# Podcast Episode 241: A Case of Scientific Self-Deception

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.

See full show notes …

# The Cognitive Reflection Test

1. A bat and a ball cost \$1.10 in total. The bat costs \$1.00 more than the ball. How much does the ball cost? _____ cents
2. If it takes 5 machines 5 minutes to make 5 widgets, how long would it take 100 machines to make 100 widgets? _____ minutes
3. In a lake, there is a patch of lily pads. Every day, the patch doubles in size. If it takes 48 days for the patch to cover the entire lake, how long would it take for the patch to cover half of the lake? _____ days

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.)

# A for Effort

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.

# Bertrand’s Problem

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.

# Moondance

What is the shape of the moon’s path around the sun? The moon orbits the earth, and the earth orbits the sun, so many of us imagine it looks something like the image on the left, a looping motion in which the moon periodically slides “backward” during its progress around the larger body.

But it’s not! The shape is closer to a 13-gon with rounded corners; there are no loops. Helmer Aslaksen, a mathematician at the National University of Singapore, writes, “I like to visualize this as follows. Imagine you’re driving on a circular race track. You overtake a car on the right, and immediately slow down and go into the left lane. When the other car passes you, you speed up and overtake on the right again. You will then be making circles around the other car, but when seen from above, both of you are driving forward all the time and your path will be convex.”

More at his page.

(Thanks, Drake.)

# For Pi Day

From the prolifically interesting Catriona Shearer: The red line is perpendicular to the bases of the three semicircles. What’s the total area shaded in yellow?

# A Thorough Anagram

This is incredible. In 2005, mathematician Mike Keith took a 717-word section from the essay on Mount Fuji in Lafcadio Hearn’s 1898 Exotics and Retrospective and anagrammed it into nine 81-word poems, each inspired by an image from Hokusai’s famous series of landscape woodcuts, the Views of Mount Fuji.

That’s not the most impressive part. Each anagrammed poem can be arranged into a 9 × 9 square, with one word in each cell. Stacking the nine grids produces a 9 × 9 × 9 cube. Make two of these cubes, and then:

• In Cube “D” (for Divisibility), assign each cell the number “1” if the sum of the letter values in the corresponding word (using A=1, B=2, C=3 etc.) is exactly divisible by 9, or “0” if it is not.
• In Cube “L” (for Length), assign each cell the number “1” if its word has exactly nine letters, or “0” if it does not.

Replace each “1” cell with solid wood and each “0” cell with transparent glass. Now suspend the two cubes in a room and shine beams of light from the top and right onto Cube D and from the front and right onto Cube L:

The shadows they cast form reasonable renderings of four Japanese kanji characters relevant to the anagram:

The red shadow is the symbol for fire.
The green shadow is the symbol for mountain.
Put together, these make the compound Kanji symbol (“fire-mountain”) for volcano.

The white shadow is the symbol for wealth, pronounced FU
The blue shadow is the symbol for samurai, pronounced JI
Put together, these make the compound word Fuji, the name of the mountain.

See Keith’s other anagrams, including a 211,000-word recasting of Moby-Dick.

# A Pretty Puzzle

I don’t know who came up with this; I found it on r/mathpuzzles. What’s the area of the red region?

# Topology

Like the Tehachapi Loop, this is a beautiful solution to a nonverbal problem. When the towpath switches to the other side of a canal, how can you move your horse across the water without having to unhitch it from the boat it’s towing?

The answer is a roving bridge (this one is on the Macclesfield Canal in Cheshire). With two ramps, one a spiral, the horse passes through 360 degrees in crossing the canal, and the tow line never has to be unfastened.

# A Bite

From the Royal Society of Chemistry’s Chemistry World blog: In 1955, when impish graduate student A.T. Wilson published a paper with his humorless but brilliant supervisor, Melvin Calvin, Wilson made a wager with a department secretary that he could sneak a picture of a man fishing into one of the paper’s diagrams. He won the wager — can you find the fisherman?