The absurd-looking 1. Re7! wins in every variation.

From A Collection of Two Hundred Chess Problems, 1866.

1. Qh3 Ke4 2. Rc4#

From Will H. Lyons, Chess-Nut Burrs: How They Are Formed and How to Open Them, 1886.

If there were 9 games in the set, then one player served 4 games and one served 5. Suppose Agassi served 4 games and Becker had k service breaks. That would mean that Agassi won 4-k of the games in which he served, and 5-k of the games in which Becker served (because there were 5 service breaks altogether). Then Agassi would have won a total of (4-k) + (5-k) = 9-2k games, an odd number. But we know that Agassi won 6 games, so this is a contradiction. Hence Agassi must really have served 5 rather than 4 games, and therefore he served the first game.

From The Inquisitive Problem Solver, 2002.

Black’s king is in check, so White must just have moved his knight from b4 to a2. But what was Black’s move before that? He cannot have moved his king, as there’s no square from which it can have come. The only other possibility is that Black moved a piece that’s not now on the board — a piece that White’s a2 knight has just captured. What can that have been? Not a knight or a pawn, as there’s no square that either of these can have come from. Nor a queen or a rook, as those could only have moved from a1, where they would have been checking the White king at the end of White’s last move. The only remaining possibility is a bishop: Black has just moved a bishop from b1 to a2, and White has responded by capturing it with a knight from b4.

None. The same amount of water is necessary to raise any boat.

One way to make this intuitive is to imagine that a 2-meter “slab” of water is inserted at the bottom of the lock chamber. This will raise a boat of any size to the level of the higher lake.

Yes. The water imparts an upward force on your finger as you immerse it, so you must exert a downward force to overcome it. That force is transferred to the glass and then to the scale, which sinks.

From Yuri B. Chernyak, The Chicken From Minsk, 1995.

White could mate in three different ways if it were his turn, but it can’t be, as there’s no legal move that Black can just have made. So Black plays Bxf6#.

It seems at first that the chance must be 1/2, as we’ve simply added a white counter and then removed it again, returning the bag to its original state.

But this is an error. After the white counter is added, the bag contains either (a) 2 white counters or (b) 1 white counter and 1 black counter. These states are equally probable. The chance of drawing a white counter from bag (a) is 1; from bag (b) it’s 1/2. So when the white counter is drawn, the chance that the bag had contained (a) 2 white counters or (b) 1 white counter and 1 black counter are proportional to (a) 1/2 × 1 (b) 1/2 × 1/2, i.e., 2 to 1. So there’s a 2/3 chance that the remaining counter is white.

In other words, the white counter that we drew is twice as likely to have come from a white-white bag than from a white-black bag.

Both were wrong, and the lady was badly cheated. She only got half the quantity that would be contained in the large bundle, and therefore should have been charged half the original price. A circle with the circumference half that of another must have its area a quarter that of the other.

(Henry E. Dudeney, “The Paradox Party,” Strand 38:228 [January 1910], 626-632.)