In 1999, University of California psychologist Nicholas Christenfeld and his colleagues reviewed thousands of state death certificates and found that males with negative initials (D.I.E., P.I.G., R.A.T.) had died 2.80 years younger than matched controls. Males with positive initials (H.U.G., W.I.N., V.I.P.) had lived 4.48 years longer.
Why? “At present, the best available explanation for these findings is that they are due to the symbolic power of one’s name. It seems unlikely that a person with initials like A.S.S. or J.O.Y. could fail to notice the negative or positive connotations.” Suicide and accidents showed the strongest differences between the positive and negative groups.
But a later study by Pomona College economist Gary Smith found no such pattern.
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.”
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).
You’re in a pitch-black room with a clock that chimes the hour and also chimes once at each quarter hour. If you hear the clock chime once, what’s the longest you’ll have to wait to be sure what time it is?
The longest interval marked by single chimes is 1 hour and 45 minutes (from 12:15 to 2:00). But in that event you’ll be able to infer at 1:45 that it will be 2:00 in 15 minutes. So the answer is an hour and a half.
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.
In a certain street are three tailors. The first to set up shop hung out this sign — ‘Here is the best tailor in the town.’ The next put up — ‘Here is the best tailor in the world.’ The third simply had this — ‘Here is the best tailor in this street.’
