In the late 1970s, Richard Feynman visited a Thai restaurant in Glendale, California, for lunch with his friend Ralph Leighton. Leighton wondered whether he should order his favorite dish, the ginger chicken, or try something new. Feynman, on the spot, scribbled out a solution: If the ginger chicken didn’t exceed a certain high threshold, Leighton ought to try a new dish. But the threshold descended over time — on Leighton’s final visit to the restaurant, for example, it would make more sense to choose a meal he knew he’d enjoy rather than to gamble on an untested candidate.

Leighton kept Feynman’s notes, but his mathematical reasoning remained undeciphered for 50 years. Now Berkeley computational cognitive scientist Brian Christian and his colleagues have established Feynman’s argument and published it in the Proceedings of the National Academy of Sciences.

They also ran an experiment with 2,520 participants to see whether people actually follow this advice. They found that “people adapt linear thresholds used in optimal stopping tasks in a way that is sensitive to the underlying distribution — a simple strategy that we show is nearly as effective as Feynman’s solution.”

On this culinary theme: The “dining philosophers problem,” a puzzle in computer science, is described memorably in Wikipedia:

Five philosophers dine together at the same table. Each philosopher has their own plate at the table. There is a fork between each pair of adjacent plates. The dish served is a kind of spaghetti which has to be eaten with two forks. Each philosopher can only alternately think and eat. Moreover, a philosopher can only eat their spaghetti when they have both a left and a right fork. Thus, two forks will only be available when their two nearest neighbors are thinking, not eating. After an individual philosopher finishes eating, they will put down both forks. The problem is how to design a regimen (a concurrent algorithm) such that any philosopher will not starve; i.e., each can forever continue to alternate between eating and thinking, assuming that no philosopher can know when others may want to eat or think (an issue of incomplete information).

(Thanks, Sharon.)

In a carnival game, you roll seven ordinary dice and then arrange them to form a 7-digit number.

• If your number is a multiple of 2, you’ll win £2.
• If your number is a multiple of 3, you’ll win £3.
• If your number is a multiple of 4, you’ll win £4.
• If your number is a multiple of 5, you’ll win £5.
• If your number is a multiple of 6, you’ll win £6.
• If your number is a multiple of 7, you’ll win £7.

The catch is that you have to announce the prize you’re attempting before you roll the dice. Which prize should you pick?

At first it seems that the £2 prize must be best. If even one of the seven dice produces an even number, you can put that at the end of string and fulfill the condition. This will happen 99.2 percent of the time.

Surprisingly, though, choosing 7 has an even higher success rate, 99.997 percent! “In fact, almost all numbers can be rearranged to make a multiple of 7,” writes James Grime. “But finding the multiple of 7 is the tricky part.” See the paper below for a strategy that will win the jackpot nearly every time.

(James Grime, “The Seven Dice Shuffle,” Recreational Mathematics Magazine 13:22 [June 2026], 95-101.)

How many distinct knots have exactly 10 crossings? By the late 20th century, mathematicians believed the number to be 166.

Then, in 1973, New York attorney and part-time mathematician Kenneth A. Perko Jr. discovered that two of these were essentially the same knot.

The correspondence had gone unnoticed for 75 years.

Origami can solve general cubic equations! The method was developed by Italian mathematician Margherita Piazzolla Beloch, who in 1936 found a way to use paper folding to construct the common tangents to two parabolas.

Given two points p₁ and p₂ and two lines l₁ and l₂, we can, whenever possible, make a single fold (dashed line) that puts p₁ onto l₁ and p₂ onto l₂ simultaneously. This fold finds a common tangent to two parabolas: one with focus p₁ and directrix l₁, the other with focus p₂ and directrix l₂.

“Now, two parabolas drawn in the plane can have at most three different common tangents, suggesting that this origami fold is equivalent to solving a cubic equation,” writes Western New England College mathematician Thomas C. Hull. “Straightedge and compass constructions, on the other hand, can only solve general quadratic equations.”

Beloch’s contribution went uncredited for decades, but it’s now receiving a fuller appreciation. See the 2011 paper below for more details.

(Thomas C. Hull, “Solving Cubics With Creases: The Work of Beloch and Lill,” American Mathematical Monthly 118:4 [April 2011], 307-315. More here.)

The 1961 GCE O-level exam included this question:

If one square yard of material costs 18 pence, what is the price of one square foot?

One student considered:

1 square yard costs 18 pence.

Therefore 1 yard costs , or 4.243, pence.

Therefore 1 foot costs 4.243 ÷ 3 = 1.414 pence.

Therefore 1 square foot costs 1.414² = 2 pence.

(Via Eureka.)

05/26/2026 UPDATE: Reader Catalin Voinescu adds:

For more dimensional fun, check out ‘ohms per square’ (symbol: capital omega divided by a literal square). The resistance of sheet material depends only on the shape of the object, not on the scale (assuming the thickness of the sheet stays the same). Any square of a given material, of any size, has the same resistance when measured between opposite edges. Longer, narrower shapes have higher resistance, and shorter, wider ones have lower resistance, but only the aspect ratio matters, not the actual dimensions. So the resistivity of conductive sheet is expressed as the resistance of a square piece of that material: ohms per square. This unit is used in electrical engineering, where thin conductive layers and foils are common, most obviously in PCB manufacturing, but also in the manufacturing of resistors, capacitors, semiconductors, batteries and solar panels.

(Thanks, Catalin.)

This is the most efficient way to pack 10 unit squares into a surrounding square.

It looks like something I would turn in after running out of time.

Cut the decimal expansion of π into parcels of 10 digits:

Mathematician John Conway points out that the chance is only about 1 in 40,000 that all of the digits 1234567890 will appear in any given parcel … and yet there they all are in the seventh.

(Via Alfred Posamentier and Ingmar Lehmann, π: A Biography of the World’s Most Mysterious Number, 2004.)