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Is It Anyone’s Birthday?

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,1 and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

Riddler Express

From James Nugent comes a puzzle that goes out to all the folks with March birthdays:

Earlier today, James’s boss was surprised to find out that not only did no one on their team have a birthday this week, but that nobody was celebrating a birthday for the entire month. With a total of 40 people on the team, the probability of this happening seemed to be miniscule.

But was that really the case? What was the probability that none of the 40 people had birthdays this month? (For the purpose of this riddle, assume that a year consists of 12 equally long months. It’s a sufficiently good approximation!)

Extra credit: What is the probability that there is at least one month in the year during which none of the 40 people had birthdays (not necessarily this month)?

The solution to this Riddler Express can be found in the following column.

Riddler Classic

From Rolfe Petschek comes a puzzle that’s good fun in snow, rain, heat or gloom of night. (Okay, maybe not in gloom of night.)

A postal worker and his customer joke about the various ways the customer could mathematically encode her post office box number.

The customer realizes that every integer greater than 1 can be encoded via at least one Fibonacci-like sequence using an ordered triple (m, n, q). The encoded number is the qth member of the sequence after the first two positive integers m and n, where each term is the sum of the previous two terms. For example, 7 has the encodings (3, 4, 1) and (1, 3, 2).

In an attempt to stump the postal worker, the customer prefers encodings with a maximal value of q. What encoding should she use for the number 81?

Extra credit: What encoding should she use for the number 179?

The solution to this Riddler Classic can be found in the following column.

Solution to last week’s Riddler Express

Congratulations to 👏 Jesus Petry 👏 of Porto Alegre, Brazil, winner of last week’s Riddler Express.

Last week, you participated in a scavenger hunt to determine who received tickets to a basketball game. The scavenger hunt had three tasks. For each task, whoever finished first got 1 point, whoever finished second got 2 points and so on. Your score was the total number of points you earned across all three tasks, and lower scores were better. While many people participated, only the top 10 finishers in the scavenger hunt received tickets.

Without knowing how anyone else did on the scavenger hunt, what was the highest score that guaranteed you were in the top 10? (Being tied for 10th was acceptable.)

Rather than ask about the highest score that still landed you in the top 10, you could equivalently look for the lowest score that still landed you in 11th place. If you did any better than that, then surely you were in the top 10.

To get a low score and still only come in 11th, you wanted the top 11 scores to all be low and also be fairly close together. Taken together, the top 11 spots in each task were worth a total of 66 points, or 198 points across all three tasks. In the event of an 11-way tie, that meant each person got 18 points. But if there had been an 11-way tie, then coming in 11th was just as good as coming in 10th — or first for that matter.

So 18 points was too good to have landed you in 11th place. It might have been good enough to guarantee a top-10 finish. But first, you had to make sure it was possible to come in 11th with 19 points. (Otherwise, that too might have guaranteed a top-10 finish.)

Several solvers, including Amy Leblang, found a set of scores where 19 points indeed landed you in 11th place:

  • 2 + 6 + 9 = 17
  • 3 + 8 + 7 = 18
  • 4 + 9 + 5 = 18
  • 5 + 10 + 3 = 18
  • 6 + 11 + 1 = 18
  • 7 + 1 + 10 = 18
  • 8 + 2 + 8 = 18
  • 9 + 3 + 6 = 18
  • 10 + 4 + 4 = 18
  • 11 + 5 + 2 = 18
  • 1 + 7 + 11 = 19

If 19 points was the lowest score that could result in 11th place, then 18 points was the highest score that guaranteed a top-10 finish.

Now that you got those basketball tickets, I hope you enjoy the madness!

Solution to last week’s Riddler Classic

Congratulations to 👏 Al Shaheen 👏 of New Haven, Indiana, winner of last week’s Riddler Classic.

Last week, we were playing a game where you had to pick four whole numbers. Then I rolled four fair dice. If any two of the dice added up to any one of the numbers you had picked, then you won! Otherwise, you lost.

For example, suppose you had picked the numbers 2, 3, 4 and 12, and the four dice came up 1, 2, 4 and 5. Then you would have won, because two of the dice (1 and 2) added up to at least one of the numbers you had picked (3).

To maximize your chances of winning, which four numbers should you have picked? And what were your chances of winning?

First off, two dice can add up to any whole number between 2 (snake eyes) and 12 (double sixes), so all four of your numbers should have come within that range. Also, whenever one of your numbers was a match, you won, so there was no point in picking the same number twice. In other words, you were picking four distinct whole numbers between 2 and 12.

Many readers knew that 7 was the most likely sum you could get from two dice, with a probability of 6/36 (or 1/6). This was followed by 6 and 8, which each had a probability of 5/36, and then 5 and 9, which each had a probability of 4/36 (or 1/9). So if you had to pick four numbers, why not pick the four most likely numbers? That meant the optimal choice of four was 5, 6, 7 and 8, or 6, 7, 8 and 9. (By symmetry, these two sets of numbers would win with equal probabilities.)

All that made sense. But it was wrong!

You could have checked it by writing out all 64 (or 1,296) cases, or preferably by checking these cases with the aid of a computer. Two dice added up to 5, 6, 7 or 8 in 1,224 of them, good for a 94.4 percent chance of victory. That sounded rather close to 100 percent, but it was possible to do even better.

The best four numbers to pick were 4, 6, 8 and 10. Two dice added up to at least one of these values in 1,264 cases, meaning your chances of victory were 97.5 percent. In the end, the most likely single sum to achieve, 7, wasn’t one of the four numbers you should have picked. What a surprising result! How was this even possible?

Your goal was to pick the four numbers that collectively covered as many of the 1,296 cases as possible. So while 7 was the single best choice, it unfortunately had significant overlap with other common sums. For example, the probability of having a pair of dice add up to 7 and a pair of dice add up to an 8 was 36.9 percent — quite a bit of overlap between those two sums. Meanwhile, the probability of having a pair add up to 6 and a pair add up to 8 was just 29.8 percent. That’s only about 7 percentage points lower, but every little bit made a difference here.

The big idea was that choosing the four most likely sums resulted in more overlap, and therefore less overall coverage of the 1,296 cases. Meanwhile, picking a set of sums that were not quite as likely individually, but more spread out across the sample space, improved your odds. Below is a sketch that is not mathematically precise, but which analogously illustrates how four smaller circles can cover a slightly greater area by having less overlap.

Two sets of four overlapping circles. On the left, the circles 6, 7, 8 and 9 overlap. 7 is the largest.

On the right, 4, 6, 8 and 10 overlap. These circles are smaller on average, but more spread out, so together they cover a greater collective area.

Solver Emma Knight took this puzzle a step further, exploring what happened when you changed the number of dice and the number of guesses. Regardless of the number of dice, if you only had one guess, you should have, of course, picked 7. With two guesses, you should have picked 6 and 7 (or, equivalently, 7 and 8). With three guesses, you should have picked 6, 7 and 8. But after that, the even numbers started to take over, leaving 7 by the wayside. As Emma pointed out, with three or more dice, you were guaranteed to have at least one even sum.

And if you were rolling at least three dice with more than six guesses, after you guessed all the even numbers, Emma suggested guessing e or 𝜋. And why not? When it comes to The Riddler, more often than not, one of those numbers shows up around here.

Want more riddles?

Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!

Want to submit a riddle?

Email Zach Wissner-Gross at riddlercolumn@gmail.com.

Footnotes

  1. Important small print: In order to 👏 win 👏, I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!

Zach Wissner-Gross leads development of math curriculum at Amplify Education and is FiveThirtyEight’s Riddler editor.

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