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How Long Will It Take To Blow Out The Birthday Candles?

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. There are two types: Riddler Express for those of you who want something bite-sized and Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,1 and you may get a shoutout (chosen at random) in next week’s column. If you need a hint, you can try asking me nicely on Twitter.

Riddler Express

From Conor McMeel, a birthday party puzzle:

It’s your 30th birthday (congrats, by the way), and your friends bought you a cake with 30 candles on it. You make a wish and try to blow them out. Every time you blow, you blow out a random number of candles between one and the number that remain, including one and that other number. How many times do you blow before all the candles are extinguished, on average?

Submit your answer

Riddler Classic

Also from Conor, an impromptu gambling problem:

You and I stumble across a 100-sided die in our local game shop. We know we need to have this die — there is no question about it — but we’re not quite sure what to do with it. So we devise a simple game: We keep rolling our new purchase until one roll shows a number smaller than the one before. Suppose I give you a dollar every time you roll. How much money do you expect to win?

Extra credit: What happens to the amount of money as the number of sides increases?

Submit your answer

Solution to last week’s Riddler Express

Congratulations to 👏 Tim Courtney 👏 of Phoenixville, Pennsylvania, winner of last week’s Express puzzle!

Sometime in the 21st century, you are about to throw out your calendar to get ready for the new year. But someone tells you, “Not so fast! You could reuse it again in the future when the days and dates realign.” In fact, they say, this will happen in 40 years. Moreover, this calendar had never had this sort of 40-year gap before. What year is it?

It’s 2072.

Why? For starters, you should look at leap years. Non-leap years are more common, so those years’ calendars will repeat too quickly — every six or 11 years. Leap-year calendars usually repeat every 28 years, except when their gap spans a year, such as 2100, that is divisible by four but is not a leap year. Then, the period of repetition increases to 40 years. The puzzle’s submitter, Ben Zimmer, provides the rest of the explanation: The year 2072 uses a leap-year calendar of 366 days beginning on a Friday (just like 2016). After 2072, that calendar will next be used in 2112. There are no previous 40-year gaps between Friday-starting leap years since the advent of the Gregorian calendar in 1582. (Before the Gregorian reform, which specified that centurial years not divisible by 400 are not leap years — 2100, specifically, is not a leap year — leap-year calendars always repeated after 28 years.)

Other 21st-century leap years with 40-year gaps — those with calendars that start on Monday, Wednesday, Thursday and Saturday — have predecessors. There are no 21st-century leap years with 40-year gaps starting on a Sunday or Tuesday, although they’ve occurred previously in the Gregorian era. Below is a list of the pairs of calendars that repeat after 40-year gaps (the 21st-century calendars are in bold):

  • 366 days, starting Monday: 1776-1816, 1872-1912, 2080-2120
  • 366 days, starting Tuesday: 1788-1828, 1884-1924
  • 366 days, starting Wednesday: 1772-1812, 2076-2116
  • 366 days, starting Thursday: 1784-1824, 1880-1920, 2088-2128
  • 366 days, starting Friday: 2072-2112
  • 366 days, starting Saturday: 1780-1820, 1876-1916, 2084-2124
  • 366 days, starting Sunday: 1888-1928

An error in last week’s Riddler Express solution

The solution I published last week for Dec. 30’s Riddler Express, about a solitaire card game, was incorrect. I assumed that the types of cards were dealt independently over the course of the game, but they were not. (There are only four aces, for example, so the odds of flipping an ace will change as you go through the deck.)

My assumption led to a \((12/13)^{52}\approx 0.01557\), or about 1.6 percent, chance of winning the game. This is a good approximation for a 52-card deck, but it is imprecise because the cards are not dealt independently. The precise, correct solution is far more complicated than I’d intended and relies on derangements. You can find a discussion of that on StackExchange. Alternatively, you could turn to computer simulation. The derangement-based solution matched many simulations that sharp, gracious readers sent me, revealing the precise answer to be about 0.01623, or also about 1.6 percent. Thank you for your astute emails, and I regret the error.

Solution to last week’s Riddler Classic

Congratulations to 👏 Donald Rauscher 👏 of Minneapolis, winner of last week’s Classic puzzle!

The game of Hip is played by two players on an N-by-N checkerboard. Each player has checkers of his or her own color, and each person takes turns placing them on empty squares on the board. If any four of a player’s checkers form a square — of any size, tilted to any angle — that player loses. What’s the largest board on which this game can end in a draw?

The largest such board is 6-by-6. Our winner, Donald, created an illustration of what a drawn board — a board where no player’s checkers form a square — looks like:

hip-n6

Donald found his solution using an integer program. Basically, he wrote an equation for the number of squares generated on the board, subject to the various constraints of the game — the checkers need to be evenly divided between the players, for example — and tried to find the lowest number of squares possible on a board of a given size. If the minimum achieved is zero, that means a draw is possible. For all boards bigger than 6-by-6, the minimum is never zero. For a board of size 7-by-7, for example, the minimum number of squares is three.

hip-n7

You can read more detail in Donald’s excellent write-up. Other solvers found the 6-by-6 draws with computational brute force. From there, you can make a sort of induction argument to prove that no larger-board solutions exist. “Because there is no way for each 6-by-6 ‘corner’ of a 7-by-7 grid to be a rotation and/or reflection of such a 6-by-6 solution, it follows that there is no 7-by-7 (or larger) solution,” solver Hector Pefo wrote.

Want to submit a riddle?

Email me at oliver.roeder@fivethirtyeight.com.

Footnotes

  1. Important small print: To be eligible, I need to receive your correct answer before 11:59 p.m. EST on Sunday. Have a great weekend!

Oliver Roeder is a senior writer for FiveThirtyEight.

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