The Three-Dice Problem

In 1620, the Grand Duke of Tuscany wrote to Galileo with a puzzling problem. In rolling three fair six-sided dice, it would seem that the sums 9 and 10 should appear with equal frequency, as there are six ways to produce each result:

10 = 6 + 3 + 1 = 6 + 2 + 2 = 5 + 4 + 1 = 5 + 3 + 2 = 4 + 4 + 2 = 4 + 3 + 3

9 = 6 + 2 + 1 = 5 + 3 + 1 = 5 + 2 + 2 = 4 + 4 + 1 = 4 + 3 + 2 = 3 + 3 + 3

But the duke had noticed that in practice 10 appears somewhat more often than 9. Why is this?

Galileo considered the problem and put his finger on the reason. What is it?

Click for Answer
| Puzzles · Science & Math

Inventory

A striking passage from Avrahm Yarmolinsky’s 1959 biography of Ivan Turgenev:

By the end of May [1840] the traveler was back in Berlin. Before he reached the capital he touched at Leghorn, Pisa, Genoa, sailed on Lago Maggiore, traveled to St. Gotthard in a sleigh, visited Lucerne, Basel, Mannheim, Mainz, Frankfort and Leipzig, all within thirteen days. In the same period he managed to lose an umbrella, a cloak, a box, a walking stick, an opera glass, a hat, a pillow, a pen knife, a purse, three towels, two neckerchiefs, two shirts, and, for a short time, his heart.

He had entered a brief affair with Mikhail Bakunin’s sister Tatyana, but passed just as quickly out of it. “I never loved any woman more than you,” he wrote her, “though I don’t love even you with a complete and steadfast love.”

| Literature

A Banner Year

Next year’s date, 2025, is remarkable:

  • It’s a square (452), the sum of two squares (272 + 362), the product of two squares (92 × 52), and the sum of three squares (402 + 202 + 52).
  • It’s the sum of the cubes of the first nine positive integers (13 + 23 + 33 + 43 + 53 + 63 + 73 + 83 + 93).
  • Equivalently, it’s the square of the sum of those integers (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)2.
  • It’s the second in a trio of square numbers in arithmetic progression (81, 2025, 3969).
  • It’s one of only three four-digit numbers whose halves can be split, summed, and squared to produce the original number: (20 + 25)2 = 2025.
  • It’s the smallest square starting with 20 and the smallest number with exactly 15 odd factors (1, 3, 5, 9, 15, 25, 27, 45, 75, 81, 135, 225, 405, 675, 2025).
  • It’s the sum of the entries in a 9×9 multiplication table.
  • July 24, 7/24/25, will be a “Pythagorean day,” because 72 + 242 = 252.

When asked his age, Augustus De Morgan used to say, “I was x years of age in the year x2.” (He was 43 in 1849.) People born in 1980 will be able to make the same cryptic response starting next year.

(Thanks to readers Chris Smith, Sam Householder, and Jim Howell.)

| Science & Math

Loaded

https://commons.wikimedia.org/wiki/File:Wreck_of_the_Central_America.jpg

“Lately in a wreck of a Californian ship, one of the passengers fastened a belt about him with two hundred pounds of gold in it, with which he was found afterwards at the bottom. Now, as he was sinking — had he the gold? or had the gold him?”

— John Ruskin, Unto This Last, 1860

| Society

Who You Know

https://commons.wikimedia.org/wiki/File:Cezanne_Ambroise_Vollard.jpg

Art dealer Ambroise Vollard was acquainted with many of the foremost artists of the early 20th century, and as a result he appears often in their work. Above are portraits by Cézanne, Renoir, and Bonnard, and he sat also for Rouault, Forain, Vallotton, Bernard, and Picasso.

Picasso wrote, “The most beautiful woman who ever lived never had her portrait painted, drawn, or engraved any oftener than Vollard.”

| Art

Checkless Chess

dawson checkless chess puzzle

In checkless chess it’s illegal to give check without giving checkmate. This changes the whole complexion of the game. T.R. Dawson published this example in Die Welt in 1951. White is to mate in two moves.

The answer, 1. f6, threatens 2. Qf5, which strangely is mate because the black king can’t move off the long light diagonal, since that would discover check by the black bishop without giving checkmate. Black can try to prevent this finish by playing 1. … Qc8, guarding f5 and thus making 1. Qf5 itself illegal. But this permits 2. Nxd6, which now is mate because the black king has nowhere to run on the long diagonal and 2. … cxd6 is illegal because this would give a mateless check.

Similarly, if Black tries to stop 1. Qf5# with 1. … Qxf6, then 2. Nc5 is mate because Black can’t capture the knight — again, this would expose White’s king to an illegal check. And if Black tries to answer White’s first move with 1. … d5, to block the long light diagonal and free his king to flee elsewhere, then White can play 2. Qe5#, an ordinary (and legal) mate.

| Puzzles

Lost Lessons

THOUGH I can never pay enough to your Grandfather’s Memory, for his tender care of my Education, yet I must observe in it this Mistake; That by keeping me at home, where I was one of my young Masters, I lost the advantage of my most docile time. For not undergoing the same Discipline, I must needs come short of their experience, that are bred up in Free Schools; who, by plotting to rob an Orchard, &c. run through all the Subtilties required in taking of a Town; being made, by use, familiar to Secresie, and Compliance with Opportunity; Qualities never after to be attained at cheaper rates than the hazard of all: whereas these see the danger of trusting others, and the Rocks they fall upon, by a too obstinate adhering to their own imprudent resolutions; and all this under no higher penalty than a Whipping: And ’tis possible this indulgence of my Father might be the cause I afforded him so poor a Return for all his Cost.

— Francis Osborne, Advice to a Son, 1656

| Society

Opaque Sets

https://commons.wikimedia.org/wiki/File:Unit_Square_Opaque_Forest_Solutions.svg

A creature living in the plane can’t see through a unit square — the square’s four line segments block its line of sight from any angle. Is there a way to achieve the same result using fewer building materials? Removing one of the square’s sides does the job — this requires only 3 units of line segment and still prevents anyone from seeing across the square’s area. The arrangement at lower left does better still, requiring only about 2.732 units. And the one at lower right requires only about 2.639 units.

Is that the shortest possible opaque set for the square? Possibly — but no one has been able to prove it.

| Science & Math