Dog Tired

http://commons.wikimedia.org/wiki/File:Canal_de_Panam%C3%A1_Construcci%C3%B3n_003.jpg

Maybe figures can’t lie, but liars can certainly figure, and that is why statistics can be made to ‘prove’ almost anything. Consider a group of ten girls, nine of them virgins, one pregnant. On the ‘average’ each of the nine virgins is ten per cent pregnant, while the girl who is about to have a baby is ninety per cent a virgin. Or, assuming that a fox terrier two feet long, with a tail an inch and a half high, can dig a hole three feet deep in ten minutes, to dig the Panama Canal in a single year would require only one fox terrier fifteen miles long, with a tail a mile and a half high.

— Stuart Cloete, The Third Way, 1947

Never the Twain

proclus paradox

A paradox attributed to Proclus Lycaeus (412-485):

Consider two nonparallel lines, AQ and BP. BP is perpendicular to AB; AQ isn’t. Find the midpoint of AB and mark AC = BD = AB/2. Now if AQ and BP are going to intersect, it can’t happen on AC or BD; if it did, say at a point R, then that would give us a triangle ARB where the sum AR + RB < AB, which is impossible. But now we can connect CD and follow the same process: CE and DF can't intersect for the same reason. EG and FH are likewise ruled out, and so on up the line forever. This seems to mean that two nonparallel lines will never intersect. That can’t be right, but where is the error?

(From Alfred Posamentier, Magnificent Mistakes in Mathematics, 2013.)

Non-Fiction

http://www.sxc.hu/photo/440835

Sherlock Holmes is an honorary fellow of the Royal Society of Chemistry.

“Holmes did not exist, but he should have existed,” society chief David Giachardi said in bestowing the award in 2002. “That is how important he is to our culture. We contend that the Sherlock Holmes myth is now so deeply rooted in the national and international psyche through books, films, radio and television that he has almost transcended fictional boundaries.”

The Wheel Cipher

http://commons.wikimedia.org/wiki/File:Jefferson%27s_disk_cipher.jpg

Thomas Jefferson, already absurdly accomplished by 1795, somehow found time to delve into cryptography, where he devised this cipher system. The letters of the alphabet are printed along the rim of each of 36 disks, which are stacked on an axle. One party rotates the disks until his message can be read along one of the 26 rows of letters, somewhat like a modern cylindrical bike lock. Now he can record the letters in any one of the other 25 rows and send that string safely to another party, who decodes it by reversing this procedure. If the message is intercepted, it’s useless even to someone who has the disks, because he must also know the order in which to stack them, and 36 disks can be stacked in 371,993,326,789,901,217,467,999,448,150,835, 200,000,000 different ways.

This is pretty robust. The cipher below, created in 1915 by U.S. Army cryptographer Joseph Mauborgne, has never been solved. “The known systems from this year (or earlier) shouldn’t be too hard to crack with modern attacks and technology,” writes NSA cryptologist Craig P. Bauer. “So, why don’t we have a plaintext yet? My best guess is that it used a cipher wheel” like Jefferson’s.

mauborgne cipher

(L. Kruh, “A 77-year-old challenge cipher,” Cryptologia, 17(2), 172-174, 1993, quoted in Bauer’s Secret History: The Story of Cryptology, 2013.)

The Traveler’s Dilemma

Two travelers are transporting identical antiques. Unfortunately, the airline smashes both of them. The airline manager proposes that each traveler write down the cost of his antique, any value from $2 to $100. If both write the same number, the airline will pay this amount to both travelers. If they write different numbers, the airline will assume that the lower number is the accurate price; the low bidder will receive this amount plus $2, and the high bidder will receive this amount minus $2. If they can’t confer, what strategy should the travelers take in deciding how to bid?

At first Traveler A might like to bid $100, the maximum allowed. If his opponent does the same then they’ll both net $100. But A can do better than this: If B bids $100 and A bids $99 then A will come away with $101.

Unfortunately if B is rational then he’ll have the same insight and also bid $99. So A had better undercut him again and bid $98.

This chain leads all the way down to $2. If both travelers are perfectly rational then they’ll both bid (and make) $2, the minimum price.

But this seems very unlikely to happen in actual practice — in real life both travelers would likely make high bids and get high (though perhaps unequal) payoffs.

“All intuition seems to militate against all formal reasoning in the traveler’s dilemma,” wrote economist Kaushik Basu in propounding the problem in 1994. “There is something very rational about rejecting (2, 2) and expecting your opponent to do the same. … The aim is to explain why, despite rationality being common knowledge, players would reject (2, 2), as intuitively seems to be the case.”