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.)