U.S. senator Alan Cranston once lost a copyright suit to Adolf Hitler. Cranston, who had begun his career in journalism, spotted an abridged translation of Mein Kampf in a New York bookstore in 1939. He had read the full text in German and was concerned that the English adaptation omitted Hitler’s anti-Semitism and ambitions to dominate Europe.

To publicize the truth, Cranston worked with a friend to publish an anti-Nazi version of the book. “I wrote this, dictated it [from Hitler’s German text] in about eight days, to a battery of secretaries in a loft in Manhattan,” Cranston told the Los Angeles Times in 1988. They produced a tabloid edition of 32 pages, reducing Hitler’s 270,000 words to 70,000 to yield a “Reader’s Digest-like version [showing] the worst of Hitler.”

At 10 cents apiece, Cranston’s version sold half a million copies in 10 days. But by that time the original was a best-seller in Germany, and the publishers sued Cranston for undercutting the market. In June the U.S. Circuit Court of Appeals in New York ordered the presses stopped. The truth had gotten out, Cranston said, but “we had to throw away half a million copies.”

Should parents be licensed? We ask teachers to study full-time for years and to pass qualifying exams before we let them educate children for six hours a day. And we carefully assess the suitability of adoptive and foster parents. But anyone has the right to become a biological parent without any training at all in child development.

Philosopher Peg Tittle writes, “How many children have been punished because they could not do what their parents mistakenly thought they should be able to do at a certain age — remember X, carry Y, say Z? How many have been disadvantaged because they grew up on junk food — for their bodies as well as their minds? How many have been neglected because their parents didn’t notice the seeds of some talent?”

Today’s children are tomorrow’s citizens, so the public has a legitimate concern in this. Psychiatrist Jack Westman writes, “The way children are parented plays a vital role in the quality of all our lives. We no longer can afford to avoid defining and confronting incompetent parenting.”

Psychologist Roger McIntire writes, “We already license pilots, salesmen, scuba divers, plumbers, electricians, teachers, veterinarians, cab drivers, soil testers, and television repairmen. … Are our TV sets and toilets more important to us than our children?”

(Peg Tittle, ed., Should Parents Be Licensed?, 2004.)

I had a growing feeling in the later years of my work at the subject that a good mathematical theorem dealing with economic hypotheses was very unlikely to be good economics: and I went more and more on the rules — (1) Use mathematics as a shorthand language, rather than an engine of inquiry. (2) Keep to them until you have done. (3) Translate into English. (4) Then illustrate by examples that are important in real life. (5) Burn the mathematics. (6) If you can’t succeed in 4, burn 3. This last I did often.

— Alfred Marshall, in a letter to A.L. Bowley, Jan. 27, 1906

In 1919 a bizarre catastrophe struck Boston’s North End: A giant storage tank failed, releasing 2 million gallons of molasses into a crowded business district at the height of a January workday. In this week’s episode of the Futility Closet podcast we’ll tell the story of the Boston Molasses Disaster, which claimed 21 lives and inscribed a sticky page into the city’s history books.

We’ll also admire some Scandinavian statistics and puzzle over a provocative Facebook photo.

Intro:

In 1888 three women reported encountering a 15-foot flying serpent in the woods near Columbia, S.C.

In 1834 the American Journal of Science and Arts reported the capture of a pair of conjoined catfish near Fort Johnston, N.C.

Sources for our feature on the Boston Molasses Disaster:

Stephen Puleo, Dark Tide: The Great Boston Molasses Flood of 1919, 2003.

Fred Durso Jr., “The Great Boston Molasses Flood of 1919,” NFPA Journal 105:3 (May/June 2011), 90-93.

Sean Potter, “Retrospect: January 15, 1919: Boston Molasses Flood,” Weatherwise 64:1 (January/February 2011), 10-11.

Please consider becoming a patron of Futility Closet — on our Patreon page you can pledge any amount per episode, and we’ve set up some rewards to help thank you for your support. You can also make a one-time donation on the Support Us page of the Futility Closet website.

Many thanks to Doug Ross for the music in this episode.

If you have any questions or comments you can reach us at podcast@futilitycloset.com. Thanks for listening!

Our legal system assumes that a defendant is innocent until proven guilty beyond a reasonable doubt. But what constitutes a reasonable doubt? Law professors Ariel Porat and Alon Harel suggest that an “aggregate probabilities principle” might help to determine whether an accused party is innocent or guilty.

Suppose we’ve decided that the evidence must indicate a probability of 95 percent guilt before we’re willing to declare a defendant guilty. Mr. Smith is accused of two separate crimes, with a 90 percent probability of guilt in each case. Under the 95 percent rule he’d be acquitted of both crimes. But Porat and Harel point out that there’s a 10 percent chance that Smith is innocent of each crime, and aggregating the probabilities gives a 0.10 × 0.10 = 0.01 chance that Smith is innocent of both — that is, there’s a 99 percent chance that he’s guilty of at least one of the offenses.

On the other hand, consider Miller, who is also accused of two different crimes. Suppose that the evidence gives a 95 percent probability that he committed each crime. Normally he’d be convicted of both offenses, but aggregating the probabilities gives a 0.95 × 0.95 = 0.9025 chance that he’s guilty of both offenses, and hence he’d be acquitted of one.

In A Mathematical Medley (2010), mathematician George Szpiro points out that this practice can produce some paradoxical outcomes. Peter and Paul are each accused of a crime, each with a 90 percent chance of being guilty. Normally both would be acquitted. But suppose that each was accused of a similar crime in the past, Peter with a 90 percent chance of guilt and Paul with a 95 percent chance. Accordingly Peter was acquitted and Paul went to prison. But historically Peter has now been accused of two crimes, with a 90 percent chance of guilt in each case; according to the reasoning above he ought to be convicted of one of the two crimes and hence ought to go to jail today. Paul has also been accused of two crimes, with a 0.95 × 0.90 = 0.855 chance that he’s guilty of both. He’s already served one prison term, so the judge ought to acquit him today.

Szpiro writes, “Thus we have the following scenario: in spite of the evidence being identical, the previously convicted Peter is acquitted, while Paul, with a clean record, is incarcerated.”

Van Helsing, who is of course famous for his part in the destruction of Dracula, has had many other encounters with supernatural creatures. In the early hours of one morning, he was woken by a loud knock at the door. “Come quickly!” cried the chief of police. “There’s been a ghastly attack at the manor house on the hill!”

Van Helsing dressed hurriedly and followed the chief. A grisly sight met him when he arrived. The front door of the house was open, and the beam of light that came from within shone on the body of a young man lying on the path. His throat had been torn out viciously, as though he had been attacked by some kind of hideous wild beast. Van Helsing looked around, but the grounds were dark, since the moon had set some time before, and he could see nothing else.

He stepped inside and found that several officers of the local constabulary were comforting a woman who appeared to be the maid. “It was horrible!” she cried. “I came down here after hearing some racket outside, and I found the young master at the door. ‘There’s something out there,’ he told me, ‘some beast, and I mean to drive it off.’ And he had in his hand the poker from the fireplace as a weapon. But when he opened the door, it was on him in a flash, a great beast, all hairy and shaggy, bigger than a man it was!”

Van Helsing stepped forward. “What was it?” he demanded.

The maid let out a little scream and gasped, “It was a werewolf!” And with that she fainted dead away.

“Could it be, Van Helsing?” said the chief, sounding worried.

Van Helsing knew it couldn’t be a werewolf because the moon was not full. A full moon is always directly opposite the sun in the sky — it sets when the sun rises and rises when the sun sets. However, if it was the early hours of the morning, the full moon would still be up, lighting up the land. Since the moon had already set, it could not possibly be full. And if the moon wasn’t full, it could not have transformed anyone into a werewolf.

Cheering news from India. My press clipping agency has sent me a letter from the correspondence column of an Indian paper about a cow that came into the bungalow of a Mr. Verrier Elwyn, who lives at Patengarth, Mandla District, and ate his copy of Carry On, Jeeves, ‘selecting it from a shelf which contained, among other works, books by Shakespeare, Thomas Hardy and Henry Fielding.’ A pretty striking tribute I look on that as.

— P.G. Wodehouse to William Townend, Sept. 3, 1929

Towns A and B are connected by two roads. Suppose that two cars connected by a rope of length 2r can travel from A to B without breaking the rope. How can we prove that two circular wagons of radius r, moving along these roads in opposite directions, will necessarily collide?

This can be solved neatly by creating a configuration space. Map each road onto a unit segment, and set these up as two sides of a square. The northern car’s progress is reflected by a point moving up the left side of the square, and the southern car’s by a point moving from left to right along the bottom. Now the motion of the two cars from A to B is represented by a continuous curve connecting (0,0) and (1,1).

The wagons start from opposite towns, so their motion is represented by a curve from (0,1) to (1,0), and it’s immediately clear that the two curves must intersect. The intersection point corresponds to the collision of the wagons.

This example, by N. Konstantinov, is reportedly common in Russian mathematical folklore; I found it in Serge Tabachnikov’s 2005 book Geometry and Billiards (of all places).

SIR, — While surveying in northern Labrador I had occasion to visit the island of Nukusustok, a few miles to seaward of the village of Nain. On the slope of a hill, and about 300ft. inland, I found a golf ball in good condition. How did the ball come to be there, and so far inland? It is possible that the ball was driven by a golfer from an Atlantic liner during practice, drifted northward past Greenland, and was finally carried ashore by the Labrador current which runs from north to south along the Labrador coast.

I have sent the ball to Dunlops, the makers, who suggest that it was probably carried so far inland by a sea bird. Perhaps some of your readers could help in explaining the mystery.

A problem from the American Mathematical Monthly, March 1930:

Two men jointly own x cows. They sell these for x dollars per head and use the proceeds to buy some sheep at $12 per head. Their income from the cows isn’t divisible by 12, so they buy a lamb with the remainder. Later they divide the flock so that each man has the same number of animals. This leaves the man with the lamb somewhat short-changed, so the other man gives him a harmonica. What’s the harmonica worth?

The men sold x cows for $x, so their income from that sale is $x^{2}. x^{2} isn’t divisible by 12, so x can’t be a multiple of 6 (if it were, then x^{2} would be divisible by 36 and hence by 12). So x is 12n ± y, where n is any integer and y is an integer from 1 to 5.

Each man ends up with the same number of animals, so the total number of animals must be even, which means there’s an odd number of sheep and one lamb. So (12n ± y)^{2} when divided by 12 should leave a remainder and have an odd integer for its quotient.

12n^{2} ± 2ny will always be even, so that last term, y^{2}/12, must provide an odd contribution toward this quotient. In order to do this, y^{2} must exceed 12, so y must be 4 or 5. If it’s 5, we get

which provides an even contribution (2) to the quotient. So y must be 4 and y^{2} = 16:

That’s what we need; the integer part of this quotient, 1, is odd. Since the remainder when 4^{2} is divided by 12 is 4, a lamb costs $4. Consequently, in dividing the flock one man got a $4 lamb and the other got a $12 sheep. To make this fair, the traded harmonica must be worth $4.

(The problem was proposed by J.H. Neelley, T.L. Smith, and P.S. Ganesa Sastri in AMM 37:3, 162; Ross Honsberger introduced the harmonica in recounting it in his Mathematical Morsels, 1978. This solution is by P.S. Ganesa Sastri of Trichinopoly, South India.)