Misleading advice Yet further ahead, London is hosting the Summer Olympics this year. Let’s hope the city’s many vibrant multicultural communities can refrain from rioting for the duration.
With a lot of first-time foreign tourists expected, the game around London dinner tables right now is “misleading advice to foreigners.” This was a hardy perennial of the old New Statesman weekly competitions. Some of the winners are preserved on the Internet, but I have a more complete set in an old book, from which:
The Monument commemorates the Great Fire of London: on reaching the top, your party is invited to take part in symbolic firefighting by urinating through the railings onto the heads of the busy fish porters below, who wear bowler hats for protection; they will respond with cheerful waves and merry shouts in their colourful language.
I’m surprised there isn’t a New York equivalent. How about:
Subway token booth clerks are always glad to engage in conversation with patrons on matters of general interest . . .
Math Corner. A reader:
I have heard it said and seen it but I can’t get my head around the math. My math brain went critical with trig.
If you stop forty people at random, you have an almost 50-50 chance that two will have the same birthday, not years. With 366 possible birthdays, how is this possible?
Easy. You stop a guy. You ask his birthday. It is whatever it is.
You stop guy number 2. You ask his birthday. The chance it is different from the first guy’s is 365/366.
You stop guy number 3. You ask his birthday. The chance it is different from the first guy’s and also different from the second guy’s is 364/366.
You stop guy number 4. You ask his birthday. The chance it is different from the birthdays of all three previous guys is 363/366.
Keep going . . .
You stop guy number N. You ask his birthday. The chance it is different from the birthdays of all N − 1 previous guys is (367 − N) / 366.
Now, if the chance of A being the case is p, and the chance of B being the case is q, and the chance of C being the case is r, and so on, then the chance of A and B and C and . . . all being the case is p × q × r × . . . (So long as A, B, C, . . . are independent events, which in this case they are.)