They’re equally likely. Imagine someone approaching the building from the opposite side. If you can see two sides, she can see three, and vice versa. Because these are the only two possibilities, the probability of each must be 1/2.

(Strictly speaking, the two probabilities are perfectly equal only if you’re approaching an abstract pentagonal building from an infinite distance. But you get the point.)

1. Rh8 Kxc2 2. Qh7#

Ask either guard, “Which door would the other guard advise me to take?” The truth-telling guard will honestly indicate the bad door, and the lying guard will falsely indicate the bad door. Either way, you’re safe in taking the opposite door.

A sequel, from John Finnemore’s Souvenir Programme:

Let C be the set of residents who drink coffee and C′ the set who don’t, and so on. So C′ contains 10 percent of the residents, T′ 20 percent, W′ 30 percent, and G′ 40 percent. Because every resident abstains from something, these four segments must cover 100 percent of the population. And because the four segments just total 100 percent, there can be no overlap among them. Hence every resident drinks three of the four beverages, and 100 percent drink liquor.

Twenty-five. He could avoid pitching in the ninth inning if his team lost. But in order to lose he must have allowed at least one run, and that would require at least one pitch. Twenty-four additional pitches produced three outs in each of eight innings, for a total of 25 pitches.

02/06/2012 UPDATE: I found this problem in the Litton Problematical Recreations series, but as many of you have pointed out, the answer is wrong. Technically there are at least two ways in which a pitcher can complete a game without throwing a single pitch:

— Rule 8.04 says that if a pitcher fails to deliver the ball within 12 seconds of receiving it, the umpire can call a ball. A very dilatory pitcher could put 27 men on base in this way and then pick them all off.
— If the pitcher plays right field, he won’t pitch at all. :)

Thanks for everyone who wrote in about this.

1. Re6! and the bishop will mate.

Well, suppose they don’t. If there are n Welshmen in the world, each can have any number of Welsh friends from 0 to n-1 (he can’t be friends with himself). If we are to arrange things so that no two Welshmen have the same number of friends, then the first must have 0 friends, the next 1 friend, and so on. But then the loneliest Welshman has no friends and the most popular is friends with every Welshman but himself. This is a contradiction. Hence there must be some overlap — at least two Welshmen must have the same number of friends.

This is essentially the same idea as in Hand Count. I think it appeared originally in the Litton Problematical Recreations series.

Pirate A picks out what he thinks is a third of the loot. Pirates B and C, in turn, have the option of returning some of it to the chest. In effect, each is making a bid as to the smallest “third” that he’s willing to accept. The last one to make a proposal (which might be Pirate A himself) keeps this first portion for himself. The other two pirates then split the remaining treasure in the conventional way: one divides it into two portions and the other chooses between them.

1. Bd1 Kxb1 2. Bb3#