering the queen. If Black captures her then 2. Nxd4 is mate, but other options are no better:

• If Black moves his knight to any other square, then 2. Qa1#. (If 1. … Nxd2 then 2. Nxd2#.)
• If he moves his bishop then the queen mates from b2.
• If the king captures the knight then 2. Qd3#.
• If the a-pawn captures the knight then 2. Qa1#.

The following year Crowley began a three-year course at Cambridge, where he later claimed to have beaten the president of the university chess club in his first year. He considering a professional career in chess and practiced two hours a day, but he gave up the idea when occultism captured his interest.

The king must visit the perimeter squares of the chessboard in order; if he skips any he’ll be unable to visit them and get home again without crossing his own path. And since the perimeter squares alternate colors, the path to each successive perimeter square must involve at least one horizontal or vertical move (diagonal moves alone won’t bring the king to a new color). There are 28 perimeter squares, so the task requires at least 28 horizontal or vertical moves. Here’s a tour that requires exactly 28:

(From John J. Watkins, Across the Board: The Mathematics of Chessboard Problems, 2004.)

1. Qf8! and 2. b5 or 2. Qe8 will mate.

From Magyar Sakkvilág, 1911.

1. Qc6! threatens 2. Qh1#. If the bishop moves to d3 or e4 to prevent this, then 2. Qc1 is mate.

1. Rd3! The king is forced to take the rook, and 2. Qf3 is mate.

This solution is by Howard Haber, who writes, “The key observation is that if two of the three integers are equal, then the person who sees the equal integers knows that his number is twice the same integer.”

Disregard the actual values for the moment and focus on their ratio instead. For A, B, C respectively, this ratio must be 5, 2, 3. To see why, consider the first round. A, who sees the numbers 2, 3, concludes that he himself must be wearing either 1 (1 + 2 = 3) or 5 (2 + 3 = 5). A can’t get any further than this in the first round and watches the play continue, adopting for the moment the hypothesis that A, B, C are 1, 2, 3. He’s not surprised when B passes his turn, then, because that’s consistent with his idea: B would have seen A wearing 1 and C wearing 3 and been unable to decide whether he himself were wearing 2 (1 + 2 = 3) or 4 (1 + 3 = 4). So he’d pass. So far, so good. Now play passes to C. If A, B, C were 1, 2, 3, then C would see A wearing 1 and B wearing 2. That would tell C that he himself must be wearing 3 (1 + 2 = 3) or 1 (1 + 1 = 2). But unlike the first two players, C would have a way to eliminate one of his candidates: He would know that he himself could not be wearing 1, because if he were, then B would have seen the numbers 1 and 1 on his turn, and thus would have realized that his own (B’s) number must have been 2 (according to the key observation above). With that insight B would have had enough information to win the game on his turn — he could have declared that A, B, C were 1, 2, 1. But B passed his turn, revealing to C that B was unable to reach that conclusion. And seeing that, C could then have concluded that the numbers could not have been 1, 2, 1 and so must have been 1, 2, 3.

But C too passes his turn. That tells A (finally!) that his starting hypothesis was wrong: A, B, C can’t be 1, 2, 3. That leaves only the second possibility he’d identified, where A, B, C are 5, 2, 3, and he announces this.

“For the problem as stated [i.e., using actual values rather than ratios], just multiply all numbers above by 10.”

1. Qh7! and the queen will mate.

From Chess Problems, 1873.