In 1903, French physicist Prosper-René Blondlot decided he had discovered a new form of radiation. But the mysterious rays had some exceedingly odd properties, and scientists in other countries had trouble seeing them at all. In this week’s episode of the Futility Closet podcast we’ll tell the story of N-rays, a cautionary tale of self-deception.
We’ll also recount another appalling marathon and puzzle over a worthless package.
Answer these questions:
The correct answers are 5 cents, 5 minutes, and 47 days, but each question also invites a quick, intuitive response that’s wrong. In order to succeed, you have to suppress your “gut” response and reflect on your own cognition deeply enough to see the error. Psychologist Shane Frederick devised the three-question test in 2005 to illustrate these two modes of thought, unreflective and reflective, which he called System 1 and System 2.
Scores on the CRT correlate with various measures of intelligence, patience, and deliberation, but cognitive ability alone isn’t strongly correlated with CRT scores: If you’re not disposed to answer impulsively then the problems aren’t hard, and if you do answer impulsively then cognitive ability won’t help you. A sample of students at MIT averaged 2.18 correct answers, Princeton 1.63, Carnegie Mellon 1.51, Harvard 1.43; see the link below for more.
(Shane Frederick, “Cognitive Reflection and Decision Making,” Journal of Economic Perspectives 19:4 [2005], 25-42.) (Thanks, Drake.)
In this original logic puzzle by the Japanese publisher Nikoli, the goal is to connect lattice points to draw a closed loop so that each number in the grid denotes the number of sides on which the finished loop bounds its cell, as above: Each cell bearing a “1” is bounded on 1 side, a “2” on 2 sides, and so on.
Here’s a moderately difficult puzzle. Can you solve it? (A loop that merely touches a cell’s corner point without passing along any side is not considered to bound it.)
So many more men seem to say that they may soon try to stay at home so as to see or hear the same one man try to meet the team on the moon as he has at the other ten tests.
This ungainly but grammatical 41-word sentence was constructed by Anton Pavlis of Guelph, Ontario, in 1983. It’s an alphametic: If each letter is replaced with a digit (EOMSYHNART = 0123456789), then you get a valid equation:
SO 31 MANY 2764 MORE 2180 MEN 206 SEEM 3002 TO 91 SAY 374 THAT 9579 THEY 9504 MAY 274 SOON 3116 TRY 984 TO 91 STAY 3974 AT 79 HOME 5120 SO 31 AS 73 TO 91 SEE 300 OR 18 HEAR 5078 THE 950 SAME 3720 ONE 160 MAN 276 TRY 984 TO 91 MEET 2009 THE 950 TEAM 9072 ON 16 THE 950 MOON 2116 AS 73 HE 50 HAS 573 AT 79 THE 950 OTHER 19508 + TEN 906 TESTS 90393
Apparently this appeared in the Journal of Recreational Mathematics in 1972; I found the reference in the April 1983 issue of Crux Mathematicorum, which confirmed (by computer) that the solution is unique.
When the Allies secured New Guinea’s Goodenough Island in October 1942, they left a small Australian occupation force to hold this important position against the Imperial Japanese. They succeeded through deception: The Australians built dummy structures (including a hospital), pointed logs at the sky to suggest anti-aircraft guns, wove jungle vines into barbed wire, lighted numerous “cooking fires” at night, and sent messages in easily broken code that suggested that a full brigade occupied the island.
It worked. The small force held the island until December 28, and a new garrison arrived the following year.
A puzzle by University College London mathematician Matthew Scroggs: A princess lives in a row of 17 rooms. Each day she moves to a new room adjacent to the last one (e.g., if she sleeps in Room 5 on one night, then she’ll sleep in Room 4 or Room 6 the following night). You can open one door each night. If you find her you’ll become her prince. Can you find her in a finite number of moves?
https://www.youtube.com/watch?v=BaCzOuHYuB8
Austrian artist Peter Kogler uses twisting lines and geometric shapes to generate dramatic illusions in ordinary spaces.
“The black-and-white grid provides a maximum contrast which has a very strong visual presence,” he says. “The structure of the image is comprehensive and completely surrounds the beholder. In a sense, you are standing in the picture, and the work can be experienced physically.”
In 1812 Percy Shelley and his wife Harriet had committed themselves to a vegetarian diet. During their residence in Ireland that March, Harriet sent a note to a friend in Dublin:
Sunday morng.
17 Grafton StreetMrs. Shelley’s comps. to Mrs. Nugent, and expects the pleasure of her company to dinner, 5 o’clock, as a murdered chicken has been prepared for her repast.
Isaac Bashevis Singer once said, “I am a vegetarian for health reasons — the health of the chicken.”
French mathematician Joseph Bertrand offered this observation in his Calcul des probabilités (1889). Inscribe an equilateral triangle in a circle, and then choose a chord of the circle at random. What is the probability that this chord is longer than a side of the triangle? There seem to be three different answers:
1. Choose two random points on the circle and join them, then rotate the triangle until one of its vertices coincides with one of these points. Now the chord is longer than a side of the triangle when its farther end falls on the arc between the other two vertices of the triangle. That arc is one third of the total circumference of the circle, so by this argument the probability is 1/3.
2. Choose a radius of the circle, choose a point on that radius, and draw a chord through that point that’s perpendicular to the radius. Now imagine rotating the triangle so that one of its sides also intersects the radius perpendicularly. Our chord will be longer than a side of the triangle if the point we chose is closer to the circle’s center than the point where the triangle’s side intersects the radius. The triangle’s side bisects the radius, so by this argument the probability is 1/2.
3. Choose a point anywhere in the circle and draw the chord for which this is the midpoint. This chord will be longer than a side of the triangle if the point we chose falls within a concentric circle whose radius is half the radius of the larger circle. That smaller circle has one-fourth the area of the larger circle, so by this argument the probability is 1/4.
Further methods yield still further solutions. After more than a century, the implications of Bertrand’s conundrum are still being discussed.
