If there were no ties, then Andy’s 6 scissors used up all of Bob’s non-scissors plays (2 rocks and 4 paper). That means that Bob’s remaining plays, 4 scissors, were used against Andy’s 3 rocks and 1 paper:

Andy won.

(By Yoshinao Katagiri, via Nob Yoshigahara.)

1. Qd6! and mate follows.

Let P be the number of pennies, D the number of dimes, and H the number of half dollars.

P + D + H = 100

and

P + 10D + 50H = 500,

so

9D + 49H = 400.

Now, we know that H is less than 10, and the only value for which (400 – 49H) is divisible by 9 is 1. So the totals are 1 half dollar, 39 dimes, and 60 pennies.

“If you take care of your peonies,” wrote Franklin P. Adams, “the dahlias will look after themselves.”

1. Second place.
2. $20.
3. February in a leap year.
4. S. (They’re the initial letters of the question.)
5. They were the same man, Grover Cleveland.

A full house with three aces will beat any other full house dealt in this hand, because only one hand can have three aces, assuming no cards are wild. But even this can still lose to a superior hand — four of a kind or a straight flush.

There’s nothing we can do to reduce the possibility that another player holds four of a kind — if we’re holding a full house then 11 varieties are possible. But by choosing carefully we can minimize the likelihood of a straight flush. Specifically, each of the hands AAA99, AAA88, AAA77, and AAA66 will prevent 13 straight flushes — each ace prevents one, and each spot card prevents five.

These are thus the best to choose; of the 40 possible straight flushes, each leaves only 40 – 13 = 27 straight flushes that can beat us. By contrast, the impressive-looking AAAKK leaves 40 – 6 = 34, or even more if we haven’t covered all four suits.

From Aaron Friedland’s Puzzles in Math and Logic, via Peter Winkler’s Mathematical Mind-Benders.

Suppose there had been just one unfaithful husband. At the mayor’s announcement, every other wife in the village would know he was the guilty party, and his own wife, aware of no other cheating husbands, would realize that her own husband was guilty and shoot him on the first day.

If there were two unfaithful husbands, Mr. X and Mr. Y, nothing would happen on the first night. The wives with faithful husbands would know of both guilty parties, and Mrs. X and Mrs. Y could each assume that the other’s husband accounted for the mayor’s announcement. But the next day, aware that no other wife has shot her husband, Mrs. X finds herself in the position described by the preceding paragraph: She knows of Mrs. Y’s cheating husband, and the fact that Mrs. Y forbore shooting him tells her that Mrs. Y is aware of another cheating husband that she herself (Mrs. X) can’t account for. That husband can only have been her own. Mrs. Y reasons in just the same way, and both shoot their husbands on the second day.

A pattern emerges. If there were three unfaithful husbands, then nothing would happen on the first day, and that very fact would leave all three wives in the position described by the preceding paragraph. All three would shoot their husbands on the third day.

So if there are 40 unfaithful husbands in all, the village would experience 39 days of peace, and then all 40 cheating husbands would be shot by their wives on the 40th day.

(I think the principle here was introduced in “The Women of Sevitan,” a problem in Herbert Gintis’ 2000 book Game Theory Evolving.)

1. c7 and the queen will mate.

It’s certain. Point A, Point B, and the planet’s center determine a plane that cuts the planet in half; Point C must lie on one side or the other of this plane (or directly on the great circle).

1. Qa2 Ke8 2. Qa8+ Kd7 3. e6#

Connect the midpoints of the triangle’s sides to make four smaller triangles. Because there are five points, two of them must fall within one of these triangles. The maximum distance between these two is 5 inches.